CHEMISTRY ISC CLASS XII - Web Educationwebéducation.com/wp-content/uploads/2019/11/Most-Likely... · 2019. 11. 2. · CHEMISTRY MOST LIKELY QUESTION BANK CATEGORY WISE & CHAPTER

CHEMISTRY

MOST LIKELY QUESTION BANKCATEGORY WISE & CHAPTER WISE

2019 EXAMINATIONCLASS XIIISC

• Covers Previous Years’ Board Questions• Solutions provided by subject experts• Comprehensive Revision of the Syllabus

In accordance with the latest syllabus of Council for the Indian School Certificate Examinations.

ISCMost Likely Question Bank

Class XII

CHEMISTRY

byA PANEL OF AUTHORS

OSWAL PUBLISHERS A1/12, Sahitya Kunj, M. G. Road, Agra-282 002

© Publishers

Edition : 2018

ISBN : 978-93-87660-12-0

OSWAL PUBLISHERSHead office : 1/12, Sahitya Kunj, M.G. Road, Agra-282 002Phone : (0562) 2527771– 4E-mail : [email protected], [email protected] : www.oswalpublishers.comFacebook link : https://www.facebook.com/oswalpublishersindiaAlso available on : amazon.in, Flipkart, snapdeal

PrefaceOswal Publishers take great pride in presenting their series of “ISC Most Likely Question Bank”,

especially for the students preparing for CISCE Board Examinations.

This book will help the students in revising the whole syllabus in a comparatively lesser time and willdevelop the aptitude for effective learning in the subject. This will enable candidates to acquireknowledge and to develop an understanding of the terms, facts, concepts, definitions, fundamental laws,principles and processes.

We hope that this book will instill a sense of confidence in the students and empower them towardsachieving their goals and scholastic excellence.

Suggestions for the improvement of the book are invited and shall be gratefully acknowledged.

—The Publisher

How to Prepare for Board Exams Using Oswal’s Most Likely Question BankCategorywise & Chapterwise

The Most Likely Question Bank is a great learning tool for CISCE Examinations as it is the mostcomprehensive and concisely composed set of questions for the students. In order to prepare well for theirexams, The students may follow the following approach.

➢ Step 1 : Prepare Your Theory

Students must go through the theory of the Chapters from their textbooks covering the syllabus asper the Council Guidelines.

➢ Step 2 : Understand the Categories

The contents have been arranged as per different ‘Category’ of Questions. These ‘Categories’ coverall the different types of questions that the Board asks in their examinations. Each Category in turncovers topics from all the chapters of the syllabus. Eg. Short Answers category will cover the shortanswer type questions from all the chapters of the syllabus in that subject.

➢ Step 3 : Complete Syllabus Revision

When students revise questions from any single category, they are revising the entire syllabus indifferent assemblages.

➢ Step 4 : Customised Learning

Students with different learning calibers can customize their preparations. They can refer theobjective questions first to get a basic understanding of the chapters and topics then they can takeup the most complex categories.

➢ Step 5 : Check Your Progress

Students can check their answers against the solutions given next to the questions. This will savetime and help in faster revisions.

Students can benefit tremendously from these series by using them optimally for their exam preparations.

CONTENTS

1. Fill in the blanks 5–14

2. Multiple Choice Questions 15–41

3. Match the Column 42–52

4. Identify the Compounds / Products / Reagents 53–81

5. IUPAC Nomenclature 82–90

6. Reasoning based Questions 91–138

7. Balance the Chemical Equations 139–163

8. Formula / Structure based Questions 164–181

9. Short Answer Questions / Definitions 182–235

10. Numerical Problems 236–278

11. Differentiate Between 279–297

12. Laws and Naming Reactions 298–309

13. Conversions 310–325

14. Mechanism of Reactions 326–333

15. Description based questions 334–384

Question

1Set Fill in the blanks

Chapter 1. Solid State(i) Both ccp and hcp are ………… packings and occupy about ………… % of the available space.(ii) The crystal of graphite is made up of ………… while that of sodium chloride is made up of

……….(iii) The unit cell of a body centred cubic system contains ……………… atoms.(iv) In diamond inter particle forces are .......................... .(v) In hcp arrangement of spheres, the coordination number is .......................... .

(vi) In ccp arrangement of atoms ........................... percent of the available space remains vacant.(vii) In bcc arrangement of atoms ................. percent of the available space is occupied by atoms.

(viii) In rock-salt structure .................... percent of the octahedral voids are occupied by cations.(ix) If the radius ratio r+/r – is 0·325, the cation would most probably be present in a/an

........................... void.(x) In CsCl structure the cations are present in ........................... voids.

(xi) Crystals of ................ and .......................... have face centered cubic lattices.(xii) The number of formula units in a CsCl unit cell is equal to .......................... .

(xiii) An atom at the corner of a unit cell makes ........................... contribution to a particular unit cell.(xiv) In zinc blende structure ........................... have ccp arrangement whereas ....................... are

present in ........................... voids.(xv) In anti-fluoride structure, cations are present in ........................... voids.

(xvi) The number of formula units per unit cell in fluoride structure is ................ .(xvii) Glass is an example of ………… solid.

(xviii) The crystal of diamond is made of …………while that of calcium chloride is made of …………(xix) In fluoride structure positive ions are arranged in ………… .(xx) Of the three cubic lattices the one that has the largest amount of empty space is the … .

Answers(i) efficient, 68% (ii) atoms, ions, (iii) two

(iv) covalent bonds (v) 12 (vi) 26

(vii) 68 (viii) 100 (ix) tetrahedral

(x) cubic (xi) CsCl, NaCl (xii) one

(xiii) 1/8 (xiv) anions, cations, tetrahedral (xv) tetrahedral

(xvi) 4 (xvii) amorphous (xviii) atoms, ions

(xix) ccp (xx) simple cubic

Chapter 2. Solutions(i) Ideal solutions obey ………… law and they ………… form azeotropic mixtures.(ii) The van’t Hoff factor of acetic acid solution is ………… than one and the value of normal

colligative property is ………… than the observed colligative property of this solution.(iii) Solutions which strictly obey ………… law are called …………… solutions.(iv) The ………… pressure of an aqueous solution of 0.1 M cane sugar is ………… than that of pure

water.(v) Molality of the solution is number of moles of the solute in ………… kg of …………

(vi) The molarity of the solution of non electrolyte at 0°C if its osmotic pressure is 17 mm is approx.…………

(vii) The boiling point of sea water at 1 atm. pressure is ………… that of distilled water.(viii) The ………………… of the boiling point of a solvent by the addition of a solute is …………

proportional to the molality of the solution.

6 ■ ISC Most Likely Question Bank, Class : XII

(ix) The evaporation of aqueous solution of glucose causes its molarity to……….(x) A solution which distills without change in composition is called …………..

(xi) Relative lowering of vapour pressure is equal to the mole fraction of the ……………(xii) For sodium chloride solution, van’t Hoff factor is ................. .

(xiii) When solvent starts flowing from ……… into ……… through semipermeable membrane, thephenomenon is termed as reverse osmosis..

(xiv) ………… and ……… are intensive properties.(xv) A solution which does not obey Raoult’s law at all range of concentration is called ……….

(xvi) For an ideal solution ΔH mixing is ……………Answers

(i) Raoult’s, do not (ii) more, less (iii) Raoult’s, ideal, (iv) Vapour, less. (v) One, solvent, (vi) 0·001 M,

(vii) greater than, (viii) elevation, directly, (ix) Increase,(x) Azeotropic, (xi) greater than 1, (xii) greater than 1,

(xiii) Solution, solvent (xiv) concentration, temperature, (xv) non-ideal,(xvi) Zero.

Chapter 3. Electrochemistry

(i) The unit of conductance is ………… and that of specific conductance is …………(ii) The more ………… is standard reduction potential of a metal, the ………… is its ability to

displace hydrogen from acids.(iii) An electrochemical cell converts ………… energy to ………… energy.(iv) The molar conductance of a solution ………… with dilution, while its specific conductance

………… with dilution.(v) In a galvanic cell, electrons flow from ………… to ………… through the connecting wires.

(vi) Zinc can displace ………… from CuSO4 solution, but cannot displace ………… from MgSO4solution.

(vii) The electrode at which there is acceptance of electrons is called ....................... .(viii) The conductance of a solution placed between two opposite faces of a cm. cube is called

................... .(ix) The solution of sugar in water is ....................... conductor of electricity.(x) Strong electrolytes give almost linear plot of Λm versus ....................... .

(xi) The resistance of a solution is measured with the help of ....................... .(xii) In a Daniel cell the copper vessel serves as ....................... .

(xiii) The E° value of a NHE is ....................... .(xiv) A galvanic cell stops after some time because ....................... .(xv) The equivalent conductance of a solution ....................... with decrease in concentration of the

electrolyte in the solution.(xvi) A device in which chemical energy of a fuel is directly converted into electrical energy is called

....................... .(xvii) In a galvanized iron, the iron is coated with a layer of ....................... metal.

(xviii) The molar conductance (Λm) of a solution is related to specific conductance (K) by the relation....................... .

(xix) In representation of an electrochemical cell, the cathode is written on ....................... .(xx) Specific conductance of an electrolyte solution ....................... with increase in dilution.

(xxi) ....................... the reduction potential ....................... is the oxidizing agent.

Answers(i) ohm–1, ohm–1 cm–1 (ii) negative, greater (iii) chemical, electrical

(iv) increases, decreases (v) anode, cathode (vi) copper, magnesium (vii) Cathode

Fill in the blanks ■ 7

(viii) Specific conductance (ix) Bad (x) C (xi) Multimeter

(xii) Cathode (xiii) Zero (xiv) Potential difference becomes zero

(xv) Increase (xvi) Fuel cells (xvii) Zinc (xviii) Λm = K × 103

M(xix) Right hand side (xx) Decreases (xxi) Higher, Stronger

Chapter 4. Chemical Kinetics

(i) When the concentration of a reactant of first order reaction is doubled, the rate becomes………… times, but for…………order reaction, the rate remains same.

(ii) Half life period of a ……… order reaction is ………… of the concentration of the reactant.

(iii) For a first order reaction, the unit of rate is…………and that of rate constant is …………….

(iv) The ……………constant of a first order reaction is proportional to the concentration of thereactant.

(v) Hydrolysis of ethyl acetate in an acidic solution is an example of ........................ order reaction.

(vi) If the activation energy of the reaction is low, it proceeds at ........................ rate.

(vii) The reaction taking place under the influence of visible light is called ................... reaction.

(viii) In a multi-step reaction, the ........................ step determines the rate of reaction.

(ix) The sum of powers to which the concentration terms are raised in the rate law is called............... of the reaction.

(x) For a first order reaction, the half life period is equal to ........................ .

(xi) The order of a reaction with respect to B, whose rate law is = K [A]3/2 [B]1/2 is .................. .

(xii) The order and molecularity of a complex reaction ....................... be same.

(xiii) In photochemical reactions, the necessary activation energy is provided by ...................... .

(xiv) The unit of rate of gaseous reaction is ...................... .

(xv) For a zero order reaction A → Products, the plot of [A] vs time has a slope ...................... .

(xvi) For the reaction xA + yB ⎯→ lL + mM, the average rate of disappearance of B = .............. .

(xvii) In the Arrhenius equation, K = A exp (–Ea/RT), A may be termed as rate constant at ................. .

(xviii) The half life period of a…………order reaction is …………on the concentration of the reactant.

(xix) A catalyst provides an alternate path for the reaction with.................... energy barrier.

(xx) A catalyst .................... the time required to establish the equilibrium.

(xxi) Heterogeneous catalysis is also called .................... catalysis.

(xxii) .................... converts starch into maltose.(xxiii) Adsorption is a .................... process.(xxiv) Chemisorption is a .................... process.(xxv) Larger surface area .................... the efficiency of the catalyst.

Answers(i) four, zero (ii) first, independent (iii) mol l–1 s–1, s–1.(iv) rate (v) first (vi) Faster(vii) photochemical (viii) slowest (ix) Order

(x)0·693

K (xi) 0·5 (xii) may or may not

(xiii) visible radiations (xiv) atm sec–1 (xv) K(xvi) –Δ [B]/Δt (xvii) T → ∞ (xviii) Second, dependent(xix) lower (xx) alters (xxi) contact(xxii) Diastase (xxiii) exothermic (xxiv) exdothermic(xxv) increases.


Chapter 5. Surface Chemistry

(i) The formation of micelles takes place above a particular concentration called …………(ii) The rate of adsorption in general is …………in the beginning and then …………till equilibrium

is attained.(iii) The size of particles of colloidal solution is in the range of …………nm to ……………nm.(iv) In benzosol, the dispersion medium is ……………(v) In whipped cream, the dispersion medium is ………and dispersed phase is ……………

(vi) Milk is an example of …………in ………emulsion.(vii) The process in which adsoption and absoption takes place simultaneously is called …………

(viii) The nature of forces involved in physical adsoption are ……………(ix) The formation of micelles takes place above a particular temperature called …………(x) The movement of colloidal particle under the influence of an electric field is called …………

(xi) The scattering of light by colloidal particle is called ……………(xii) The swelling of gel in water is called …………

(xiii) Colloidal solution of gold in Tin is called …………Answers

(i) critical micelle concentration (ii) high, decreases (iii) 1,1000(iv) benzene (v) Liquid, gas (vi) liquid, liquid (vii) Sorption

(viii) vander Waal’s forces (ix) Kraft temperature (x) electrophoresis

(xi) Tyndall effect (xii) imbibition (xiii) purple of cassius

Chapter 6. General Principles and Processes of Isolation of Elements

(i) Levigation generally used in froth floatation process is ……………(ii) Calcination and roasting process are carried out in ……………furnace.

(iii) In the mond’s process ………is used to purify impure nickel.(iv) Flux combines with infusible impurities to form …………(v) In electrolytic refining impure metal is made …………while pure metal act as …………

(vi) The formula of the sylvine ore is …………(vii) Limonite is the ore of …………having chemical formula ………………

(viii) The two basic fluxes are …………and ……………(ix) Feldspar is a …………ore of …………and …………(x) Baeyer’ process is used when …………ore contain higher percentage of …………

(xi) Two acidic fluxes are …………and …………(xii) Red bauxite is purified by …………method.

(xiii) The process used for refining of Aluminium is known as …………(xiv) Invar contain iron and …………(xv) The actual reducing agent of haematite in blast furnance is …………

(xvi) The purest from of iron is …………(xvii) The frothing agent used in froth floatation process is …………

Answers(i) oxide (ii) reverberatory (iii) CO

(iv) slag (v) anode, cathode (vi) KCl(vii) iron, FeO(OH).nH2O (viii) lime (CaO), limestone (CaCO3)(ix) Silicate, potassium, aluminium (x) bauxite, iron oxide(xi) Sand (SiO2), borax (Na2B4O7.10H2O) (xii) Baeyer’s (xiii) Hoope’s process

(xiv) Nickel (xv) CO (xvi) Wrought iron(xvii) pine oil


Chapter 7. p-Block Elements

(i) Halogens are strong ………… agents because of their high …………

(ii) …………… the Swedish chemist, was the first to prepare oxygen.

(iii) Oxygen family is also called ………… .

(iv) …………… is most electronegative among halogens.(v) Among the hydrides of halogen ……… is most acidic.

(vi) AX3 type of inter halogen compounds show …………… type of hybridisation.(vii) The shape of AX7 type of inter halogen is …………… .

(viii) XeOF4 shows ………… hybridisation and is ………… in shape.(ix) ……………a noble gas is used in the treatment of cancer.(x) The first compound of noble gases prepared by N. Bartlett was …………… .

(xi) The halogen which does not show variable oxidation state is …………… .(xii) On boiling sulphur in NaOH solution …………… are produced.

(xiii) Aqua regia is a mixture of …………… and …………… in the ratio of 3 : 1.(xiv) Na2O2 + …… ⎯→ Na2SO4 + ……(xv) Cl2 + ……⎯→ O2 + ……

(xvi) Sulphuric acid is a …… molecule having hexavalent ……… atom. It has a ……… structure.(xvii) O3 condenses to deep ………… liquid (b.p. 161.2 K) and to ………… solid (mp 80·6 K).

(xviii) In contact process for the manufacture of H2SO4, the SO 3 is absorbed in ………… .(xix) The two series of salts that sulphuric acid can produce are ……… and ……… .(xx) The most volatile halogen acid …………… .

(xxi) The oil of vitriol is …………… .(xxii) At –76°C ……… solidifies into snow like solid.

(xxiii) With NH3 …………… gives white dense fumes,(xxiv) Boiling of ……… is low among the hydrides of halogen.(xxv) Oxygenated water is ……………

(xxvi) In the preparation of ozone ……… electric ……… produces ……… heat and prevents the local……… in temperature.

(xxvii) H2SO4 reacts with H2S to form ………, H2O and …………Answers

(i) oxidizing, electronegativity (ii) Carl Scheele (iii) Chalcogen (iv) Fluorine

(v) HI (vi) sp3d (vii) pentagonal bipyramidal

(viii) sp3d2, square pyramidal (ix) Radon (x) Xe+ [Pt F6]–

(xi) Fluorine (xii) Na2S and Na2S2O3

(xiii) Concentrated hydrochloric acid, concentrated nitric acid.(xiv) Na2O2 + H2SO4 ⎯→ Na2SO4 + H2O2 (xv) Cl2 + H2O2 ⎯→ O2 + 2HCl(xvi) covalent, S, tetrahedral (xvii) blue, violet black

(xviii) H2SO4 (xix) Sulphate and bisulphates (xx) HCl(xxi) H2SO4 (xxii) SO2 (xxiii) HCl

(xxiv) HCl (xxv) Hydrogen Peroxide(xxvi) silent, discharge, less, rise (xxvii) S, SO2

Chapter 8. d-and f-Block Elements(i) The transition metals show………… character because of the presence of unpaired electrons

and Cu+ is………… because its electronic configuration is [Ar]3d10.(ii) The most abundant transition metal is …………… .

(iii) In acidic medium KMnO4 is reduced to ………… .(iv) The formula of pyrolusite is …………… .


(v) Among transition element, the element having lowest melting point is ……… .(vi) The common oxidation state of lanthanoids is …………… .

(vii) The common oxidation state of actinoids is …………… .

(viii) The ferrous metals are …………… .(ix) The last element of lanthanoid series is …………… .(x) The acidic oxide of manganese is …………… .

(xi) CrO42– ion has ………… geometry.

(xii) The metal having highest melting point is …………… .(xiii) f-block elements are also known as …………… .

Answers(i) paramagnetic, diamagnetic (ii) Iron (iii) Mn2+

(iv) MnO2 (v) Mercury (vi) + 3

(vii) + 3 (viii) Fe, Co, Ni (ix) Lutetium

(x) Mn2O7 (xi) Tetrahedral (xii) Tungsten

(xiii) Inner-transition elements

Chapter 9. Coordination Compounds(i) The number of chelating ligands indicate the ………… of the ligand.(ii) A bidentate ligand is ………… if the two coordinating atoms are the same.

(iii) Glycinato is a ………… dentate ligand.(iv) π acid ligands are also known as ………… ligands.(v) The portion present outside the square brackets is called ………… sphere.

(vi) Species in the coordination sphere are ………… .(vii) In …………… complexes metal is bound to only one kind of donor groups.

(viii) The oxidation number of cobalt in [CoBr2(en)2]+ is ……… .(ix) The EAN of central metal atom in K4 [Fe(CN)6] is ………… .(x) The formula of the coordination compound dichlorobis (ethane–1, 2–diammine) cobalt (III) is

………… .(xi) [Cr(NH3)6]3+ is an ………… orbital complex and ………… in nature.

(xii) The complex [Co (NH3)6]3+ has a ………… geometry and is ………… magnetic in nature.(xiii) Grignard reagent is an ………… .(xiv) For [Rh(NH3)3Cl] the possible number of geometrical isomers is/are ………… .

Answers(i) denticity, (ii) symmetrical (iii) bi

(iv) π–acceptor (v) Ionisation (vi) non-ionisable

(vii) homoleptic (viii) +3 (ix) 36

(x) [CoCl2(en)2]+ (xi) Inner, paramagnetic (xii) Octahedral, dia,

(xiii) organometallic compound (xiv) zero.

Chapter 10. Haloalkanes and Haloarenes

(i) The well known refrigerant freon has the structure ……… .

(ii) Preparation of chlorobenzene from benzene diazonium chloride and aqueous HCl is known as……… reaction.

(iii) Formation of phenol from chlorobenzene is an example of ……… aromatic substitution.

(iv) RCH2OH ?⎯⎯→ RCH2Br ?

⎯⎯→ RCH2CN ?⎯⎯→ RCH2COOH

(v) Butane nitrile can be prepared by heating ……… with alcoholic KCN.


(vi) Vinyl chloride on reaction with dimethyl copper gives ………

(vii) Ethyl bromide on reaction with moist silver gives ……… as the main product.

(viii) The trade name of carbon tetrachloride is ………

(ix) In SN1 mechanism ……… are involved as intermediate species.

(x) Phenyl isocyanide is formed when chloroform is treated with ……… in the presence ofalcoholic KOH.

(xi) Vinyl chloride is ……… reactive than aryl chloride.

(xii) BHC is commercially called ………… .Answers

(i) CCl2F2 (ii) Gattermann (iii) Nucleophilic

(iv) HBr, KCN, Dil. HCl (v) n-propyl chloride (vi) Propene, CH3CH = CH2

(vii) Ethanol (viii) Pyrene (ix) Carbocation

(x) Aniline (xi) Less (xii) Gammaxene

Chapter 11. Alcohols, Phenols and Ethers

(i) ………… is an example of trihydric alcohol and …………… is an example of dihydric alcohol.

(ii) Amongst three isomers of nitrophenol, the one that is least soluble in water is ……… .

(iii) The common name of 1, 2, 3-trihydroxy-benzene is ………… .

(iv) IUPAC name of cumene is ……

(v) Ethyl bromide on reaction with moist silver oxide gives …… as the main product.

(vi) Phenol reacts with sodium liberating …… gas.(vii) IUPAC name of picric acid is ……… .

(viii) Phenol is ……… reactive than chlorobenzene towards electrophilic substitution reactions.(ix) In case of ……… alcohols, the cloudiness appears immediately while performing Lucas test.(x) During acid catalysed dehydration of alcohols, the intermediate species involved are ……… .

(xi) Ethers behave as weakly ………… substances because of the presence of ……… of electrons onthe oxygen atom.

Answers(i) Glycerol, glycol (ii) o-nitrophenol(iii) pyrogallol (iv) 2-Phenylpropane(v) ethanol (vi) hydrogen (or H2)(vii) 2, 4, 6-Trinitrophenol (viii) more (ix) tertiary(x) carbocations (xi) basic, lone pair

Chapter 12. Aldehydes, Ketones and Carboxylic Acids

(i) Benzaldehyde undergoes ………… reaction on treatment with concentrated sodium hydroxidebecause it has………… atom.

(ii) Benzaldehyde when treated with an alcoholic solution of ………… forms ………… .

(iii) When benzaldehyde reacts with ………… it forms ………… and POCl3.

(iv) Acetaldehyde reacts with HCN to give ………… which on hydrolysis gives ………… .

(v) A ………precipitate is obtained on adding iodine and sodium hydroxide to ………… .

(vi) A 40% solution of formaldehyde in water is known as …………… .

(vii) The hybrid state of carbon atom in carbonyl group is …………… .(viii) The addition reactions of aldehydes and ketones are initiated by ………… attack on carbonyl

carbon.


(ix) The reaction of acetic anhydride with LiAlH4 followed by subsequent acidic hydrolysisproduces …………… .

(x) Fehling ‘A’ consists of an aqueous solution of CuSO4 while Fehling ‘B’ consists of an alkalinesolution of …………… .

(xi) Trimer of acetaldehyde is named as …………… .(xii) Meta formaldehyde is a ……… of formaldehyde.

(xiii) Hydrolysis of methyl propanoate gives …………… and …………… .(xiv) Acetic acid forms ........................ with NH3 which on heating gives ....................... .(xv) Soda lime on decarboxylation with sodium propionate produces ....................... .

(xvi) Ethanamide on heating with P2O5 gives ....................... .(xvii) Carbonation of Grignard’s reagent involves reaction of ........................ with .............. .

(xviii) Trimethyl acetic acid is ........................ than trichloroacetic acid.(xix) The reaction of acetic anhydride with LiAlH4 followed by subsequent acid hydrolysis produces

....................... .(xx) The Hoffmann bromide reaction involves the treatment of amides with ....................... .

(xxi) Addition of water to acetylenic compounds is catalyzed by...................... and ................... .(xxii) The IUPAC name of succinic acid is ....................... .

(xxiii) Benzoic acid reacts with alcohol to give ....................... .(xxiv) Carboxylic groups attached directly to the benzene are called ....................... .(xxv) Alkaline hydrolysis of ester is called ……………………… .

Answers(i) cannizzaro, no α-hydrogen (ii) potassium cyanide, benzoin(iii) PCl5, benzal chloride. (iv) Acetaldehyde cyanohydrin, Lactic acid(v) Yellow, Acetone (vi) formalin(vii) sp2 (viii) nucleophilic(ix) ethanol (x) Rochelle salt(xi) para aldehyde (xii) Trimer.(xiii) methanol, propanoic acid (xiv) Ammonium acetate, acetamide(xv) Ethane (xvi) Ethanenitrile (xvii) CO2, Grignard’s reagent(xviii) Weaker acid (xix) Ethanol (xx) Br2/KOH(xxi) dil. H2SO4, HgSO4 (xxii) Butanedioic acid (xxiii) Ester(xxiv) Aromatic Acid (xxv) Saponification

Chapter 13. Organic Compounds Containing Nitrogen(i) Ethyl isocyanide, on hydrolysis with dilute sulphuric acid, gives ………… and ………… .(ii) When acetamide is treated with bromine and caustic soda, it gives …………… as the main

product and the reaction is called…………….(iii) The boiling point of isocyanides is .................... than the corresponding isomeric cyanides.(iv) The alkyl nitrites and nitro alkanes are ............................. of each other.(v) Direct nitration is ............................. suitable to get pure nitroalkanes.

(vi) Alkyl nitrite can be considered to be an ............................. of alcohol and nitrous acid.

(vii) Aniline is .................. basic than . .NH3 whereas ethylamine is ................ basic than

. .NH3.

(viii) N, N-dimethyl aniline on treatment with HNO2 gives ............................. .(ix) Schotten Baumann reaction involves reaction between ............................. .(x) Sulphanilic acid exists as ............................. structure.

(xi) C6H5OH + C6H5 N+2Cl

– pH(9–10)⎯⎯⎯→ ......................... .

(xii) Stephan’s reduction converts alkyl cyanide into ............................ .(xiii) Benzene diazonium chloride reacts with ............................. to give benzene.(xiv) Electrolytic reduction of nitrobenzene in strongly acidic medium produces ....................... .


Answers(i) ethylamine, formic acid (ii) methylamine, Hoffmann’s degradation(iii) Less (iv) Isomers(v) Not (vi) Ester(vii) Less, more (viii) p-nitroso-2, 4-dimethyl aniline(ix) Aniline and benzoylchloride (x) Zwitter ion

(xi) —N = N— —OH (xii) Aldoximes

(xiii) hypophosphorous acid (xiv) 4-Aminophenol

Chapter 14. Biomolecules

(i) Insulin is secreted by …………… .

(ii) The three parts of a nucleotide are : a phosphate group, a base and a ………… .

(iii) The linkage holding monosaccharide units in maltose is …………… .

(iv) Primary structure of proteins refers to ……… .

(v) The sugar in RNA is …………… .

(vi) The helical structure of proteins is stabilised by …………… .

(vii) The three dimensional structure of DNA was proposed by …………… .

(viii) All enzymes are …………… .

(ix) The chemical changes in DNA molecule that could lead to synthesis of proteins with an alteredamino acid sequence is called …………… .

(x) …………… is a monosaccharide which is sweeter than sucrose.

(xi) The chemical name of vitamin C is …… .

(xii) Human body can synthesize …………… from Carotene.

(xiii) Vitamin B-complex and vitamin ………… are synthesised by micro-organisms present in theintestinal tract.

(xiv) Biotin is neither soluble in …………… nor in water.

(xv) Vitamin ………… causes impaired clotting of blood.

(xvi) Neutral amino acids have …………… in the pH range of 5·6—6·3.

(xvii) At isoelectric point the amino acids have ………… solubility in water.

(xviii) Solubility is exploited in the…………… of different α-amino acids obtained by hydrolysis of…… .

(xix) Condensation of two molecules of amino acids gives……… involving elimination of ………molecule.

(xx) Condensation of more than ten molecules of amino acids, product is …………… .

(xxi) Globular proteins function as ……………, regulates ………… processes and act as …………which protects body from diseases.

Answers(i) pancreas (ii) sugar (iii) glycoside linkage

(iv) sequence of amino acids (v) D-ribose (vi) H-bonds

(vii) James Watson and Francis Crick (viii) Globular proteins (ix) mutation

(x) Fructose (xi) ascorbic acid (xii) Vitamin A

(xiii) K (xiv) fat (xv) K(xvi) isoelectric point (xvii) least (xviii) separation, protein(xix) dipeptide, water (xx) polypeptide(xxi) enzymes, metabolic, antibodies


Chapter 15. Polymers(i) On the basis of their origin polymers are classified as ............. and........... polymers.(ii) The three natural fibres are ......................., ............ and .............

(iii) Natural rubber is a polymer of ....................... .

(iv) Phenol formaldehyde resin is commonly called ....................... .

(v) The monomer unit of PAN is ....................... .

(vi) The monomer unit of PMMA is ....................... .

(vii) Lucite is a polymer of ....................... .

(viii) The starting material of PCTEF is ....................... .

(ix) Protein is a polymer of ....................... and cellulose is a polymer of ....................... .Answers

(i) Natural, synthetic (ii) Wool, cotton and silk(iii) Isoprene (iv) Bakelite(v) Acrylonitrile (vi) Methylmethacrylate(vii) Methylmethacrylate (viii) Chlorotrifluoroethylene(ix) α-amino acids, β-glucose

Chapter 16. Chemistry in Everyday Life(i) Morphine may be used as an ………but its use should be avoided because it is ……………(ii) The antiseptic usually added to soap is …………

(iii) Aspirin act as ……………and ……………(iv) Bithional is added to soap to impart them ……………properties.(v) The first antibiotic …………was discovered in ………………

(vi) The sulpha drug usually used in pneumonia is …………(vii) Equanil is a……………

(viii) Penicillin was discovered by ……………(ix) Saccharin is a well known artificial …………agent. It is a cyclic imide of …………acid.(x) A detergent can be represented by the general formula…………where R is a large alkyl group

containing …………to …………carbon atoms.(xi) Chloramphenicol is a ……………

(xii) The birth control pills are called …………drugs.Answers

(i) analgesics, habit forming (ii) bithional(iii) antipyretics, analgesics (iv) antiseptic (v) Penicillin, 1929(vi) Sulphapyridine (vii) tranquilizer (viii) Alexander fleming

(ix) Sweetening, O-sulpho benzoic (x) ROSO2O–Na+, 12,18(xi) Broad spectrum antibiotic (xii) antifertility

❐

Question

2Set Multiple Choice Questions

Chapter 1. Solid State

1. Copper has the face centred cubic structure. The coordination number of each ion is :

(a) 4 (b) 12

(c) 14 (d) 8

2. In a crystal, the atoms are located at the position of :

(a) Maximum P.E. (b) Minimum P.E.

(c) Zero P.E. (d) Infinite P.E.3. Designation of the pattern as AB, AB, AB…… etc., of successive vertical layers of identical atoms

gives the arrangement called as :(a) Hexagonal close packing (hcp) (b) Cubic close packing (ccp)(c) Face centered cubic (fcc) (d) Body centered cubic (bcc).

4. Cubic close packing arrangement is also known as :(a) Hexagonal close packing (b) Face centered cubic

(c) Body centered cubic (d) None of these.5. In a rock salt structure each Cl– ion is surrounded by :

(a) 4 Na+ ions (b) 6 Na+ ions

(c) 8 Na+ ions (d) 12 Na+ ions.6. CsCl has which type of lattice ?

(a) SC (b) FCC(c) BCC (d) HCP

7. Which of the following is an example of covalent crystal solid ?(a) Si (b) Al(c) Ar (d) NaF

8. The solid NaCl is a bad conductor of electricity since :(a) In solid NaCl there are no ions

(b) Solid NaCl is covalent

(c) In solid NaCl there is no movement of ions

(d) In solid NaCl there are no electrons.

9. In a solid lattice the cation has left a lattice site and is located at an interstitial position, the latticedefect is :

(a) Interstitial defect (b) Valency defect

(c) Frenkel defect (d) Schottky defect.

10. Piezoelectric crystals are used in :(a) Radio (b) T.V.

(c) Record player (d) Refrigerator.

11. Schottky defect in crystal is observed when :

(a) An ion leaves its normal site and occupies the interstitial site.

(b) Equal number of cations and anions are missing from the lattice.

(c) Unequal number of cations and anions are missing from the lattice.(d) Density of the crystal is increased.


12. How many kinds of space lattice are possible in a cubic crystal ?(a) 23 (b) 7(c) 30 (d) 14

13. A solid has a structure in which ‘W’ atoms are located at the corners of a cubic lattice, ‘O’ atomsat the centre of edges and ‘Na’ atoms at the centre of the cube. The formula for the compound is :(a) NaWO2 (b) NaWO3

(c) Na2WO3 (d) NaWO4

14. In a compound, atoms of Element Y forms ccp lattice and those of Element X occupy 23 rd of

tetrahedral voids, the formula of the compound will be :(a) X4X3 (b) X2X3

(c) X2Y (d) X3X4

15. Which of the following is an example of paramagnetic solid ?(a) NaCl (b) KF

(c) TiO2 (d) CuO.16. A substance AxBy crystallizes in a face centered cubic (fcc) lattice in which atoms ‘A’ occupy each

corner of the cube, atom ‘B’ occupy the centres of each face of the cube. Identify the correctcomposition of the substance AxBy :(a) AB3 (b) A4B3

(c) A3B (d) Composition cannot be specified

Chapter 2. Solutions

1. The molal freezing point constant of water is 1·86 K kg mol–1. Therefore, the freezing point of0·1M NaCl solution in water is expected to be :(a) – 1·86°C (b) – 0·372°C(c) – 0·186°C (d) + 0·372°C

2. Of the following terms used for denoting concentration of a solution, the one which does not getsaffected by temperature is :(a) Molarity (b) Molality

(c) Normality (d) Formality3. Out of the following solutions, the one having the highest boiling point will be :

(a) 0·1 M NaCl (b) 0·1 M BaCl2

(c) 0·1 M KNO3 (d) 0·1 M K4 [Fe(CN)6]

4. For a dissociated solute in solution the value of van’t Hoff factor is :(a) Zero (b) One(c) Greater than one (d) Less than one

5. An example of intensive property is :(a) Number of moles (b) Mass(c) Volume (d) Density

6. The solubility of a gas varies directly with pressure of the gas, is based upon :(a) Raoult’s Law (b) Henry’s Law(c) Nernst’s Distribution Law (d) None of these

7. The relative lowering of vapour pressure of a solvent by the addition of a solute is :(a) Proportional to the molarity of the solution (b) Proportional to the molality of the solution(c) Equal to the mole fraction of the solute (d) Equal to the mole fraction of the solvent

8. Which solution is isotonic to the blood ?(a) 0·75% by weight of NaCl approximately (b) 0·99% by weight of NaCl approximately(c) 0·85% by weight of NaCl approximately (d) None of these

Multiple Choice Questions ■ 17

9. The osmotic pressure of equimolar solution of NaCl, BaCl2 and glucose will be in the order of :(a) NaCl > BaCl2 > Glucose (b) BaCl2 > NaCl > Glucose(c) Glucose > NaCl > BaCl2 (d) NaCl > Glucose > BaCl2

10. Which of the following is not a colligative property ? (a) Depression in freezing point (b) Elevation in boiling point(c) Osmotic pressure (d) Modification of refractive index

11. The molecular weight of sodium chloride determined by measuring the osmotic pressure of itsaqueous solution is :(a) Double the theoretical value (b) Same as the theoretical value(c) Half the theoretical value (d) Three times the theoretical value

12. Determination of correct molecular mass from Raoult’s law is applicable to :(a) An electrolyte in solution (b) A non-electrolyte in dilute solution(c) A non-electrolyte in conc. solution (d) An electrolyte in a liquid solvent.

13. The number of moles of solute present in 1000 gm of the solvent is known as :(a) Molarity (b) Molality(c) Normality (d) Mole fraction.

14. Which of the following 0·1 M aqueous solution will have the lowest freezing point ?(a) Potassium sulphate (b) Sodium chloride(c) Urea (d) Glucose.

15. Colligative properties of the solution depends on :(a) Nature of solute (b) Nature of solvent(c) Number of particles present in the solution (d) Number of moles of solvent only.

16. Solution that obey’s Raoult’s law :(a) Normal (b) Molar(c) Ideal (d) Saturated

17. A liquid pair of benzene-toluene shows :(a) Positive deviation from Raoult’s law. (b) Negative deviation from Raoult’s law.(c) Practically no deviation from Raoult’s law (d) Irregular deviation from Raoult’s law.

18. When 0·1 mole urea is dissolved in 9·9 mole water, then the vapour pressure is :(a) Increased by 1% (b) Increased by 10%(c) Decreased by 1% (d) Decreased by 10%.

19. 12 gm of urea is dissolved in 1 litre of water and 68·4 gm of sucrose is dissolved in 1 litre ofwater. The lowering of vapour pressure of first reaction is :(a) Equal to second (b) Greater than second(b) Less than second (d) Double that of second.

20. Molecular weight of non-volatile solute can be determined by :(a) Victor-Mayer’s method (b) Graham’s law of diffusion(c) Gay Lussac’s law (d) Raoult’s law.

21. The molal elevation constant is the ratio of the elevation in boiling point to :(a) Molarity (b) Molality(c) Mole fraction of solute (d) Mole fraction of solvent

22. Osmotic pressure of a dil. solution is given by :(a) P = P0x (b) ππππV = nRT(c) π = VRT (d) None.

23. For a dissociated solute in solution the value of van’t Hoff factor is :(a) Zero (b) One(c) Greater than one (d) Less than one.

24. The elevation in boiling point of a solution of 13·44 gm of CuCl2 in 1kg of H2O using thefollowing information will be : (Mol. wt. of CuCl2 = 134·4 and Kb = 0·52 K g. mol–1)(a) 0·16 (b) 0·05(c) 0·1 (d) 0·2



1. Electrochemical equivalent is the amount of substance which gets deposited from its solution onpassing electrical charge equal to :

(a) 96,500 coulomb (b) 1 coulomb(c) 60 coulomb (d) 965 coulomb

2. A current liberates 0·50 g of hydrogen in 2 hours. The weight of copper (at. wt. = 63·5) depositedat the same time by the same current through copper sulphate solution is :(a) 63·5 g (b) 31·8 g(c) 15·9 g (d) 15·5 g

3. The quantity of electricity required to deposit 1·15 g of sodium from molten NaCl (Na = 23,Cl = 35·5) is :

(a) 1 F (b) 0·5 F

(c) 0·05 F (d) 1·5 F

4. When zinc granule is dipped into copper sulphate solution, copper is precipitated because :

(a) Both, copper and zinc have a positive reduction potential(b) Reduction potential of copper is higher than that of zinc(c) Reduction potential of zinc is higher than that of copper(d) Both, zinc and copper have a negative reduction potential

5. The unit of equivalent conductance is :(a) ohm–1cm2equiv -1 (b) ohm–1cm2gm–1

(c) ohm cm2equiv–1 (d) ohm–1mole–1

6. The number of Faradays required to reduce one mol of Cu+2 to metallic copper is :(a) One (b) Two

(c) Three (d) Four7. Conductivity of a solution is directly proportional to :

(a) Dilution (b) Number of ions

(c) Current density (d) Volume of the solution8. Electrolysis involves oxidation and reduction respectively at :

(a) Anode and cathode (b) Cathode and anode(c) At both the electrode (d) None of the above

9. All galvanic cell do not contain :(a) A cathode (b) An anode(c) Ions (d) A porous plate

10. The reaction is spontaneous if the cell potential is :(a) Positive (b) Negative(c) Zero (d) Infinite

11. Which of the following cells can convert chemical energy of H2 and O2 directly into electricalenergy ?(a) Mercury cell (b) Daniel cell(c) Fuel cell (d) Lead storage cell

12. When lead storage battery discharges ?(a) SO2 is evolved (b) PbSO4 is consumed(c) Lead is formed (d) Sulphuric acid is consumed

13. The electrode Pt, Hg (g) /HCl is reversible with respect to :(a) Cl– ions (b) HCl(c) H+ ions (d) Both H+ and Cl– ions


14. Saturated solution of KNO3 is used to make salt bridge because :(a) Velocity of K+ is greater than that of NO3

–

(b) Velocity of NO3– is greater than that of K+

(c) Velocity of both K+ and NO3– are nearly the same

(d) KNO3 is highly soluble in water15. Zn/Zn2+ (a = 0·1M) ||Fe2+ (a = 0·1M)/Fe. The e.m.f. of the above cell is 0·290 V. Equilibrium

constant for the cell reaction is :(a) 100·32/0·0991 (b) 100·32/0·0295

(c) 100·26/0·0295 (d) e0·32/0·295

16. The standard electrode potentials of four elements A, B, C and D are – 3·05, 1·66, – 0·40 and 0.80volts respectively. The highest chemical activity will be shown by :(a) A (b) B(c) C (d) D


1. For a first order reaction the rate constant for decomposition of N2O5 is 6 × 10– 4 sec– 1. The half-life period for the decomposition in seconds is :(a) 11·55 (b) 115·5(c) 1155 (d) 1·155.

2. In a plot of log k vs 1/T, the slope is :(a) – Ea/2·303 (b) Ea/2·303 R(c) Ea/2·303 (d) – Ea/2·303 R

3. 75% of a first order reaction was completed in 32 minutes. When was 50% of the reactioncompleted ?(a) 24 minutes (b) 16 minutes(c) 8 minutes (d) 4 minutes

4. The reaction between X and Y is first order with respect to X and second order with respect to Y.If the concentration of X is halved and the concentration of Y is doubled, the rate of the reactionwill be :(a) Same as the initial value (b) Three times the initial value(c) Double the initial value (d) Half the initial value.

5. The rate of a chemical reaction :(a) Increases as the reaction proceeds(b) Decrease as the reaction proceeds

(c) May increase or decrease during the reaction(d) Remains constant as the reaction proceeds.

6. The correct order indicating against the rate of reaction A + B K

⎯⎯→ is :

(a)d [c]dt = k [A] (b) –

d [c]dt = k [B]

(c)– d [c]

dt = k [A] [B] (d) –

d [A]dt = k [A].

7. The one which is unimolecular reaction is :(a) 2HI → H2 + I2 (b) N2O5 →→→→ N2O4 + 1/2 O2

(c) H2 + Cl2 → 2HCl (d) PCl3 + Cl2 → PCl58. If the concentration is expressed in moles per litre the unit of the rate constant for a first order

reaction is :(a) Mole litre– 1 sec– 1 (b) Mole litre– 1

(c) sec– 1 (d) Mole– 1


9. The rate of a chemical reaction is double for every 10°C rise in temperature because of :(a) Increase in the activation energy(b) Decrease in the activation energy(c) Increase in the number of molecular collisions(d) Increase in the number of activated molecules.

10. The minimum energy required by reacting molecules to permit a reaction is :(a) Internal energy (b) Threshold energy(c) Activation energy (d) Free energy

11. A quantitative relationship between the temperature and rate constant of a reaction is given by :(a) Nernst equation (b) Arrhenius equation(c) Van’t Hoff equation (d) Henderson equation

12. For the first order reaction with rate constant K, which expression gives the half-life period ?

(a)ln2

k (b)1

ka

(c)0·693

k(d)

32ka2

13. The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at25°C are 3·0 × 10– 4 s– 1, 104·4 kJ mol– 1 and 6·0 × 1014 s– 1 respectively. The value of the rateconstant as T → ∞ is :(a) 2·0 × 1018 s– 1 (b) 6·0 × 1014 s– 1

(c) Infinity (d) 3·6 × 1030 s– 1

14. The rate constant for the reaction 2N2O5 → 4NO2 + O2, is 3·0 × 10– 5 s– 1. If the rate is 2·40 × 10– 5

mol L– 1 s– 1, then the concentration of N2O5 (in mol L– 1) is :(a) 1·4 (b) 1·2(c) 0·04 (d) 0·8

15. For the reaction 2SO2 + O2 2SO3, the unit of equilibrium constant is :(a) L mol– 1 (b) J mol– 1

(c) mol L– 1 (d) [L mol– 1]2

16. Consider a reaction aG + bH → products. When concentration of both the reactants G and H isdoubled, the rate increases by eight times. However, when concentration of G is doubled keepingthe concentration of H fixed, the rate is doubled. The overall order of the reaction is :

(a) 0 (b) 1(c) 2 (d) 3

17. A catalyst :

(a) Increase the free energy change in the reaction

(b) Decrease the free energy change in the reaction

(c) Does not increase or decrease the free energy change in the reaction

(d) Can either increase or decrease the free energy change depending on what catalyst we use.

18. The effect of a catalyst in a chemical reaction is to change the :

(a) Activation energy (b) Equilibrium concentration

(c) Heat of reaction (d) Final products

19. An example of autocatalytic reaction is :

(a) The decomposition of nitroglycerine

(b) Thermal decomposition of KClO3 and MnO2 mixture

(c) Break down of 14C

(d) Hydrogenation of vegetable oil using nickel catalysts


20. In which of the following commercial process a catalyst is not used ?

(a) Haber’s process (b) Deacon’s process

(c) Solvay process (d) Lead Chamber process

21. A biological catalyst is :

(a) An amino acid (b) A carbohydrate

(c) The nitrogen molecule (d) An enzyme

Chapter 5. Surface Chemistry1. Blue colour of water in sea is due to :

(a) Refraction of blue light by impurites in sea water(b) Scattering of light by water(c) Refraction of blue sky by water(d) None of these

2. The coagulation power of an electrolyte for arsenious sulphide sol decreases in the order :

(a) Na+ > Al3+ > Ba2+ (b) PO43– > SO4

2– > Cl–

(c) Cl– > SO4

2– > PO43–

(d) Al3+

> Ba2+ > Na+

3. Volume of a colloidal particle Vc as compared to the volume of a solute particle in a true solution,VS could be :

(a)VC

VS ≈ 1 (b)

VC

VS ≈ 1

(c)VC

VS ≈ 10–3

(d)VC

VS ≈≈≈≈ 10

3

4. Lyophilic sols are :(a) Reversible sols(b) They are prepared from inorganic compound(c) Coagulated by adding electrolytes(d) Self-stabilizing

5. A plot of log x/m versus log P for the adsorption of a gas on a solid gives a straight line withslope equal to :(a) n (b) 1/n(c) log K (d) – log K

6. On adding 1 mL of solution of 10% NaCl to 10 mL of gold sol in the presence of 0·25 g of starchthe coagulation is just prevented the gold number of starch is :(a) 0·25 (b) 0·025(c) 250 (d) 25

7. Hair cream is an example of :(a) Gel (b) Sol(c) Foam (d) Emulsion

8. Which of the following impurities present in colloidal solution cannot be removed byelectrodialysis ?(a) Sodium chloride (b) Potassium sulphate(c) Urea (d) Calcium chloride

9. The ion that is more effective for coagulation of As2S3 sol in :(a) Ba2+ (b) Na+

(c) Al3+ (d) SO4

2–

10. The most adsorbed gas on activated charcoal is :(a) N2 (b) H2

(c) CO2 (d) CH4


11. The charge of Fe (OH)3 sol is due to :(a) Adsorption of hydroxyl ion (b) Adsorption of hydroxyl ion(c) Absorption of ferric acid (d) Adsorption of ferric acid

12. Which one of the following does not involve coagulation ?(a) Formation of delta region(b) Peptization(c) Treatment of drinking water by potash alum(d) Clotting of blood by the use of ferric chloride

13. The dispersed phase and dispersion medium in soap lather are respectively :(a) Gas and liquid (b) Liquid and gas(c) Solid and gas (d) Solid and liquid

14. Milk is an example of :(a) W/O type of emulsion (b) O/W type of emulsion(c) W/W type of emulsion (d) O/O type of emulsion

15. The disease kala azar is cured by :(a) Colloidal antimony (b) Milk of magnesia(c) Argyrols (d) Colloidal gold

16. According to Freundlich adsorption isotherm, which of the following is correct ?

(a)xm

∝ P′ (b)xm

∝ P1/n

(c)xm

∝ P°

(d) All the above are correct for different ranges of pressure

17. Identify the positively charged sol. :(a) Haemoglobin (blood) (b) Clay(c) As2S3 (d) Gold sols.

Chapter 6. General Principles and Processes of Isolation of Elements1. The ore chromite is :

(a) FeCr2O4 (b) CoCr2O3

(c) CrFe2O4 (d) FeCr2O3

2. When lime stone is heated, CO2 is given off. The metallurgical operation is :(a) Smelting (b) Reduction(c) Calcination (d) Roasting

3. The process of zone refining is used in the purification of :(a) Al (b) Ge(c) Cu (d) Ag

4. Identify the alloy containing a non-metal as a constituent in it :(a) Invar (b) Steel(c) Bell metal (d) Bronze

5. Aluminium is extracted from alumina (Al2O3) by electrolysis of a molten mixture of :(a) Al2O3 + HF + NaAlF4 (b) Al2O3 + CaF2 + NaHF4

(c) Al2O3 + Na3AlF6 + CaF2 (d) Al2O3 + KF + Na3AlF6

6. The ore magnetite is :(a) Fe3O4 (b) ZnCO3

(c) CuCO3.Cu(OH)2 (d) FeS2

7. Which of the following is not a sulphide ore ?(a) Magnetite (b) Iron pyrites(c) Copper glance (d) Galena


8. Duralumin is used in aircraft industry for its light weight and high strength. It is an alloy of :(a) Al, Cu, Mg and Mn (b) Al, Zn, Fe and Sn(c) Al, Ti, Ce and Fe (d) Al, Fe, Zn and Sn

9. Which of the following pair of metals is purified by Van-Arkel method ?(a) Ga and In (b) Zr and Ti(c) Ag and Au (d) Ni and Fe

10. Which of the following element is present as the impurity to the maximum extent in the pig ion ?(a) Manganese (b) Carbon(c) Silicon (d) Phosphorus

11. Which one of the following is an oxide ore ?(a) Malachite (b) Copper glance(c) Haematite (d) Zinc blende

12. In the equation 4M + 8CN– + 2H2O + O2 ⎯⎯→ 4 [M(CN)2]– + 4OH– the metal M is :(a) Copper (b) Iron(c) Gold (d) Zinc

13. For which ore of the metal, froth floatation process is used for concentration ?(a) Horn silver (b) Bauxite(c) Cinnabar (d) Haematite

14. Which of the following metal is leached by cyanide process ?(a) Ag (b) Na(c) Al (d) Cu

15. Which of the following ore is best concentrated by froth floatation process ?(a) Magnetite (b) Cassiterite(c) Galena (d) Malachite

16. Which one contains both iron and copper ?(a) Cuprite (b) Chalcocite(c) Malachite (d) Copper pyrites

17. In aluminothermic process, Al is used as :(a) Reducing agent (b) Oxidising agent(c) Catalyst (d) Electrolyte

18. The temperature of the slag zone in the metallurgy of iron using blast furance is :(a) 1500-1600 °C (b) 400-700 °C(c) 800-1000 °°°°C (d) 1200-1500 °C

19. Heating Cu2O and Cu2S will give :(a) Cu + SO2 (b) Cu + SO3

(c) CuO + CuS (d) Cu2SO3

20. Δ°G vs T plot in the Ellingham’s diagram sloped downward for the reaction :

(a) Mg + 12 O2 ⎯→ MgO (b) 2Ag +

12 O2 ⎯→ Ag2O

(c) C + 12 O2 ⎯⎯⎯⎯→→→→ CO (d) CO +

12 O2 ⎯→ CO2

Chapter 7. p-Block Elements1. The geometry of XeF6 molecule and the hybridization of Xe atom in the molecule is :

(a) Distorted octahedral and sp3d3 (b) Square planar and sp3d2

(c) Pyramidal and sp3 (d) Octahedral and sp3d3

2. Among the following halogens, the one which does not forms an oxyacid is :(a) Fluorine (b) Chlorine(c) Bromine (d) Iodine


3. The tendency of group 16 elements to form catenated compounds is greatest in case of :(a) Oxygen (b) Sulphur

(c) Selenium (d) Tellurium4. The minimum bond angle in hydrides of group 16 elements is in :

(a) H2O (b) H2Te

(c) H2Se (d) H2S5. Which of the following has lowest reducing character ?

(a) H2O (b) H2S(c) H2Te (d) H2Se

6. Which of the following mainly exhibits – 2 oxidation state ?(a) S (b) O

(c) Se (d) Te7. Oxygen molecule is :

(a) Paramagnetic (b) Diamagnetic(c) Ferromagnetic (d) Ferrimagnetic

8. The halogen with highest electron affinity :(a) F (b) Cl

(c) Br (d) I

9. The halide ion easiest to oxidise is :

(a) F – (b) Cl–

(c) Br– (d) I–

10. The reaction, 3ClO– (aq) → ClO–3 (aq) + 2Cl– (aq), is an example of :

(a) Oxidation reaction (b) Reduction reaction

(c) Disproportionation reaction (d) Decomposition reaction

11. The low bond energy is best explained by :

(a) The attainment of noble gas configuration (b) The low electron affinity of F

(c) Repulsion by electron pairs on F (d) The small size of F

12. The oxo-acid of halogen with maximum acidic character is :

(a) HClO4 (b) HClO3

(c) HClO2 (d) HClO

13. Which of the following reaction will not occur spontaneously ?

(a) F2 + 2Cl– ⎯→ 2F– + Cl2 (b) I2 + 2Br– ⎯⎯⎯⎯→→→→ 2I

– + Br2

(c) Br2 + 2I– ⎯→ 2Br– + I2 (d) 2I– + Cl2– ⎯→ 2Cl– + I2

14. The high viscosity and high boiling point of HF is due to :

(a) Low dissociation energy of F2 molecule

(b) Associated nature due to hydrogen bonding

(c) Ionic character of HF

(d) High electronegativity of fluorine

15. The most powerful oxidising agent is :

(a) Fluorine (b) Chlorine

(c) Bromine (d) Iodine

16. Shape of ClF3 is :

(a) Trigonal planar (b) Tetrahedral

(c) T-Shaped (d) Distorted octahedral


17. Which of the following is not an interhalogen compound ?

(a) ICl4– (b) ClF5

(c) IPO4 (d) ClF3

18. Which noble gas was discovered in chromosphere ?(a) He (b) Ar(c) Xe (d) Rn

19. XeF2 molecule is :

(a) Trigonal planar (b) Square planar

(c) Linear (d) Pyramidal

20. The noble gas used in the treatment of Cancer is :

(a) Argon (b) Xenon

(c) Radon (d) Helium

21. In XeF2, Xenon involves the hybridisation :

(a) sp (b) sp2

(c) sp3d (d) sp3

22. Which one of the following displaces Bromine from an aqueous solution of Bromine ?

(a) Cl2 (b) Cl–

(c) I2 (d) I3–

23. When SO2 gas is passed through acidified K2Cr2O7 solution, the colour of the solution changesto :(a) Red (b) Black(c) Orange (d) Green

24. Which of the following reagent does not give O2 gas on reaction with Ozone ?(a) KMnO4 (b) SnCl2/HCl(c) FeSO4/H2SO4 (d) PbS

25. Aqua Regia is a mixture of :(a) Conc HNO3 and conc. H2SO4

(b) Conc HCl and conc. H2SO4 in the ratio of 3 : 1(c) Conc. HCl and conc. HNO3 in the ratio of 3 : 1(d) None of these

26. Tailing of mercury is due to the formation of :(a) Hg2O (b) HgO(c) Hg(OH)2 (d) HgCl2

27. Which one is called oleum ?(a) Liq. NH3 (b) H2SO4 + SO3

(c) Conc. HNO3 (d) Dilute solution of H2O2

28. Which one absorbs U.V. radiation in stratosphere ?(a) CO2 (b) N2

(c) O3 (d) H2

Chapter 8. d-and f-Block Elements1. The general outer electronic configuration of transition elements is :

(a) (n – 1) d1-10 ns1 (b) (n – 1) d10 ns2

(c) (n – 1) d1-10 ns1 – 2 (d) (n – 1) d5 ns1

2. Which one of the following does not show different oxidation states ?(a) Iron (b) Copper(c) Zinc (d) Manganese


3. Which of the following is not a condition for complex salt formation ?(a) Small size (b) Higher nuclear charge(c) Availability of vacant d-orbitals (d) Variable oxidation states

4. Which of the following metal is used in incandescent lamps ?(a) Chromium (b) Tungsten(c) Zirconium (d) Molybdenum

5. The outer electronic configuration of chromium is :(a) 4s1 3d5 (b) 4s2 3d4

(c) 4s0 3d6 (d) 4s2 3d 5

6. Which ion gives coloured solution ?(a) Cu+ (b) Zn2+

(c) Ag+ (d) Fe2+

7. Paramagnetism is a property of :(a) Completely filled electronic sub-shells (b) Unpaired electrons

(c) Non-transition elements (d) Melting point and boiling point of elements.8. Which transition metal has the highest density ?

(a) Os (b) Zn(c) Sc (d) La

9. Which transition metal shows highest oxidation state ?(a) Sc (b) Ti(c) Os (d) Zn

10. In KMnO4 oxidation number of Mn is :(a) + 2 (b) + 4(c) + 6 (d) + 7

11. The most common oxidation state of lanthanoids is :(a) + 4 (b) + 3(c) + 6 (d) + 2

12. Which element among the lanthanoids has the smallest atomic radius ?(a) Cerium (b) Lutetium

(c) Europium (d) Gadolinium13. Which of the following oxides of chromium is amphoteric in nature ?

(a) CrO (b) Cr2O3

(c) CrO3 (d) CrO5

14. Which compound of chromium is widely used in tanning of leather ?(a) CrCl3 (b) Cr2O3

(c) CrO2Cl2 (d) K2SO4·Cr2(SO4)3·24H2O

15. Which of the following element belongs to actinoid series ?(a) La (b) Gd(c) Lu (d) Th

16. Number of moles of K2Cr2O7 reduced by one mole of Sn2+ ions is :(a) 1/3 (b) 3(c) 1/6 (d) 6

17. The lanthanoid contraction is responsible for the fact that :(a) Zr and Y have about the same radius (b) Zr and Nb have similar oxidation state(c) Zr and Hf have about the same radius (d) Zr and Zn have the same oxidation state

18. Which lanthanoid is most commonly used ?(a) Lanthanum (b) Nobelium(c) Thorium (d) Cerium


19. Ammonium dichromate is use in some fireworks. The green coloured powder blown in air is :(a) CrO3 (b) Cr2O3

(c) Cr (d) CrO(O2)2

20. Which of the following statement is not correct ?(a) La(OH)3 is less basic than Lu(OH)3

(b) La is actually an element of transition series rather than lanthanoids(c) Atomic radii of Zr and Hf are same because of lanthanoid contraction(d) In lanthanid series the ionic radius of Lu3 + is smallest

Chapter 9. Coordination Compounds1. In the complexes [Fe(CN)6]3– and [Pt(en) (H2O)2(NO2) (Cl)]2+ the respective oxidation numbers of

central metal atoms are :(a) + 3 and + 4 (b) + 6 and + 4(c) + 6 and + 3 (d) + 3 and + 3

2. The complex ion [Ni (CN)4]2– is :(a) Square planar and diamagnetic (b) Tetrahedral and paramagnetic(c) Square planar and paramagnetic (d) Tetrahedral and diamagnetic

3. Among the following coordination compounds, the one giving a white precipitate with BaCl2solution is :(a) [Cr(H2O)5Br]SO4 (b) [Cr(H2O)5SCN](c) [Co(NH3)5SO4]Br (d) [Pt(NH3)6]Cl4

4. The hybridization of the iron atom in [Fe(CN)6]-3 complex is :(a) sp3 (b) d2sp3

(c) sp3d2 (d) dsp2

5. Which of the following statement about primary and secondary valencies is true ?(a) Both primary and secondary valency are ionisable(b) Both primary and secondary valencies are non-ionisable(c) Primary valencies are ionisable while secondary are non-ionisable(d) Primary valencies are non-ionisable while secondary are ionisable

6. What is the oxidation number of central atom in Na [Hg (CN)2] ?(a) +4 (b) +2(c) 0 (d) +1

7. What is the coordination number of central metal atom in [Pt (NH3)2Cl2] Cl ?(a) 6 (b) 4(c) 3 (d) 7

8. Which of the following is π acid ligand ?(a) NH3 (b) CO

(c) F– (d) Ethylene diamine9. Which of these compound does not show paramagnetism ?

(a) [Cu(NH3)4]Cl2 (b) [Ag(NH3)2]Cl

(c) NO (d) NO2

10. The geometry of Ni(CO)4 and [Ni(CN)4]2– are :(a) Both square planar(b) Tetrahedral and square planar respectively(c) Both tetrahedral(d) Square planar and tetrahedral respectively

11. Which of the following compounds show optical isomerism ?(a) [Co(CN)6]3– (b) [Cr (C2O4)3]3–

(c) [ZnCl4]2– (d) [Cu(NH3)4]2+


12. The correct order of hybridization of the central atom in the following species NH3, [PtCl4]2–,PCl5 and BCl3 is :(a) dsp2, dsp3, sp2 and sp3 (b) sp3, dsp2, dsp3, sp

2

(c) dsp2, sp2, sp3 dsp3 (d) dsp2, sp3, sp2, dsp3

13. Which is not true about the coordination compound [Co(en)2Cl2]Cl ?(a) Exhibits geometrical isomerism (b) Exhibits optical isomerism(c) Exhibits ionization isomerism (d) Is an octahedral complex

14. In Fe(CO)5, the Fe–C bond possesses :(a) π–character only (b) Both σσσσ and ππππ characters

(c) Ionic character (d) σ–character only15. The IUPAC name for the complex [Co(NO2) (NH3)5]Cl2 is :

(a) Nitrito–N–pentaammine cobalt (III) chloride(b) Nitrito–N–pentaammine cobalt (II) chloride(c) Pentaammine nitrito–N–cobalt (II) chloride(d) Pentaammine nitrito–N– cobalt (III) chloride


1. Which of the following poisonous gas is formed when chloroform is exposed to light and air ?

(a) Mustard gas (b) Carbon monoxide

(c) Phosgene (d) Chlorine

2. What is the IUPAC name of CH3—

CH3|C—|CH3

CH2Cl ?

(a) 2-dimethylchloropropane (b) 1-chloro-2-dimethyl-pentane(c) 2, 2,-dimethyl-chlorobutane (d) 1-chloro-2, 2-dimethyl propane

3. Halogenation of alkane gives :(a) Only required alkyl halide(b) Alkyl halide and unreacted halogen(c) A mixture of mono-, di-, tri- and tetra-halogen derivatives

(d) Alkyl halide and unreacted alkane

4. Which of the following compound has been suggested as causing depletion of the ozone layer inthe upper stratosphere ?(a) CH4 (b) CCl2F2

(c) CF4 (d) CH2Cl2

5. Which of the following reagent cannot be used to prepare an alkyl chloride from an alcohol ?

(a) HCl + ZnCl2 (b) SOCl2

(c) NaCl (d) PCl5

6. Alkyl halides undergo :

(a) Electrophilic substitution reactions (b) Electrophilic addition reactions

(c) Nucleophilic substitution reactions (d) Nucleophilic addition reactions

7. Carbylamine test involves heating a mixture of :

(a) Alcoholic KOH, methyl iodide, and sodium metal

(b) Alcoholic KOH, methyl iodide, and primary amine

(c) Alcoholic KOH, chloroform, and primary amine

(d) Alcoholic KOH, methyl alcohol, and primary amine


8. When chloroform is heated with aqueous NaOH, it gives :

(a) Formic acid (b) Sodium formate

(c) Acetic acid (d) Sodium acetate

9. Which alkyl halides react most readily by nucleophilic substitution ?

(a) CH3CH2Cl (b) CH3CH2I

(c) CH3CH2Br (d) CH3CH2F

10. The action of sodium on alkyl halide to form an alkane is called :

(a) Grignard reaction (b) Wurtz coupling reaction

(c) Isocyanide reaction (d) Halogenation reaction

11. Conversion of ethyl bromide to ethylene is an example of :

(a) Hydrohalogenation (b) Intramolecular dehydrohalogenation

(c) Dehydration (d) Hydration

12. Which of the following compound is an organometallic compound ?

(a) CH3COOAg (b) CH3MgI

(c) MgCl2 (d) CH3—O—Na

13. The reaction,

2C2H5Br + 2Na dry ether

⎯⎯⎯⎯⎯→ C2H5—C2H5 + 2NaBr is an example of :(a) The Wurtz reaction (b) Sandmeyer’s reaction(c) Aldol condensation (d) Williamson’s reaction

14. Grignard’s reagent is prepared by the action of magnesium metal on :(a) Alcohol (b) Phenol(c) Alkyl halide (d) Benzene

15. The major product of the following reaction is :

Cl

+ CH3COCl Anhydrous

AlCl3

⎯⎯⎯⎯⎯→

(a)

ClCOCH3

(b)COCH3

Cl

(c)

Cl

COCH3

(d)

ClCOCH3

COCH3

16.

CH3

ClCH3

Cl

2

Cl

+ 2HCl+ 2CH3Cl +Anhyd. AlCl3⎯⎯⎯⎯⎯→

The above reaction is known as :(a) Wurtz-Fittig reaction (b) Friedel Craft’s reaction

(c) Sandmeyer’s reaction (d) Swaits reaction17. p-p-dichlorodiphenyl trichloroethane is used as :

(a) Insecticide (b) Anaesthetic(c) Antiseptic (d) Refrigerant


18. The following compound is called :

Cl CH

CCl3

Cl

(a) Chloral (b) DDT

(c) Lindane (d) BHC

19. DDT is prepared by the reaction of chlorobenzene with (in the presence of conc. H2SO4) :

(a) Chloral (b) Chlorine

(c) Chloroform (d) Carbon tetrachloride


1. When acetaldehyde is treated with Grignard reagent, followed by hydrolysis the product formedis :

(a) Primary alcohol (b) Secondary alcohol

(c) Carboxylic acid (d) Tertiary alcohol.

2. When oxalic acid is heated with glycerol we get :

(a) Formic acid (b) Acetic acid

(c) Lactic acid (d) Tartaric acid

3. Ethanol on heating with conc. H2SO4 at 445 K gives :

(a) Diethyl sulphate (b) Ethylene, C2H4

(c) Diethyl ether, (C2H5)2O (d) Ethyl hydrogensulphate, C2H5HSO4

4. Which of the following is most acidic ?

(a) H2O (b) CH3OH

(c) C2H5OH (d) CH3CH2CH2OH

5. Which of the following has highest boiling point ?

(a) CH3CH2CH2OH (b) CH3CH2CH2CH2OH

(c)CH3 CH3

CH—CH2OH (d) CH3—

CH3|C—|CH3

OH

6. C6H5Cl NaOH (aq)

⎯⎯⎯⎯⎯⎯→623 K‚ 300 atm

A. Here, A is :

(a) Phenol (b) Sodium phenoxide(c) Benzene (d) Cyclohexyl chloride

7. Which one of the following will produce a primary alcohol by reacting with CH3MgI ?(a) Acetone (b) Methyl cyanide(c) Ethylene oxide (d) Ethyl acetate

8. In the sequence HO SO3H Br2⎯⎯→

H2O X, X is :

(a) 2-Bromo-4-hydroxybenzene sulphonic acid

(b) 3, 5-Dibromo-4-hydroxybenzene sulphonic acid

(c) 2-Bromophenol (d) 2, 4, 6-Tribromophenol

9. Phenol can be distinguished from ethyl alcohol by all reagents except :

(a) NaOH (b) FeCl3

(c) Br2/H2O (d) Na


10. Alcohols can be obtained from all methods except :(a) Hydroboration-oxidation(b) Oxymercuration-demercuration(c) Reduction of aldehyde/ketones with Zn-Hg/HCl(d) By fermentation of starch

11. Chlorine reacts with ethanol to give :(a) Diethyl chloride (b) Chloroform(c) Acetaldehyde (d) Chloral

12. Which of the following alcohol is least soluble in water ?(a) N-Butyl alcohol (b) Iso-Butyl alcohol(c) Tert-Butyl alcohol (d) Sec-Butyl alcohol

13. Glycerol on heating with potassium bisulphate yields :(a) Acetone (b) Glyceraldehyde(c) Acrolein (d) Propanol.

14. The reaction of Lucas reagent is fastest with :(a) (CH3)3COH (b) (CH3)2CHOH(c) CH3(CH2)2OH (d) CH3CH2OH

15. The ionization constant of phenol is higher than that of ethanol because :(a) Phenoxide ion is a stronger base than ethoxide ion(b) Phenoxide ion is stabilized through delocalization(c) Phenoxide ion is less stable than ethoxide ion(d) Phenoxide ion is bulkier than ethoxide ion

16. The correct order of boiling points for primary (1°), secondary (2°) and tertiary alcohol (3°) is :(a) 1° > 2° > 3° (b) 3°°°° > 2°°°° > 1°°°°

(c) 2° > 1° > 3° (d) 2° > 3° > 1°.17. Which of the following is the most suitable method for removing the traces of water from

ethanol ?(a) Heating with Na metal (b) Passing dry HCl gas through it

(c) Distilling Cl

_(d) Reacting with Mg

18. Phenol is heated with CHCl3 and alcoholic KOH when salicylaldehyde is produced. This reactionis known as :(a) Rosenmund’s reaction (b) Reimer-Tiemann reaction(c) Friedel-Crafts reaction (d) Sommelet reaction

19. Lucas test is used for distinction of :(a) Alcohols (b) Phenols(c) Alkyl halides (d) Aldehydes

20.OH

+ CHCl3 + NaOH ⎯→

CHO

O Na– +

The electrophile involved in the above reaction is :

(a) Dichloromethyl cation (+CHCl2) (b) Dichlorocarbene (: CCl2)

(c) Trichloromethyl anion (–CCl3) (d) Formylcation (

⊕CHO)

21. When phenol is treated with excess of Bromine water, it gives :(a) m–bromophenol (b) o and p–bromophenol(c) 2, 4–dibromophenol (d) 2, 4, 6–tribromophenol

22. Which of the following is simple ether ?

(a) C2H5OCH3 (b) CH3OCH3

(c) C6H5OCH3 (d) All are simple ethers.


Chapter 12. Aldehydes, Ketones and Carboxylic acids

1. The compound which gives a positive Haloform test and a positive Fehling solution test is :(a) Acetone (b) Acetaldehyde

(c) Formaldehyde (d) Diethyl ether2. Benzaldehyde, when heated with an alcoholic solution of potassium cyanide, forms :

(a) Benzyl alcohol (b) Benzoin

(c) Hydrobenzamide (d) Benzoic acid3. Which of the following reagent can be used to prepare ketone from acid chloride ?

(a) Grignard’s reagent (b) LiAlH4

(c) Dimethyl cadmium (d) Cadmium chloride4. Which of the following reaction cannot be used for the reduction of

R R

C = O ⎯⎯→ R R

CH2

(a) Clemmensen reaction (b) Wolff-Kishner reaction

(c) Wurtz reaction (d) HI and red phosphorus

5. Benzaldehyde can be prepared by the hydrolysis of :

(a) Benzyl chloride (b) Benzal chloride

(c) Benzotrichloride (d) Benzo nitrite

6. The reaction, C6H5COCl + H2 Pd

⎯⎯→BaSO4

C6H5CHO + HCl, is called–

(a) Rosenmund’s reaction (b) Sandmeyer’s reaction

(c) HVZ reaction (d) Cannizzaro’s reaction

7. The end product ‘C’ in the following sequence of chemical reaction is :

CH3COOH CaCO3⎯⎯⎯→ A

heat⎯⎯⎯→ B

NH2OH⎯⎯⎯→ C

(a) Acetaldehyde oxime (b) Formaldehyde oxime(c) Methyl nitrate (d) Acetoxime

8. Which of the following statements about benzaldehyde is/are true ?(a) Reduces Tollen’s reagent

(b) Undergoes aldol condensation(c) Undergoes Cannizzaro reaction(d) Does not form an addition compound with sodium hydrogen sulphite

9. Which one of the following aldehyde gives Cannizzaro reaction when heated with strongalkali ?(a) Benzaldehyde (b) Acetaldehyde(c) Propionaldehyde (d) All the above

10. In the Cannizzaro reaction given below,

2Ph—CHO OH

–⎯⎯⎯→ Ph—CH2OH + PhCO2

–,the slowest step is :(a) The attack of —OH at the carbonyl group(b) The transfer of hydride ion to the carbonyl group

(c) The abstraction of proton from the carboxylic acid(d) The deprotonation of Ph—CH2OH.

11. Acetaldehyde cannot show :(a) Iodoform test (b) Lucas test

(c) Benedict’s test (d) Tollen’s test


12. Iodoform test is not given by :(a) 2-Pentanone (b) 3-Pentanone

(c) Ethanal (d) Ethanol13. The reaction in which sodium cyanide is used

(a) Perkin condensation (b) Reimer-Tiemann reaction(c) Benzoin condensation (d) Rosenmund’s reduction

14. Which one of the following reaction is a method for the conversion of a ketone into ahydrocarbon ?

(a) Aldol condensation (b) Reimer-Tiemann reaction

(c) Cannizzaro reaction (d) Wolff-Kishner reduction

15. The IUPAC name of the compound having the formula, Cl3CCH2CHO is :(a) 3, 3, 3-Trichloropropanal (b) 1, 1, 1-Trichloropropanal(c) 2, 2, 2-Trichloropropanal (d) Chloral

16. When ethanal is treated with Fehling’s solution, it gives precipitate of :(a) Cu (b) CuO(c) Cu2O (d) Cu2O + Cu2O3

17. A new carbon-carbon bond formation is possible in :(a) Cannizzaro reaction (b) Friedel-Crafts reaction(c) Clemmensen reduction (d) Reimer-Tiemann reaction

18. The reduction of benzoyl chloride with Pd and BaSO4 produces :(a) Benzoyl chloride (b) Benzaldehyde

(c) Benzoic acid (d) None of above19. Which of the following cannot reduce Fehling’s solution ?

(a) Formic acid (b) Acetic acid(c) Formaldehyde (d) Acetaldehyde

20. During reduction of aldehydes with hydrazine and potassium hydroxide, the first is formationof :(a) R—C ≡ N (b) R—CO—NH2

(c) R—CH = NH (d) R—CH = N—NH2

21. Reduction of >C = O to >CH2 can be carried out with—(a) Catalytic reduction (b) Na/C2H5OH(c) Wolff-Kishner reduction (d) LiAlH4

22. Identify the compounds A and B in the following reaction sequence.

(CH3)2C = O NaCN

⎯⎯⎯→HCl

A H3O+

⎯⎯⎯→Heat

B

(a) A = CH3CO2H, B = (CH3CO)2O(b) A = (CH3)2C(OH)CN, B = (CH3)2C(OH)CO2H(c) A = CH3CHO, B = CH3CO2H(d) A = (CH3)2C(OH)CN, B = (CH3)2C = O

23. Acetone on heating with conc. H2SO4 mainly gives :(a) Mesitylene (b) Mesityl oxide(c) Toluene (d) Xylene

24. The end product of the reaction is : HC ≡ CH 1% HgSO4⎯⎯⎯⎯⎯→20% H2SO4

A CH3MgX

⎯⎯⎯⎯→ B [H2O]⎯→

(a) Acetic acid (b) Isopropyl alcohol(c) Acetone (d) Ethanol

25. Which one of the following undergoes reaction with 50% NaOH to give the correspondingalcohol and acid ?(a) Phenol (b) Benzoic acid(c) Butanal (d) Benzaldehyde


26. Which of the following does not undergo Cannizzaro’s reaction ?(a) Benzaldehyde (b) 2-methylpropanal(c) p-methoxybenzaldehyde (d) 2, 2-dimethyl propanal

27. A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysisforms a racemic mixture of α-hydroxy acid. The carbonyl compound is :(a) Formaldehyde (b) Acetaldehyde(c) Acetone (d) Diethyl ketone

28. In the equation CH3COOH +Cl2 Red P

⎯⎯→–HCl

A, the compound A is :

(a) CH3CH2Cl (b) ClCH2COOH(c) CH3Cl (d) CH3COCl

29. When acetic acid is reacted with calcium hydroxide and the product is distilled dry, thecompound formed is :(a) Calcium acetate (b) Acetone(c) Acetaldehyde (d) Acetic anhydride

30. HVZ reaction is used to prepare :(a) β-haloacid (b) αααα-haloacid(c) α, β-unsaturated acid (d) none of these

31. In the reaction CH3COOH ⎯⎯⎯→P4‚ Cl2 A ⎯⎯→

NH3

excess B, the product B is :

(a) Acetamide (b) Glycine(c) Ammonium acetate (d) Methane

32. The weakest acid amongst the following is :(a) ClCH2 COOH (b) FCH2CH2COOH(c) HCOOH (d) CH2(I)COOH

33. Phenol and benzoic acid can be distinguished by reaction with :(a) aq. NaOH (b) aq. NaHCO3

(c) Neutral FeCl3 (d) aq. NH3

34. The carboxylic acid that does not undergo HVZ reaction is :(a) CH3COOH (b) (CH3)2CHCOOH(c) CH3CH2CH2CH2COOH (d) (CH3)3C COOH

35. The correct order of reactivity is :(a) CH3COCl > CH3COOC2H5 > CH3CONH2

(b) CH3COOC2H5 > CH3COCl > CH3CONH2

(c) CH3CONH2 > CH3COCl > CH3COOC2H5

(d) CH3COCl > CH3CONH2 > CH3COOC2H5

36. When sodium formate is heated with sodalime, it forms :(a) Hydrogen (b) Ethylene(c) Ethane (d) Methane

37. The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B),acetic acid (C), and formic acid (D) is :(a) A > B > C > D (b) A > C > B > D(c) B > A > D > C (d) B > D > C > A

38. Lower carboxylic acids are soluble in water due to :(a) Lower Mol wt. (b) Hydrogen bonding(c) Dissociation into ions (d) Easy hydrolysis

39. Among acetic acid, phenol and n-hexanol, which of the compounds will react with NaHCO3solution to give sodium salt and carbon dioxide ?(a) Acetic acid (b) n-Hexanol(c) Acetic acid and phenol (d) Phenol


40. Which of the following is the strongest acid ?(a) CH3OH (b) CH3CH2OH(c) C6H5SO3H (d) C6H5COOH

41. Which reaction is suitable for the preparation of α-chloroacetic acid ?(a) Hell-Volhard Zelinsky reaction (b) Nef reaction(c) Stephen’s reaction (d) Perkin condensation

42. When 2-hydroxybenzoic acid is distilled with zinc dust, it gives :(a) Phenol (b) Benzoic acid(c) Benzaldehyde (d) A polymeric compound

43. Benzoic acid may be converted into ethyl benzoate by reaction with(a) Ethyl chloride (b) Dry HCl—C2H5OH(c) Ethanol (d) Sodium ethoxide

44. A carboxylic acid is converted into its anhydride using :(a) Thionyl chloride (b) Sulphur chloride(c) Sulphuric acid (d) Phosphorus pentoxide

45. The acid which do not contain a —COOH group is :(a) Propionic acid (b) Picric acid(c) Tartaric acid (d) Lactic acid

46. Which of the following statements is not true about HCOOH ?(a) It is a stronger acid than CH3COOH(b) It forms formyl chloride with PCl5

(c) It gives CO and H2O on heating with conc. H2SO4

(d) It reduces Tollen’s reagent47. Phenol and benzoic acid may be distinguished by their reaction with :

(a) Aqueous NaOH (b) Aqueous NaHCO3

(c) Neutral FeCl3 (d) Aqueous NH3

Chapter 13. Organic Compounds Containing Nitrogen1. The product formed when aniline is warmed with chloroform and caustic potash is :

(a) Phenyl chloride (b) Methyl isocyanide(c) Phenyl isocyanide (d) Nitro phenol

2. The formula of acrylonitrile is :(a) CH3CH = CHCN (b) CH2 = CHCN

(c) NC–CH = CH–CN (d) HC ≡ C–CN3. Lower amines are soluble in water due to :

(a) Low molecular mass(b) Formation of complexes(c) Formation of hydrogen bonds with water

(d) Affinity with water4. Which one of the following amines gives an alcohol with nitrous acid ?

(a) CH3NH2 (b) (CH3)2NH(c) (CH3)3N (d) C6H5NH2

5. Reaction of R—

O||C—NH2 with a mixture of Br2 and KOH gives RNH2 as main product. The

intermediate involved in the reaction are :

(a) R—

O||C—NHBr (b) R—N = C = O

(c) R—NH—Br (d) H—CO—NBr2


6. Activation of benzene ring in aniline can be decreased by treating with :(a) Dil. HCl (b) Ethyl alcohol

(c) Acetic acid (d) Acetyl chloride7. Basic hydrolysis of alkyl cyanide produces :

(a) Carboxylic acid (b) Salt of ammonia(c) Salt of carboxylic acid (d) NaNH2

8. Which of the following does not reacts with Hinsberg reagent ?(a) Ethyl amine (b) (CH3)2NH(c) (CH3)3N (d) Propane–2–amine

9. Dehydration of an amide with phosphorus pentoxide yields :(a) Ammonia (b) Alkyl cyanide

(c) Alkyl isocyanide (d) Alkyl amine10. Which of the following when heated with a mixture of ethanamine and alcoholic potash gives

ethyl isocyanide ?(a) 2-chloropropane (b) 2, 2–dichloropropane(c) Trichloromethane (d) Tetrachloromethane

11. Which of the following pairs of species will yield carbylamine ?(a) CH3CH2Br and KCN (b) CH3CH2Br and NH3 (excess)(c) CH3CH2Br and AgCN (d) CH3CH2 NH2 and HCHO

12. The main product of reaction of alcoholic silver nitrite and ethyl bromide is :(a) Ethane (b) Ethylnitrite(c) Ethylisocyanide (d) Nitroethane

13. Which of the following will not give primary amine ?(a) Dehydration of amides (b) Acidic hydrolysis of alkyl cyanides

(c) Reduction of amides (d) Reduction of alkyl cyanides14. The reagents used in Hoffmann’s mustard oil reaction are :

(a) Mustard oil and 1° amine (b) CS2 and aniline in HgCl2

(c) Nitrobenzene and CS2 (d) S and RNC15. The hybrid state of N in R2NH is :

(a) sp3 (b) sp2

(c) sp (d) dsp2

16. Arrange the following :(I) CH3NH2, (II) (CH3)2NH, (III) C6H5NH2 and (IV) (CH3)3Nin increasing order of basicity in aqueous medium.

(a) II < I < IV < III (b) III < IV < I < II

(c) I < II < III < IV (d) II < III < I < IV17. Identify the term C in the series :

CH3CN Na/C2H5OH

⎯⎯⎯→ A HNO2⎯→ B

Cu⎯⎯→573°K C

(a) CH3COOH (b) CH3CH2NHOH

(c) CH3CONH2 (d) CH3CHO

18. In the following reaction identify ‘Y’ :

CH3CN + 2H ⎯→ X Boiling⎯⎯→H2O Y

(a) Acetone (b) Ethylamine(c) Acetaldehyde (d) Dimethyl amine


19. Which is the end product in the following equation ?

C2H5NH2 HNO2⎯→ A

PCl5⎯→ B NH3⎯→ C

(a) Ethylcyanide (b) Methylamine(c) Ethylamine (d) Acetamide

Chapter 14. Biomolecules1. The deficiency of vitamin D causes :

(a) Rickets (b) Gout(c) Scurvy (d) Night blindness.

2. Which of the following is an example of aldohexose ?(a) Ribose (b) Fructose(c) Sucrose (d) Glucose

3. The linkage which holds various amino acid units in primary structures of proteins is :(a) Glycoside linkage (b) Peptide linkage(c) Ionic linkage (d) Hydrogen bond

4. Maltose on hydrolysis gives :(a) αααα-D-glucose (b) α and β−D-glucose(c) Glucose and fructose (d) Fructose only

5. The amino acids are the end products of the digestion of :(a) Lipids (b) Fats(c) Proteins (d) Enzymes

6. α-helix refers to :(a) Primary structure of proteins (b) Secondary structure of proteins(c) Tertiary structure of proteins (d) Quaternary structure of proteins

7. Which of the following bases is not present in DNA ?(a) Adenine (b) Guanine(c) Uracil (d) Cytosine

8. The relation between nucleotide triplets and the amino acids is called :(a) Transcription (b) Duplication(c) Genetic code (d) Gene

9. Nucleic acids are polymers of :(a) Nucleotides (b) Nucleosides(c) Nuclei of heavy metals (d) Proteins

10. The non-proteinous substances with certain enzymes require for their activity are called :(a) Catalysts (b) Inhibitors(c) Co-enzymes (d) Epimers

11. Which of the following α-amino acid is not optically active ?(a) Alanine (b) Glycine

(c) Phenylanine (d) All are optically active12. Glucose on treatment with NH2OH undergoes :

(a) Condensation (b) Reduction(c) Hydrolysis (d) Oxidation

13. Niacin is vitamin :(a) B1 (b) B2

(c) B12 (d) B4

14. Which of the following is a ketohexose ?(a) Fructose (b) Maltose(c) Glucose (d) Ribose


15. The linkage that holds monosaccharide units together in a polysaccharide is called

(a) Peptide linkage (b) Glycoside linkage

(c) Ester linkage (d) Ionic linkage

16. A nucleoside is made up of :

(a) A base and sugar (b) A base and phosphoric acid

(c) A sugar and phosphoric acid (d) A sugar, a base and phosphoric acid

17. The disease albinism is caused by the deficiency of enzyme :

(a) Trypsin (b) Tyrosinase

(c) Phenylalanine hydroxylase (d) None of these

18. Which of the following base is a purine ?

(a) Thymine (b) Uracil

(c) Cytosine (d) Adenine

19. Which of the following is a pyrimidine base ?

(a) Adenine (b) Gyanine

(c) Uracil (d) None of these

20. Which of the following is an example of globular protein ?

(a) Myosin (b) Collagen

(c) Keratin (d) Haemoglobin

21. Which of the following is an example of fibrous protein ?

(a) Insulin (b) Haemoglobin

(c) Fibroin (d) Glycogen

22. α−Amino acids behave as crystalline ionic solids and have high melting point due to thepresence of :

(a) —NH2 group (b) —COOH group

(c) Both —NH2 and —COOH (d) None of these

23. The main structural feature of proteins is :

(a) Ether linkage (b) Ester linkage

(c) Peptide linkage (d) All the three above

24. Enzymes are :

(a) Fatty acids (b) Vitamins

(c) Proteins (d) None of these

25. Enzymes belong to which class of compounds ?

(a) Polysaccharides (b) Polypeptides

(c) Polynitro heterocyclic compounds (d) Hydrocarbons

26. Which of the following base is not present in RNA ?

(a) Thymine (b) Uracil

(c) Adenine (d) Guanine

27. The sequence of bases on m-RNA molecule, synthesized on the GCATA strand of DNA is :

(a) CGUAU (b) CGTAT

(c) TACGC (d) ATCGC

28. Amino acids are least soluble in water :

(a) at pH = 7 (b) at pH > 7

(c) at pH < 7 (d) at isoelectric point


29. —

O||C—NH— (peptide bond). Which statement is incorrect about peptide bond ?

(a) C—N bond length in proteins is longer than usual bond length of C—N bond(b) C—N bond length in proteins is smaller than usual bond length of C—N bond(c) Spectroscopic analysis shows planar nature of —C

||

O

—NH— group

(d) None of these30. Which of the following is correct about H-bonding in nucleotide ?

(a) A-T, G-C (b) A-G, T-C,(c) G-T, A-C (d) A-A, T-T

31. ATP represents :(a) Abnormal temperature and pressure (b) Atmospheric temperature(c) Adenosine triphosphate (d) None of these

32. Which one of following vitamin checks k night blindness ?(a) A (b) B(c) C (d) D

33. Which one of the following is synthesized in our body by sun rays ?(a) Vitamin D (b) Vitamin B(c) Vitamin K (d) Vitamin A

34. The helical structure of protein is stabilized by :(a) Hydrogen bonds (b) Ether bonds(c) Peptide bonds (d) Dipeptide bonds

35. The enzyme which hydrolyses triglycerides to fatty acids and glycerol is called :(a) Lipase (b) Zymase(c) Pepsin (d) Maltase

36. The nucleic acid base having two possible binding sites is :(a) Thymine (b) Cytosine(c) Guanine (d) Adenine

Chapter 15. Polymers

1. Natural rubber is a :(a) Polyester (b) Polyamide(c) Polyisoprene (d) Polysaccharide

2. The fibre obtained by the condensation of hexamethylene diamine and adipic acid is :(a) Nylon-6 6 (b) Dacron(c) Teflon (d) Polyester

3. Polymers are :(a) Micromolecules (b) Macromolecules(c) Sub-macromolecules (d) None of these

4. Bakelite is :(a) Addition polymer (b) Elastomer(c) Thermoplastic (d) Thermosetting

5. The repeating units of PTFE are :(a) Cl2CH—CH3 (b) F2C = CF2

(c) F3C—CF3 (d) FClC = CF2

6. Which of the following is not a condensation polymer ?(a) Glyptal (b) Nylon-6 6(c) Dacron (d) PTFE


7. Which of the following polymer is a copolymer ?(a) Polypropylene (b) Nylon-6 6(c) PVC (d) Teflon

8. Which of the following types of polymer has the strongest inter particle forces ?(a) Elastomers (b) Thermoplastics(c) Fibers (d) Thermosetting polymers

9. Synthetic polymer prepared by using ethylene glycol and terephthalic acid is known as :(a) Teflon (b) Terylene(c) Nylon (d) PVC

10. The fibre obtained by the condensation of hexamethylene diamine and adipic acid is :(a) Nylon-6 6 (b) Dacron(c) Teflon (d) Polyester

11. Which is not a macromolecule ?(a) DNA (b) Starch(c) Palmitate (d) Insulin

12. Copolymer of acrylonitrile (40%) and vinyl chloride (60%) is called :(a) Orlon (b) Dacron(c) Dynel (d) Perlon

13. Copolymer of vinyl chloride (90%) and vinyl acetate (10%) is called :(a) Saran (b) Teflon(c) Vinyon (d) All of the above

14. Polymers have :(a) Absolute molecular weight (b) Average molecular weight(c) Low molecular weight (d) Absolute melting point

15. Chemical name of melamine is :(a) 2, 4-diamino–1, 3, 5-triazine (b) 2-amino–1, 3, 5-triazine(c) 2, 4, 6-triamino–1, 3, 5-triazine (d) 1, 3, 5-triamino-2, 4, 6-triazine

16. Synthetic human hair wigs are made from a copolymer of vinyl chloride and acrylonitrile and iscalled :(a) PVC (b) Polyacrylonitrile(c) Cellulose (d) Dynel

17. Plexiglass (PMMA) is a polymer of :(a) Acrylic acid (b) Methyl acrylate(c) Methyl methacrylate (d) None of the above

18. Of the following which is a step growth polymer ?(a) Bakelite (b) Polyethylene(c) Teflon (d) PVC

Chapter 16. Chemistry in Everyday Life

1. An Antibiotic with a broad spectrum :

(a) Kills the antibodies (b) Act on a specific antigen

(c) Act on different antigens (d) Act on both the antigen and antibodies2. Penicillin was first discovered by :

(a) A-Fleming (b) Tence and Salke(c) S. A. Waksna (d) Lewis Pasteur

3. An example of psychedelic agent is :(a) DNA (b) LSD

(c) DDT (d) TNT4. Acetoxy benzoic acid in :

(a) Antiseptic (b) Aspirin(c) Antibiotic (d) Mordent dyes


5. Antiseptic chloroxylenol is :(a) 4-chloro-3,5-dimethyl phenol (b) 3-chloro-4,5 dimethyl phenol(c) 4-chloro-2,5-dimethyl phenol (d) 5-chloro 3,4-dimethyl phenol

6. The bacteriostatic antibiotic among the following :(a) Erythromycin (b) Penicillin(c) Aminoglyconside (d) Ofloxacin

7. Which one of the following is employed as a transquilizer ?(a) Equanil (b) Naproxen(c) Tetracycline (d) Chlorpheninamine

8. Aspirin is also known as :(a) Methyl salicylic acid (b) Acetyl salicylic acid(c) Acetyl salicylate (d) Methyl salicylate

9. Which one of the following pair is the strongest pesticides ?(a) Chloroform and benzene hexachloride (b) D. D. T. and 666(c) 666 and ether (d) Isocyanides and alcohol

10. The drug used as an antidepressant is :(a) Luminel (b) Tofranil(c) Mescaline (d) Sulphadiazine

11. Dettol is the mixture of :(a) Chloroxylenol and Bithionel (b) Chloroxylenol and Torpineol(c) Phenol and Iodine (d) Terpineol and Bithionol

12. Phenacetin is used as :(a) Antipyretics (b) Antiseptic(c) Antimalarial (d) Analgesics

13. Substance used for bringing down temperature in high fever are called :(a) Pyretics (b) Antipyretics(c) Antibitics (d) Antiseptic

14. The correct structure of drug paracetamol is :

(a)

OH

NHCOCH3

(b)

OH

CONH2

(c)

OH

CONH2

(d)

Cl

COCH3

15. Which is correct about Saccharin ?

(a) It is

O||C

NHSO2

(b) It is 600 times sweeter than sugar

(c) It is used as sweetening agent (d) All of the these

16. Bithional is an example of :(a) Disinfectant (b) Antiseptic(c) Antibiotic (d) Analgesies

17. Indigo is a or an :(a) Organic dye (b) Organic polymer(c) Detergent (d) Pesticide

❐

Question

3Set Match the Column


Q. Match the items of column A to those given in column B.Column A Column B

1. Colour in crystals Tetrahedral arrangement of atoms2. Diamond F-centre3. Hexagonal close packing Co-ordination number of 84. CsCl crystal BCC5. 68% occupancy of space ABAB type of close packing6. Metallic crystal Diamond7. Covalent crystal Malleable and ductile

Ans. Column A Column B1. Colour in crystals F-centre2. Diamond Tetrahedral arrangement of atoms3. Hexagonal close packing ABAB type of close packing4. CsCl crystal Co-ordination number of 85. 68% occupancy of space BCC6. Metallic crystal Malleable and ductile7. Covalent crystal Diamond


(a) Column A Column B1. Molal depression constant Henry’s law2. Colligative property Osmotic pressure3. Dilute solution Relative lowering of vapour pressure4. Elevation of boiling point K Kg mol–1

5. Solubility of gas in liquid Colligative property

Ans. Column A Column B1. Molal depression constant K Kg mol–1

2. Colligative property Relative lowering of vapour pressure3. Dilute solution Osmotic pressure4. Elevation of boiling point Colligative property5. Solubility of gas in liquid Henry’s Law

(b) Column A Column B1. Colligative property Raoult’s law2. Dilute solution Osmotic pressure3. Cottrell’s Ebullioscopic constant4. Temperature Solutions having same osmotic pressure5. Isotonic solution Intensive property

Ans. Column A Column B1. Colligative property Osmotic pressure2. Dilute solution Raoult’s law3. Cottrell’s Ebullioscopic constant4. Temperature Intensive property5. Isotonic solution Solutions having same osmotic pressure

Match the Column ■ 43


(a) Column A Column B

1. Molar conductivity Siemen cm– 1

2. Dry cell Nernst equation

3. Zn/Zn2+ (0·1M) || Zn2+ (0·5M)/Zn Maximum work obtainable from cell

4. Specific conductivity Electrochemical cells

5. E = E° + 0·59

n log [Mn+][M]

Ohm– 1 cm2 mol– 1

6. Nernst Equation Primary cell

7. E° cell Concentration cell

8. 96500 C mol– 1 Faraday’s constant

Ans. Column A Column B

1. Molar conductivity Ohm– 1 cm2 mol– 1

2. Dry cell Primary cell

3. Zn/Zn2+ (0·1M) || Zn2+ (0·5M)/Zn Concentration cell

4. Specific conductivity Siemen cm– 1

5. E = E° + 0·59

n log [Mn+][M]

Nernst equation

6. Nernst Equation Electrochemical cells

7. E° cell Maximum work obtainable from cell

8. 96500 C mol– 1 Faraday’s constant

(b) Column A Column B1. Conductivity Sm2 mol–1

2. Cell constant m–1

3. S Sm–1

4. Molar conductivity Conductance


1. Conductivity Sm–1

2. Cell constant m–1

3. S Conductance

4. Molar conductivity Sm2 mol–1

(c) Column A Column B

1. Dry cell Potassium hydroxide

2. Nickel-cadmium cell Aqueous H2SO4

3. Lead storage cell Zinc chloride


1. Dry cell Zinc chloride

2. Nickel-cadmium cell Potassium hydroxide

3. Lead storage cell Aqueous H2SO4




1. Rate of reaction Arrhenius equation2. Activation energy mol l–1 sec–1

3. Molecularity Short interval of time4. Half life of 1st order reaction Arrhenius equation5. Instantaneous rate Specific reaction rate6. Mathematical expression for rate of

reactionSum of powers to which concentration ofreactants are raised

7. Order of a complex reaction is determinedby

Is independent of initial concentration ofreactants

8. Average rate Rate law9. k = Ae–Ea/RT Molar concentration10. Active mass Order of slowest step11. Rate of a reaction when molar concentra-

tion of reactant is unityCannot be a fraction or zero

12. Order of a reaction Long duration of time13. Threshold energy Orientation barriers14. Rate constant varies with Energy is given out15. To form products, reactants must cross Energy of activated complex16. Exergonic Temperature

Ans. Column A Column B1. Rate of reaction mol l–1 sec–1

2. Activation energy Arrhenius equation3. Molecularity Cannot be a fraction or zero4. Half life of 1st order reaction Is independent of initial concentration of

reactants5. Instantaneous rate Short interval of time6. Mathematical expression for rate of

reactionRate law

7. Order of a complex reaction is determinedby

Order of slowest step

8. Average rate Long duration of time9. k = Ae–Ea/RT Arrhenius equation10. Active mass Molar concentration11. Rate of a reaction when molar concentra-

tion of reactant is unitySpecific reaction rate

12. Order of a reaction Sum of powers to which concentration ofreactants are raised

13. Threshold energy Energy of activated complex14. Rate constant varies with Temperature15. To form products, reactants must cross Orientation barriers16. Exergonic Energy is given out

(b) Column A Column B1. A positive catalyst Catalyst to catalyst2. Catalyst works best at Adsorption theory3. Promoter Provides an alternative path with low energy

barrier4. A negative catalyst Optimum temperature5. Ni as catalyst Provides an alternative path with high energy

barrier6. Dehydration of ethanol using conc. H2SO4. Intermediate compound formation theory.


Ans. Column A Column B1. A positive catalyst Provides an alternative path with low energy

barrier2. Catalyst works best at Optimum temperature3. Promoter Catalyst to catalyst4. A negative catalyst Provides an alternative path with high energy

barrier5. Ni as catalyst Adsorption theory5. Dehydration of ethanol using conc.

H2SO4.Intermediate compound formation theory.


(a) Column A Column B1. Coagulation Tyndall Effect2. Colloidal sol formation Dialysis3. Purification Peptisation4. Cleansing action of soap Emulsification5. Scattering of light Electrophoresis


1. Coagulation Electrophoresis

2. Colloidal sol formation Peptisation

3. Purification Dialysis

4. Cleansing action of soap Emulsification

5. Scattering of light Tyndall Effect

(b) Column A Column B1. Foam Solid in liquid2. Gel Liquid in solid3. Sol Liquid in liquid4. Emulsion Gas in liquid

Ans. Column A Column B1. Foam Gas in liquid2. Gel Liquid in solid3. Sol Solid in liquid4. Emulsion Liquid in liquid


(a) Column A Column B1. Cyanide Process Dressing of ZnS2. Froth Floatation process Ultra pure Ge3. Electrolytic reduction Extraction of Al4. Zone refining Purification of Ni

Ans. Column A Column B1. Cyanide Process Dressing of ZnS2. Froth Floatation process Extraction of Al3. Electrolytic reduction Ultra pure Ge4. Zone refining Purification of Ni


(b) Column A Column B1. Blistered Cu 2 Cu2O + Cu2S → 6Cu + SO2

2. Blast Furnace Aluminium3. Reverberatory Furnace Iron4. Hall-Herault process FeO + SiO2 → FeSiO3

Ans. Column A Column B1. Blistered Cu 2 Cu2O + Cu2S → 6Cu + SO2

2. Blast Furnace Iron3. Reverberatory Furnace FeO + SiO2 → FeSiO3

4. Hall-Herault process Aluminium

(c) Column A Column B1. Pendulum NaCN2. Malachite Al2O3

3. Calamine Nickel steel4. Cryolite Na3AlF6

5. Corundum CuCO3.Cu(OH)2

6. Depressant ZnCO3

Ans. Column A Column B1. Pendulum Nickel steel2. Malachite CuCO3.Cu(OH)2

3. Calamine ZnCO3

4. Cryolite Na3AlF6

5. Corundum Al2O3

6. Depressant NaCN


(a) Column A Column B1. Frasch process Sulphur2. Noble gas For respiration by deep divers3. O2 NaCl4. Common salt N. Bartlett5. ICl2 Inter halogen6. He Paramagnetic

Ans. Column A Column B1. Frasch process Sulphur2. Noble gas N. Bartlett3. O2 Paramagnetic4. Common salt NaCl5. ICl2 Inter halogen6. He For respiration by deep divers

(b) Column A Column B1. CFC HCl, HNO3

2. Ozone Acid rain3. Ozonide Depletion4. Na2SO4.10H2O Ethylene5. Aqua regia Glauber’s Salt6. SO2 Chloro fluoro carbon



1. CFC Chloro fluoro carbon2. Ozone Depletion3. Ozonide Ethylene4. Na2SO4.10H2O Glauber’s Salt5. Aqua regia HCl, HNO3

6. SO2 Acid rain

Chapter 8. d-and f Block Elements


1. Five unpaired electrons Diamagnetic2. First Lanthanoid element Amphoteric oxide3. Cu (I) salts Sc3+

4. NiO An acidic oxide5. Colourless Scandium6. Mn2O7 A basic oxide7. ZnO Mn2+

8. Lightest transition metal Cerium

Ans. Column A Column B1. Five unpaired electrons Mn2+

2. First Lanthanoid element Cerium3. Cu (I) salts Diamagnetic4. NiO An acidic oxide5. Colourless Sc3+

6. Mn2O7 An acidic oxide7. ZnO Amphoteric oxide8. Lightest transition metal Scandium

(b) Column A Column B1. Zieglar Natta catalyst Ni in presence of hydrogen2. Haber’s process Cu2Cl2

3. Contact process V2O5

4. Vegetable oil or ghee Finally divided iron5. Sandmeyer reaction TiCl4 + Al (CH3)3


1. Zieglar Natta catalyst TiCl4 + Al (CH3)3

2. Haber’s process Finally divided iron3. Contact process V2O5

4. Vegetable oil or ghee Ni in presence of hydrogen5. Sandmeyer reaction Cu2Cl2


1. Production of iron alloy Lanthanoid oxide2. Television screen Lanthanoid3. Petroleum cracking Misch metal4. Lanthanoid metal + iron Magnesium based alloy5. Bullets Mixed oxides of Lanthanoid are employed



1. Production of iron alloy Lanthanoid2. Television screen Lanthanoid oxide3. Petroleum cracking Mixed oxides of Lanthanoid are employed4. Lanthanoid metal + iron Misch metal5. Bullets Magnesium based alloy

Chapter 9. Coordination Compounds


1. Low spin complex, d2sp3 Werner’s theory2. EDTA Hexaamine cobalt (III) ion3. Coordination compounds Hexadentate

Ans. Column A Column B1. Low spin complex, d2sp3 Hexaamminecobalt (III) ion2. EDTA Hexadentate3. Coordination compounds Werner’s theory

(b) Column A Column B

1. Rhodium Chlorophyll2. Cobalt Blood pigment3. Iron Wilkinson catalyst4. Magnesium Vitamin B12


1. Rhodium Wilkinson catalyst2. Cobalt Vitamin B12

3. Iron Blood pigment4. Magnesium Chlorophyll


1. dsp2 [Cr(H2O)6]3+

2. sp3d2 [Co(CN)4]2–

3. d2sp3 [Ni(NH3)6]2+

4. sp3 [MnF6]4–


1. dsp2 [MnF6]4–

2. sp3d2 [Ni(NH3)6]2+

3. d2sp3 [Co(CN)4]2–

4. sp3 [Cr(H2O)6]3+

Chapter 10. Haloalkanes and Haloarenes(a) Column A Column B

1. Ammonical silver nitrate F-centre2. DDT Insecticide3. Freon RMgX4. Iodoform Refrigerant5. Grignard Reagent Antiseptic


Ans. Column A Column B1. Ammonical silver nitrate Tollen’s reagent2. DDT Insecticide3. Freon Refrigerant4. Iodoform Antiseptic5. Grignard Reagent RMgX


(a) Column A Column B1. Anhydrous ZnCl2 + conc.HCl Hexaamminecobalt(III)ion2. Phenol Hexadentate3. Fermentation Tertiary alcohol4. Dynamite Phenol5. Enzyme Mixture of petrol and ethyl alcohol6. Lucas test Ethyl alcohol7. Coupling test Nitro glycerine8. Methylated spirit 1, 2, 3-propane triol9. Power alcohol Polar10. Glycerine 100% ethyl alcohol11. O—H Bond Invertase12. Absolute alcohol Methyl alcohol

Ans. Column A Column B1. Anhydrous ZnCl2 + conc.HCl Lucas reagent2. Phenol Reimer Tiemann reaction3. Fermentation 1, 2, 3-propane triol4. Dynamite Nitro glycerine5. Enzyme Invertase6. Lucas test Tertiary alcohol7. Coupling test Phenol8. Methylated spirit Methyl alcohol9. Power alcohol Mixture of petrol and ethyl alcohol10. Glycerine 1, 2, 3-propane triol11. O—H Bond Polar12. Absolute alcohol 100% ethyl alcohol


(a) Column A Column B1. —COOH Tollen’s reagent2. —CHO Sodium bisulphite

3. CH3—

O||C—

Sodium bicarbonate

4. —

O||C—

Sodium hypoiodite

5. —COOCH3 Sodium hydroxide6. Clemmensen reduction α, β-unsaturated compound7. Cannizzaro’s reaction Red brown precipitate8. Aldol condensation Aldehyde + Zn—Hg/HCl9. Fehling solution Aldehydes containing no α-hydrogen10. Claisen-Schmidt reaction Carbonyl compounds having α-hydrogen


Ans. Column A Column B1. —COOH Sodium bicarbonate2. —CHO Tollen’s reagent

3. CH3—

O||C—

Sodium hypoiodite

4. —

O||C—

Sodium bisulphite

5. —COOCH3 Sodium hydroxide6. Clemmensen reduction Aldehyde + Zn—Hg/HCl7. Cannizzaro’s reaction Aldehydes containing no α-hydrogen8. Aldol condensation Carbonyl compounds having α-hydrogen.9. Fehling solution Red brown precipitate10. Claisen-Schmidt reaction α, β-unsaturated compound

(b) Column A Column B1. HVZ reaction Aromatic acid2. Benzoic acid Reducing agent3. Formic Acid White Vinegar4. Acetic Acid Ethanoic anhydride5. Carbamide NH2—CO—NH2

6. Acetic Anhydride Formalin7. Dilute solution of formic Acid Hell Volhard Zelinsky


1. HVZ reaction Hell Volhard Zelinsky2. Benzoic Acid Aromatic acid3. Formic Acid Reducing agent4. Acetic Acid White Vinegar5. Carbamide NH2—CO—NH2

6. Acetic Anhydride Ethanoic anhydride7. Dilute solution of formic Acid Formalin

Chapter 13. Organic Compounds Containing Nitrogen(a) Column A Column B Column C

1. Nitro ethane Benzonitrite Sulphonation of aniline.2. Aniline Stronger base than ethyl

amine.Caused Bhopal tragedy in1984.

3. C6H5CN Reduction product of C6H5NO2. Gives nitrolic acid with HNO2.4. Diethyl amine Zwitter ion Can not be used to prepare 2°

and 3° amines.5. Phenyl hydroxyl amine C2H5I + AgNO2 Gives benzoic acid.6. p-amino benzene

sulphonic acidMIC Benzylamine

7. Phthalamide C6H5CH2NH2 Gives orange dye withdiazonium chloride.

8. CH3 NC C is in sp hybrid state Does not react withCHCl3/KOH.

9. HCN Gabriel synthesis Gives silver mirror withTollen’s reagent.

10. Aryl alkyl amine 1° amine Prussic acid


Ans. Column A Column B Column C

1. Nitro ethane C2H5I + AgNO2 Gives nitrolic acid with HNO2.2. Aniline 1° amine Gives orange dye with

diazonium chloride.3. C6H5CN Benzonitrile Gives benzoic acid.4. Diethyl amine Stronger base than ethyl

amine.Does not react withCHCl3/KOH.

5. Phenyl hydroxyl amine Reduction product of C6H5NO2.

Gives silver mirror withTollen’s reagent.

6. p-amino benzenesulphonic acid

Zwitter ion Sulphonation of aniline.

7. Phthalamide Gabriel synthesis Can not be used to prepare 2°and 3° amines.

8. CH3 NC MIC Caused Bhopal tragedy in1984.

9. HCN C is in sp hybrid state Prussic acid10. Aryl alkyl amine C6H5CH2NH2 Benzylamine


Column A Column B1. Disaccharide Lucas reagent2. Starch Polarimeter3. Purine Urea4. Phospholipids Linkage in carbohydrates5. Nucleotide Linkage in proteins6. Peptide linkage A sugar and heterocyclic base combination7. Glycoside linkage Major constituent of cell membranes8. Nucleoside The monomeric unit in nucleic acids9. Maltase Polysaccharide10. Starch Mg2+, Mn2+, CO2+

11. Prosthetic group Triosinase12. Albinism Dissolves blood clots13. Streptokinase Maltose into glucose


1. Disaccharide Sucrose2. Starch Polysaccharide3. Purine DNA4. Phospholipids Major constituent of cell membranes5. Nucleotide The monomeric unit in nucleic acids6. Peptide linkage Linkage in proteins7. Glycoside linkage Linkage in carbohydrates8. Nucleoside A sugar and heterocyclic base combination9. Maltase Maltose into glucose10. Starch Polysaccharide11. Prosthetic group Mg2+, Mn2+, CO2+

12. Albinism Triosinase13. Streptokinase Dissolves blood clots


Chapter 15. PolymersColumn A Column B

1. Dacron Condensation polymer2. Buna–S Nylon-6 63. Elastomers Wool4. Addition co-polymer Synthetic rubber5. Thermosetting polymer Intermolecular forces are H-bonds6. A polymer of hexamethylene diammine Saran7. A natural polymer Bakelite

Ans. Column A Column B1. Dacron Condensation polymer2. Buna–S Synthetic rubber3. Elastomers Intermolecular forces are H-bonds4. Addition co-polymer Saran5. Thermosetting polymer Bakelite6. A polymer of hexamethylene diammine Nylon-6 67. A natural polymer Wool

Chapter 16. Chemistry in Everyday Life(a) Column A Column B

1. Inhibit the growth of microorganism,given orally.

Transquillizers

2. Treatment of stress Analgesics3. Pain killing effect Antacids4. Applied to diseased skin surfaces Antiseptic5. Treatment of Acidity Antibiotic

Ans. Column A Column B1. Inhibit the growth of microorganism,

given orally.Antibiotic

2. Treatment of stress Transquillizers3. Pain killing effect Analgesics4. Applied to diseased skin surfaces Antiseptic5. Treatment of Acidity Antacids

(b) Column A Column B1. CH3(CH2)16 COO(CH2CH2O)nCH2CH2OH Cationic detergent2. C17H35COO–Na+ Anionic detergent

3. CH3—(CH2)10—CH2SO–3Na+ Non-ionic detergent

4.CH3—(CH2)15—N—CH3

–

CH3

Br–

+CH3

Soap

Ans. Column A Column B1. CH3(CH2)16 COO(CH2CH2O)nCH2CH2OH Non-ionic detergent2. C17H35COO–Na+ Soap

3. CH3—(CH2)10—CH2SO–3Na+ Anionic detergent

4.CH3—(CH2)15—N—CH3

–

CH3

Br–

+CH3

Cationic detergent

❐

Question

4Set

Identify the Compounds/Products/Reagents


Q. 1. A colourless inorganic salt A decomposes at about 250° C to give only two products B and Cleaving no residue. The oxide C is a liquid at room temperature and is neutral to litmus paperwhile B is a neutral oxide. White phosphorus burns in excess of B to produce strong dehydratingagent. Give balance equation for above process.

Ans. A = NH4NO3 (Ammonium Nitrate), B = N2O (Nitrous oxide), C = H2OReaction :

NH4NO3 250°C

⎯⎯⎯→ N2O + 2H2O

Ammonium Nitrous Water

Nitrate Oxide (C)

(A) (B)

10 N2O + P4 ⎯⎯→ 10N2 + P4O10

(B) Phosphorus pentaoxide

Q. 2. An Element ‘A’ exist as a yellow solid in a standard state. It forms a volatile hydride ‘B’ which isa foul smelling gas and is extensively used in qualitative analysis of salts. When treated withoxygen ‘B’ forms an oxide ‘C’ which is a colourless, pungent smelling gas. This gas when passedthrough acidified KMnO4 solution, decolourises it. ‘C’ gets oxidised to another oxide ‘D’ in thepresence of a heterogeneous catalyst. Identify ‘A’ ‘B’, ‘C’, ‘D’ and also give chemical equation ofreaction of C with acidified KMnO4 solution and for conversion of ‘C’ to ‘D’.

Ans. ‘A’ = Sulphur, ‘B’ = H2S gas, ‘C’ = SO2 gas, ‘D’ = SO3 gasReaction :

MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O ] × 2

SO2 + 2H2O ⎯⎯→ SO42–

+ 4H+ + 2e– ] × 5

2MnO4– + 5SO2 + 2H2O ⎯⎯→ 2Mn2+ + 5SO4

2– + 4H

+

2SO2 (g) + O2 (g) V2O5

⎯⎯⎯→ 2SO3 (g)

Sulphur Sulphur trioxide

dioxide (D)

(C)

Q. 3. On reaction with Cl2, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowishwhite powder but halide ‘B’ is a colourless oily liquid. Identify A and B and write the formula oftheir hydrolysis product.

Ans. A = PCl5 (Yellowish white powder)

P4 + 10Cl2 ⎯⎯→ 4PCl5

B = PCl3 (Colourless oily liquid)

P4 + 6Cl2 ⎯⎯→ 4PCl3

Hydrolysis product are formed as follows :

PCl3 + 3H2O ⎯⎯→ H3PO3 + 3HCl

PCl5 + 4H2O ⎯⎯→ H3PO4 + 5HCl


Q. 4. When conc. H2SO4 was added into an unknown salt present in a test tube, a brown gas (A) wasevolved. This gas intensified when copper turnings were also added into this test tube. Oncooling the gas (A) it changed into a colourless gas (B).

(i) Identify the gases A and B.

(ii) Write the equation for reaction involved.

A = NO2 (g) B = N2O4 (g)

MNO3 + H2SO4heat

⎯⎯⎯→ MHSO4 + HNO3

4HNO3heat

⎯⎯⎯→ 4NO2 + 2H2O + O2Nitrogen dioxide(Brown gas)

Cu + 4HNO3 heat

⎯⎯⎯→ Cu (NO3)2 + 2H2O + 2NO2Copperturning

2NO2

CoolN2O4

(Brown gas) (Colourless)

Q. 5. Concentrated sulphuric acid is added followed by heating to each of the following test tubelabelled (i) to (v)

Identify in which of the given test tube the following change will be observed. Support youranswer with the help of a chemical equation.

(a) Formation of black substance.

(b) Evolution of brown gas.

(c) Evolution of colourless gas.

(d) Formation of brown substance which on dilution becomes blue.

(e) Disappearance of yellow powder along with evolution of colourless gas.

Ans. (a) Test tube (i) C12H22O11 ⎯⎯→ 12C + 11H2O(White) Black

Substance

(b) Test tube (ii) 2NaBr + 2H2SO4 ⎯⎯→ Br2 ↑ + Na2SO4 + SO2 + 2H2OBrowngas

(c) Test tube (v) 2KCl + H2SO4 ⎯⎯→ 2HCl ↑ + K2SO4

colourless gas

(d) Test tube (iii) Cu + 2H2SO4 ⎯⎯→ CuSO4 + SO2 + 2H2Obluesubstance

(e) Test tube (iv) S + 2H2SO4 ⎯⎯→ 3SO2 ↑ + 2H2OColourlessgas

Identify the Compounds/Products/Reagents ■ 55

Q. 6. X2 is a greenish yellow gas with pungent smell and is used in purification of water. On dissolvingwater it gives a solution which turns blue litmus red. When it is passed through NaBr solutionBr2 is obtained(i) Identify the gas(ii) What are the products obtained when X2 reacts with ammonia ? Give chemical equation.(iii) What happens when X2 reacts with cold and dil NaOH solution ? Write chemical equationand give the name of reaction.

Ans. (i) X2 = Cl2

(ii) 8NH3 + 3Cl2 ⎯⎯→ 6NH4Cl + N2

NH3 + 3Cl2 ⎯⎯→ NCl3 + 3HCl

excess (oxidation)

(iii) 2 NaOH + Cl2 ⎯⎯→ 2Na + NaOCl + H2O

(cold and dil) (Disproportionation reaction)

Q. 7. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treatedwith 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic innature. Gas C on further oxidation in moist condition gives a compound (D) which is a part ofacid rain. Identify compound (A) to (D) and also give necessary equation of all the steps involved.

Ans. A = NH4NO2, B = N2 C = NH3 D = HNO3

Equation of all the steps involved

(i) NH4NO2 ⎯⎯→ N2 + 2H2O

(A) (B)

(ii) N2 + 3H2 ⎯⎯→ 2NH3

(B) (C)(iii) 4NH3 + 5O2 ⎯⎯→ 4NO + 6H2O

(C)

(iv) 2NO + O2 ⎯⎯→ 2NO2

3NO2 + H2O ⎯⎯→ 2HNO3 + NO

(D)

Q. 8. On heating lead (II) nitrate gives a brown gas ‘A’. The gas ‘A’ on cooling changes to colourlesssolid ‘B’. Solid ‘B’ on heating with NO changes to a blue solid ‘C’. Identify A, B, C and also writethe reactions involved and draw structure of B and C.

Ans. 2 Pb (NO3)2Δ

⎯⎯→673 K

2PbO + 4NO2 + O2

(A)

(Brown colour)

2NO2

on cooling

on heating N2O4

(B)(Colourless solid)

2NO + N2O4Δ

⎯⎯→250 K

2N2O3

(B) (C)

(Blue solid)

Structure of N2O4

N — N

ON — N

O

OOO O

O O

→←


Structure of N2O3

N — N

O

O

O

O

O

N — N

O

→←

Q. 9. A translucent white waxy solid (A) on heating in an inert atmosphere is converted to itsallotropic form (B). Allotrope (A) on reaction with very dilute aqueous KOH liberates a highlypoisonous gas (C) having rotten fish smell with excess of chlorine it forms (D) which hydrolysesto compound (E). Identify compound (A) to (E).

Ans. A = White phosphorus, B = Red Phosphorus, C = Phosphine (PH3), D = Phosphoruspentachloride (PCl5), E = Phosphoric acid (H3PO4)Reaction :

P4 (S)Δ

⎯⎯⎯→inert gas

P4 (S)

white Red phosphorus

phosphorus (B)

(A)

P4(s) + 3KOH + 3H2OΔ

⎯→ 3KH2PO2 + PH3

White phosphorus Potassium Phosphine

(A) hypophosphite (C)

PH3 (g) + 4Cl2(g) ⎯⎯→ PCl5 (g) + 3HCl

Phosphine Phosphorus

(C) pentachloride

(D)PCl5(s)

pentachloride

Phosphorus

+ 4H2O ⎯⎯→ H3PO4

(acid)

Phosphoric

+ 5HCl

(D) (E)

Chapter 8. The d and f Block Elements

Q. 1. A mixed oxide of iron and chromium FeOCr2O3 is fused with sodium carbonate in the presenceof air to form a yellow coloured compound (A). On acidification the compound (A) forms anorange coloured compound (B), which is a strong oxidising agent

(i) Identify the compound (A) and (B).

(ii) Write balanced chemical equation of each step.

Ans. (i) Compound (A) sodium chromate (Na2Cr2O4)

Compound (B) sodium dichromate (Na2Cr2O7)

(ii) (a) 4FeOCr2O3 + 8Na2CO3 + 7O2 ⎯⎯→ 8Na2CrO4 + 2Fe2O3 + 8CO2

(A)

(b) 2NaCrO4 + 2H+ ⎯⎯→ Na2Cr2O7 + 2Na+ + H2O

(B)

Q. 2. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas takenin excess, reacts with NH3 to give an explosive compound (C). Identify compound A, B and C.

Ans. A = MnO2, B = Cl2, C = NCl3

MnO2 + 4HCl ⎯→ MnCl2 + Cl2 + 2H2O

(A) (B)

NH3 + 3Cl2 ⎯⎯→ NCl3 + 3HCl (excess)


Q. 3. A voilet compound of manganese (A) decolourises on heating to liberate oxygen and compound(B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence ofpotassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl,chlorine gas is liberated and a compound D of manganese along with other products is found.Identify compound A to D and also explain the reactions involved.

Ans. A = KMnO4, B = K2MnO4, C = MnO2, D = MnCl2

2KMnO4Δ

⎯→ K2MnO4 + MnO2 + O2

(A) (B) (C)

2MnO2 + 4KOH + O2 ⎯⎯→ 2K2MnO4 + 2H2O

(C) (B)

MnO2 + 4NaCl + 4H2SO4 ⎯⎯→ MnCl2 + 4NaHSO4 + 2H2O + Cl2

(C) (D)

Q. 4. When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent anddissolved in water it gives a dark solution of compound (B). Compound (B) disproportionates inneutral or acidic solution to give purple compound (C). An alkaline solution of compound (C)oxidises potassium iodide solution to a compound (D) and compound (A) is also formed.Identify compound A to D and also explain the reaction involved.

Ans. A = MnO2, B = K2MnO4, C = KMnO4, D = KIO3

2MnO2 + 4KOH + O2 ⎯⎯→ 2K2MnO4 + 2H2O

(A) (B)

3MnO42– + 4H

–⎯⎯→ 2MnO4

– + MnO2 + 2H2O

(C)

2MnO4– + H2O + KI ⎯⎯→ 2MnO2 + 2OH

– + KIO3

(C) (A) ( D)

Q. 5. (a) A blackish brown coloured solid ‘A’ when fused with alkali metal hydroxides in presenceof air produces a dark green coloured compound ‘B’ which on electrolytic oxidation inalkaline medium gives a dark purple coloured compound (C). Identify A, B and C andwrite the reaction involved.

(b) What happens when in an acidic solution the green compound (B) is allowed to stand forsome time ? Give the equation involved. What is this type of reaction called ?

Ans. (a) A = MnO2 B = K2MnO4 C = KMnO4

Pyrolusite ore Potassium manganate Potassium permanganateReaction involved :

2MnO2 + 4KOH + O2 ⎯⎯→ 2K2MnO4 + 2H2O

(A) (B)

MnO42–

Electrolytic oxidation⎯⎯⎯⎯⎯⎯⎯⎯→ MnO4

– + e–

Manganate in alkaline solution Permanganate ion (Purple)

ion (Green)

(b) In acidic medium K2MnO4 changes to give purple coloured compound along with blackprecipitate

3MnO42– + 4H

+⎯⎯→ 2MnO4

– + MnO2 + 2H2O

Green Purple Blackcompound compound

This type of reaction is called disproportionation reaction.



Q. 1. Fe3+ SCN

–

⎯⎯⎯→excess

A F

–

⎯⎯⎯→excess

B . What are A and B ? Give IUPAC names of A and B.

Ans. A = Fe (CSN)3 B = [FeF6]3–

Trithiocyanato iron (III) Hexafluorido ferrate (III) ionQ. 2. CoSO4Cl. 5NH3 exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with AgNO3 to give

white precipitate, but does not reacts with BaCl2. Isomer ‘B’ gives white precipitate with BaCl2

but does not reacts with AgNO3. Answer the following questions.(i) Identify ‘A’ and ‘B’ and write their structural formulae.(ii) Give the IUPAC names of ‘A’ and ‘B’.

Ans. (i) A = [Co (NH3)5 SO4]ClB = [Co (NH3)5 Cl]SO4

(ii) A = Pentaammine sulphatocobalt (III) chlorideB = Pentaamminechloridocobalt (III) sulphate.


Q. 1. Identify A, B, C, D, E and F in the following :

⎯⎯⎯→dryether

⎯⎯→H2O

Br + Mg A B

⎯⎯⎯→dryether

⎯⎯→D2O

C CH3—CH—CH|(D)

R – Br + Mg

|CH3

CH3

CH3—C—X ⎯→Mg

E ⎯⎯→H2O

F|

3

Ans. Mg Br A = B =

C = RMgBr where R = CH3CHCH3

E = H3C—C—MgX|CH3

CH3

F = H3C—C—H|CH3

|CH3

|

Q. 2. Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of thereaction depends upon the concentration of the compound ‘A’ only. When another opticallyactive isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction wasfound to be dependent on concentration of compound with KOH.

(i) Write down the structural formula of both compounds ‘A’ and ‘B’.

(ii) Out of these two compounds, which one will be converted to the product with invertedconfiguration.

Ans. (i) Compound (A) : CH3—C—CH3

|Br

|CH3

(ii) Compound (B) : CH3—CH2—CH—CH3|Br


Q. 3. Hydrocarbon C5H10 does not reacts with chlorine but gives a single monochloro compound,C5H9Cl in bright sunlight. Identify the hydrocarbon.

Ans. (i) The hydrocarbon with molecular formula C5H10 can be either a cycloalkane or an alkene.(ii) Since the hydrocarbon does not reacts with Cl2 in the dark, it cannot be an alkene but must bea cycloalkanes.(iii) As the cyclo alkane reacts with Cl2 is the presence of bright sunlight, to give a single monochloro compound C5H9Cl therefore all the ten hydrogen atoms of the cycloalkane must beequivalent thus, the cycloalkane is cyclopentane.

No reaction ⎯⎯⎯→Sunlight←⎯⎯

Cl2 Cl

cyclopentane(C5H10)

monochlorocyclopentane

(C5H9Cl)

Cl2Dark

Q. 4. Write the structures of A, B and C in the following reactions :

(i) C6H5NO2 Sn/HCl

⎯⎯⎯⎯→ A NaNO2 + HCl

⎯⎯⎯⎯⎯⎯→273

B H2O

⎯⎯→Δ

C

(ii) CH3Cl KCN

⎯⎯⎯→ A LiAlH4

⎯⎯⎯⎯→ B HNO2

⎯⎯⎯→273K

C

Ans. (i)

NH2 N+ ≡ NCl–

A = B = C =

OH

Aniline PhenolBenzene diazoniumchloride

(ii) A = CH3—C ≡ N B = CH3—CH2—NH2 C = CH3CH2—OH

Ethane Nitrile Ethanamine Ethanol

Q. 5. Write the major organic product in each of the following reactions.

(a) CH3CH2CH2Cl + NaI Acetone

⎯⎯⎯→Heat

………

(b) (CH3)3CBr + KOH Ethanol

⎯⎯⎯→Heat

………

(c) CH3CH(Br) CH2CH3 + NaOH water

⎯⎯⎯→ ………

(d) CH3CH2Br + KCN aq. ethanol

⎯⎯⎯⎯⎯→ ………

(e) C6H5ONa + C2H5Cl ⎯⎯⎯⎯⎯→ ………

(f) CH3CH2CH2OH + SOCl2 ⎯⎯⎯⎯⎯→ ………

(g) CH3CH2CH = CH2 + HBr Peroxide

⎯⎯⎯⎯→ ………

(h) CH3CH = C (CH3)2 + HBr ⎯⎯⎯→ ………

Ans. (a) CH3CH2CH2Cl + NaI Acetone‚heat

⎯⎯⎯⎯⎯⎯⎯→Finkelstein reaction

CH3CH2CH2I1-Iodopropane

+ NaCl,

(b) (CH3)3CBr + KOH Ethanol ‚ Δ

⎯⎯⎯⎯⎯⎯⎯⎯→Dehydrohalogenation

CH3|

CH3—C = CH2 + KBr + H2O 2-Methyl propene

(c) CH3 CH2CH3 + NaOH water

⎯⎯⎯⎯⎯⎯→Hydrolysis|

Br

CH3—CH—CH2—CH3|OH

Butan-2-ol

CH

(d) CH3CH2Br + KCN aq. ethanol

⎯⎯⎯⎯⎯→ CH3CH2CN + KBr

Propane nitrile


(e) C6H5O–

Na+

+ C2H5Cl Williamson’s

⎯⎯⎯⎯⎯→Synthesis

C6H5OCH2CH3 + NaClPhenetole

(f) CH3CH2CH2OH + SOCl2 ⎯⎯⎯⎯⎯→ CH3CH2CH2Cl + HCl + SO21-Chloropropane

(g)But-1-ene

CH3CH2CH = CH2 + HBr Peroxide

⎯⎯⎯⎯→ CH3CH2CH2CH2Br1-Bromobutane

(h) CH3CH = C—CH3 + HBr ⎯⎯⎯⎯⎯⎯→ |CH3

Markovnikov’srule

CH3CH2—C—CH3

CH3

Br2-Bromo-2-methyl butane

Q. 6. Identify the reagents A, B, C and D.

C2H5Br A

⎯⎯→ C2H4 B

⎯⎯→ C2H5Cl C

⎯⎯→ C2H5OH D

⎯⎯→ C2H4

Ans. A ⎯→ KOHalc, B ⎯→ HCl, C ⎯→ KOHaq., D ⎯→ Conc. H2SO4 170°C.

Q. 7. During the conversions,

C6H5CH2CH3 (a)

⎯⎯→ X (b)

⎯⎯→ C6H5CH = CH2

Find out the reagents (a) and (b).

Ans. C6H5CH2CH3 SO2Cl2⎯⎯→ X

alc. KOH⎯⎯⎯⎯→ C6H5CH = CH2

Q. 8. An aromatic compound (A) C8H9Br reacts with CH2(COOC2H5)2 in the presence of C2H5ONa to

give (B). Compound (B) on refluxing with dil H2SO4 gives (C) which on vigrous oxidation gives

(D). The compound (D) is dibasic but on heating does not gives an anhydride. It forms a mono

nitro derivative (E) in which all the substituents are equidistant from one another. Give structure

of A to E.

Ans. (A)

CH2Br

CH3

(B)

CH2CH (COOC2H5)2

CH3

(C)

CH2CH2COOH

CH3

(D)

COOH

COOH(E)

NO2

COOH

COOH


Q. 1. An organic compound (A) C4H10O reacts with HI to give a compound (B) C4H9I which onreduction gives a normal hydrocarbon having four carbon atoms. On oxidation, (A) givescompound (C) C4H8O and then an acid (D) C4H8O2. Deduce the structures of A, B, C and D.

Ans. A = C4H9OH

B = C4H9I

C = C3H7CHO

D = C3H7COOH


Q. 2. Compound A (C4H10O) undergoes oxidation to give (B) C4H8O. B forms an oxime but does notgives Tollen’s test (B) reacts with Iodine and potassium hydroxide to give Iodoform. Deduce thestructure of A and B.

Ans. A = 2-ButanolB = 2-Butanone

(A)

OH

2-Butanol(A)

(B)

O

H3CCH3

2-Butanone(B)

Q. 3. An organic compound (A) on treatment with acetic acid in presence of sulphuric acid producesan ester (B). A on mild oxidation gives (C). (C) with 50% potassium hydroxide followed byacidification with dilute hydrochloric acid generates (A) and (D). (D) when treated withphosphorus pentachloride followed by reaction with ammonia gives (E). (E) on dehydrationproduces hydrochloric acid. Identify the compounds A, B, C, D and E.

Ans. A = CH3OHB = CH3COOCH3

C = HCHOD = HCOOHE = HCONH2

Q. 4. A compound A (C6H14O) liberates hydrogen gas with sodium metal. A does not reacts withsodium hydroxide and gives a positive lucas test immediately. When A is treated with PBr5compound B is formed (C6H13Br). When B is treated with alcoholic KOH, compound C and Dboth having formula C6H12 are formed. C is a major product while D is a minor product. When Cis treated with ozone followed by hydrolysis, only single ketone is formed. The ketone can beshown to be identical with the compound produced by hydration of propyne in the presence ofsulphuric acid and Hg+. Deduce the structural formula of the compound.

Ans. (A) (CH3)2CH — (CH3)2|OH

(B) (CH2)2CH — C — (CH3)2|Br

(C) (CH3)2 C = C (CH3)2 (D) (CH3)2 CH — CH = CH2|CH3

Q. 5. An optically active alcohol (C6H10O) absorbs two moles of hydrogen per mole of A upon catalytichydrogenation and gives a product B. The compound B is resistant to oxidation by CrO3 anddoes not shows any optical activity. Deduce the structures of A and B.

Ans. (A) HC ≡ C — C — OH|CH3

|C2H5(A)

(B) CH3CH2 — C — OH|C2H5

|CH3

(B)Q. 6. An organic compound (A) composed of C, H and O gives characteristic colour with ceric

ammonium nitrate treatment of A with PCl5 gives (B) which reacts with KCN to form (C). Thereduction of (C) with warm Na/C2H5OH produces (D) which on heating gives (E) with evolutionof ammonia Pyridine is obtained on treatment of E with nitrobenzene. Give structures of (A) to(E).

Ans. (A) CH2 —CH2—CH2| |OH OH

(B) ClCH2CH2CH2Cl

(C) NCCH2CH2CH2CN (D) H2NCH2CH2CH2CH2NH2

(E)N

H


Q. 7. Compound (A) C10H 12O gives off hydrogen on treatment with sodium metal and also

decolourises Br2 in CCl4 to give (B). (A) on treatement with I2, NaOH gives Iodoform and an acid

(C) after acidification. Give structures (A) to (C).

Ans.

⎯→

⎯→

CH = CHCOCH3

I2, NaOH

Br2,CCl4

CH = CHCHOHCH3

⎯⎯⎯→

(A) (B)

(C)

+ CHI3

CH Br CH Br CH OH CH3

H+CH = CHCOOH

Q. 8. An organic compound ‘A’ having molecular formular C3H6 on treatment with aqueous H2SO4

gives ‘B’ which on treatment with HCl/ZnCl2 gives ‘C’. The compound C on treatment with

HCl/ZnCr2 gives ‘C’. The compound C on treatment with ethanolic KOH gives back the

compound ‘A’. Identify the compounds A, B, C.

Ans. A = CH3 — CH = CH2 Propene

(B) = CH3 — CH — CH3|OH

Propane-2-ol

(C) CH3—CH—CH3|Cl

2-Chloropropane

CH3 —CH = CH2aq. H2SO4

HydrationCH3 — CH—CH3

|OH

Propene(A)

Propan-2-ol(B)

2-Chloropropane(C)

CH3 — CH—CH3|Cl

Ethanolic KOH CH3 —CH = CH2

Propene

←⎯⎯⎯⎯⎯⎯⎯⎯

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→HCl/ZnCl2

Q. 9. An aromatic compound ‘A’ on treatment with CHCl3/KOH gives two compounds ‘B’ and ‘C’.Both B and C gives the same product D when distilled with zinc dust. Oxidation of D gives Ehaving molecular formula C7H6O2. The sodium salt of E on heating with sodalime gives F whichmay also be obtained by distillating A with zinc dust. Identify A to F.

Ans. OH OHCHO

CHO

A = B = C =

OH

Phenolp-Hydroxy

benzaldehyde

o-hydroxybenzaldehyde

CHO COOH

Benzene

D = E = F =

Benzaldehyde


Reaction :

OH OH

(A) (B) (C)

CHO

CHO

+

OH

Phenol

p-Hydroxybenzaldehyde

o-hydroxy benzaldehyde

COOH

630K

CHO

Benzene(F) (E) (D)

Benzoic acid Benzaldehyde

⎯⎯⎯⎯⎯→CHCl3/KOH

(O) ⎯→

⎯⎯

→

Zn (dust)distillation

⎯⎯

→

Zn (dust)distillation

←⎯⎯⎯⎯⎯(i) NaOH

(ii) NaOH/CaO

Q. 10. An organic compound A having molecular formula C6H6O gives a characteristic colour withaqueous FeCl3 solution. A on treatment with CO2 and NaOH at 400 K under pressure gives Bwhich on acidification gives a compound C. The compound C reacts with acetyl chloride to giveD which is a popular pain killer. Deduce the structures of A, B, C and D.

Ans. OH OH

OCO CH3

(A) (B) (C)

COONa

COOH

COOH

Salicylic acidSodiumsalicylate

OH

Phenol

2-Acetoxy benzoic acid(Aspirin)

⎯⎯⎯⎯⎯→CO2/NaOH

410K ⎯⎯⎯→+H2O/H

Under pressure

⎯⎯

→ CH3—C—Cl

O

Q. 11. A compound ‘A’ is an optically active compound, on mild oxidation it gives a compound ‘B’ buton vigorous oxidation gives another compound ‘C’. C along with D is also formed from B byreaction with iodine and alkali. Deduce the structure of A, B, C and D.

Ans.

CH3—CH2—CH—CH3

|OH

(O)Mild oxidation

CH3—CH2—C—CH3

Sec. butyl alcohol Ethyl methyl ketone(B)

Vigrous oxidation

CH5—CH2—COOH(C)

Propanoic acidCH3—CH2—COOH + CHI5

Propanoic acid (D)Iodoform

⎯⎯⎯⎯⎯⎯→

←⎯

⎯

O

(O)

←⎯

⎯

I / NaOH2

(C)

(A)

Q. 12. Identify A, B, C, D in the following :

CH3CH2CH2OH PBr3

⎯⎯⎯→ A alc. KOH⎯⎯⎯→ B

HBr⎯⎯⎯→ C

NH3⎯⎯⎯→ D

Ans. A → CH3CH2CH2Br, B → CH3CH = CH2, C → CH3CHBrCH3, D → CH3CHCH3

NH2


Q. 13. C2H5OH PCl5⎯⎯→ A

KCN⎯⎯→ B

H3O+⎯⎯→ C2H5COOH

NH3⎯⎯→ Δ

C

Ans. C2H5OH PCl5⎯⎯⎯→ CH3CH2Cl (A)

(A) CH3CH2Cl KCN

⎯⎯⎯→ CH3CH2CN (B)

ethyl chloride ethyl cyanide

(B) CH3CH2CN H3O+

⎯⎯⎯→ C2H5COOH

C2H5COOH NH3⎯⎯⎯→Δ

C2H5COONH4 Δ

⎯⎯⎯→–H2O C2H5CONH2Propanamide

(C)

Q. 14. Identify the reagents A, B, C, D, E and F required for the following conversion :

C6H5NO2 A→ C6H5NH2

B→ C6H5N2

+ Cl– C→ C6H5Cl

D→ C6H5OH

H2SO4⎯⎯→ E + F

Ans. A ⎯→ Sn + conc. HCl.

B ⎯→ NaNO2 + conc. HCl

C ⎯→ Cu2Cl2 + HCl

D ⎯→ NaOH

E ⎯→ 2-hydroxy benzene sulphonic acid.

F ⎯→ 4-hydroxy benzene sulphonic acid.Q. 15. An organic compound A with molecular formula C3H8O3 reacts with oxalic acid at 110°C to give

a monocarboxylic acid B. B gives a silver mirror with Tollen’s reagent and reduces acidifiedpotassium permanganate solution. Identify A and B and give the reaction of B with acidifiedKMnO4 solution.

Ans. A is glycerol (CH2OH·CHOH·CH2OH).B is formic acid (HCOOH).When B reacts with acidified KMnO4 solution, the later gets decolourised.2KMnO4 + 3H2SO4 + 5HCOOH → K2SO4 + 2MnSO4 + 8H2O + 5CO2.

Q. 16. Identify A, B, C and D.

CH3CHOHCH3 Conc H2SO4⎯⎯⎯⎯→

Δ A ⎯⎯⎯→

B2H6 B

(O)⎯⎯→ C

C2H5OH⎯⎯⎯⎯→ D.

Ans. A ⎯→ CH3CH = CH2, B ⎯→ CH3CH2CH2OH, C ⎯→ CH3CH2COOH,D ⎯→ CH3CH2COOC2H5.

Q. 17. Identify compounds A, B, C and D.

C6H5HSO3 HOH

⎯⎯⎯→

D

ZnA

Br2C

↑

↓

CH3COCl⎯⎯⎯⎯→ B

Ans.A ⎯→ C6 H5OH, B ⎯→ C6H5OCOCH3, C ⎯→

D ⎯→ C6H6

OHBr

BrBr

Q. 18. In the following equations identify A and B :

(i)

OH

Zn dust

⎯⎯⎯→ A H2SO4⎯⎯⎯→

SO3 B (ii) A

Sn/HCl⎯⎯⎯→ B

(i) NaNO2/HCl⎯⎯⎯⎯⎯⎯→

(ii) H2O

OH

(iii) Cl2⎯⎯→

AlCl3 A

aq. NaOH⎯⎯⎯⎯⎯⎯→

575 K‚ 300 atm. B.


Ans. (i) A = B =

SO3H

(ii) A =

NO2

B =

NO2

Benzene Benzene Nitrobenzene Anilinesulphuric acid

(iii) A =

Cl

B =

OH

Chlorobenzene PhenolQ. 19. Name the reagents used in the following reactions :

(i) Oxidation of primary alcohol to carboxylic acid.(ii) Oxidation of primary alcohol to aldehyde.(iii) Bromination of phenol to 2,4,6-tribromo phenol.(iv) Benzyl alcohol to benzoic acid.(v) Dehydration of propan-2-ol to propene.(vi) Butan-2-one to butan-2-ol

Ans. (i) Acidified K2Cr2O7 or KMnO4

(ii) Pyridinium chlorochromate (PCC) in CH2Cl2 or pyridinium dichromate (PDC) in CH2Cl2.(iii) Aqueous Br2 i.e. Br2/H2O

(iv) Acidified or alkaline KMnO4 followed by hydrolysis with dil. H2SO4.

(v) Conc. H2SO4 at 443 K.

(vi) NaBH4 or Ni/H2.

Q. 20. Write the names of reagents and equation for the preparation of the following ethers byWilliamson’s synthesis.(i) 1-Propoxy propane (ii) Ethoxybenzene(iii) 2-Methoxy-2-methyl propane (iv) 1-methoxy ethane(v) 2-Methoxy-2-Methyl propane

Ans. (i) ⎯→

ONa

Sodiumphenoxide

Ethoxybenzene

+ C2H5Br

O2C2H5

+ NaBr

(ii) CH3CH2ONa + CH3Br ⎯⎯⎯→ CH3OCH2CH2 + NaBrSodium Bromo 1-Methoxy ethaneethoxide methane

(iii) CH3—C—ONa

CH3

|

|CH3

+ BrCH3 ⎯⎯→ CH3 —C—OCH3 + NaBr|CH3

|CH3

Sodium 2- methyl2-propoxide

2-Methyl-2-methoxypropane

Bromo-methane

(iv) CH3CH2CH2ONa + CH3CH2CH2 Br ⎯⎯→ CH3CH2CH2OCH2CH2CH3Sodium 1-Bromo propane 1-Propoxy propane

propoxide

(v) CH3—C—CH2ONa + CH3CH2Br ⎯⎯→|CH3

|CH3

CH3 — C — CH2OCH2—CH3|CH3

|CH3

Sodium 2, 2,-dimethyl

propoxide

Bromoethane

1-Ethoxy-2,2-

dimethyl propane



Q. 1. Write the balanced equation for the conversion of A to B.

Ans. A

CH ≡ CH + H2O H2SO4 + HgSO4⎯⎯⎯⎯⎯→

BCH3CHO

⎯⎯⎯⎯⎯⎯→HgSO2, dil H2SO4CH ≡ CH + H2O

(Ethyne) 33K⎯⎯⎯⎯⎯⎯→TautomerisationCH — OH

||CH2

(Unstable)

CHO

CH3(Ethanal)

Q. 2. Identify the products A, B, C and D.

C6H6 CH3Cl

Anhy.AlCl3⎯⎯⎯→ A

CrO2Cl2

CCl4⎯⎯⎯→ B

Conc.

NaOH⎯⎯⎯→ C + D

Ans.

C6H6 CH3 Cl

Anhy. AlCl3 C6H5CH3

CrO2 Cl2

CCl 4

Benzene Toluene BenzaldehydeA B

+ Benzyl alcohol Sod. Benzoate

C D

CHO

CH2OH||O

C – +

⎯⎯⎯⎯→ ⎯⎯⎯→

O Na→

conc. NaOH⏐↓

Q. 3. Compound (A) C4H10 O, is found to be soluble in H2SO4 acid. (A) does not reacts with Na-metalor KMnO4. When (A) is heated in excess of HI, it is converted into a single alkyl iodide. What isthe structural formula of (A) ?

Ans. (A) = CH3CH2—O—CH2CH3.

Q. 4. Give balanced equations for reaction of diethyl ether with

(i) Cl2 (ii) PCl5

Ans. (i) CH3CH2 — O—CH2CH3 + Cl2 ⎯→ CH3CHCl—O—CH2CH3 + HCl

(ii) CH3CH2O—CH2CH3 + PCl5 ⎯→ 2CH3CH2Cl + POCl3

Q. 5. C6H5COOH SOCl2⎯⎯→ A

NH3⎯⎯→ B Br2/KOH

⎯⎯⎯⎯→ C

Ans.

C6H5COOH

CONH2COCl NH2

⎯⎯⎯→

(BenzoylchlorideBenzoic

acid )

SOCl2 ⎯⎯→⎯⎯→Br2 /KOHNH3

(Benzamide) (Aniline)(A) (B) (C)

Q. 6. CH3COCH3 Conc.HNO3

(O)⎯⎯⎯⎯→ A

SOCl2⎯⎯→ B NH3⎯→ C

LiAlH4⎯⎯⎯→ D HNO2⎯⎯→ E

CH3COCl⎯⎯⎯⎯→ F

Ans. CH3COCH3

CH3COOC2H5

CH3COOH CH3COCl CH3CONH2

SOCl2Conc. HNO3

[O]

NH3

A

E

F

B

CH3CH2 NH2CH3CH2OH

D

LiAlH4

C

HNO2CH3COCl


A—Acetic acid (CH3COOH)

B—Acetyl chloride (CH3COCl)

C—Acetamide (CH3CONH2)

D—Ethyl amine (CH3CH2NH2)

E—Ethyl alcohol (CH3CH2OH)

F—Ethyl acetate (CH3COOC2H5)Q. 7. An organic compound A with molecular formula C2H7N on reaction with nitrous acid gives a

compound B. B on controlled oxidation gives compound C. C reduces Tollen’s reagent to givesilver mirror and D. B reacts with D in the presence of concentrated sulphuric acid to give sweetsmelling compound E. Identify A, B, C, D and E. Give the reaction of C with ammonia.

Ans.A

HNO2⎯⎯⎯→ B

[O]⎯⎯→ C

Tollen’s

reagent⎯⎯⎯⎯→ D + Ag↓

(C2H7N) H+D

E

A : C2H5NH2 (ethylamine)B : C2H5OH (ethanol)C : CH3CHO (ethanal)D : CH3COOH (ethanoic acid)E : CH3COOC2H5 (ethyl acetate)

CH3 C|H

= O + NH3

Ammonia

Acetaldehyde ammoniaAcetaldehyde (C)

OH|

CH3 —C—NH 2|H

⎯→

Q. 8. An organic compound A with molecular formula C7H8 on oxidation by chromyl chloride in thepresence of CCl4 gives a compound B which gives positive Tollen’s test. The compound B ontreatment with NaOH followed by acid hydrolysis gives two products C and D. C on oxidationgives B which on further oxidation gives D. The compound D on distillation with sodalime givesa hydrocarbon E. Below 60°C, concentrated nitric acid reacts with E in the presence ofconcentrated sulphuric acid forming a compound F. Identify the compounds A, B, C, D, E and F.

Ans. A → C6H5CH3 B → C6H5CHO C → C6H5CH2OH Toluene Benzaldehyde Benzyl alcoholD → C6H5COOH E → C6H6 F → C6H5NO2

Benzoic acid Benzene NitrobenzeneQ. 9. Identify the compounds A, B, C, D, E and F :

HC ≡ CH [A]

[E]

dil.H2SO4

Hg2+

[O]

NH2OH C6H5OH

[B]SOCl2

[C]

[F]

ΔΔ

[D]CH3COONa

Ans. A is CH3CHO ⎯→ Acetaldehyde

B is CH3COOH ⎯→ Acetic acid

C is CH3COCl ⎯→ Acetyl chloride

D is CH3CO.O.COCH3 ⎯→ Acetic anhydride

E is CH3CH=NOH ⎯→ Acetaldoxime

F is C6H5.O.COCH3 ⎯→ Phenyl acetate.


Q. 10. An organic compound [A] having molecular formula C2H7N on treatment with nitrous acid givesa compound [B] having molecular formula C2H6O. [B] on treatment with an organic compound[C] gives a carboxylic acid [D] and a sweet smelling compound [E]. Oxidation of [B] withacidified potassium dichromate also gives [D].

(i) Identify [A], [B], [C], [D] and [E].

(ii) Write balanced chemical equation of [D] with chlorine in the presence of red phosphorus andname the reaction.

Ans. (i) A is C2H5NH2 ⎯→ EthylamineB is C2H5OH ⎯→ EthanolC is CH3CO.O.CO.CH3 ⎯→ Acetic anhydrideD is CH3COOH ⎯→ Acetic acidE is CH3COOC2H5 ⎯→ Ethyl acetate

(ii) CH3COOH + Cl2 Red P

⎯⎯⎯→– HCl

CH2ClCOOHmonochoro acetic acid

This reaction is called Hell-Volhard-Zelinsky (HVZ) reaction.Q. 11. Identify the products A, B and C :

Alk.KMnO4⎯⎯⎯⎯→ A

ΔNaOH‚ CaO

(soda lime)⎯⎯⎯⎯→

Cl2, FeCl3

CH3

⎯⎯

→

C

B

Ans. A → Benzoic acid, B → Benzene, C → 3, 5 dichloro benzoic acid.CH3

alk KMnO4⎯⎯⎯⎯→

COOH

[Benzoic acid]

ΔNaOH‚ CaO

⎯⎯⎯⎯→ [Benzene]

Cl

COOH

Cl

Cl2, FeC

⎯⎯

→

[3, 5–dichloro benzoic acid]Q. 12. Identify A to F :

A LiAIH4⎯⎯→ C2H5OH

PBr3⎯⎯→ B KCN

⎯⎯→ C D

⎯→ C3H7NH2 HNO2⎯⎯→ E

[O]K2Cr2O7/H+⎯⎯⎯⎯→ F

Ans. A → CH3CHO, B → C2H5Br, C → C2H5CN, D → LiAlH4, E → C3H7OH, F → C2H5COOH.Q. 13. An organic compound A has the molecular formula of C7H6O. When A is treated with NaOH

followed by acid hydrolysis, it gives two products, B and C. When B is oxidised, it gives A. WhenA and C are each treated separately with PCl5, they give two different organic products D and E.

(i) Identify A to E.

(ii) Give the chemical reaction when A is treated with NaOH and name the reaction.

Ans. (i) A → C6H5CHO, B → C6H5CH2OH, C → C6H5COOH

D → C6H5CHCl2, E → C6H5COCl.(ii) 2C6H5CHO + NaOH ⎯→ C6H5COONa + C6H5CH2OH

This is Cannizzaro’s reaction.


Q. 14. Identify the products A, B, C and D

CH3COOH PCl3⎯⎯→ A

NH3⎯⎯→ B NaOH

⎯⎯→ C + DAns. A → CH3COCl (Acetyl chloride) B → CH3CONH2 (Acetamide)

C → CH3COONa (Sodium acetate) D → NH3 (Ammonia)Q. 15. Identify the reagents W, X, Y and Z

C6H6 W

⎯→ C6H5SO3H X

⎯→ C6H5ONa Y

⎯→ C6H5OH Z

⎯→ C6H5OCOCH3Ans. W → Conc. H2SO4, X → NaOH, Y → HCl, Z → CH3COCl.

Q. 16. Identify the compounds A, B, C, D

CH3CN H+

HOH A

P2O5 B C6H5NH2 C + A

Cl2, PD

⎯⎯→ ⎯⎯⎯⎯→⎯⎯⎯⎯→↓

Ans. A ⎯→ CH3COOH, B ⎯→ CH3CO.OCOCH3,C ⎯→ C6H5NHCOCH3, D ⎯→ CH2Cl COOH

Q. 17. Identify the reagents A, B and C.CH2OH

CH2OH

A⎯→

COOH

COOH

↓ CCO2 + H2O

B⎯→

COOC2H5

COOC2H5

Ans. A ⎯→ Conc. HNO3 , B ⎯→ C2H5OH, C ⎯→ acidified KMnO4

Q. 18. A compound with molecular formula C7H6O2 when treated with SOCl2 gives compound Awhich on treatment with ammonia forms compound B and reacts with Br2/KOH to form C.compound C7H6O2 also gives a fruity smell when heated with conc. H2SO4 and ethanol.IdentifyA, B, and C.

Ans. A ⎯→ C6H5COCl, B ⎯→ C6H5CONH2, C ⎯→ C6H5NH2

Q. 19. Identify products A, B, C and D required for the following conversions.

(i) CH3COOH Ca (OH)2⎯⎯⎯→ A

Dry⎯⎯⎯→Distillation

B NH2OH

⎯⎯⎯→ C KOH

⎯⎯→ D

(ii) CH3COOH A

⎯→ CH3COCl B

⎯→ CH3CONH2 C

⎯→ CH3NH2 D

⎯→ CH3OH

(iii) CH3CH2COOHP/Br2⎯⎯→ (A)

(i) alc. KOH⎯⎯⎯⎯→

(i) H+ (B)

(iv) CH3COOH Ca(OH)2⎯⎯⎯→ (A)

Heat⎯⎯⎯→ (B)

NH2.NH2⎯⎯⎯→ (C)

(v) CH3COCH3 LiAlH4⎯⎯⎯→ (A)

PCl5⎯⎯→ (B) alc.KOH

⎯⎯⎯→ (C)

Ans. (i) CH3COOH Ca(OH)2

⎯⎯⎯→ (CH3COO)2CaCalcium acetate

dry

⎯⎯⎯⎯→distillation

CH3COCH3Acetone

NH2OH

⎯⎯⎯⎯→ CH3 CH3

C = NOH

Acetone oxime

(A) (B) (C)

KOH ⎯→ CH3—

OH|C—CH2|CH3

—

O||C—CH3

4 Hydroxy-4-methyl pentane-2-one

(D)

(ii) CH3COOH PCl5⎯⎯→(A) CH3COCl

NH3⎯⎯→(B) CH3CONH2 Br2 + KOH

⎯⎯⎯⎯→(C) CH3NH2 HNO2⎯⎯⎯→(D) CH3OH

A—Phosphorus pentachlorideB—AmmoniaC—Bromine and alkaliD—Nitrous acid


(iii) CH3CH2COOHPropionic acid

P/Br2⎯⎯→HVZ α-Bromopropionic (A)

CH3CHCOOH|Br

(i) alc. KOH⎯⎯⎯⎯→

(i) H+ CH2 = CH–COOHAcrylic acid (B)

(iv) CH3COOHAcetic acid

Ca(OH)2⎯⎯⎯→ (CH3COO)2Ca

Calcium acetate(A)

heat⎯→ CH3COCH3

Acetone(B)

NH2.NH2⎯⎯⎯⎯→ (CH3)2C = N.NH2

Acetone hydrazone

(C)

(v) CH3COCHAcetone

3 LiAlH4⎯⎯⎯→ CH3.C

|OH

H.CH3

Propane-2-ol(A)

PCl 5⎯⎯→ CH3.C|

Cl

H.CH3

2-chloropropane(B)

alc. CH2 = CH—CH 3

Propene

(C)

⎯→KOH

Q. 20. An organic compound ‘A’ on treatment with ethyl alcohol gives carboxylic acid ‘B’ andcompound ‘C’. Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ withKMnO4 also gives B. B on heating with Ca (OH)2 gives E with molecular formula C3H6O. E doesnot gives Tollen’s test or reduces Fehling’s solution but forms 2, 4-Dinitro phenyl hydrazone.Identify A, B, C, D and E.

Ans. Given that E does not gives Tollen’s test or Fehlings test, hence, it should be a Ketone. Thesequence of reaction is—

CH3—

O||C—O—

O||C—CH3 + C2H5OH ⎯→ CH3—

O||C—OH + C2H5—O—

O||C—CH3

H2O/H+

⎯⎯⎯→ C2H5OHAcetic anhydride Ethanol Acetic acid Ethylacetate Ethanol (A) (B) (C) (D)

KMnO4⎯⎯→

[O] CH3COOH

(B)

(i) Ca(OH)2⎯⎯⎯⎯⎯→(ii) Distillation

CH3—

O||C—CH3

Acetone (E)

Q. 21. Compound A (C6H12O2) on reaction with LiAlH4 yields two compounds B and C. The compoundB on oxidation gives D, which on treatment with aqueous alkali and subsequent heatingfurnished E and E on catalytic hydrogenation gave C. D was oxidized to from F, which ismonobasic carboxylic acid with molecular mass 60. Deduce the structure of A, B, C and D.

Ans. Let the monobasic carboxylic acid, F be CnH2nO2.

–14 n + 32 = 60or 14n = 60 – 32

or n =2814 = 2

Hence, the acid is –C2H4O2 or CH3COOHThus F is CH3COOH (Ethanoic acid)Thus, the structures deduced are

A—Ethylbutanoate — CH3CH2CH2

O||C —O—C2H5

B—Ethanol — C2H5OHC—1-Butanal — CH3CH2CH2CH2OHD—Ethanal — CH3CHOE—2-Butenal — CH3CH = CHCHOF—Ethanoic acid — CH3COOH

CH3CH2CH2—

O||C—OC2H5

LiAlH4⎯⎯→ C2H5OH + CH3CH2CH2OHEthyl butanoate Ethanol 1-Butanol(A) (B) (C)

H2/Ni (Both – CHO and C = C are reduced)

CH3COOH [O]

←⎯⎯⎯Oxidation

CH3CHOEthanal

(i) Aq. NaOH⎯⎯⎯⎯→

(ii) Δ CH3CH = CHCHO

2-Butanal

⎯⎯→

←⎯⎯⎯⎯

[O]

(E) (D)(F) Aldol condensation


Q. 22. A compound which reduced Fehlings solution, on oxidation with acidified potassiumpermangnate, forms a compound Y having the same number of carbon atoms as X, which reactswith aqueous sodium carbonate generating carbon dioxide. Y on reaction with ethanol in thepresence of sulphuric acid forms a pleasant smelling compound Z, of molecular mass 88. Identifythe compounds X, Y and Z by name. Explain how you named compound Z. What is thecompound that is isomeric with Z and has similar properties ?

Ans. Because compound X reduces Fehling solution and oxidizes acidified KMnO4 into a compoundof same carbon atoms, it must be an aldehyde. Y reacts with aqueous Na2CO3 and ethanol inpresence of H2SO4, thus it must be an acid (R–COOH). Y on treatment with ethanol in presence ofH2SO4 gives a pleasant smelling compound, Z which should be an ester (RCOOC2H5) ofmolecular mass 88. Now the molecular mass of –COOC2H5 is 72, thus the molecular mass of Rwill be 15 (i.e., 88 – 72 = 16). Thus the –R will be –CH3 (mol. mass 15).

CH3CHOEthanol

KMnO4/H+⎯⎯⎯→ CH3COOH

Acetic acid

C2H5OH⎯⎯⎯→conc. H2SO4

CH3COOC2H5Ethylethanoate

(X) (Y) (Z)Hence,

X = Ethanol or AcetaldehydeY = Ethanoic acid or Acetic acid

Z = Ethylethanoate or Ethylacetate

Z is an ester. As per the IUPAC system, the names of the alkyl group attached with the oxygen ofcarboxylate group comes before the root word followed by the name of carboxylate ion. Thus thename of Z would be Ethyl acetate or in IUPAC system it would be Ethylethanoate.

The isomeric compound of Z having similar properties would be methyl propanoate i.e.C2H5COOCH3.

Q. 23. An organic compound ‘A’ with molecular formula C5H8O2 is reduce to n-pentane on treatmentwith Zn-Hg/HCl. ‘A’ forms a dioxime with hydroxyl amine and gives a positive Iodoform testand Tollen’s test. Identify the compound A and deduce its structure.

Ans. As ‘A’ gives positive iodoform test. So, it has CH3—C—group

O

As ‘A’ gives positive Tollen’s test so it must have —CHO group So A is CH3—C—CH2CH2—CHO

O4-Oxopentenal

Q. 24. An organic compound ‘A’ with molecular formula C8H8O gives positive DNP and Iodoformtests. It does not reduce’s Tollen’s or Fehling’s reagent, does not decolourised bromine wateralso. On oxidation with chromic acid (H2CrO4), It gives a carboxylic acid (B) with molecularformula C7H6O2. Deduce the structure of A and B.

Ans. As ‘A’ does not gives Fehling’s solution or Tollen’s test, so it does not have —CHO group but itgives positive iodoform test and DNP test so it has CH3—C—group

O

.

So, as ‘A’ is COCH3

Acetophenone

B is carboxylic acid obtained by oxidation of A with H2CrO4.So, ‘B’ is COOH

Benzoic acid


Q. 25. An alkene ‘A’ with molecular formula (C5H10) on ozonolysis gives a mixture of two compounds‘B’ and ‘C’. Compound ‘B’ gives positive Fehlings test and also reacts with iodine and NaOHsolution compound ‘C’ does not gives Fehlings test but forms iodoform. Identify the compounds‘A’, ‘B’ and ‘C’ giving suitable explanation and write the reaction of ozonolysis and iodoformformation from either ‘B’ or ‘C’.

Ans. CH3 — CH = C — CH3|CH3

(i) O3⎯⎯⎯⎯→(ii) Zn/H2O CH3 —CHO + O = C — CH3

|CH3

(A)

(B)Acetaldehyde

Acetone(C)

2-Methyl but-2-ene

CH3—CHO + 3I2 + 4NaOHIodoform

reaction HCOONa + CHI3 + 3NaI + 3H2O⎯⎯⎯⎯⎯→

Other isomers of ‘A’ will not give product corresponding to the given test.Q. 26. A compound ‘X’ (C2H4O) on ozonolysis gives Y (C2H4O2). ‘X’ undergoes haloform reaction on

treatment with HCN. ‘X’ forms a product ‘Z’ which on hydrolysis gives -2-hydroxy propanoicacid.(i) Write down structure of ‘X’ and ‘Y’(ii) Name the product when X reacts with dil. NaOH(iii) Write down the equation for the reaction involved.

Ans. (i) X = CH3CHO, Y = CH3COOH(ii) 3-Hydroxybutanol

(iii) CHI3 I2/NaOH

[Haloform test]CH3CHO

[O]⎯⎯→ CH3COOH HCN

CH —CH—OH|COOH

H2O/H+

CH3—CH—OH|CN(Z)

←⎯⎯⎯⎯⎯⎯

3 ←⎯⎯⎯⎯

⎯⎯→[X] [Y]

Q. 27. An organic compound ‘A’ on treatment with ethyl alcohol gives carboxylic acid ‘B’ andcompound C. Hydrolysis of C under acidic conditions gives ‘B’ and D. oxidation of ‘D’ withKMnO4 also give ‘B’. B on heating with Ca(OH)2 gives E. With molecular formula C3H6O. E doesnot gives Tollen’s test or reduces Fehling’s solution but forms 2,4-dinitro phenyl hydrozoneIdentify A, B, C, D and E.

Ans. A = CH3—C

O

O

CH3—C—OH

Acetic anhydride

Acetic acid CH3—C

B =

OO

C = CH3 — C — O — CH2 —CH3Ethylacetate

D = CH3CH2—OHEthyl alcohol

E = CH3 — C—CH3Acetone

O

O

Q. 28. An organic compound (A) having molecular formula C9H10O forms an orange red precipitate (B)with 2,4-DNP reagent. Compound (A) gives a yellow precipitate C when heated in the presenceof Iodine and NaOH along with a colourless compound (D). (A) does not reduce’s Tollen’sreagent or Fehlings solution nor does it decolourizes bromine water. On drastic oxidation of (A)with chromic acid, a carboxylic acid E of molecular formula C7H6O2 is formed. Deduce thestructure of the organic compounds A to E.


Ans. (A) CH2—C—CH3

O

(B)CH2—C = N—NH—

H NO3

NO3

(C) CHI3 (Yellow precipitate)

(D)CH2COO–Na+

(Colourless compound)

(E)COOH

Q. 29. An unknown aldehyde ‘A’ on dissolving in alkali reacts with it to give a β-Hydroxy aldehydewhich loses water molecule to form an unsaturated aldehyde 2-butanol. Another Aldehyde ‘B’undergoes disproportionation reaction in the presence of conc. alkali to form product C and D. Cis an aryl alcohol with the formula C7H8O.

(i) Identify A and B.

(ii) Write the sequence of reaction involved.

(iii) Name the product when B reacts with zinc amalgam and hydrochloric acid.

Ans. (i) A is CH3CHO (ethanal)

B is C6H5CHO (benzaldehyde)

(ii) 2CH3CHO NaOH CH3 —CH—CH2—CH3

– H2 O⎯⎯⎯→ CH3—CH = CH—CHO

But-2 enal(A)

OH

CHO CH2OH

2

(B) (C)

COO–

Alkali⎯⎯⎯→ +

(iii) Toluene

Q. 30. An organic compound ‘A’ (C3H4) on hydration in presence of H2SO4/HgSO4 gives compound B(C3H6O). Compound ‘B’ gives white crystalline product (D) with sodium hydrogen sulphide. Itgives negative Tollens test and positive Iodoform test. On drastic oxidation ‘B’ gives compoundC (C2H4O2) along with formic acid. Identify compounds ‘A’, ‘B’ and ‘C’ and explain all thereactions.

Ans.

CH3 — C ≡ CH Hg2+ / H2SO4 CH3—C—CH3

O

(A)

(B)

(C)

(D)

NaHSO4

CH3

CH3

C

OSO2–Na+

OH

[O]

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→

CH3COOH + HCOOH↓

Compound B (propanone) is a Ketone therefore Fehling’s Test and Tollen’s test are negative.


Q. 31. A ketone (A) which gives haloform reaction, on reduction gives compound (B). The compound onheating with conc. H2SO4 gives a compound (C) which forms monoozonide. The decompositionof this monoozonide gives only acetaldehyde. Identify (A), (B) and (C).

Ans. (A) = CH3COCH2CH3 (B) CH3CH—CH2—CH3 (C) CH3CH —— CH CH3|

OHQ. 32. Compound (A) C5H10O, forms a derivative with hydrazine and gives yellow ppt. with iodine and

sodium hydroxide solution. It does not reacts with Tollen’s reagent. Identify the compound A.

Ans. 2-Pentanone.

O

Q. 33. A compound having molecular formula C3H6O forms a crystalline white precipitate with sodiumbisulphite and gives a red precipitate with Fehling’s solution. Write structure formula, commonname and IUPAC name of the compound.

Ans. Structure : CH3CH2CHO

Common name : Propionaldehyde

IUPAC name : Propanal

Q. 34. An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and acompound ‘C’. Hydrolysis of C under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ withKMnO4 also gives ‘B’. B upon heating with Ca(OH)2 gives E (C3H6O). E does not gives Tollen’stest and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenyl hydrozone. Identify A,B, C, D and E.

Ans. (A) = (CH3CO)2O (B) = CH3COOH

(C) = CH3COO C2H5 (D) = C2H5OH

(E) = CH3COCH3

Q. 35. An organic acid (A) C5H10O2 reacts with Br2. In the presence of phosphorus to give (B).Compound B contains an asymmetrical carbon atom and yields (C) on dehydrobromination.Compound (C) does not shows geometrical isomerism and on decarboxylation gives an alkene(D) which upon ozonolysis gives (E) and (F). Compound (F) gives positive Schiff’s test but (F)does not. Give structures of (A) to (F).

Ans.CH3

CH3

CHCH2COOHBr2,P CH3

CH3

CH—CH—COOH

Br

– HBr C = CHCOOH(C)

– CO2

HCHO

(F)

+ C = O ozonolysis C = CH2

(D)

⎯⎯⎯→ ⎯⎯⎯→CH3

CH3

(B)(A)

CH3

CH3

CH3

CH3

(E)

←⎯⎯⎯⎯⎯⎯

⎯⎯→

Q. 36. An aromatic compound ‘A’ of molecular formula C7H6O2 undergoes a series of reaction as shownbelow. Write the structures of A, B, C, D and E of the following reactions.

(C7H6O2) A B

LiAlH4/Ethen Br2 (aq)

CC6H5—CONH2Heat/NH3

⎯⎯⎯→

⎯⎯⎯→

⎯⎯⎯⎯⎯→(CH3Cl) 2O⎯⎯⎯⎯⎯→

Br2 + NaOH⎯⎯⎯⎯⎯→

D E


Ans.

Benzoicacid

Aniline

Benzylamine

2, 4, 6-Tribromoaniline

COOH NH2

CH2NH2

NH—C—CH3

A = B = C =

D =

O

N-phenylethanamide

NH2Br Br

Br

E =

Q. 37. Write the structures of A, B, C and D in the following reaction :

AC6H5—COCl

CH3MgBr/H3O+

D

B + C⎯⎯

⎯→

H2/Pd-BaSO4⎯⎯⎯⎯⎯⎯→

Conc. NaOH⎯⎯⎯⎯⎯⎯→

Ans.

Benzaldehyde Sodiumbenzoate

CHO COONa CH2OH

A = B or C= C or B =

Benzoylalcohol

CH—CH3

D =

OH

1-phenyl ethanol

Chapter 13. Organic Compounds Containing NitrogenQ. 1. Identify A, B, C and D.

N2Cl

Cu/HCl

⎯⎯⎯→

Na

A

D↓

Mg⎯⎯⎯→ B

HOH⎯⎯⎯→ C

Ans.

A →

Cl

, B →

MgCl

C → , D

Q. 2. Identify the reagents X, Y and Z.

C2H5Cl X

⎯→ C2H5CN Y

⎯→ C2H5CH2NH2 Z

⎯→ C2H5CH2NHCOCH3

Ans. X → alcoholic KCN, Y → LiAlH4, Z → CH3COCl.

Q. 3. An organic compound A with molecular formula C2H7N on reaction with nitrous acid gives acompound B. B on controlled oxidation gives a compound C. C reduces Tollen’s reagent to givesilver mirror and D. B reacts with D in the presence of concentrated sulphuric acid to give a


sweet smelling compound E. Identify A, B, C, D and E. Give the reaction of C with ammonia andname the product.

Ans. CH3CH2NH2

A

Ethyl Amine

HNO2⎯⎯→ CH3CH2OH

B

Ethyl alcohol

[O]

Controlled⎯⎯⎯⎯→ CH3CHO

C

Acetaldehyde

Tollen’s

Reagent⎯⎯⎯→ CH3COOH

D

Acetic acid

CH3COOH + CH3CH2OH

CH3CHO + NH3

H2 SO4

CH3—C—OH|H

|NH2

Acetaldehyde ammonia

conc.

⎯⎯⎯⎯→

Q. 4. State the reagents for the following conversions.

Benzene A

⎯→ Nitrobenzene B

⎯→ Aniline C

⎯→ Aniline hydrochloride D

⎯→ Benzene diazoniumchloride

Ans. A is conc. HNO3 B is Sn and HCl acidC is conc. HCl D is HNO2

Q. 5. Give the name and formula of each A, B, C, D, E and F in the following conversion reactions :

[A] HNO3 Conc.H2SO4 Conc.

[B] Sn/HClHeat

[C] CHCl3/KOH

Heat [D]

H2/Pt

[E]

[F]

NHCOCH3

+ HCl

⎯→

⎯⎯→⎯⎯⎯⎯⎯→⎯⎯⎯→⎯⎯⎯⎯⎯⎯→

Ans. A → Benzene (C6H6)

B → Nitrobenzene (C6H5NO2)

C → Aniline (C6H5NH2)

D → Phenyl Isocyanide (C6H5NC)

E → Methyl phenyl amine (C6H5NHCH3)

F → Acetyl chloride (CH3COCl)

Q. 6. Identify the reagents A, B, C and D

C2H5—Cl A

⎯→ C2H5CN B

⎯⎯→ C2H5CH2NH2 C

⎯⎯→ C2H5CONH2 D

⎯⎯→ C2H5NH2

Ans. A ⎯→ KCNalc, B ⎯→ LiAlH4, C ⎯→ H2O (H+), D ⎯→ Br2/KOH

Q. 7. Identify A, B, C, D and E.

C6H6 Conc. HNO3 Sn/HCl C6H5COCl

⎯⎯⎯⎯⎯⎯→ B

Cl2 + AlCl3

NaNO2 + Conc. HCl

E D

⎯⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯⎯⎯→

↓

A

↓

C

Ans.

, B ⎯→ , C ⎯→A ⎯→

, E ⎯→D ⎯→

NO2 NH2 NHCOC6H5

N2Cl NO2


Q. 8. A compound A having a molecular formula C2H7N on treatment with nitrous acid gave acompound B having a molecular formula C2H6O. The compound B on treatment with acetylchloride gave a compound C having the molecular formula C4H8O2.

(i) Identify the compounds A, B and C and write their structural formulae.

(ii) Write balanced equations for the formation of B and C.

Ans. Removal of nitrogen atom from compound A, on treatment with HNO2 means that compound Ais a primary amine, hence A is C2H5NH2. Primary amine with nitrous acid forms alcohol (B)which on treatment with acetyl chloride forms esters (C).

(i) A = CH3CH2NH2 Ethanamine

B = CH3CH2OH Ethanol

C = CH3CH2O—C||O

—CH3 Ethylethanoate

(ii) CH3CH2NH2HNO2⎯→ CH3CH2OH

CH3COCl⎯⎯⎯→ CH3CH2—O—C

||O

—CH3

(A) (B) (C)

Q. 9. An aromatic compound (A) C7H8 on nitration gives (B) as a major product. (B) on reduction withSn/HCl gives (C). (C) on treatment with NaNO2/HCl at 273 K followed by subsequent treatmentwith CuCl/HCl gives (D). When (D) is oxidized, it forms ortho substituted monocarboxylic acidwith formula C7H5O2Cl. Predict the structures of A, B, C and D.

Ans. D is an ortho substituted monocarboxylic acid. It means it contains —COOH and a chloro groupat the ortho position.

Therefore its formula is :

COOH

Cl

2-chlorobenzoic acid

Hence starting compound must contain a benzene ring and its formula could be C6H5—CH3 i.e.,Toluene.

⎯⎯⎯⎯⎯⎯→

⎯⎯

⎯→

COOH

Cl Cl

CH3

NO2

CH3

HNO3/H2SO4

–H2O

←⎯⎯⎯[O]

Oxidation←⎯⎯⎯⎯CuCl/HCl

–N2

⎯⎯⎯⎯⎯⎯⎯→273 K NaNO2/HCl

–H2O⎯⎯⎯→Sn/HCl

4 [H]

(C)

(D)

(A) (B)o-nitro toluene

o-chlorobenzoic acid

CH2

o-chlorotoluene

N2+Cl–

CH2

o-toluenediazonium chloride

CH3

NH2

o-toluidineToluene

Q. 10. Write the structure of A, B, C, D and E in the following reaction :

C6H5NH2 CH3COCl

Pyridine⎯⎯⎯⎯→ A

Br2/CH3COOH⎯⎯⎯⎯⎯⎯→ B

|||||↓

CHCl3 + KOH

|||||↓

HNO3 + H2SO4

|||||↓

H+

D E C


Ans.

Acetanilide

p-bromoacetanilide

A = B =

NH—C—CH3

O

Br

NH—C—CH3

O

p-nitro acetanilideNO2

NH—C—CH3

O

p-bromoaniline

C =

NH2

Br

phenylisocyanide

D =

N+ ≡ NC–

E =

Q. 11. An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions asshown below :Write the structure of A, B, C, D and E in the following reactions :

(C7H7ON) A Br2 + KOH C6H5NH2NaNO2 + HCl

273 K

CaCN

CHCl3 + NaOH H2O

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→ ⎯⎯⎯→B⎯⎯

⎯→

C

D E

⎯⎯

⎯→

Ans.

BenzamideA = C6H5CONH2

phenylisocyanide

D =

N ≡ C

phenol

E =

OH

Benzenediazoniumchloride

B =

N+ ≡ NCl–

Benzonitrile

C =

C ≡ N

Q. 12. An aromatic compounds ‘A’ of molecular formula C7H7ON undergoes a series of reaction asshown below. Write the structures A, B, C, D and E in the following reactions :

(C7H7ON) A Br2 + KOH C6H5NH2NaNO2 + HCl

273 K

KI

CH3 + NaOH⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→ B⎯⎯

⎯→

⎯⎯⎯⎯⎯→ C

D E

CHCl3 + NaOH

⎯⎯

⎯→

Ans.

Benzamide

C—NH2

O

A =

Benzenediazonium

chloride

N+ ≡ NC–

B =

Benzene

C =

phenylisocyanide

N+ NC–

D =

Iodobenzene

I

E =

=→


Q. 13. Give the structures of A, B and C in the following reactions :

(i) CH3CH2I NaCN

⎯⎯⎯→ A OH

–

⎯⎯⎯⎯⎯⎯⎯→Partial hydrolysis

B NaOH + Br2

⎯⎯⎯⎯⎯→ C

(ii) CH3CH2Br KCN


⎯⎯⎯⎯⎯→ B HNO2

⎯⎯⎯→0°C

C

(iii) CH3COOH NH3

⎯⎯⎯→Heat

A NaOBr

⎯⎯⎯→ B NaNO2/HCl

⎯⎯⎯⎯⎯⎯→ C

Ans. (i) CH3CH2 I NaCN

⎯⎯⎯→ CH3CH2CN OH

–

⎯⎯⎯⎯⎯⎯⎯→Partial hydrolysis

CH3—CH2—C—NH2

O

Iodoethane Propanenitrile Propanamide

(A) (B)NaOH + Br2

⎯⎯⎯⎯⎯→ CH3—CH2—NH2

Ethanamine (C)

(ii) CH3CH2Br KCN

⎯⎯⎯→ CH3CH2CN LiAlH4

⎯⎯⎯⎯⎯→(Reduction)

CH3—CH2—CH2NH2

Propane nitrile Propan-1-amine

(A) (B)

HNO2 0°C

⎯⎯⎯⎯→ [CH3CH2CH2N+ ≡ NCl–]

H2O⎯⎯→ CH3CH2CH2OH

Propanol (C)

(iii) CH3COOH NH3

⎯⎯⎯→Heat

CH3CONH2 NaOBr

⎯⎯⎯→ CH3NH2 NaNO2/HCl

⎯⎯⎯⎯⎯⎯→ CH3OH

Ethanoic acid Ethanamide Methamine Methanol

(A) (B) (C)Q. 14. Give the structures of A, B and C in the following reactions :

(i) C6H5N2Cl– CuCN

⎯⎯⎯→ A H2O/H+

⎯⎯⎯⎯→ B NH3

⎯⎯⎯→Δ

C

(ii) C6H5NO2 Fe/HCl

⎯⎯⎯⎯⎯→ A NaNO2 + HCl

⎯⎯⎯⎯⎯⎯→273 K

B H2O/H+

⎯⎯⎯⎯⎯→ C

(iii) C6H5NO2 Fe/HCl

⎯⎯⎯⎯→ A HNO2

⎯⎯⎯→273 K

B C6H5OH

⎯⎯⎯⎯→ C

Ans. (i) C6H+5N2Cl

–

CuCN⎯⎯⎯→ C6H5CN

H2O/H+

⎯⎯⎯⎯→ C6H5COOH NH3

⎯⎯→ C6H5—COO–NH4+

Benzenediazonium Cyano hydrolysis Benzoic acid

chloride benzene (A) (B)

|||↓Δ Heat

C6H5—CONH2

Benzamide

(C)

(ii)Nitrobenzene

C6H5NO2 Fe/HCl

⎯⎯⎯⎯⎯→ Aniline (A)C6H5NH2

NaNO2 + HCl⎯⎯⎯⎯⎯→

273 K BenzenediazoniumC6H5N

+ ≡ NCl

–

H2O/H+

⎯⎯⎯⎯→Heat

Phenol (C)C6H5OH

Chloride(B)

(iii) C6H5NO2 Fe/HCl

⎯⎯⎯⎯→ C6H5NH2 NaNO2/HCl

⎯⎯⎯⎯⎯⎯→273 K

C6H5—N ≡ NCl

Nitrobenzene Aniline Benzenediazonium(A) chloride (B)

C6H5OH

⎯⎯⎯⎯→ N N OH=

p-Hydroxyazobenzene

(C)


Q. 15. Identify A and B in the following reactions :O

ClAKCN

⎯⎯→H2/Pd

⎯⎯⎯→ B

Ans.

(A)

KCN

OCl

OCH2NH2

⎯⎯→H2/Pd

⎯⎯⎯→(B)

OCN

Q. 16. Write the main products of the following reactions :

(i) C6H5N+2 Cl–

H2PO2 + H2O⎯⎯⎯⎯⎯⎯→ ?

(ii)

NH2

?Br2(aq)

⎯⎯⎯→

(iii) CH3 —C—NH2

O

Br2 + NaOH⎯⎯⎯⎯→ ?

Ans. (i) C6H5N+2 Cl–

H3PO2 + H2O⎯⎯⎯⎯⎯⎯→ C6H6 + N2 + H3PO3 + HCl

(ii)

NH2

Br2(aq)⎯⎯⎯→

NH2Br Br

Br

(iii) CH3—C—NH2

O

Br2 + NaOH⎯⎯⎯⎯⎯→ CH3—NH2 + Na2CO3 + 2NaBr + 2H2O.

Q. 17. Give the structures of A, B and C in the following reactions :

(i) CH3Br KCN


⎯⎯⎯⎯→ B HNO2

⎯⎯⎯→273 K

C

(ii) CH3COOH NH3

⎯⎯⎯→Δ

A Br2 + KOH

⎯⎯⎯⎯→ B CHCl3 + NaOH⎯⎯⎯⎯⎯→ C

Ans. (i) CH3Br KCN

⎯⎯⎯→ CH3CN LiAlH4

⎯⎯⎯⎯→ CH3CH2NH2 HNO2

⎯⎯⎯→273 K

CH3CH2OH

Bromomethane Ethane nitrile Ethanamine Ethanol(A) (B) (C)

(ii) CH3COOH NH3

⎯⎯⎯→Δ

CH3CONH2 Br2 + KOH

⎯⎯⎯⎯→ CH3NH2 CHCl3 + NaOH⎯⎯⎯⎯⎯→ CH3NC

Ethanoic Ethanamide Methanamine Methyl isocyanideacid (A) (B) (C)

Q. 18. Write structure of reagents or organic compounds (A to F) in the following sequence of reaction :

AHNO 3 (conc.)

H2SO4 (conc.)Sn/HCl (conc.) CHCl3/KOH

Δ

H2/Pt F⎯⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯→⎯⎯⎯⎯⎯⎯→B C D ENHCOCH3

+ CH3COOH

Ans.

Benzene Nitrobenzene

Phenylisocyanide

N-methylaniline

NO2 NH2

NC

A = B = C =

D =

Aniline

NHCH3

E =


Q. 19. Write the structure of reagents/organic compounds (A to P) in the following sequence ofreaction.

A NaOH/Br2

⎯⎯⎯⎯⎯→ B NaNO2/HCl

O°C⎯⎯⎯⎯⎯⎯→ C

Δ⎯→ E

Cl2/Fe⎯⎯⎯→ F

H2/Pt

⎯⎯

⎯→

CH2NH2

Ans.

Benzamide Aniline Benzenediazonium

chloride

Benzene Chlorobenzene

A = B = C =

E = F =

Cl

N ≡ NClCONH2 NH2

D = H3PO2/H2O

Q. 20. Predict the reagent or the product in the following reaction sequences :

NaNO2/HCl←⎯⎯⎯⎯⎯

5←⎯

1⎯→

⎯⎯→

CH3

NHCOCH3

2

4

3

CH3

NO2

CH3

NO2

CH3

NH2

(CH3CO)2O⎯⎯⎯⎯⎯→

PyridineHNO3⎯⎯⎯→H2SO4

CH3

NH2

NO2

Ans. 1. Sn—HCl 2.

CH3

NHCOCH3

NO23. H2O/H+ 4.

CH3

N2ClNO2

5. H3PO2/H2O

Q. 21.(Optically active)

C5H13N2 Aq. NaNO2/HCl⎯⎯⎯⎯⎯⎯→

– N2 3°°°° Alcohol

Y + other product

(i) Identify (X) and (Y) (ii) Is (Y) optically active ?Ans. (i) Since X give N2 gas on treatment with NaNO2/HCl therefore it must be a primary amine

CH3—CH2—CH—CH2—NH22-Methyl butanamine

(Optically active)

|CH3

(X)

(ii)CH3—CH2—C—CH3

|OH

|CH3

2-Methyl butan-2-ol

(Optically inactive)(Y)

❐

Question

5Set IUPAC Nomenclature

Chapter 9. Coordination CompoundsQ. 1. Write the IUPAC names of the following :

(i) [Co(NH3)4SO4] NO3

(ii) K[Pt (NH3) Cl3]Ans. (i) Tetraamminesulphatocobalt(III) nitrate

(ii) Potassiumaminetrichloroplatinate(II)Q. 2. Write the IUPAC names of the following coordination compounds :

(i) [Cr(NH3)4(H2O)2]Cl3

(ii) [PtCl2(NH3)4] [PtCl4]Ans. (i) Tetraamminediaquachromium (III) chloride

(ii) Tetraamminedichloroplatinum (IV) tetrachloroplatinate (II)Q. 3. Give the IUPAC names of the following coordination compounds :

(i) K2[Zn(OH)4](ii) [Co(NH3)5(CO3)]Cl.

Ans. (i) Potassiumtetrahydroxozincate (II)(ii) Pentaamminecarbonatocobalt (III) chloride

Q. 4. Give the IUPAC names for the following :(i) Na3[AlF6] (ii) [Co(NH3)6]Cl3

Ans. (i) Sodium hexafluroaluminate (III).(ii) Hexaamminecobalt (III) chloride.

Q. 5. Write the IUPAC names for the following coordination compounds :(a) [Co(NH3)6]Cl3 (b) [Co(NH3)5Cl]Cl2

(c) K3[Fe(CN)6] (d) K2 [Zn (OH)4](e) [Co(NH3)5 (CO3)] Cl.

Ans. (a) Hexaamminecobalt (III) chloride.(b) Pentaamminechlorocobalt (III) chloride.(c) Potassium hexacyanoferrate (III)(d) Potassium tetrahydroxozincate (II)(e) Pentaamminecarbonatocobalt (III) chloride.

Q. 6. Write the formulas of the following coordination compounds :(a) Sodium tetrafluorooxochromate (IV)(b) Sodium hexafluorosilicate (IV)(c) Bis (cyclopentadienyl) iron (II)(d) Tetracarbonylnickel (0)(e) Potassium dicyanoargentate (I)

Ans. (a) Na2 [CrOF4] (b) Na2 [SiF6](c) [Fe (C5H5)2] (d) [Ni (CO)4](e) K [Ag(CN)2]

Q. 7. Using IUPAC norms write the formula for the following :(a) Hexamminecobalt (III) chloride.(b) Tetramine chloronitrocobalt (III) chloride.(c) Hexaminenickel (II) chloride.(d) Diamminechloro (methylamine) platinum (II) chloride.

IUPAC Nomenclature ■ 83

(e) Hexaquamanganese (II) ion(f) Tris (1, 2-ethane diamine) cobalt (III) ion(g) Hexaquatitanium (III) ion(h) Tetrachloronickelate (II) ion(i) Tetracarbonyl nickel (O).(j) Potassium tetracyanonickel (0)(k) Triamminetrinitrocobalt (III)

Ans. (a) [Co(NH3)6]Cl3 (b) [CoCl (NO2) (NH3)4]Cl(c) [Ni (NH3)6]Cl2 (d) [PtCl (NH2CH3) (NH3)2]Cl(e) [Mn (H2O)6]2+ (f) [Co (en)3]3+

(g) [Ti (H2O)6]3+ (h) [NiCl4]2–

(i) [Ni(CO)4] (j) K4[Ni(CN)4](k) [Co(NH3)4(NO2)3]

Q. 8. Using IUPAC norms write the systematic names of the following :(a) [Zn (OH)4]2– (b) [Co (NH3)6]2 (SO4)3

(c) K2[PdCl4] (d) K3[Cr(C2O4)3](e) [Pt (NH3)2Cl2] (f) [Pt (NH3 )6]4+

(g) K2 [Ni (CN)4] (h) [CuBr4]2–

(i) [Co(NH3)5 (ONO)]2+.

Ans. (a) Tetrahydroxozincate (II) (b) Hexaamminecobalt (III) sulphate

(c) Potassium tetrachlorodopalladate (II) (d) Potassium trioxalato chromate (III)

(e) Diamminedichlorodoplatinum (II) (f) Hexaammineplatinum (IV)

(g) Potassium tetracyanonickelate (II) (h) Tetrabromidocuprate (II)

(i) Pentaamminenitrito-o-cobalt (III).Q. 9. Give the name of the following :

(i) [Co (NH3)5Cl] Cl2

(ii) [K2 [Ni (CN)4](iii) [Pt (NH3)2 Cl (NO2)](iv) K3 [Cr (C2O4)3](v) [Co (NH3)5 (CO3)] Cl

Ans. (i) Pentaamminechlorido cobalt (III) chloride(ii) Potassium tetracyanonickelate (II)(iii) Diammine chlorido nitrito–N–platinum (II)(iv) Potassium trioxalato chromate (III)(v) Pentaammine carbanato cobalt (III) chloride

Q. 10. Using IUPAC norms write the systematic names of the following :(i) [Co (NH3)6]Cl3

(ii) [Co (NH3)4 Cl(NO2)]Cl(iii) [Ni (NH3)6]Cl2

(iv) [Mn (H2O)6]2+

(v) [Co(en)3]3+

(vi) [Ti (H2O)6]3+

(vii) [NiCl4]2–

(viii) [Ni (CO)4](ix) [CO(NH3)6]Cl3

(x) [CO (NH3)5Cl]Cl2

(xi) K3 [Fe (CN)6]Ans. (i) Hexaammine cobalt (III) chloride


(ii) Tetraammine chlorido nitrito-N-cobalt (III) chloride(iii) Hexaamine nickel (II) chloride(iv) Hexaaqua manganese (II) ion(v) Tris (ethane – 1, 2-diamine) cobalt (III) ion(vi) Hexaaquatitanium (III) ion(vii) Tetrachlorido nickelate (II) ion(viii) Tetracarbonyl nickel (0)(ix) Hexaammine cobalt (III) chloride(x) Pentaammine chlorido cobalt (III) chloride(xi) Potassium hexacyanoferrate (III).

Q. 11. Write IUPAC name of the following coordination compounds :(i) K3 [Fe(C2O4)3](ii) K2 [PdCl4](iii) [Pt (NH3)2Cl(NH2CH3)]Cl

Ans. (i) Potassium trioxalatoferrate (II)(ii) Potassium tetrachloridopalladate (II)(iii) Diammine chlorido (methyl amine) platinum (II) chloride

Q. 12. Write down the IUPAC name of each of the following complexes :(i) K4[Mn(CN)6](ii) [Co(NH3)5Cl]Cl2

(iii) Cs[FeCl4](iv) [Cr(NH3)4Cl2]+

(v) [CrCl2(H2O)4]ClAns. (i) Potassium hexacyanomanganate (II)

(ii) Pentaaminechlorocobalt (III) chloride(iii) Caesiumtetrachloroferrate (III)(iv) Tetra amminedichlorido chromium (III) ion(v) Tetra aquadichlorido chromium (III) chloride

Q. 13. Write the IUPAC name of Fe4 [Fe (CN)6]3.Ans. Iron (III) hexa cyanoferrate (III)

Q. 14. Write down the IUPAC nomenclature of the following compounds :(i) [Pt (NH3)2ClNO2](ii) [CoCl2(en)2Cl](iii) K3[Fe(C2O4)3](iv) [Ag(NH3)2] [Ag(CN)2](v) CrCl3(Py)3

Ans. S. No. Formula Name

(i) [Pt (NH3)2ClNO2] Diammine chloride nitrito–N–platinum (II)(ii) [CoCl2 (en)2Cl] Dichloridobis (ethane – 1, 2–diamine) cobalt (III) chloride(iii) K3[Fe(C2O4)3] Potassium trioxalato ferrate (III)(iv) [Ag(NH3)2] [Ag (CN)2] Diammine silver (I) dicyano argentate (I)(v) CrCl3 (Py)3 Trichlorido pyridine chromium (III)

Q. 15. Write down the IUPAC name of the following compounds :(i) [Co (NH3)4 (H2O) Cl] Cl2 (ii) K2 [Ni (CN)4]

Ans. (i) Tetrammineaquachlorido cobalt (III) chloride(ii) Potassium tetra cyanonickelate (II)

Q. 16. Write down IUPAC name of the following complexes :(i) [Co(NH3)4(H2O)2]Cl3 (ii) K2[PdCl4](iii) [Cr (en)3]Cl3


Ans. (i) Tetraammine diaquacobalt (III) chloride(ii) Potassium tetrachlorido palladate (II)(iii) Tris (ethane–1, 2–diamine) chromium (III) chloride.

Q. 17. Write down the name of the following complexes according to IUPAC norms :(i) [Pt(NH3)BrCl(NO2)]–

(ii) [PtCl2(en)2](NO3)2

(iii) [Fe4[Fe(CN)6]3

Ans. (i) Amminebromidochloridonitrito –N–platinate (II)(ii) Dichloridobis (ethane–1, 2–diammine) N-nitrito platinum (IV) nitrate(iii) Iron (III) hexacynoferrate (II)

Chapter 10. Haloalkanes and HaloarenesQ. 1. Write the structure and IUPAC name of the following compounds :

(i) tert-butyl bromide (ii) neo-pentyl chloride(iii) o-bromotoluene (iv) p-dichlorobenzene

Ans. (i) CH3—C—Br|CH3

|CH3

(ii) CH3—C—CH2Cl|CH3

|CH3

2-Bromo-2-methyl propene 1-Chloro-2, 2-dimethyl propane

(iii) Br CH3

(iv) ClCl

2-bromomethyl benzene 1, 4–dichlorobenzeneQ. 2. Write IUPAC name and structure of the major product of the following reactions :

(i) CH2CH3 + Cl2 hν

⎯⎯→

(ii) Cl + CH3CH2Cl Na

ether⎯⎯⎯→

(iii) Cl Ni/Al alloy

NaOH⎯⎯⎯⎯⎯→

Ans. (i)

Cl|CH—CH3

1-Chloro-1-phenyl ethane

(ii) CH2CH3

Ethyl benzene

(iii)

BenzeneQ. 3. Give the IUPAC names of the following compounds :

(i) ClCH2C ≡ C CH2Br(ii) (CCl3)3CCl(iii) CH3C (p-Cl C6H4)2 CH(Br)CH3

(iv) (CH3)3CCH = Cl C6H4 I – pAns. (i) 1-Bromo-4-chlorobut-2-yne

(ii) 2-(Trichloromethyl)-1, 1, 1, 2, 3, 3, 3 - hepta chloropropane(iii) 2-Bromo-3, 3-bis (4-chlorophenyl) butane(iv) 1-chloro-1-(4-iodophenyl)- 3, 3,-dimethyl but-1-ene


Q. 4. Write the structure of the following organic halogen compounds :(i) p-Bromochlorobenzene(ii) 1-chloro-4-ethyl cyclohexane(iii) Perfluorobenzene(iv) 1, 4-Dibromobut-2-ene.

Ans. (i)

Cl

Br

(ii)

Cl

C2H5

(iii)

FF

FF

F

F(iv)

1 2 43BrCH2CH = CH CH2Br

Q. 5. Give the IUPAC names of following :

(i) CH3—CH—CH—CH3 |Cl

|Br

(ii) CH3—CH—CH2—CH—CH2—CH3 |CH3

|I

(iii)

Cl

ClCH2—CH26

21

122 3

4

Ans. (i) 2–Bromo–3–chlorobutane (iii) 4 – Iodo–2–methyl hexane

(iii) 1–(2–chlorophenyl) 2-(4 chlorophenyl) ethane

Q. 6. Write down the IUPAC name of following :

(i) FCH2—

CH3|C—CH3|F

(ii)I

Br

(iii) CH3—C = CHCH2Cl|CH3

Ans. (i) 1, 2 –Difluoro–2–methyl propane

(ii) 1–Bromo–2–iodocyclobut –1–ene

(iii) 1–chloro–3–methyl but–2–ene

Chapter 11. Alcohols, Phenols and EthersQ. 1. Name the following alcohols according to IUPAC system :

(i) CH3CH2CH2—OH (ii) CH3—

CH3

|

CH—OH (iii) CH2 = CH—CH2—OH

(iv) C6H5CH2CH2COOCH(CH3)2

(v) CH3OOCCH2CH2COOCH3

(vi) m-BrC6H4CH2COOCH2CH2CH3

Ans. (i) 1-propanol (ii) 2-propanol (iii) prop-2-ene-1-ol

(iv) Isopropyl-3-phenyl propanoate

(v) Dimethyl butanedioate

(vi) Propyl (3-bromophenyl) ethanoate


Q. 2. Write the IUPAC names of the following :

(i) CH3—CH2—CH—CH—CH—CH3||

|CH2OH

CH3 CH2Cl

(ii)| |

|CH2OH

CH3

CH3—CH—CH —CH2—CH—CH—CH3

OH

2

(iii)

OH

Br(iv) CH2 = CH—CH—CH2—CH2—CH3

|OH

(v) CH3—C = |CH3

C |Br

— CH2OH (vi) C6H5O(CH2)6CH3

Ans. (i) 3-chloromethyl-2-isopropylpentan-1-ol(ii) 2, 5-Dimethylhxan – 1, 3 – diol(iii) 3–Bromocyclohexanol(iv) Hex–1–en–3–ol(v) 2–Bromo–3–methyl but–2–en–1–ol(iv) 1–phenoxy heptane

Q. 3. Write IUPAC name of following :

OHH1

2

34

CH3HO CH3

CH2OH

(i) (ii) (iii)Ans. (i) Cyclohex–2–ene–1–ol

(ii) 4, 4–Dimethyl cyclohex–2–en–1–ol(iii) (1–cyclo pentyl) methanol

Q. 4. Give the IUPAC name of following :

(i) CH3—CH|OH

—CH2—CH2—OH (ii) HOCH2—CH2—CH2—CH2

(iii) CH2 = CH—CH|OH

—CH3 (iv) CH3—C—OH

CH3

(v) CH2|OH

—CH|OH

—CH2|OH

Ans. (i) Butan – 1, 3–diol (ii) 4–phenyl butan–1–ol(iii) But – 3 – en – 2 – ol (iv) 2–phenyl propan –2–ol(v) Propane – 1, 2, 3 – triol.


(i) CH3CH2O CH2 CH2 Cl (ii) CH3—O—CH—CH2

|OCH3

—CH3

(iii) CH3CH2—O—C—CH3C|CH3

|CH3

(iv) O—CH2CH3

(v)O CH2CH3

H


Ans. (i) 1–chloro – 2–ethoxy ethane(ii) 1, 1 –Dimethoxy propane(iii) 2– Ethoxy – 2–methylpropane(iv) Ethoxy benzene(v) Ethoxy cyclohexane.

Chapter 12. Aldehydes, Ketones and Carboxylic acidsQ. 1. Name the following compounds according to IUPAC system of nomenclature :

(a) CH3CH(CH3)CH2CH2CHO (b) CH3CH = CHCHO(c) CH3CH(CH3)CH2C(CH3)2COCH3 (d) OHCC6H4CHO-p(e) CH3CH2COCH(C2H5)CH2CH2Cl (f) CH3COCH2COCH3

(g) (CH3)3CCH2COOH (h) (CH3)2CHCH(CH3)COCl

(i) [(CH3)2CHCH2CO]2O (j) CH3CH(CH3)CONH2

(k) CH3CH(Br)CH2CONHCH3 (l) CH3CH|Cl

COCl

(m) CH3(CH2)3COOH

Ans. (a) 4-methyl pentanal (b) But-2-enal

(c) 3, 3, 5-trimethyl hexan-2-one (d) Benzene-1, 4-dicarbaldehyde

(e) 6-chloro-4-ethyl hexan-3-one (f) Pentane-2, 4-dione

(g) 3, 3-dimethyl butanoic acid (h) 2, 3-dimethyl butanoylchloride

(i) Bis-(3-methyl butanoic) anhydride (j) 2-methylpropanamide

(k) 3-bromo-N-methyl butanamide (l) α-chloropropionyl chloride

(m) Pentanoic acid

Q. 2. Write IUPAC names of the following :

(i) CH3CH = CH COOH (ii) CH3CH2CH2CH2COOH

(iii) C6H5CH2CH2COOH (iv) (CH3)3C COOH

(v) CH3CH2COCH2COOH (vi) (CH3)3 CCH2COOH

Ans. (i) 2–Butanoic acid

(ii) Pentanoic acid

(iii) 3–phenylpropanoic acid

(iv) 2, 2–Dimethyl propanoic acid

(v) 3–Oxopentanoic acid

(vi) 3, 3–Dimethyl butanoic acid

Q. 3. Give the structure of the following :

(i) Trimethyl acetic acid (ii) Iso-valeric acid

(iii) Adipic acid (iv) Iso-butyric acid

Ans. (i) (CH3)3CCOOH

(ii) (CH3)2 CHCH2COOH

(iii) HOOC(CH2)4COOH

(iv) CH3—CH—COOH|CH3

Q. 4. Write IUPAC name of the following compounds :

(i) CH3—CH2—CH—|CHO

CH2—CH3 (ii) CH3CH = CH CH2CHO


(iii)

CH—CH2—CHO

CH3

(iv ) CH2—C—C—CH3

OO

(v)OCH3

Ans. (i) 2 Ethyl butanal (ii) Pent – 3–enal(iii) 3–phenyl butanal (iv) Butane – 2, 3 –dione(v) 3–methyl cyclohexanone.

Q. 5. Give the IUPAC name of following :

(i)CH3

O

(ii)CHOH

(iii) OHC—CH2—CH2—CH—CH2

|CHO

—CHO

(iv) C6H5—CH2—C ||O

—C ||O

—CH2—CH3

(v) CH3—CH—CH2

|CH2CHO

—CHOAns. (i) 3–methyl cyclopent – 2–ene –1–one (ii) cyclohexane carbaldehyde

(iii) Butane – 1, 2, 4 – tri carbaldehyde (iv) 1–phenyl pentane – 2, 3 dione(v) 3– (formyl methyl) hexane dial.

Q. 6. Give IUPAC name of following :(i) HOOC—(CH2)8—COOH

(ii) H—C ||O

—COOH

(iii)COOHH

(iv) CH3—C ||

O

—CH2—CH2—CH2COOH

(v) C6H5CH2COOH.

Ans. (i) Decanedioic acid (ii) Formyl methanoic acid

(iii) Cyclohexane carboxylic acid (iv) 5–oxo pentanoic acid

(v) Phenyl ethanoic acid.


(i) CH3—C ||

O

—CH2—CH2CH2COOH

(ii) HOOC—CH2—CH2—CH—CH2—

|COOH

CH2—COOH

(iii) COOH

H H

O14

5 6

3 2CHO

Ans. (i) 5–oxo hexanoic acid

(ii) pentane – 1, 3, 5 – tricarboxylic acid

(iii) 4 – formyl – 2–oxo cyclo hexane –1–carboxylic acid.


Chapter 13. Organic Compounds Containing Nitrogen

Q. 1. Write the IUPAC name of the following :

(a) CH3CN and CH3NC (b) CH3CH2NO2

(c) (i) CH3NH2 (ii) CH3—NH—CH3 (iii) CH3 CH.|NH2

CH3

(iv) CH3CONH|CH3

(v) CH3CH2CONH2 (vi) CH3CONH|CH3

(vii) CH3CH2CONH2

Ans. (a) Ethane nitrile and methyl carbylamine. (b) Nitro ethane.

(c) (i) Methanamine . (ii) N-methylmethanamine.

(iii) Propane-2-amine.

(iv) N-methylacetamide (v) Propionamide

(vi) N-methylacetamide (vii) Propionamide

Q. 2. Give the IUPAC name of following compound :

(i) N ≡ C—CH2—CH2—COOH (ii)C ≡ NH

(iii) N ≡ C—CH2—CH2—CH2—C ≡ N

(iv) N ≡ C—CH2—CH2—CH—CH2—C ≡ N|CH2—C ≡ N

(v) CH3CH2CH2CH2 C ≡ N

Ans. (i) 3–cyano propanoic acid

(ii) Cyclohexane carbonitrile

(iii) Pentane – 1, 5, – dinitrile

(iv) 3–(cyanomethyl) hexane dinitrile

(v) Pentane nitrile.

Q. 3. Write IUPAC name of the following compound :

(i) (CH3)2CHNH2 (ii) CH3—(CH2)2—NH2 (iii) CH3NHCH(CH3)2

(iv) (CH3)2CNH2 (v) C6H5—NHCH3 (vi) (CH3CH2)2NCH3

(vii) m–BrC6H4NH2.

Ans. (i) Propan-2-amine (ii) Propan-1-amine (iii) N-methyl propan-2-amine

(iv) 2-methyl propan-2-amine (v) N-methyl aniline (vi) N-ethyl-N-methyl ethamine

(vii) 3-Bromo aniline or 3-Bromobenzamine.

Q. 4. Write down the IUPAC name of following compound :

(i) CH3CH2NH2 (ii) NH2CH2CH2NH2 (iii) CH3CH2CH2CH2NHCH3

(iv) CH3CH2N(CHClCH3)2 (v) ClCH2—CH2—NH—CH2—CH2—CH3

Ans. (i) Ethanamine (ii) Ethane-1, 2 diamine (iii) N-methyl butan-1-amine

(iv) N, N-bis (1-chloroethyl) ethamine (v) N-(2-chloro ethyl) propan-1-amine.❐

Question

6Set Reasoning Based Questions

Chapter 1. Solid StateQ. 1. Graphite is anisotropic with respect to conduction of electric current. Explain.Ans. Graphite is a hexagonal crystal, that is carbon atoms in it are arranged in layers parallel to the

plane. The distance between adjacent layers is 3·4 Å. Carbon atoms are arranged in differentlayers and each atom is covalently bonded to three of its neighbouring atoms in the same layer.

Q. 2. Why does sodium chloride on heating with sodium vapours acquire yellow colour ?Ans. Sodium chloride on heating with sodium vapours acquires yellow colour because sodium

chloride crystal suffers metal excess defect when heated with sodium vapours. Due to electronictransitions in the excited state of sodium atom, it appears yellow.

Q. 3. Ionic solids conduct electricity in molten state, but not in solid state. Why ?Ans. To conduct electricity ions of a substance should be free to move to oppositely charged

electrodes. In solid state of a substance (ionic compound) ions are not free to move but in molten(liquid) state they are free to move.

Q. 4. Why is Frenkel defect found in AgCl ?

Ans. This is because the cation Ag+ and anion Cl– differ in their size due to a larger extent, therefore

cations occupy voids due to smaller size.Q. 5. Why is glass considered a super cooled liquid ?Ans. Glass is an amorphous solid. Like liquid it has tendency to flow due to this the glass panes in the

window of old building are invariably found to be highly thicker at the bottom than at the topi.e., why glass is consider a super cooled liquid.

Q. 6. Why the window glass of old building looks milky ?Ans. Glass is not a true solid but a super cooled liquid of high viscosity (called pseudo-solid) so it has

property to flow like liquid.Q. 7. Why Frenkel defect does not change the density of AgCl crystals ?Ans. Because of the Frenkel defect, no ion is missing from the crystal therefore there is no change in

the density of AgCl crystal.Q. 8. Will hcp and ccp structure for a given element have same density. Why ?Ans. Yes, because in both cases coordination number is same and hence the packing fraction is the

same.Q. 9. What are F-centres in an ionic crystal ?Ans. F-Centres : In ionic compounds having non-stoichiometric defects anionic vacancies are

occupied by electrons. These are called F-centres. These electrons remain delocalised and areresponsible for imparting colours to crystals.

Fig. Ionic Crystal with F Centre

Q. 10. Why crystals having F-centres are paramagnetic ?Ans. Free, unpaired electrons occupying anionic vacancies undergo spin in a particular direction. This

produces a magnetic moment and therefore, ionic crystals with F-centres shows paramagneticbehaviour.


Q. 11. Explain, why diamond is hard and graphite is soft ?Ans. The crystal of diamond is a network of large number of carbon atoms attached with each other in

a tetrahedral manner. The force present among the carbon atoms is very strong and it is verydifficult to break this network. Due to this reason diamond crystal is very hard. On the otherhand graphite has many layers of sp2 hybridized carbon atoms, which are arranged at the cornersof regular hexagons. Two successive layers are held together by weak van der Waal’s forces ofattraction hence one layer can slip over the other. Due to this reason graphite is soft.

Q. 12 . Excess of potassium in KCl makes the crystal appear violet. Explain, why ?Ans. When KCl is heated in an atmosphere of K metal vapour, the metal K deposits on the surface of

the KCl crystal. The chloride ions diffuse into the surface and combines with K atoms. Theelectrons produced by the ionization of K atoms then diffuses into the crystals and are trapped inanion vacancies called F–centres. The excess of K in KCl makes the crystal appear violet.

Q. 13. Why molecular solids are generally soft and easily compressible ?Ans. Molecular solids have weak van der Waal’s forces among the molecules hence they are soft and

easily compressible.Q. 14. Why the energy required to vapourize one mole of copper is smaller than that of energy required

to vapourize one mole of diamond ?Ans. Copper is a metallic solid having metallic bonds while diamond is a covalent solid having

covalent bonds. Metallic bonds are weaker than covalent bonds thus lesser amount of energy isrequired to break metallic bonds than covalent bonds.

Q. 15. All metals have metallic bonds but some metals are soft having low melting points and some arehard having high melting points. Why ?

Ans. The strength of metallic bond depends upon the number of valence electrons of the metal ion andthe magnitude of forces binding these electron to the nuclei. Thus, the strength of metallic bondin different metals may be weaker or strong hence their nature and melting points may bedifferent.

Q. 16. Covalent crystals like diamond and silicon carbide are quite hard and difficult to break., why ?Ans. In covalent crystals the shared pair of electrons gives rise to very strong forces between atom,

therefore covalent solids like diamond and silicon carbide are quite hard and difficult to breakand shows very high melting points.

Q. 17. Explain how vacancies are introduced in an ionic solid when a cation of higher valency is addedas an impurity in it.

Ans. When a cation of higher valency is added as an impurity in an ionic solid, some of the sites of theoriginal cations are occupied by the cations of higher valency. Each cation of higher valencyreplaces two or move original cation and occupies the site of one original cation and other sitesremains vacant.Cationic vanancies produced = [Number of cations of higher valency × Difference in valencies of

original cation and cation of higher valency]Q. 18. Ionic solids, which have anionic vacancies due to metal-excess defect, develop colour. Explain

with the help of a suitable example.Ans. In ionic solid with anionic vacancies due to metal excess defect, when the metal atoms deposit on

the surface, they diffuse into the crystal and after ionisation the metal ion occupies cationicvacancy while electron occupies anionic vacancy. Such ionic sites occupied by an electron areknown as F-centres. These electrons get excited to higher energy level by absorption of suitablewavelength from the visible white light and therefore appears coloured when Na vapours arepassed over NaCl crystal such defect is created and the crystal appears yellow due to excess Na+

and presence of F-centres.Q. 19. How would you account for the following ?

(i) Frenkel defect is not found in alkali metal halides.(ii) Schottky defect lowers the density of related solids.(iii) Impurity doped silicon is a semiconductor.

Ans. (i) This is because alkali metal ions have larger size which cannot fit into interstitial sites.

Reasoning based Questions ■ 93

(ii) As the number of ions decreases as a result of Schottky defect, the mass decreases whereasthe volume remains the same.(iii) This is due to addition of electron or creation of an electron hole on doping with impurity.Creation of electron hole results in p-type semiconductor whereas additional electron results inthe n-type semiconductor.

Q. 20. Non-stoichiometric cuprous oxide, Cu2O can be prepared in the laboratory. In this oxide copperto oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor ?

Ans. The ratio less than 2 : 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by

cupric (Cu2+

) ion. For maintaining electrical neutrality every two Cu+

ion will be replaced by oneCu

2+ ion thereby creating a hole. As conduction will be due to the presence of these positive

holes, hence it is a p-type semiconductor.

Chapter 2. SolutionsQ. 1. Water boils below 100°C by the addition of NaCl.Ans. Water boils above 100°C by the addition of NaCl.Q. 2. The molecular weights of sodium chloride and glucose are determined by the Depression of

freezing point method. Compared to their theoretical molecular weights, what will be theirobserved molecular weights when determined by the above method ? Justify your answer.

Ans. Sodium chloride is a strong electrolyte therefore dissociates in solution to give twice the actualnumber of particles, therefore the molecular weight determined by colligative property will behalf of the actual weight. The molecular weight of the non-electrolyte glucose will be same as thetheoretical molecular weight.

Q. 3. When and why is molality preferred over molarity in handling solution in chemistry ?Ans. Molality is preferred because molality depends on masses of solvent which do not change with

temperature.Q. 4. Why vapour pressure of solution of glucose in water is lower than that of water ?Ans. This is due to decrease in the escaping tendency of the water molecules from the surface of the

solution as of the surface area is occupied by non-volatile solute, i.e., glucose particles.Q. 5. Why does a solution of ethanol and cyclohexane shows positive deviation from Raoult’s law ?Ans. On adding cyclohexane to ethanol, its molecule get in between the molecule of ethanol thus

breaking the hydrogen bond and reducing ethanol -ethanol interaction. This will increase thevapour pressure of the solution and results in positive deviation from Raoult’s law.

Q. 6. Why is osmotic pressure of 1 M KCl higher than 1 M urea solution ?

Ans. This is because KCl dissociates to give K+and Cl– ion while urea being a molecular solid does not

dissociates into ions in the solution.Q. 7. Why does sodium chloride solution freezes at a lower temperature than water ?Ans. When a non-volatile solute is dissolved in a solvent, the vapour pressure decreases. As a result,

the solvent freezes at a lower temperature.Q. 8. The dissolution of ammonium chloride in water is an endothermic process but still it dissolves in

water readily. Why ?Ans. This is because the entropy changes. In this case, ΔS = positive

NH4Cl (aq) ⎯⎯⎯→ NH4+ + Cl

–

The ion that were held together in crystalline solid are free and moves in all positive directions.Its entropy increases and this makes TΔS > ΔH i.e. ΔG = negative.

Q. 9. Why are aquatic species more comfortable in cold water in comparison to warm water ?Ans. At a given pressure the solubility of oxygen in water increases with decrease in temperature.

Presence of more oxygen at lower temperature makes the aquatic species more comfortable incold water.

Q. 10. Why is liquid ammonia bottle first cooled in ice before opening it ?Ans. At room temperature, the vapour pressure of ammonia is very high, on cooling, the vapour

pressure decreases and hence liquid ammonia will not splash out.


Q. 11. Cutting onion taken from the fridge is more comfortable than cutting onion lying at roomtemperature. Explain why.

Ans. At lower temperature, the vapour pressure is low. Less vapour of tear producing chemicals areproduced.

Q. 12. Why does pressure cooker reduces cooking time ?Ans. The weight over lid does not allow the steam to go out. As a result, pressure, inside the cooker

becomes high. Higher the pressure, higher is the boiling point and faster is the cooking.Q. 13. Why glycol and water mixture is used in car radiators in cold countries ?

Ans. Ethylene glycol lowers the freezing point of water. Due to this coolant in radiator will not freezeotherwise radiator will burst due to freezing of coolant (water).

Q. 14. 0·1 M urea solution shows less depression in freezing point than 0·1 M MgCl2 solution. Explain.Ans. Depression in freezing point is a colligative property which depends upon the number of

particles present in the solution.MgCl2 when dissolved in water gives three particles per molecule

MgCl2 Δ

⎯→ Mg2+ + 2Cl– (3 particles on dissociation)while, urea remains undissociated.

Q. 15. The elevation in boiling point produced by dilute equimolal solutions of three substances are inthe order A > glucose > B. Suggest a reason for this observation.

Ans. This observation suggests that substance A dissociates while substance B associates in thesolution because elevation in boiling point depends upon number of particles of the solutepresent in the solution and not on the molal concentration.

Q. 16. Addition of one liquid to other causes decrease in its vapour pressure. Explain, why ?Ans. Addition of liquid decreases surface area of previous liquid which decreases number of

molecules of vapour as well as vapour pressure of liquid.Q. 17. Solution A is prepared by dissolving 2 mole of glucose in 1 litre of water and solution B is

prepared by dissolving 1 mole of sodium chloride in 1 litre of water. Will the osmotic pressure ofsolution A be higher, lower or equal to that of B ? Give reasons for your answer.

Ans. The osmotic pressure of the two solutions A and B will be equal because both the solutionscontain the same number of particles. Osmotic pressure is a colligative property, hence itdepends upon the number of particles of solute in solution and not upon the nature of theparticles. Glucose is a non electrolyte solute therefore, on dissolving two mole of glucose in onelitre of water it will not dissociate. While sodium chloride will dissociate into sodium ions andchloride ions because it is a strong electrolyte.

NaCl ⎯→ Na+ + Cl–

(1 mole) (1 mole) (1 mole)

Due to dissociation one mole of NaCl will form two mole of particles (one mole Na+ and onemole Cl–) hence the number of particles in both the solutions A and B will be the same whichresults in the same osmotic pressure of A and B.

Q. 18. The relative molecular masses of two substances known to have the same molecular masseswhen determined by the elevation of boiling point method were found to be different. One wasalmost twice as that of the other. Give your comments on the results.

Ans. Elevation in boiling point is a colligative property i.e., it depends upon the number of soluteparticles and not upon the nature of the solute particles. Suppose the two substances say XY andWZ have same relative molecular masses. If there is an association in WZ molecules its particlesbecomes half and when the molecular mass is determined with the help of elevation in boilingpoint boiling (colligative property) it would be found just double of the actual molecular massbecause colligative property (ΔTb) is inversely proportional to the molecular mass.

2WZ Association⎯⎯⎯→ (WZ)2

(Number of particles remains half hence the colligative property remains half )

Colligative property (ΔTb) ∝ 1

Molar mass of solute (Thus the molar mass becomes double)


In this way, the molar mass of WZ (where no association or dissociation is observed) is founddouble than that of the molar mass of XY.

Q. 19. The molecular weights of potassium chloride and glucose are determined by the depression infreezing point method. As compared to their theoretical molecular weights, what do you expectabout their molecular weights determined by this experiment ? Why ?

Ans. When potassium chloride is dissolved in water it is dissociated into K+ and Cl–. Thus the numberof particles becomes double (because one KCl molecule forms two ions). The depression infreezing point (ΔTf) is a colligative property which depends upon the number of solute particlesand not upon their nature. Thus ΔTf becomes double due to the dissociation of KCl. Themolecular weight of solute is inversely proportional to the colligative property. Thus due todouble the value of colligative property (ΔTf) the observed molar mass of KCl will be half of itsactual value.On the other hand glucose molecules will not undergo any association or dissociation, thereforethe number of particles will remain same. Thus, there will be no change in colligative propertiesand molar mass and the actual value and observed value with the help of ΔTf will be same in caseof glucose.

Q. 20. What care is generally taken during intravenous injection and why ?Ans. During intravenous injection the concentration of the solution should be approximately same as

that of blood plasma so that they are isotonic. If solution to be injected is hypertonic, it will causecells to shrink. On the other hand if the solution to be injected is hypotonic, it will cause cells toburst.

Q. 21. Define osmotic pressure. Arrange the following in increasing order of osmotic pressure and givereasons in support of your answer :(i) 34·2 g per litre of sucrose (MW = 342).(ii) 90·0 g per litre of glucose (MW = 180).(iii) 5·85 g per litre of sodium chloride (MW = 58·5).

Ans. Osmotic pressure of a solution is the pressure required to prevent osmosis when the solution isseparated from pure solvent by a semi-permeable membrane.As osmotic pressure is a colligative property so it depends on the no. of solute particles.

(i) Mole of Sucrose =34·2342 = 0·1.

(ii) Mole of Glucose =90180 = 0·5.

(iii) Mole of NaCl =5·8558·5 = 0·1.

1 mole = 6·023 × 10 23 particles.As Sucrose and Glucose are undissociatedSo, Sucrose = 6·023 × 10 22 particles

Glucose = 30·115 × 10 22 particles.NaCl is dissociated to give two ions so

NaCl = 12·046 × 10 22 particles.Sucrose < NaCl < Glucose.

Q. 22. Correct the following statements :(i) Freezing point of a solution is directly proportional to its molality.(ii) The molality of a solution depends on the temperature whereas the mole fraction of a

solution does not changes with temperature.(iii) The relative lowering of vapour pressure of a solvent by a solute is proportional to the

molarity of the solution.(iv) Colligative properties of dilute solutions containing non-volatile solute depends upon the

nature of solute.(v) Molarity of a solution is independent of temperature.

(vi) Osmotic pressure and boiling points are colligative properties.(vii) Addition of sodium chloride lowers the boiling point and freezing point of water.


Ans. (i) Depression in freezing point of a solution is directly proportional to its molality.(ii) Molality as well as mole fraction are independent of temperature.(iii) The relative lowering of vapour pressure of a solvent by a solute is proportional to the

mole fraction of the solute.(iv) Colligative properties of dilute solution containing non-volatile solute does not depend

upon the nature of the solute dissolved but depends only upon the number of particles ofsolute present in the solution.

(v) Molarity of a solution varies with temperature.(vi) Osmotic pressure and elevation in boiling point are colligative properties.(vii) Addition of sodium chloride elevates the boiling point and decreases the freezing point of

water.Q. 23. Give reasons for the following :

(i) At higher altitudes, people suffer from a disease called anoxia. In this disease they becomeweak and cannot think clearly.(ii) When mercuric iodide is added to an aqueous solution of KI, the freezing point is raised.(iii) A person suffering from high blood pressure is advised to take minimum quantity ofcommon salt.

Ans. (i) At higher altitudes, partial pressure of oxygen is less than that at ground level, so thatoxygen concentration becomes less in blood or tissue. Hence people suffer from anoxia.(ii) Due to the formation of complex K2[HgI4] the number of particles in the solution decreasesand hence the freezing point is raised.(iii) Osmotic pressure is directly proportional to the concentration of solutes. Our body fluidcontains a number of solutes. On taking a large amount of common salt Na+ and Cl

– ions enter

into the body fluid thereby raising the concentration of solutes. As a result, osmotic pressureincreases which may rupture the blood cells.


Q. 1. Why does the conductivity of a solution decreases with dilution ?Ans. Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On

dilution, the number of ions per unit volume decreases. So, the conductivity also decreases.Q. 2. Why on dilution the λm of CH3COOH increases drastically while that of CH3COONa increases

gradually ?Ans. In the case of CH3COOH which is a weak electrolyte the number of ions increases on dilution

due to increase in degree of dissociation.

CH3COOH + H2O ⎯⎯→ CH3COO– + H3O+

Q. 3. Why is alternating current used for measuring resistance of an electrolytic solution ?Ans. Alternating current is used to prevent electrolysis so that concentration of ions in the solution

remains constant.Q. 4. Why electrolysis of NaBr and NaI gives Br2 and I2 respectively while that of NaF gives O2 instead

of F2 ?

Ans. Br– and I – ion have higher oxidation potentials than water. Hence, they are more easily oxidised.

But F– ions have lower oxidation potential than H2O. So, H2O is easily oxidised to give O2 gas.

Q. 5. Why does an aqueous solution of NaCl on electrolysis gives H2 gas at the cathode and notsodium metal ?

Ans. This is because of the fact that standard reduction potential of water is greater than that ofsodium.

Q. 6. Value of standard electrode potential for oxidation of Cl– ions is more positive than that of water,

even then in the electrolysis of aqueous sodium chloride. Why is Cl– oxidised at anode instead of

water ?Ans. On electrolysis of aqueous sodium chloride oxidation of water at anode requires over potential

hence Cl– is oxidised instead of water.


Q. 7. Why does a dry cell becomes dead after a long time even if it has not been used ?Ans. Even though not in use, a dry cell becomes dead after some time because the acidic NH4Cl

corrodes the zinc container.Q. 8. Why does a galvanic cell becomes dead after some time ?Ans. As the reaction proceeds, concentration of ions in anodic half cell keeps on increasing while in the

cathodic half it keeps on decreasing. Hence, their electrode potentials also keeps on changinguntil they become equal and then e.m.f. of the cell becomes zero.

Q. 9. Why is it not possible to measure the single electrode potential ?Ans. Oxidation or reduction cannot takes place alone. Moreover electrode potential is a relative

tendency and can be measured with respect to a reference electrode only.

Q. 10. How would you determine the standard electrode potential Mg2+

| Mg ?Ans. Set up an electrochemical cell consisting of Mg/MgSO4 (1 M) as one electrode by dipping a

magnesium rod in 1 M MgSO4 solution and standard hydrogen Pt, H2 (1 atm) | H+ (1 M) as the

second electrode as shown in given figure.

Measure the EMF of the cell and also note the direction of deflection in the voltmeter. Thedirection of deflection shows that electrons flow from magnesium electrode to hydrogenelectrode. Thus the cell may be represented as follows :

Mg | Mg2+ (1M) || H+ (1M) | H2, (1 atm), Pt (s)

E°cell = E°H+

/H2 – E°Mg

2+/Mg

But E°H+

/H2= 0

Hence E°Mg2+

/Mg = – E°cell

Q. 11. Explain how rusting of iron is envisaged a setting up of an electrochemical cell.Ans. The water layer present on the surface of iron dissolves acidic oxides of air like CO2 to form acid

which dissociates to give H+ ions.

H2O + CO2 ⎯⎯→ H2CO3 2H+ + CO3

2–

In the presence of H+ ions, iron starts losing electrons at some spot to form ferrous ions. Hence,

this spot acts as the anode :

Fe(s) ⎯⎯→ Fe2+

(aq) + 2e–

The electrons thus released through the metal moves to reach another spot where H+ ions and the

dissolved oxygen gains those electrons and reduction reaction takes place. Hence this spot acts asthe cathode.

O2(g) + 4H+ (aq) + 4e

– ⎯→ 2H2O (l)

The overall reaction is given as : Ferrous ion are further oxidised by the atmospheric oxygen toferric ions which combines with water molecule to form hydrated ferric oxide Fe2O3.xH2O whichis rust.

Q. 12. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed ?Ans. The pH of the solution will increase as NaOH is formed in electrolytic cell.


Q. 13. Can you store copper sulphate solution in a zinc pot ?Ans. For this we have to check whether the following reaction will take place or not.

Zn(s) + CuSO4 (aq) ⎯⎯⎯→ ZnSO4 (aq) + Cu(s)E°cell = E°Cu2+/Cu – E°Zn

2+/Zn = 0·34 – (–0·76)

= 1·10 VE°cell is positive, the reaction will take place there. Therefore we cannot store copper sulphate inzinc pot.

Q. 14. Account for the following :(i) Alkaline medium inhibits the rusting of iron.(ii) Iron does not rust even if the zinc coating is broken in a galvanised iron pipe.

Ans. (i) The alkalinity of the solution prevents the availability of H+ ions.

(ii) Zinc has lower reduction potential than iron. Therefore zinc coating acts as anode and theexposed iron portion acts as cathode. If zinc coating is broken zinc still undergoes oxidation,protecting iron from rusting. No attack occurs on iron till all the zinc is corroded.

Q. 15. Why a salt bridge or a porous plate is not needed in a lead storage battery ?Ans. The half cells in a cell must be separated only if the oxidising and reducing agents can migrate to

the other half cell. In lead storage cell the oxidising agent PbO2 and the reducing agent Pb as wellas their oxidation and reduction product PbSO4 are solids. Therefore there is no need to separatehalf cells.

Q. 16. Blocks of magnesium are often strapped to the steel hulls of ocean going ships. Why ?Ans. Magnesium prevents the oxidation of steel by transferring the excess of electrons to the steel.

Thus rusting of steel is protected and it is called cathodic protection.Q. 17. Will nickel displace copper from a 1 M solution of copper sulphate ? Justify your answer.

[ ]Eo

Ni+2/Ni = – 0·25V‚ Eo

Cu+2/Cu = + 0·34V

Ans. Nickel will displace Cu2+ ions from 1M copper sulphate solution because E°red for nickel is less

than that of copper.

Ni(s) + Cu2+(aq) → Ni2+

(aq) + Cu(s) ↓

Q. 18. Zinc displaces hydrogen from acid solution [E°Zn+2/Zn = – 0·76 volts]Ans. Zn ⎯→ Zn2+ + 2e– E° = 0·76 V (Oxidation potential)

2H+ + 2e– ⎯→ H2 E° = 0·0 V (Reduction potential)

Zn + 2H+ ⎯→ Zn2+ + H2 ; E° = 0·76 V (Electromotive force)

It is clear from the above example that zinc displaces hydrogen from acid solution because theelements having positive oxidation potential can replace hydrogen from acids due to zeroelectrode potential of hydrogen. Electromotive force (E°) of the complete reaction will be positivein this case.

Q. 19. Specific conductance decreases with dilution whereas equivalent conductance increases withdilution. Explain.

Ans. Specific conductance decreases with dilution as the number of current carrying particles, i.e., ionspresent per cm3 of solution becomes less and less.

Equivalent conductane increases with dilution because the degree of dissociation of an electrolyteincreases on dilution.

Q. 20. Explain precipitation of silver when copper rod is dipped in silver nitrate solution.

Ans. When copper rod is dipped in silver nitrate solution, silver is precipitated because copper metallies above in the electrochemical series. That is why it replaces silver from silver nitrate solutionor the reduction potential of silver is +0·80V while that of Cu is + 0·34V and a metal with lowerreduction potential displaces the metal with higher reduction potential.


Q. 21. Out of the following, which will gain electron most easily ?Mg2+, K+, H+, Na+.

Ans. H+ can gain electron most easily because the electrode potential of H+ is higher in comparison ofother metal ions.

Q. 22. When a zinc piece is added to CuSO4 solution, copper gets precipitated, why ?Ans. Standard electrode potential of zinc is lesser than that of copper hence, zinc metal gets oxidized

to Zn+2 ions and Cu+2 ions gets reduced to copper metal.Q. 23. Fluorine cannot be prepared from fluorides by chemical oxidation. Why ?Ans. The standard electrode potential of fluorine is maximum hence fluoride cannot be oxidized to

fluorine by any chemical oxidation.Q. 24. Zinc displaces hydrogen from acids while hydrogen displaces silver from its solutions. Explain

this on the basis of standard potentials. Ans. Lower the value of the standard electrode potential of a metal more easily it can lose electrons

hence greater is its reactivity. Therefore, a metal with lower reduction potential can displace themetals with higher reduction potential from their salt solution. The order of standard reductionpotential of H, Ag and Zn is-

Zn (–0·76 V) < H (0·0V) < Ag (+0·8 V)Hence, Zn can replace hydrogen while hydrogen can replace Ag from its solution.

Q. 25. Why salt bridge is not required in a lead storage cell ?Ans. In a cell the partition is needed to separate the oxidizing agent and reducing agent. In lead

storage cell, the oxidant PbO2, reductant Pb and their oxidation and reduction product PbSO4 areall solids. Thus, half cell need not be in separate vessels and as such no partition is needed toseparate them.

Q. 26. Can 1M solution of ferrous sulphate be stored in a nickel vessel ?(E°Fe2+/Fe = – 0·44V, E° Ni2+/Ni= – 0·25V)

Ans. From the values of standard reduction potential, it is clear that iron is more reactive than nickel.Consequently, nickel can not displace iron from 1M ferrous sulphate solution. Hence ferroussulphate solution can be stored in a nickel vessel.

Q. 27. Blue colour of copper sulphate solution is discharged slowly when zinc is dipped in it.Given E° Cu2+/Cu = +0·34 V and E° Zn2+/Zn = – 0·76 V

Ans. Zinc has lower value of standard electrode potential than copper, hence Cu2+ ion can be reducedby Zn metal, when they come in contact with it. Due to the displacement of Cu2+ ion by Zn2+

ions, colour of the solution is discharged.Cu+2(aq.) + Zn(s) ⎯→ Zn+2(aq.)


Q. 1. The rate constant of a first order reaction is proportional to the concentration of the reactant.Ans. The rate of a first order reaction is proportional to the concentration of the reactant.Q. 2. Why does the rate of a reaction increase with rise in temperature ?Ans. At higher temperature larger fraction of colloiding particles can cross the energy barrier (i.e.

activation energy) which leads to faster rate.Q. 3. Why the probability of reaction with molecularity higher than three is very rare ?Ans. The probability of more than three molecules colliding simultaneously is very small. Hence

possibility of molecularity being three is very low.Q. 4. Why does the rate of any reaction generally decreases during the course of the reaction ?Ans. The rate of reaction depends upon the concentration of reactants. As the reaction proceeds the

concentration of reactant decreases hence rate of chemical reaction decreases.Q. 5. Can a reaction have zero activation energy ?Ans. If Ea = 0, then according to Arrhenius equation K = Ae–Ea/RT = Ae° = A. This implies that every

collision results into a chemical reaction which cannot be true. Hence a reaction cannot have zeroactivation energy.


Conc. of N2O5

Rat

e

Q. 6. Consider the reaction 2N2O5(g) → 4NO2(g) + O2(g). When a graphis plotted between the rate of reaction against the concentration ofN2O5 following figure is obtained. What would be the order ofreaction ?

Ans. As a straight line is obtained in a graph of rate vs. concentration.therefore, it would be the first order reaction, it means ratedepends upon one exponent of reactant concentration.

Q. 7. The rate law for the reaction,RCl + NaOH ⎯→ ROH + NaCl; where rate = K [RCl]

What happens to the rate and rate constant of the reaction when :(i) Concentration of RCl is doubled ?(ii) Concentration of NaOH is doubled ?(iii) Temperature is increased ?

Ans. The rate law for this reaction is rate = K [RCl](i) It is clear from the rate law that order with respect to RCl is one, thus on making the

concentration of RCl doubled the rate also becomes doubled.

(ii) NaOH is not present in rate law hence on making the concentration of NaOH doubled itwould not affect the rate.

(iii) Increasing the temperature, the reaction rate will also increase due to the increase in thefraction of molecules having effective collision and also the increase in collision frequency.

Q. 8. Why highly active solids with large surface area are prepared for use as catalyst ?

Ans. This is to increase the adsorption of the reacting molecules on the surface of the catalyst.

Q. 9. How do temperature and presence of a catalyst brings about an increase in the rate of a reaction ?

Ans. Increase in temperature increases the kinetic energy of the molecules thus a larger fraction ofmolecules have energy equal to or more than the threshold energy. Hence the rate of reactionincreases. Presence of a catalyst provides an alternate path for the reaction to proceed which haslower energy barrier therefore more reacting molecules have sufficient energy to cross it.

Q. 10. Why is it that instantaneous rate of reaction does not changes when a part of the reacting solutionis taken out ?

Ans. Instantaneous rate is measured over a very small interval of time. Hence it does not changeswhen a part of solution is taken out.

Q. 11. Why does the rate of a reaction increases with rise in temperature ?

Ans. At higher temperature, larger fraction of colliding particles can cross the energy barrier (i.e., theactivation energy) which leads to faster rate.

Q. 12. Thermodynamic feasiblity of the reaction alone cannot decide the rate of the reaction explainwith the help of an example.

Ans. Thermodynamically the conversion of diamond to graphite is highly feasible but this reaction isvery slow because its activation energy is high.

Q. 13. In some cases, it is found that a large number of colloiding molecules have energy more thanthreshold value, yet the reaction is slow why ?

Ans. The colloiding molecules may not have a proper orientation at the time of collision.

Q. 14. Why is molecularity applicable only for elementary reaction and order is applicable forelementary as well as complex reaction ?

Ans. A complex reaction proceeds through several elementary reactions. Number of moleculesinvolved in each elementary reaction may be different i,.e. the molecularity of each step may bedifferent. Therefore discussion of molecularity of overall complex reaction is meaningless. On theother hand order of a complex reaction is determined by the slowest step in its mechanism and isnot meaningless even in the case of complex reaction.


Q. 15. For a zero order reaction will the molecularity be equal to zero ? Explain.

Ans. No, the molecularity can never be equal to zero or a fractional number. Molecularity is thenumber of molecules involve in each elementary reaction which may be different i.e. themolecularity of each step may be different.

Q. 16. For a reaction A + B ⎯⎯→ Product the rate law is Rate = K[A] [B]3/2. Can the reaction be anelementary reaction ? Explain.

Ans. During an elementary reaction, the number of atoms or ions colliding to react is refered to asmolecularity. If the given reaction is an elementary reaction the order of reaction with respect toB would be 1. But in the give rate law it is 3/2. This indicates that the reaction is not anelementary reaction.


Q. 1. Why are substance like platinum and palladium often used for carrying out electrolysis ofaqueous solution ?

Ans. Substance like platinum and palladium are often used for carrying out electrolysis of aqueoussolution because they are chemically inert.

Q. 2. Why is it necessary to remove CO when ammonia is obtained by Haber’s process.Ans. CO acts as catalytic poison and must be removed.Q. 3. How does catalyst works ?Ans. Catalyst provides an alternate path involving lower activation energy for the reactants.Q. 4. In what way is a sol different from a gel ?Ans. Colloidal system in which solid is dispered in liquid is called sol and that in which liquid is

dispersed in solid is called gel.Q. 5. Why is a colloidal sol stable ?Ans. All the particles in a colloidal sol carry the same charge and hence they keep on repelling each

other and cannot aggregate together to form bigger particles.Q. 6. Why does colloidal solution exhibit Tyndall effect ?Ans. Colloidal solution exhibits Tyndall effect because the size of particle (1 nm-100nm) is such that

they can scatter light.Q. 7. How can a lyophilic sol be coagulated ?Ans. This can be done (i) by adding an electrolyte (ii) by adding suitable solvent.Q. 8. Why are some medicines more effective in the colloidal form ?Ans. Medicines are more effective in the colloidal form because of large surface area and are easily

assimilated in this form.Q. 9. How does emulsifying agent stabilizes the emulsion ?Ans. The emulsifying agent forms an interfacial layer between suspended particles and the dispersion

medium there by stabilising the emulsion.Q. 10. How does a solid catalyst enhances the rate of combination of gaseous molecules ?

Ans. When gaseous molecules comes in contact with the surface of a solid catalyst adsorption ofgaseous molecules take place at the surface of the catalyst. It increases the concentration ofreactants on the surface. Different molecules adsorbed side by side have better chance to reactand form new molecules. This enhances the rate of reaction. Also, Adsorption is an exothermicprocess. The heat released in the process of adsorption is utilised in exchanging the reaction rate.

Q. 11. How does BF3 acts as a catalyst in industrial process ?Ans. It is a because BF3 is an electron deficient compound and helps to generate electrophile.

Q. 12. Does the vital functions of the body such as digestion gets affected during fever ? Explain youranswer.

Ans. The optimum temperature range for the activity of enzymes is 298-310 K. On either side of thistemperature range enzymatic activity gets affected. Thus during fever, when temperature risesabove 310 K. The activity of enzymes may be affected.


Q. 13. Why is it essential to wash the precipitate with water before estimating it quantitatively ?Ans. Some amount of the electrolyte mixed to form the precipitate remains absorbed on the surface of

the particles of the precipitate. Hence it is essential to wash the precipitate with water to removethe sticking electrolyte (or any other impurities) before estimating it quantitatively.

Q. 14. Why a finely divided substance is more effective as an adsorbent ?Ans. Powdered subtance have greater surface area as compared to their crystalline form. Greater the

surface area greater is the adsorption.Q. 15. Why is the ester hydrolysis slow in the beginning and becomes faster after some time ?

Ans. The ester hydrolysis takes place as follows :RCOOR′ + H2O ⎯⎯→ RCOOH + R′OH

Ester Water Acid Alcohol

The acid produced in the reaction acts as catalyst for the reaction. Hence, the reaction becomesfaster after some time.

Q. 16. Is it possible to cause artificial rain by spraying silver iodide on the clouds. Comment.Ans. Yes, clouds are colloidal in nature and carry charge. Spray of silver iodide an electrolyte, results

in coagulation leading to rain.Q. 17. Why does leather get hardened after tanning ?

Ans. Animal hide is colloidal in nature and has positively charged particles. When it is soaked in taninwhich has negatively charged colloidal particles, it results in mutual coagulation.

Q. 18. Why is ferric chloride preferred over potassium chloride in case of cut leading to bleeding ?Ans. Fe3+ ion has greater coagulation power than K+ ion as ferric ion has higher charge.

Q. 19. Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy stillit is a spontaneous process. Why ?

Ans. According to the equationΔG = ΔH – TΔS

For a process to be spontaneous, ΔG should be negative, Even though ΔS is negative here, ΔG isnegative because reaction is highly exothermic, i.e. ΔH is negative.

Q. 20. Explain what is observed when :(i) An electric current is passed through a colloidal sol.(ii) When a beam of light is passed through a colloidal solution.

Ans. (i) On passing direct current, colloidal particles move towards the oppositely charged electrodewhere they loose their charge and gets coagulated.(ii) Scattering of light by the colloidal particle takes place and the path of light becomes visible(Tyndall effect)

Q. 21. Explain the following observations :(i) Sun looks red at the time of setting.(ii) Cottrell’s smoke precipitator is fitted at the mouth of chimney used in factories.(iii) Physical adsorption is multilayered while chemical adsorption is mono layered.

Ans. (i) At the time of setting, the sun is at horizon. The light emitted by sun has to travel a relativelylonger distance through the atmosphere. As a result blue part is scattered away by the particulatein the atmosphere causing red part to be visible.(ii) Cottrell’s smoke precipitator, neutralizes the charge on unburnt carbon particle coming outof chimney and they get precipitated and settle down at the floor of the chamber.(iii) Physical adsorption involves van der waal’s forces, so any number of layer be formed oneover the other on the surface of the adsorbent. Chemical adsorption takes place as a result ofreaction between adsorbent and adsorbate when the surface of adsorbent is covered with onelayer no further reaction can takes place.

Q. 22. SnO2 forms a positively charged colloidal solution in acidic medium and a negatively charged solin the basic medium why ? Explain.

Ans. SnO2 is amphoteric in nature. It reacts with acid such as HCl to form SnCl4 in the solution. Thecommon Sn4+ ions are adsorbed on the surface of SnO2 particles to give a positively chargedcolloidal sol.


SnO2 + 4HCl ⎯⎯→ SnCl4 + 2H2OSnO2 + Sn4+ ⎯⎯→ (SnO2) : Sn4+

(positively charged sol.)

Similarly, with base like NaOH, it forms sodium stannate (Na2SnO3). The stannate ions getadsorbed on the surface of SnO2 to give negatively charged colloidal sol.

SnO2 + 2NaOH ⎯⎯⎯→ Na2SnO3

Sodium stannate

SnO2 + SnO32–

⎯⎯⎯→ (SnO2) : SnO32–

(Negatively charged sol)Q. 23. Which one of the following electrolytes is the most effective for the coagulation of Fe (OH)3 sol

and why ?NaCl, Na2SO4, Na3PO4.

Ans. Fe (OH)3 is a positively charged sol. According to Hardy-Schulze rule, greater the charge on theoppositely charged ions of electrolyte added, more effective it is in bringing about thecoagulation. Hence Na3PO4 (containing PO4

3–) is the most effective.Q. 24. Explain the following observations ?

(a) Lyophilic colloid is more stable than lyophobic colloid.(b) Coagulation takes place when sodium chloride is added to a colloidal solution of ferrichydroxide.(c) Sky appears blue in colour.

Ans. (a) This is because the stability of lyophobic colloid is due to the presence of charge on thecolloidal particles. On the other hand, stability of lyophillic colloid is due to charge as well assolvation of colloidal particles.

(b) Fe (OH)3 sol is positively charged which is coagulated by negatively charged Cl– present in

sodium chloride solution.(c) Sky appears blue in colour due to scattering of light by dust particles. This is known asTyndall effect.

Q. 25. In which of the following does adsorption takes place and why ?(i) Silica gel placed in the atmosphere saturated with water.(ii) Anhydrous CaCl2 placed in the atmosphere saturated with water.

Ans. (i) In silica gel, adsorption takes place due to attraction and retention of water molecule on thesurface.(ii) Anhydrous CaCl2, undergoes absorption because it combines with CaCl2, and undergoesadsorption since it combines with water molecule to form hydrated calcium chloride CaCl2.2H2O.

Q. 26. Comment on the statement that “colloid is not a substance but state of a substance.”Ans. The given statement is true. This is because the same substance may exist as a colloid under

certain conditions and as a crystalloid under some other conditions. For example : NaCl in waterbehaves as a crystalloid while in benzene it behaves as a colloid. It is the size of the particleswhich matter i.e. the state in which the substance exists. If the size of the particles lies in therange of 1 nm to 1000 nm, it is in the colloidal state.

Q. 27. Why do soaps not work in hard water ?Ans. Hard water contains calcium and magnesium salts. In hard water soap gets precipitated as

calcium and magnesium salts which being insoluble sticks to the clothes as gummy mass.Therefore, soap do not work in hard water.

Q. 28. How do size of particles of adsorbent, presence of gas and prevailing temperature influences theextent of adsorption of a gas on a solid ?

Ans. They influence in the following ways :(i) Smaller the size of particles of the adsorbent, greater is the surface area and greater is theadsorption.(ii) At constant temperature, adsorption first increases with increase of pressure and thenattains equilibrium at a high pressure.(iii) In physical adsorption, it decreases with increase of temperature but in chemisorption, firstit increases then decreases.


Chapter 6. General Principles and Processes of Isolation

Q. 1. Why it is that only sulphide ores are concentrated by Froth Floatation process ?Ans. This is because the sulphide ore particles are preferentially wetted by oil and gangue particles are

preferentially wetted by water.Q. 2. Why copper matte is put in silica lined converter ?Ans. Copper matte consist of Cu2S and FeS. When a blast of hot air is passed through molten matte

taken in a silica lined converted FeS present in matte is oxidised to FeO which combines withsilica (SiO2) to form FeSiO3, slag.

2FeS + 3O2 ⎯→ 2FeO + 2SO2

FeO + SiO2 ⎯→ FeSiO3

(slag)

When whole iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2Owhich reacts with more Cu2S to form copper metal.

2Cu2S + 3O2 ⎯→ 2Cu2O + 2SO2↑

2Cu2O + Cu2S ⎯→ 6Cu + SO2

Thus copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO3,slag.

Q. 3. Why is an external emf of more than 2·2 V required for the extraction of Cl2 from brine ?Ans. For the reaction :

2Cl– + 2H2O ⎯→ 2OH

– + H2 + Cl2

value of ΔG° is + 422 kJ. Using the equation ΔG° = – nFE°, the value of E° comes out to be – 2·2 VTherefore extraction of Cl2 from brine will require an external emf, greater than 2·2 V.

Q. 4. Why is it advantageous to roast a sulphide ore to the oxide before reduction ?Ans. The standard free energies (ΔfG°) of formation of most of the sulphide ore greater than those of

CS2 and H2S (CS2 is infact, an endothermic compound). Therefore neither carbon nor hydrogencan reduce metal sulphide to metal. In contrast, the standard free energies of formation of oxidesare much lower that that of CO2 and hence oxidation of metal sulphide to metal oxides isthermodynamically favourable. Therefore it is the common practice to roast sulphide ore to theoxide before reduction.

Q. 5. Why is the extraction of copper from pyrites more difficult than that from its oxides ore throughreduction ?

Ans. The standard free energy (ΔfG°) of formation of Cu2S is more negative than those of CS2 and H2S,so neither carbon nor hydrogen can reduce Cu2S to Cu metal.On the other hand ΔfG° of Cu2O is much less negative than that of CO and hence carbon caneasily reduce Cu2O to Cu.

Cu2O + C ⎯→ 2Cu + COThus the extraction of copper from pyrites is more difficult than from its oxide ore throughreduction.

Q. 6. Out of C and CO, which is better reducing agent at 673 K ?

Ans. At 673 K (low temperature), the ΔG° (CO, CO2) line lies below ΔG° (C, CO2) line in the Ellinghamdiagram therefore, at 673K, CO is a better reducing agent. On the other hand, at highertemperature ΔG° (CO, CO2) line lies below ΔG° (CO,CO2) line and hence at higher temperaturecarbon is better reducing agent than CO.

Q. 7. Out of C and CO, which is better reducing agent for ZnO ?

Ans. The free energy of formation (ΔfG°) of CO and C becomes lower at temperature above 1120 Kwhereas that of CO2 from C becomes lower above 1323 K than ΔfG° of ZnO. However ΔfG° ofCO2 from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO.Therefore, out of C and CO ‘C’ is better reducing agent than CO for ZnO.

Q. 8. Copper and silver lie below in the electrochemical series yet they are found in the combined stateas sulphide in nature. Comment.

Ans. Due to high polarising power of Cu and Ag ions, their sulphides are even more stable than themetals.


Q. 9. Zinc acts as reducing agent in the extraction of silver. Comment.

Ans. Zinc acts as reducing agent in the extraction of silver. It reduces Ag+ to Ag and itself getsoxidised to Zn2+.

2Na [Ag (CN)2] + Zn ⎯⎯→ Na2 [Zn(CN)4] + 2Ag ↓Q. 10. Carbon and hydrogen are not used as reducing agent at high temperature. Explain.

Ans. Because at high temperature carbon and hydrogen reacts with metal to form carbides andhydrides respectively.

Q. 11. Although thermodynamically feasible in practice magnesium metal is not used for the reductionof alumina in the metallurgy of aluminium. Why ?

Ans. Temperature below the point of intersection of Al2O3 and MgO curves, magnesium can reducealumina. But magnesium is a much costlier metal than aluminium and hence the process will beuneconomical.

Q. 12. Which reducing agents are employed to get copper from the leached low grade copper ore ?Ans. Scrap iron Cu2+ (aq) + Fe(s) ⎯⎯→ Cu(s) + Fe2+ (aq)

or H2 gas Cu2+ (aq) + H2(g) ⎯⎯→ Cu(s) + 2H+ (aq)Q. 13. Although carbon and hydrogen are better reducing agents but they are not used to reduce

metallic oxide at high temperature. Why ?Ans. At high temperature carbon and hydrogen reacts with metals to form carbides and hydrides

respectively.Q. 14. The reaction Cr2O3 + 2Al ⎯→ Al2O3 + 2Cr (ΔG° = – 421 kJ) is thermodynamically feasible as is

apparent from the Gibb’s energy value. Why it does not take place at room temperature ?Ans. Certain amount of energy of activation is required even for such reaction which are

thermodynamically feasible, therefore heating is required.Q. 15. Why is limestone added to the ore in the metallurgy of iron ?

Ans. In the metallurgy of iron, limestone (CaCO3) acts as a flux and forms calcium silicate with SiO2.

CaCO3Δ

⎯⎯→ CaO + CO2

CaO + SiO2 ⎯⎯→ CaSiO2

Flux Gangue Slag


Q. 1. Fluorine gives only one oxide but chlorine gives a series of oxides. Explain.

Ans. Fluorine has only two orbits. In 2nd orbit it has s2p5 configuration with no vacant ‘d’ orbitals. So itcan not form more than one oxide i.e., OF2.Chlorine, on the other hand, has 3 orbits in its atoms and 3d is lying vacant in it. Therefore it canshow variable covalency and gives a series of oxides. Like Cl2O, ClO2 etc.

Cl atom in ground state → ↑↓ ↑↓ ↑↓ ↑

3s 3p 3dCl atom in excited state → ↑↓ ↑↓ ↑ ↑ ↑

3s 3p 3d

⎩⎪⎪⎨⎪⎪⎧

↑↓ ↑ ↑ ↑ ↑ ↑3s 3p 3d↑ ↑ ↑ ↑ ↑ ↑ ↑

3s 3p 3dQ. 2. Interhalogen compounds are more reactive than their constituent elements. Explain.Ans. Interhalogen compounds are more reactive than their constituent elements because X-Y bonds

present in them are weaker than X-X, or Y-Y bonds.Q. 3. SF6 exists but OF6 does not, though both oxygen and sulphur belong to the same group in the

periodic table.Ans. 16S32 lies in third period of the periodic table. It has 6e– in its valence shell and 3d sub-shell is

lying vacant. Therefore it can show a maximum covalency of 6.


16S (ground state) ↑↓ ↑↓ ↑ ↑

3s 3p 3d

16S (excited state) ↑ ↑ ↑ ↑ ↑ ↑

3s 3p 3dand form molecules like SF6.

8O16 lies in 2nd period of the periodic table. It also has 6e– in its valence shell but there is no d sub-shell present. So its maximum covalency remains 2 and it does not forms OF6.

8O16 ↑↓ ↑↓ ↑ ↑

2s 2pQ. 4. Why the melting and boiling points of group 16 increases down the group ?Ans. It is because as we move down the group the molecular size increases. As a result, the magnitude

of van der Waal forces increases and hence the melting and boiling points also increases.Q. 5. Why H2S is acidic while H2O is neutral ?Ans. The S—H bond is weaker than O—H bond because size of S–atom is greater than of O–atom.

Hence, H2S can dissociate to give H+ ions in aqueous solution.Q. 6. SF6 is known but SH6 is not known. Why ?Ans. In the highest oxidation state, sulphur can combine only with highly electronegative element like

fluorine.Q. 7. Why the compounds of fluorine with oxygen are called fluorides of oxygen and not oxides of

fluorine ?Ans. This is because fluorine is more electronegative than oxygen.Q. 8. Oxygen molecule has the formula O2 while sulphur is S8. Explain why ?Ans. Oxygen atom being small in size has the tendency to form multiple bonds while sulphur atom,

being large in size forms single bonds with another sulphur atom. The puckered ring structure S8is most stable.

Q. 9. Give reasons for the late discovery of fluorine.Ans. (a) It is so reactive that it reacts with all the materials of apparatus.

(b) Greater stability and non-conducting nature of hydrofluoric acid.Q. 10. Except HF all other hydrides are gases (e.g., HCl, HBr, HI). Explain

Ans. HF is a liquid because of intermolecular hydrogen bonding.

–H—F………H—F……H—F……The bond between hydrogen and halogen is covalent in all the cases.

Q. 11. Halogens have maximum negative electron gain enthalpy in the respective periods of theperiodic table, why ?

Ans. Due to small size and high negative charge.

Q. 12. Why do noble gas forms compounds with fluorine and oxygen only ?

Ans. Fluorine and oxygen are small atoms with high value of electronegativity. Fluorine is also highlyreactive in nature. It is for this reason they form compounds with noble gases.

Q. 13. Why Iodine is more soluble in KI than water ?

Ans. Iodine combines with KI to form soluble complex, KI3 (KI + I2 ⎯→ KI3).

Q. 14. Explain why oxygen exists in a diatomic gaseous state at room temperature while other elementsare solids ?

Ans. Due to its small size, oxygen is capable of forming pπ-pπ bond and forms O2 (O = O) molecule.The inter molecular forces in oxygen are weak van der-Waal’s forces, due to which it is a gas. Therest of the elements are not able to form pπ-pπ bond due to their size and do not exist as M2molecule. On the other hand they exist as puckered ring or chain polymers in which the atomsare linked by covalent bonds, hence their boiling points being high, they exist as solid at ordinarytemperature.


Q. 15. Hydrides of group 16 are acidic, their acidic nature increase from H2O to H2Te. Give reason.Ans. The increase in strength of acid on moving down the group is due to increase in size of the atoms.

The distance between central atom and hydrogen increases as we move from O to Te whichfavours the release of hydrogen as protons.

H2O < H2S < H2Se < H2Te.Q. 16. Why oxygen is different from all other elements of its group ?

Ans. Oxygen is different because(i) It has a small size.(ii) High electronegativity.(iii) High ionisation potential.(iv) Non-availability of d-orbitals in its valence shells.(v) Ability to form hydrogen bond.(vi) The maximum covalency is two.

Q. 17. The acidic strength varies in the order HF < HCl < HBr < HI. Give reasons.Ans. All the halogen acids ionise to give H+ ion and halide ion, X–.

HX ⎯→ H+ + X– (where X– = F–, Cl–, Br–, I–). The above order of acidic strength can beexplained in terms of H—X bonds, which is in the order HI < HBr < HCl < HF since HI is theweakest therefore HI is the strongest acid. On the other hand HF bond is strongest, hence it is theweakest acid among all the halogens.

Q. 18. Why are halogens strong oxidising agents ?Ans. Halogens act as strong oxidising agents because of their high electronegativity values and high

electron affinity values. The oxidising power of the halogens is comparable in terms of theirreduction potential values given below.Element : F Cl Br I

Red. Potential (E° Volt.) : + 2·87 + 1·36 + 1·06 + 0·54

Oxidising nature decreases

As the reduction potential values decreases from fluorine to iodine, the oxidising power alsodecreases. The reduction potential value depends upon various energy terms as shown below :

12 X2 (s) ΔƒH

⎯→12 X2 (l)

ΔevH⎯⎯→

12 X2 (g) ΔDissH⎯⎯⎯→ X (g)

red pot . Electron gain(Δeg H) enthalpy

⎯⎯→

⎯⎯→

X–(aq) ΔHhyd←⎯⎯ X–(g)

E° = 12 ΔfH +

12 Δev H +

12 Δdiss H– Electron gain enthalpy – Δ Hhyd.

Q. 19. Explain why fluorine forms only one oxoacid HOF ?Ans. Fluorine is known to form only one oxoacid, HOF which is highly unstable. Other halogens

forms oxoacids of the type HOX, HXO2, HXO3, HXO4 (X = Cl, Br, I). Fluorine due to its small sizeand high electronegativity cannot act as central atom in higher oxoacids and hence do not formhigher oxoacids.

Q. 20. Explain why in spite of nearly the same electronegativity, oxygen forms hydrogen bonding whilechlorine does not ?

Ans. Oxygen atom can form H bonds whereas chlorine does not because the tendency for H bondingdepends upon the :(i) small size (ii) high electronegativity values.

Although the electronegativity of O and Cl is nearly same yet chlorine does not forms H bonddue to its larger size (99 pm) as compared to oxygen (66 pm).


Q. 21. Why are halogens coloured ?

Ans. All the halogens are coloured. The colour of halogens deepens with rise in atomic number fromfluorine to iodine.

Element : F Cl Br I

Colour : Light yellow Yellow green Reddish Brown Deep violet

The colour is due to absorption of energy from visible light by their molecules for excitation ofouter electrons of higher energy levels (the gap of energy between valence shell of halogens andhigher energy shells is less). Fluorine absorbs violet portion of the light and appears yellow whileiodine absorbs yellow and green portions of the light and thus appears violet.

Change in colour on moving from F to I is called blue shift or bathochromic shift.

Q. 22. Why do noble gases have comparatively large atomic sizes ?

Ans. The atomic size, in the case of noble gases, is expressed in terms of van der Waal’s radii whereasthe atomic size of other members of the period is either metallic radii or covalent radii. van derWaal’s radii is larger than both metallic as well as covalent radii, therefore the atomic size ofnoble gas is quite large. Among the noble gases, the atomic size increases down the group due toaddition of new electronic shells.

Q. 23. Bond dissociation energy of F2 is less than Cl2. Why ?

Ans. Bond energy of F–F is smaller due to greater repulsive interactions between the lone pair ofelectrons of one F atom with those of other. The repulsive interaction arises due to greaterconcentration of electron density on each F atom because of its small size.

Q. 24. Sulphur dioxide acts as an oxidizing agent as well as a reducing agent. Give one reaction each toshow its oxidizing nature and its reducing nature.

Ans. The structural formula of sulphur dioxide is SO2, and its a covalent planar molecule which is V-shaped with an angle of 120 degrees.Sulphur dioxide as oxidising agent : (Hydrogen sulphide)Claus Process : SO2 + 2H2S ⎯→ 3S + 2H2O

sulphur dioxide elemental sulphur water

Sulphur dioxide as reducing agent :SO2 + Cl2 ⎯→ SO2Cl2

sulphuryl chloride

Q. 25. Why sulphuric acid is always added to H2O for dilution ?

Ans. To avoid spurting, because lots of heat is produced when water is added to conc. sulphuric acid.

Q. 26. How is O3 estimated quantitatively ?

Ans. When ozone reacts with an excess of potassium iodide solution buffered with a borate solution(pH 9·2) iodine is liberated which can be titrated against a standard solution of sodiumthiosulphate. This is a quantitative method for estimating ozone gas.

Q. 27. Sea is the greatest source of halogens comment.

Ans. Sea water contains chlorides, bromides and iodides of Na, K, Mg and Ca but mainly sodiumchloride solution (2·5% by mass) certain forms of marine life contains iodine in their systems, Forexample, sea weeds contain up to 0·5% of iodine as sodium iodide.

Q. 28. Why has it been difficult to study the chemistry of radon ?

Ans. Radon is radioactive with very-short half life (3·82 days) which makes the study of chemistry ofradon difficult.

Q. 29. In the preparation of H2SO4 by contact process why is SO3 not absorbed directly in water to formH2SO4 ?

Ans. Acid fog is formed, which is difficult to condense due to this reason SO3 is not absorbed directlyin water to form H2SO4.


Q. 30. How is SO2 an air pollutant ?Ans. (i) SO2 dissolves in moisture present in air to form H2SO3 which damages building materials

especially marble.CaCO3 + H2SO3 ⎯⎯⎯→ CaSO3 + H2O + CO2

(ii) Even at a low concentration of 0·03 ppm, SO2 has a damaging effect on plants. If exposed fora long time, it slows down the formation of chlorophyll resulting in injury to the leaves includingloss of green colour. This is called chlorosis.(iii) SO2 is strongly irritating to the respiratory tract SO2 at a concentration of 5 ppm causesthroat and eye irritation resulting in cough, tear and redness in eyes. It causes breathlessness. andaffects voice box.

Q. 31. Why are pentahalides more covalent than trihalides ?Ans. Higher the positive oxidation state of the central atom more will be its polarising power which in

turn increases the covalent character of bond formed between the central atom and other atom.As, in pentahalides, the central atom is in +5 oxidation state while in trihalides it is in +3oxidation state. Therefore pentahalides are more covalent than trihalides.

Q. 32. Explain each of the following :(i) H3PO4 is tribasic.(ii) H2O is a liquid and H2S is a gas.

Ans. (i)

O||P

HO HO OHThere P—OH groups are present in the molecule of H3PO4. Therefore its

basicity is three.(ii) Due to small size and high electronegativity of oxygen, molecules of water are associatedthrough hydrogen bonding resulting in its liquid state. On the other hand H2S molecules are notassociated through H-bonding. Hence it is a gas.

Q. 33. Give reason for the following :(i) ICl is more reactive than I2

(ii) Helium is used in diving apparatus.Ans. (i) ICl is more reactive than I2 because I-Cl bond is weaker than I—I bond, consequently ICl

breaks easily to form halogen atom which readily brings about the reactions.(ii) Helium is used as a diluent for oxygen in modern driving apparatus because of its very lowsolubility in blood.

Q. 34. Account for the following :(i) Ozone acts as a powerful oxidising agent.(ii) Noble gases have comparatively large atomic size.

Ans. (i) Due to the ease with which it liberates atom of nascent oxygen, it acts as a powerfuloxidising agent.

O3 ⎯→ O2 + O (nascent oxygen)(ii) Noble gases have only van der waal’s radii while others have covalent radii. As van derwaal’s radii are larger than covalent radii, hence noble gases have comparatively large atomicsizes.

Q. 35. Give reason for the bleaching action of Cl2.Ans. In the presence of moisture or in aqueous solution Cl2 liberates nascent oxygen.

Cl2 + H2O ⎯→ 2HCl + O

Nascent oxygen

This nascent oxygen brings about the oxidation of coloured substance present in vegetable andorganic matter to colourless substances.Coloured substance + O ⎯→ colourless substance.Thus the bleaching action of Cl2 is due to oxidation.


Q. 36. The HNH angle value is higher than HPH, HAsH, HSbH angles. Why ?Ans. As we move from N to P to As to Sb, the electronegativity of the central atom goes on decreasing.

Bond pairs of electrons, lies much away from the central atom. In other words, force of repulsionbetween the adjacent bond pairs goes on decreasing and hence the bond angle keep ondecreasing from NH3 to SbH3.

Q. 37. Explain why NH3 is basic while BiH3 is only feebly basic.Ans. Since the atomic size of N is much smaller than that of Bi, electron density on the N-atom is much

higher than that on Bi-atom. As a result the tendency of N in NH3 to donate its lone pair ofelectron is much higher than that of Bi in BiH3. Thus NH3 is much more basic than BiH3.

Q. 38. Nitrogen exist as diatomic molecule and phosphorus as P4. Why ?Ans. Nitrogen because of its small size and high electronegativity forms pπ-pπ multiple bonds. Thus it

exists as a diatomic molecule having a triple bond between the two N-atoms. Phosphorus due toits larger size and lower electronegativity usually does not form pπ-pπ multiple bond with itself.Instead it prefers to form P—P single bond and hence it exist as tetrahedral P4 molecule.

Q. 39. Why does nitrogen shows catenation property less than phosphorus ?Ans. The property of catenation depends upon E—E bond strength of the element. Since the N—N

(159 kJ mol–1) bond strength is much weaker than P—P (213 kJ/mol) bond strength, hencenitrogen shows less catenation properties than phosphorus.

Q. 40. Why is dioxygen a gas but sulphur is solid ?Ans. Oxygen forms pπ–pπ multiple bonds. Due to small size and high electronegativity. Oxygen exist

as diatomic (O2) molecules thus molecules are held together by weak van der Waal’s forces.Hence O2 is a gas at room temperature.Sulphur because of its bigger size and lower electronegativity, prefers to form S—S single bondfurther because of S—S single bond than O—O single bonds. Sulphur has a much greatertendency for catenation than oxygen. Thus sulphur because of its higher tendency for catenationand lower tendency for pπ-pπ multiple bonds forms octa-atomic (S8) molecule having eight-membered pucked ring structure due to bigger size the force of attraction holding the S8 moleculetogether are much stronger. Hence sulphur is a solid at room temperature.

Pucked 8-membered ring structure of sulphur molecule

Q. 41. Why are halogens considered strong oxidising agents.Ans. Due to low bond dissociation enthalpy, high electronegativity and large negative electron gain

enthalpy, halogens have a stronger tendency to accept electrons and gets reduced

X + e– ⎯⎯⎯→ X

–

Thus halogens act as strong oxidising agents. Their oxidising power however, decreases from F2to I2.

Q. 42. Explain why in spite of nearly the same electronegativity, oxygen forms hydrogen bonding whilechlorine does not.

Ans. Although O and Cl have about the same electronegativity, yet their atomic size are muchdifferent O = 66 pm and Cl = 99 pm. Thus-electron density per unit volume on oxygen atom ismuch higher than that of chlorine atom. Hence oxygen forms hydrogen bonds while chlorinedoes not though both have nearly same electronegativity.

Q. 43. Account for the following :(i) PCl5 is more covalent than PCl3.(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.(iii) The two O—O bond length in the ozone molecule are equal.

Ans. (i) The oxidation state of central atom i.e. phosphous is + 5 in PCl5 whereas it is + 3 in PCl3 .


Higher the positive oxidation of central atom. More will be its polarising power which in turnincreases the covalent character of bond formed between the central atom and the atomsurrounding it.(ii) Iron reacts with HCl to form FeCl2 and H2

Fe + 2HCl ⎯⎯→ FeCl2 + H2

H2 thus produced prevents the oxidation of FeCl2 to FeCl3.(iii) Ozone is a resonance hybrid of the following two main structure :

O

O O

O

O O˙˙: . .

˙˙

: . .:

˙˙

˙ :: :→←

As a result of resonance, the two O—O bond lengths in O3 are equal.Q. 44. Account for the following :

(i) Bi (V) is a stronger oxidising agent than Sb(V).(ii) N—N single bond is weaker than P—P single bond.(iii) Noble gases have very low boiling points.

Ans. (i) Due to inert pair effect + 3 oxidation state of Bi is more stable than its + 5 oxidation statewhile + 5 oxidation state of Sb is more stable than its + 3 oxidation state. Therefore Bi (V) canaccept a pair of electron to form more stable Bi (III) more easily than Sb (V). Hence Bi (V) is astronger oxidising agent than Sb (V).(ii) N—N single bond is weaker than P—P single bond due to large inter electronic repulsionbetween the lone pairs of electron present on the N atom of N—N bond having small bondlength.(iii) Noble gases being monoatomic have no interatomic forces except weak London or dispersionforces and therefore they are liquified at very low temperature hence they have very low boilingpoints.

Q. 45. Account for the following :(i) Sulphur in vapour form exhibits paramagnetic behaviour.(ii) SnCl2 is more covalent than SnCl2.(iii) H3PO2 is a stronger reducing agent than H3PO3.

Ans. (i) In vapour form sulphur partly exist as S2 molecules which have two unpaired electrons in theantibonding π* molecular orbital like O2 molecule and hence exhibits paramagnetism.(ii) The oxidation state of the central atom Sn is + 4 in SnCl4 while it is + 2 in SnCl2. Higher thepositive oxidation state of central atom more will be its polarising power which in turn increasesthe covalent character of bond formation between the central atom and atoms around it.(iii) Acids which contain P—H bond have reducing character since. H3PO2 contains two P—Hbonds while H3PO3 contains only one P—H bond therefore H3PO2 is a stronger reducing agentthan H3PO3.

Q. 46. Give reasons for the following :

(a) CN– ion is known but CP– ion is not known.(b) NO2 dimerises to form N2O4.

Ans. (a) Nitrogen being smaller in size forms pπ-pπ multiple bonding with carbon so CN– ion is

known but phosphorous does not forms pπ-pπ bond as it is larger in size.(b) This is because NO2 is an odd electron molecule and therefore gets dimerised to stable N2O4.

Q. 47. Account for the following :(i) NH3 is stronger base than PH3.(ii) Sulphur has a greater tendency for catenation than oxygen.

Ans. (i) NH3 is a stronger base than PH3. This is because of lone pair of electron on N—atom in NH3is directed and not diffused as it is in PH3 due to larger size of phosphorus and hence moreavailable for donation.(ii) Sulphur has a greater tendency for catenation than oxygen because S—S bond is strongerthan O—O bond due to less inter electronic repulsion.


Q. 48. Explain the following situations :(i) In the structure of HNO3 molecule the N—O bond (121 pm) is shorter than N—OH bond(140 pm).(ii) SF4 is easily hydrolysed where as SF6 is not easily hydrolysed.(iii) XeF2 has a straight linear structure and not a bent angular structure.

Ans. (i)

H—O—N

O

O←→ HO—N

O

O

˙˙

˙˙

:

:˙˙

˙˙

:˙˙˙˙

:

+–

+

–

As a result of resonance N—O bond length is average of single bond and double bond whereasN—OH bond as purely single bond character. Therefore N–O bond is shorter than N–OH bondin HNO3.(ii) S atom in SF4 is not sterically protected as it is srrrounded by four F atom, so attack of H2Omolecule can takes place easily and hence hydrolysis takes place easily. In contrast to this in SF6,S is sterically protected by six f atom therefore does not allow H2O molecule to attack S atom. Asa result of this SF6 does not undergo hydrolysis.(iii) In XeF2, Xe is sp3d hybridised having 2 bond pairs and 3 lone pairs of electron. The presenceof 3 lone pair of electron in XeF2 are at equidistance, to have minimum repulsion is responsiblefor its linear structure.

Q. 49. Account for the following :(a) Chlorine water has both oxidising and bleaching properties.(b) H3PO2 and H3PO3 act as good reducing agents while H3PO4 does not.(c) On addition of ozone gas to KI solution, violet vapours are obtained.

Ans. (a) Chlorine water produces nascent oxygen which is responsible for bleaching action andoxidation.

Cl2 + H2O ⎯→ 2HCl + [O](b) Both H3PO2 and H3PO3 has P—H bonds so they act as reducing agents. H3PO4 has no P—Hbond but has O—H bonds so it cannot act as reducing agents.(c) Ozone gas acts as a strong oxidising agent, so it oxidises iodide ions to iodine.

2I– (aq) + H2O (l) + O3 (g) ⎯→ 2OH

– (aq) + I2(g) + O2 (g)

I2 vapours evolved have violet colour.

Chapter 8. The d and f-block ElementsQ. 1. Why do transition metal ions possess a great tendency to form complexes ?Ans. The transition metal ions possess a great tendency to form complexes because of their small size,

high ionic charge and availability of partially filled d-orbitals for bond formation. These partiallyfilled d-orbitals can easily accommodate ligands, electrons and consequently transition metal ionsform complexes. Examples [Fe(CN6)]3– , Ni(CO)4 etc.

Q. 2. The paramagnetic character in 3d-transition series elements increases upto Mn and thendecreases. Why ?

Ans. In the 3d-transition series as we move from Sc (21) to Mn (25), the number of unpaired electronsincreases and hence paramagnetic character increases. After Mn, the pairing of electrons in the d-orbital starts and the number of unpaired electrons decreases and hence, paramagnetic characterdecreases.

Q. 3. Zn2+ salts are white but Cu2+ salts are blue in colour. Why ?Ans. Zinc and copper are transition metals and their ions are generally coloured due to d-d transition

of unpaired electron. But in case of Zn2+ 3d is completely filled and no unpaired electrons areavailable for transition. So light falling on these ions gets completely transmitted and their saltsappear white. However in Cu2+ 3d has 9 electrons. The unpaired electron absorbs particular


wavelengths in visible region of light and the transmitted light shows the complementary colour‘blue’.

Zn2+ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

3d10

Cu2+ ↑↓ ↑↓ ↑↓ ↑↓ ↑

3d9

Q. 4. Transition elements form coloured compounds. Why ?Ans. Transition metal ions generally possess one or more unpaired electrons. When visible light falls

on a transition metal compound or ion, the unpaired electrons present in the lower energy d-orbital gets promoted to high energy d-orbitals, called d-d transition, due to the absorption ofvisible light. Since, the energy involved in d-d transition is quantised, only a definite wavelengthgets absorbed, remaining wavelengths present in the visible region gets transmitted. Therefore,transmitted light shows some colour complementary to the absorbed colour.

Q. 5. Cu+ is diamagnetic but Cu2+ is paramagnetic. (Z = 29).

Ans. Cu+ is diamagnetic because it has no unpaired electrons in its core.

29Cu ⎯→ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑↓

3d 4s 4p

29Cu+ ⎯→ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

3d 4s 4pOn the other hand Cu2+ is paramagnetic because it has an unpaired electron in its core.

29Cu2+ ⎯→ ↑↓ ↑↓ ↑↓ ↑↓ ↑

3d 4s 4pQ. 6. Explain why transition metals form complex compounds ?Ans. Transition metals are d-block elements with valence shell configuration ns2(n – 1) d1–10. These

metals and their ions easily coordinate to a number of negative ions or neutral molecules havinglone pair of electrons to form complex. This is due to :(i) Small size and high nuclear charge of these metals.

(ii) Availability of vacant d-orbitals of suitable energy to accommodate lone pair of electronsdonated by the ligands.

Q. 7. Iron is ferromagnetic in nature. Explain why ?Ans. Ferromagnetic substances have a larger number of unpaired electrons and are attracted strongly

in a magnetic field.26Fe has 4s23d6 valence shell configuration with four unpaired electrons in 3d . So it isferromagnetic when all these four unpaired electrons gets aligned in the same direction in agiven magnetic field.

unpaired electrons

26Fe atom’s valence shell ↑↓ ↑ ↑ ↑ ↑ ↑↓3d 4s 4p

Q. 8. In a given transition series, there is no significant change in the atomic radii of elements withincrease in atomic number. Explain why ?

Ans. In a transition series the electrons are filled in inner d orbitals so there is no increase in thenumber of shells but there is an increase in the nuclear charge and also an increase in therepulsion between inner orbital electrons. The two opposite forces i.e., increased attraction by thenucleus and increased repulsion between electrons counter balance each other, as a result there isno significant change in atomic radii of transition elements.

Q. 9. Zn2+ compounds are white in colour but Cu2+ compounds are coloured, though both zinc andcopper are d-block elements. Explain.

Ans. Both zinc and copper are d-block elements but zinc has no unpaired e- in its (n – 1) d sub-shell. Soits ion Zn2+ is colourless.Copper, on the other hand has one unpaired e– in its (n – 1) d sub-shell when it forms Cu2+

compounds. This unpaired electron can absorb a definite quantum of energy from the visiblespectrum and jumps on to the higher sub-shell. Wavelength transmitted shows a complimentarycolour.


Q. 10. Why Cd salts are white ?Ans. They have fully filled d-orbitals, hence d-d transitions are not possible.

Q. 11. Why tungsten is used as a filament ?Ans. Due to high melting point and high resistance, tungsten is used as a filament.

Q. 12. Why K2Cr2O7 is generally preferred to Na2Cr2O7 in volumetric analysis although both areoxidising agents ?

Ans. K2Cr2O7 is not hygroscopic like Na2Cr2O7 and it also fulfills the requirements of primarystandard.

Q. 13. Why chromium is used for electroplating iron ?Ans. Due to formation of inert layer of oxide.

Q. 14. Why Ti4+ complexes are diamagnetic ?Ans. It has no unpaired electrons

Ti4+(→)1s2, 2 s2, 2 p6, 3 s2, 3 p6.Q. 15. Why transition elements have high melting and boiling points ?

Ans. The melting and boiling points of transition elements are generally high. This is due to strongmetallic bonds and the presence of half filled d-orbitals in them. Due to these half filled orbitalssome covalent bonds exist between atoms of transition elements, because of strong inter atomicbonding, they have high melting and boiling points.

Q. 16. Why transition metals and their compounds are widely used as catalysts ?Ans. Transition metals and their compounds are extensively used as catalysts due to following

reasons :(i) Their partially empty d-orbitals provides surface area for the reactant molecules.(ii) They combines with reactant molecules to form transition state and lowers the activation

energy.(iii) They show multiple oxidation states and by giving electrons to reactants they form

complexes and lowers their energies.Q. 17. Explain why :

(i) Zinc, cadmium and mercury are generally not considered as transition metals.(ii) Sc3+ salts are diamagnetic(iii) Ti3 + salts are coloured whereas Ti4 + salts are white

Ans. (i) Zn, Cd and Hg have completely filled d-orbitals in their atoms and in their commonoxidation states. They do not exhibit characteristic properties of transition elements so theyare not considered as transition elements.

(ii) In Sc3+ all the electrons are paired.(iii) Ti3+ has one electron in 3d-subshell whereas Ti4+ has no electron in 3d-subshell. Therefore in

Ti3+ (3d1) d-d-transitions are possible whereas in Ti4+ (3d°) d-d transitions are not possible.Hence, Ti3+ salts are coloured whereas Ti4+ salts are white.

Q. 18. How would you account for the following d4 species :(a) Cr2+ is strongly reducing while manganese (III) is strongly oxidising.(b) The d1 configuration is very unstable in ions.

Ans. (a) For chromium (III) state is more stable as compared to II state. Therefore, Cr2+ readilychanges into Cr3+ and behaves as a strong reducing agent. On the other hand for manganese, IIstate is more stable than III state. Hence, Mn3+ readily changes into Mn2+ (3d5) by gaining anelectron and behaves as a strong oxidising agent.(b) The ions in d1 configuration have great tendency to acquire more stable d° configuration bylosing the lone-d-electron and acts as a reducing agent.

Q. 19. Explain why cuprous chloride (CuCl) is colourless while cupric chloride (CuCl2) is blue.Ans. In CuCl, Cu+ has fully-filled d-subshell hence it cannot undergo-d-d transitions. Therefore, it is

colourless. In CuCl2, Cu2+ has 3d9 configuration. Due to the presence of one half filled d- orbital, itcan absorb energy for d-d transitions. Hence, it is coloured.


Q. 20. Explain why La(OH)3 is a stronger base than Lu(OH)3 ?Ans. Lu3+ is smaller in size than La3+ due to lanthanoid contraction. Due to smaller size of Lu3 +, Lu–O

bond is stronger than La–O bond in the respective hydroxides. Due to weaker La–O bond,La(OH)3 behaves as stronger base.

Q. 21. Copper can be extracted by hydrometallurgy but not zinc. Explain.Ans. This is because the E° value of Cu2+/Cu is more than that of hydrogen while E° value of

Zn2+/Zn is less than that of zinc. Therefore, Cu2+ can be reduced to Cu by H2 but not zinc.Cu2+

(aq.) + H2(g) ⎯→ Cu(s) + 2H+ (aq.)

Q. 22. Silver ores have to be leached with cyanides. Give a reason for this.Ans. Silver forms water soluble complexes with alkali metal cyanides (NaCN, KCN) from which pure

metal can be precipitated by the addition of more electropositive metal like zinc. Therefore, silvermetal is extracted by leaching with metal cyanides.

Q. 23. Explain why, melting point of transition metals first increases to maximum and then decreasesregularly towards the end of the period.

Ans. The strength of inter atomic bonds in transition elements is roughly related to the number of halffilled d-orbitals. In the beginning the number of half filled d-orbitals increases till the middle ofthe period causing increase in strength of interparticle bonds. But thereafter, the pairing ofelectrons in d-orbitals occurs and the number of half-filled orbitals decreases which also causesdecrease in the melting point.

Q. 24. Why are transition elements named so ?Ans. Transition elements are named so because their properties are in between those of s and p-block

elements.Q. 25. In what way is the electronic configuration of transition elements different from that of the non-

transition elements ?Ans. Transition elements contain incompletely filled d-subshell i.e. their electronic configuration is

(n-1) d1–10 ns1–2 whereas non-transition elements have no d-subshell or their subshell is completelyfilled and have ns1–2 or ns2 np1–6 electrons in their outermost shell.

Q. 26. Why does a transition series contain 10 elements ?Ans. There are five d-orbitals in an energy level and each orbital can contain two electrons. As we

more from one element to the next, an electron is added for complete filling of the five d-orbitals10 electrons are required.

Q. 27. Why are transition elements known as d-block elements ?Ans. The last electron enters (n-1) d-orbital i.e. d orbital of the penultimate shell. Hence these are

known as d-block elements.Q. 28. Why do transition elements shows similarities along the horizontal period ?

Ans. All transition elements contain incompletely filled d-subshell whereas outershell electronicconfiguration remains the same.

Q. 29. Reactivity of transition element decreases almost regularly from Sc to Cu. Explain.

Ans. It is due to regular increase in ionisation enthalpy.

Q. 30. How would you account for the irregular variation of ionisation enthalpies (first and second) infirst series of the transition elements.

Ans. Irregular variation of ionisation enthalpy is mainly attributed to varying degree of stability ofdifferent 3d configuration (e.g., d°, d5, d10 are exceptionally stable).

Q. 31. Why copper does not replace hydrogen from acids ?

Ans. Since Cu has positive value of E° thus it does not replaces H2 from acids.

Q. 32. Which of the 3d series of the transition metal exhibits the largest number of oxidation state andwhy ?

Ans. Manganese (Z = 25) as its atom has the maximum number of unpaired electron.Thus it showsoxidation states from + 2 to + 7 (+ 2, + 3, + 4, + 5, + 6 and + 7) which is the maximum number.

Q. 33. Why do Zr and Hf exhibit almost similar properties ?Ans. Zr and Hf have similar ionic size, due to which they exhibit almost similar properties.


Q. 34. Which divalent metal ion has maximum paramagnetic character among the first transitionseries ? Why ?

Ans. Mn2+ has the maximum paramagnetic character because of the maximum number of unpairedelectrons i.e.; 5.

Q. 35. Why are lanthanoids called f-block elements.Ans. Lanthanoids are called f-block elements because the last electron in them enters into f-orbital.

Q. 36. On what basis can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not ?Ans. On the basis of incompletely filled 3d-orbitals in case of scandium atom in its ground state (3d1) it

is regarded as a transition element on the other hand, zinc atom has completely filled d-orbitals(3d10) in its ground state as well as in its oxidised state. Hence it is not regarded as transitionelement.

Q. 37. The E° (M2+/M) value for copper is positive (+ 0·34V). What is possibly the reason for this ?

Ans. E° (M2+/M) for any metal is related to the sum of the enthalpy changes taking place in thefollowing steps :

M(s) + ΔaH ⎯⎯→ M(g)

M(g) + ΔiH ⎯⎯→ M2+

(g)

M2+ (g) + aq ⎯⎯→ M2+(aq) + Δhyd H

Copper has high enthalpy of ionisation and relatively low enthalpy of hydration. So E° (Cu2+/Cu) ispositive. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydrationenthalpy.

Q. 38. For the first row transiton metal the E° values are

E° V Cr Mn Fe Co Ni Cu

(M2+/M) – 1·18 – 0·91 – 1·18 – 0·44 – 0·28 – 0·25 + 0·34Explain irregularity in the above values.

Ans. The E° ( M2+/M) values are not regular which can be explained from the irregular variation ofionisation enthalpies (ΔiH1 + ΔiH2) and also the sublimation enthalpies which are relatively muchless for manganese and vanadium.

Q. 39. Account for the following :(i) Silver atom has completely filled d-orbitals (4d10) in its ground state yet silver is a transitionelement.(ii) Highest oxidation state of a metal exhibited in its oxide or fluoride only.

Ans. (i) Silver (Z = 47) can exhibit + 2 oxidation state where in it will have a incompletely filledd-orbitals) 4f hence a transition element.(ii) Due to small size and high electronegativity oxygen or fluorine can oxidise a metal to itshighest oxidation state.

Q. 40. Give reason for the following :(i) Zn, Cd and Hg normally not regarded as transition metals.(ii) The metallic radii of the third (3d) series of transition metals are virtually the same as thoseof the corresponding group members of the second (4d) series.

Ans. (i) Transition elements are those (atoms or ions) which have partly filled d-orbitals. Sinceneither the atom nor ions of Zn, Cd and Hg have partly filled d-orbitals they are normally notregarded as transition metals.(ii) This is due to filling of 4f orbital which have poor shielding effect.

Q. 41. How would you account for the following :(i) Mn (III) undergoes disproportionation reaction easily.(ii) Co (II) is easily oxidised in the presence of strong ligands.

Ans. (i) Mn3+ is less stable and changes to Mn2+ which is more stable due to half filled d-orbitalsconfiguration. That is why Mn3+ undergoes disproportionation reaction.


(ii) Co (II) has electronic configuration 3d7, 4s° i.e.; it has three unpaired electrons. In thepresence of strong ligands, two unpaired electrons in 3d-subshell pair up and third unpairedelectron shifts to higher energy subshell from where it can be easily lost and hence oxidised toCo(III).

Q. 42. Explain briefly how + 2 state becomes more and more stable in the first half of the first rowtransition elements with increasing atomic number.

Ans. As the atomic number increases from 21 to 25 the number of electrons in the 3d orbital alsoincreases from 1 to 5, + 2 oxidation state is attained by the loss of the two 4s electron by thesemetals, so does not exhibit + 1 oxidation state. As the number of d-orbitals in + 2 state increasesfrom Ti to Mn. The stability of + 2 state increases (d-orbital gradually becoming half filled) Mn(+ 2) has d5 electrons which is highly stable.

Q. 43. To what extent do the electronic configuration decides the stability of oxidation state in the firstseries of transition elements ? Illustrate your answer with examples.

Ans. The stability of oxidation state in the first series of the transition elements are related to theirelectronic configuration. The first five elements in the first transition series up to Mn in which the3d-sub shell is not more than half-filled, the minimum oxidation is given by the sum of the outer sand d-electrons. For example Sc does not show + 2 oxidation state. Its electronic configuration is4s2, 3d1. It loses all the three electrons to form Sc3+. + 3 oxidation state is very stable as by loosingall three electrons, it attains the stable configuration of Argon. For, Mn2+ oxidation state is verystable after losing two 4s electrons, the d-orbitals become half-filled.

Q. 44. Account for the following :

(i) Of the d4 species Cr2+ is strongly reducing while manganese (III) is strongly oxidising.

(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it iseasily oxidised.

(iii) The d1 configuration is very unstable in ions.

Ans. (i) E° value for Cr3+/Cr2+ is negative (– 0·41V) whereas E° value for Mn3+/Mn2+ is positive(+ 1·57V). Thus Cr2+ ion can easily undergo oxidation to give Cr3+ ions and therefore acts asstrong reducing agent. On the other hand Mn3+ ions, can easily undergo reduction to give Mn2+

and hence acts as an oxidising agent and therefore act as strong reducing agent.

(ii) Co (III) has greater tendency to form coordination complexes than Co(II). Thus in thepresence of ligands, Co(II) changes to Co(III) i.e. is easily oxidised.

(iii) The ions with d1 configuration have the tendency to loose the only electron present in d-subshell to aquire stable d° configuration. Therefore they are unstable and undergo oxidation ordisproportionation.

Q. 45. Give examples and suggest reason for the following features of transition metal chemistry.

(i) The lowest oxide of transition metal is basic, the highest is acidic.

(ii) A transition metal exhibit higher oxidation states in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxo-anions of metal.

Ans. (i) The lowest oxide of transition metal is basic because the metal atom has low oxidation state.This means that it can donate valence electron which are not involved in bonding to act like abase whereas the highest oxide is acidic due to the highest oxidation state as the valence electronsare involved in bonding and are unavailable. For example, MnO is basic whereas Mn2O7 is acidic.

(ii) The transition metal exhibits higher oxidation states in oxides and fluorides because oxygenand fluorine are highly electronegative elements, small in size and strongest oxidising agents. Forexample–Osmium shows an oxidation of + 6 in OsF6 and vanadium shows an oxidation of + 5 inV2O5.

(iii) Oxo metal anions have higher oxidation state e.g. Cr in Cr2O72– has an oxidation state of + 6

whereas Mn in MnO4– has an oxidation state of + 7. This is again due to the combination of the

metal with oxygen which is highly electronegative and oxidising element.


Q. 46. Give reason for the following :(i) The halides of transition element becomes more covalent with increasing oxidation state ofthe metal.(ii) Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U.

Ans. (i) As the oxidation state increases, size of the ion of transition element decreases, As perFajan’s rule as the size of the metal ion decreases covalent character of the bond formed increases.(ii) It is because in the beginning, when 5-f orbitals begins to be occupied they will penetrateless into inner core of electrons. The 5-f electron will therefore be more effectively shielded fromthe nuclear charge than 4-f electron of the corresponding lanthanoids. Therefore, outer electronsare less firmly held and they are available for bonding in the actinoids.

Q. 47. Account for the following observations :(i) First ionisation enthalpy of Cr is lower than that of Zn.(ii) Radius of Fe2+ is less than that of Mn2+.

Ans. (i) Ionisation enthalpy of Cr is lower due to stability of d5 and the value of Zn is higher becauseits electron comes out from 4s orbital.

(ii) Fe2+

has lesser radius than Mn2+

because effective nuclear charge is more in Fe+ 2

ascompared to Mn

2+.

Q. 48. Assign reason for each of the following :

(i) Transition metals generally form coloured compounds.

(ii) Manganese exhibits the highest oxidation state of + 7 among the 3d-series of transitionelements.

Ans. (i) This is due to d-d transition. When visible (white) light falls on transition metal compoundsthey absorb certain radiation of visible light and transmits the remaining ones. The colourobserved corresponding to complementary colour of the light absorbed.

(ii) As manganese has maximum number of unpaired electron (5) in 3d-subshell in addition to 2electrons in the 4s-subshell.

Q. 49. How do the oxides of transition elements in the lower oxidation state differ from those in higheroxidation state in the nature of metal-oxygen bonding and why ?

Ans. In the lower oxidation state the transition metal oxides are basic and they are acidic if the metal isin higher oxidation state. The oxides are amphoteric when the metal is in intermediate oxidationstate. For example.

+ 3 + 4 + 7Mn2O3 MnO2 Mn2O7

Basic Amphoteric Acidic

Q. 50. Transition metals can act as catalysts because they can change their oxidation states. How doesFe(III) catalysis the reaction between iodide and per sulphate ions.

Ans. Reaction between iodide and per sulphate ion’ :

2I– + S2O8

2–Fe(III)

⎯⎯→ I2 + 2SO42–

Role of Fe (III) ions.

2Fe3+ + 2I–

⎯⎯→ 2Fe2+

+ I2

2Fe2+ + S2O82–

⎯⎯→ 2Fe3+ + 2SO42–

Chapter 9. Coordination CompoundsQ. 1. How does K2 [Pt Cl4] get ionized when dissolved in water ? Will it form precipitate when AgNO3

solution is added to it ? Give a reason for your answer.Ans. On addition of AgNO3 to K2 [PtCl4]

K2PtCl4 + 2AgNO3 → 2KCl (aq) + PtCl2 (AgNO3) (s)Grey precipitate


Q. 2. Explain, why an aqueous solution of potassium hexacyanoferrate(II) does not give the test forferrous ion ?

Ans. Potassium hexacyanoferrate (II) does not gives ferrous ion test, because it is a coordinationcompound and coordination compounds do not ionise in aqueous solution, thus does not giveferrous ion test.

Q. 3. In a coordination complex, donation of electron pair takes place from the central metal atom tothe ligands.

Ans. In a coordination complex, donation of electron pair takes place from the ligand to the centralmetal atom.

Q. 4. Why is CO a stronger ligand than Cl– ?

Ans. Because CO forms π-bonds.Q. 5. Why are low spin tetrahedral complexes not found ?Ans. Because for tetrahedral complexes, the crystal field stabilisation energy is lower than paring

energy.Q. 6. Why do compounds having similar geometry have different magnetic moment ?Ans. It is due to the presence of weak and strong ligands in complexes. If CFSE is high, the complex

will show low value of magnetic moment and vice-versa eg- [COF6]3+ the former is para magneticand the later is diamagnetic.

Q. 7. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ but CuSO4solution mixed with aqueous ammonia 1 : 4 molar ratio does not give the test of Cu2+ ion. Explainwhy ?

Ans. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio forms double saltFeSO4 (NH4)2SO4. H2O which ionise in the solution to give Fe2+ ion. Hence it gives the test of Fe2+

ion.CuSO4 solution mixed with aqueous ammonia 1 : 4 molar ratio forms a complex with the formula[Cu(NH3)4]SO4. The complex ion, [Cu(NH3)4]2+ does not ionise to give Cu2+ ion. Hence it does notgives the test of Cu2+ ion.

Q. 8. Tetrahedral complexes do not show geometrical isomerism. Why ?Ans. It is because the relative position of the ligands attached to the central metal atom are the same

with respect to each other.Q. 9. The transisomer of complex CoCl2 (en)2 is optically inactive. Why ?Ans. It is because the trans isomer has a plane of symmetry and can be divided into two equal halves.

Q. 10. Why chelate complexes are more stable than unchelated complexes ?Ans. When a chelating ligand attaches to the central metal atom the process is accompanied by the

increase in entropy resulting in the formation of a stable complex.Q. 11. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital

complex.Ans. In [Co(NH3)6]3+ inner d-orbitals are used in hybridization whereas in [Ni(NH3)6]2+ outer d-

orbitals of valence shell are used in hybridization.Q. 12. Explain why K3 [Fe(CN)6] is more stable than K4[Fe(CN)6] ?

Ans. It is because the stability of complex depends upon the charge density (i.e., charge/radius ratio)on central ion. More is the charge density greater is the stability.

Q. 13. Why inner orbital octahedral complexes are called low spin complexes ?Ans. Inner orbital octahedral complexes are also called low spin complexes because of the pairing of

electrons in them they can have either zero or maximum of one unpaired electron and hence theirmagnetic moment can’t exceed 173 B.H.

Q. 14. Why outer orbital octahedral complexes are called high spin complexes ?Ans. Outer octahedral complexes are also called high spin complexes because as no pairing occurs in

these complexes, they can have many unpaired electrons (from 1 to 5) and hence have large valueof magnetic moment.

Q. 15. [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why ?Ans. In both the above complexes nickel undergoes sp3 hybridization and hence these have tetrahedral

geometry. In [Ni(CO)4], oxidation state of nickel is zero whereas in [NiCl4]2– it is +2. Due to the


strong ligand field of the CO group the unpaired electrons of nickel gets paired but the weakligand field of Cl– does not pair the electrons. Hence [Ni(CO)4] is diamagnetic while the [NiCl4]2–

is paramagnetic.Q. 16. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.

Ans. The ground state configuration of Ni (Z = 28) is 3d84s2. In Ni2+ ion the configuration becomes 3d8.Water (H2O) provides a weak ligand field and as a result Ni undergoes sp3d

2 hybridization information of complex [Ni(H2O)6]2+ ion.

3d 4s 4p 4dNi (H2O)6]2+ ·· ·· ·· ·· ·· ··

sp3d2 hybridization

Due to presence of unpaired electrons in d-orbitals there are d-d-transitions and hence bluecomponent of light is absorbed and the complex appears green in colour (as green is thecomplementary colour of blue).In case of [Ni(CN)4]2– ion, the cyanide ion provides a strong ligand field which forces unpairedelectrons in 3d orbitals, no d-d transition would occur and hence complex is colourless.

3d 4s 4pNi (CN)4]2– ·· ·· ·· ··

dsp2 hybridization

Chapter 10. Haloalkanes and HaloarenesQ. 1. (a) Why does p-dichlorobenzene have a higher m.p. than its o-and m-isomers ?

(b) Why is (±) – Butan-2-ol is optically inactive ?Ans. (a) It is due to symmetry of p-dichlorobenzene which fits in crystal lattice better than o-and m-

isomers.(b) (±) Butan-2-ol exist in two enantiomeric forms which rotate the plane polarised light in

opposite direction in equal amounts. Therefore, canceling each other and becoming opticallyinactive.

Q. 2. Ethyl iodide undergoes SN2 reaction faster than ethyl bromide. Why ?

Ans. Since I– ion is a better leaving group than Br- ion, hence CH3I reacts faster than CH3Br in SN2

reaction with OH– ion.

Q. 3. (±) 2-Butanol is optically inactive. Why ?Ans. (±) 2-Butanol is a racemic mixture i.e. there are two enantiomers in equal proportion. The rotation

by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixtureoptically inactive.

Q. 4. C—X bond length in halobenzene is smaller than C–X bond length in CH3—X. Why ?Ans. In CH3—X the carbon atom is sp2 hybridised while in halobenzene the carbon atom is sp3

hybridised. The sp 2 hybridised carbon is more electronegative due to greater s-character and holdthe electron pair of C—X bond tightly than sp3 hybridized carbon with less s-character. Thus C—Xbond length in C—X is larger than C—X in halobenzene.

Q. 5. Explain why dipole moment of chlorobenzene is lower than that of cyclo-hexyl-chloride ?Ans. Because of the greater s-character an sp3-hybrid carbon is more electronegative than an sp2-hybrid

carbon. Thus, the sp2–hybrid carbon of C—Cl bond in chloro benzene has less tendency to releaseelectron to Cl than an sp3–hybrid carbon of cyclo-hexyl-chloride.

Q. 6. Alkyl halide, though polar, are immiscible with water. Why ?Ans. Alkyl halides though polar, are immiscible in water because they are unable to form hydrogen

bonds with water molecules.Q. 7. Grignard reagent should be prepared under anhydrous conditions. Why ?Ans. Grignard’s reagent are very reactive. They react with moisture present in the apparatus or the

starting material to give hydrocarbons. R — Mg — X + H — OH ⎯⎯⎯→ R — H + Mg (OH) XGrignard reagentHence Grignard reagent must be prepared under anhydrous conditions.


Q. 8. Chloro ethane is insoluble in water. Explain.Ans. Chloro ethane is unable to form hydrogen bond with water. Hence it is insoluble in water.Q. 9. Thionyl chloride is preferred for preparing alkyl chloride from alcohols.Ans. The byproducts of the reaction i.e. SO2 and HCl being gases escape into the atmosphere leaving

behind alkyl chloride in almost pure state.Q. 10. Why is thionyl chloride considered as the best reagent to convert alcohol into alkyl chlorides ?

Ans. Thionyl chloride is considered the best reagent because both the byproducts (SO2 and HCl) aregaseous which escape from the reaction mixture. This makes the purification of the chlorocompounds easy.

Q. 11. Why can benzene not be iodinated with I2 directly ?Ans. Iodination with I2 is slow (F2 > Cl2 > Br2 > I2) and somewhat reversible.

Q. 12. Why para-dichlorobenzene has a higher melting point than ortho dichlorobenzene ?Ans. Para-dichlorobenzene is more symmetrical and fits better in the crystal lattice of the solid form

than the ortho dichlorobenzene.Cl

ClMelting point = 323 K

Cl

ClMelting point = 256 K

Q. 13. Chloroform is stored in dark coloured bottles.

Ans. Chloroform in the presence of air gives a poisonous phosgene gas. The reaction is catalyzed bylight. To slow down this reaction, CHCl3 is stored in dark coloured bottles.

CHCl3 + 12 O2 ⎯→ COCl2 + HCl Chloroform Phosgene

gas

Q. 14. A small amount of ethanol is added to chloroform bottles. WhyAns. A small amount of ethanol is added to convert poisonous COCl2 into a non-poisonous diethyl

carbonate

COCl2 + 2C2H5OH ⎯→ (C2H5)2CO3 + 2HCl

diethyl carbonate

Q. 15. Iodoform is obtained by the reaction of acetone with hypoiodite ion but not iodite ion. Explain.

Ans. Hypoiodite ion acts as an oxidizing agent but iodite ion does not.

Q. 16. Chloroform is a chlorine compound but it does not give white precipitate with silver nitratesolution. Explain.

Ans. Chloroform (CHCl3) ionizes to give H+ and CCl –3 and not Cl–. Therefore, no precipitate with

AgNO3 solution is obtained.

Q. 17. Explain with reason the relative order of reactivity of 1°/2°/3° alkyl halides by SN1 mechanism.

Ans. In alkaline hydrolysis of an alkyl halide by SN1 mechanism, the formation of carbocation as an

intermediate product is involved.

The increasing order of stability of carbocation is,

(1°) primary < (2°) secondary < (3°) tertiary

(less stable)

CH3—+CH2 < CH3—

+CH—CH3 <

(most stable)

CH3 —+

CH3|C|CH3

The stability order for carbocation is 3° > 2° > 1°.Therefore the increasing order of reactivity by SN

1 mechanism of alkyl halides is(1°) primary < (2°) secondary < (3°) tertiary.


Q. 18. Give reason for the following :(i) Haloalkanes easily dissolves in organic solvents.(ii) Halogen compounds used in industry as solvent are alkyl chlorides rather than bromidesand iodides.

Ans. (i) Haloalkanes dissolve in organic solvent because the new intermolecular attraction betweenhaloalkanes and organic solvent molecule have much the same strength as ones being broken into separate the haloalkanes and solvent molecules.(ii) Because alkyl chlorides are more stable and more volatile than bromides and iodides.

Q. 19. Which one of the following compound is more easily hydrolysed by KOH and why ?CH3CHClCH2CH3 or CH3CH2CH2Cl

Ans. Due to + I effect of alkyl group the 2° carbonium ion CH3+CHCH2CH3 derived from secondary

butyl chloride is more stable than the 1° carbonium ion CH3CH2CH2+

derived from n-propylchloride.Therefore secondary butyl chloride gets hydrolysed more easily than n-propyl chlorideunder SN

1 reaction.Q. 20. Give reason :

(i) tert-Butyl chloride reacts with aqueous sodium hydroxide by SN1 mechanism while n-butyl

chloride reacts with SN2 mechanism.

(ii) Vinyl chloride is unreactive in nucleophilic substitution reaction.(iii) 3-Bromocyclohexane is more reactive than 4-bromo cyclohexene in hydrolysis with aqueousNaOH.

Ans. (i) tert-Butyl chloride reacts with aqueous sodium hydroxide by SN1 mechanism because the

heterolytic cleavage of C—Cl bond gives 3° carbocation which is highly stable while n-butylchloride reacts by SN

2 mechanism of C of C—Cl bond is less crowded and favourable fornucleophile to attack from back side resulting in the formation of transition state.(ii) Vinyl chloride is unreactive in nucleophilic substitution reaction because of the resonanceeffect resonance in Vinyl chloride gives rise to partial double bond character to carbon halogenbond making it stronger and therefore more difficult to cleave it than the C—Cl single bond

CH2 = CH—Cl ←⎯⎯ CH2—CH = Cl˙˙ :˙˙

⊕

(iii) 3-bromocyclo hexane

Br

is an allylic halide and high reactivity of this is due to

resonance stabilisation of carbocation.

⊕

⊕

⊕

⎯→

Q. 21. Give reason for the following :(i) Ethyl iodide undergoes SN

2 reaction faster than ethyl bromide.(ii) (±) 2– Butanol is optically inactive.

Ans. (i) Since I– ion is better leaving group than Br– ion hence CH3I reacts faster then CH3Br in SN2

reaction with OH– ion.

(ii) (±) 2-Butanol is a racemic mixture i.e. there are two enantiomers in equal proportion . Therotation by one enantiomer will be cancelled by the rotation due to the other isomer, making themixture optically inactive.(iii) In CH3—X the carbon atom is sp2 hybridised while in halobenzene the carbon atom is sp3

hybridised. The sp2 hydridised carbon is more electronegative due to greater s-character andholds the electron pair of C–X bond more tightly than sp3 hybridised carbon atom with less s-character thus C—X bond length in CH3—X is bigger than C—X in halobenzene.

Q. 22. (a) Why Alkyl halide, through polar, are immiscible with water.(b) Grignard reagents should be prepared under anhydrous condition ?


Ans. (a) Alkyl halide, through polar, are immiscible in water because they are unable to formhydrogen bond with water molecules.(b) Grignard reagent are very reactive. They react with moisture present in apparatus or thestarting material to give hydrocarbons

δ–

δ+

δ–

R—Mg —X + H—OH ⎯→ R —H + Mg (OH) X

Grignard reagentHence Grignard reagent must be prepared under anhydrous conditions.

Chapter 11. Alcohols, Phenols and EthersQ. 1. Glycerol (propane 1, 2, 3 triol) is more viscous than ethylene glycol (ethane 1, 2, diol). Explain.Ans. Viscosity depends on the amount of hydrogen bonding between molecules and presence of more

– OH groups, glycerol involves intramolecular hydrogen bonding thus glycerol is more viscousthan ethylene glycol.

Q. 2. Lower alcohols are soluble in water, higher alcohols are not.Ans. Lower alcohols can form H-bond with water whereas higher alcohols cannot due to larger

hydrocarbon part.Q. 3. Account for the following :

(i) Rectified spirit cannot be converted into absolute alcohol by simple distillation.(ii) Phenols do not undergo substitution of –OH groups like alcohols.

Ans. (i) Rectified spirit containing 95% ethyl alcohol and 5% water forms an azeotropic mixturewhich distills at a constant temperature of 351·13 K.(ii) The C–O bond in phenols has some double bond character due to resonance and hencecannot be easily cleaved by nucleophile. In contrast, the C–O bond in alcohol is a pure singlebond and hence can be easily cleaved by nucleophile.

Q. 4. Explain why does propanol has higher boiling point than that of the hydrocarbon, butane ?Ans. Propanol has higher boiling point than butane because it has stronger interparticle forces. In

propanol intermolecular H-bonding is present whereas in butane intermolecular forces are weakvan der Waal’s forces. A lot of heat is required to break intermolecular H-bonding amongpropanol molecules.

Q. 5. Alcohols are comparatively more soluble in water than the hydrocarbons of comparable mole-cular masses. Explain this fact.

Ans. Because of the presence of O—H group in them, alcohols are capable of forming H-bonds withwater molecules whereas hydrocarbons cannot form H-bonds with water. As a result, alcohols aremore soluble in water than the hydrocarbons of comparable molecular masses.

Q. 6. Explain why phenol has smaller dipole moment than methanol.Ans. In case of phenol, the electron withdrawing inductive effect of oxygen is opposed by electron

releasing resonance effect. Hence, phenol has smaller dipole moment. In case of methanol onlyelectron withdrawing inductive effect is operative. Hence, it has higher dipole moment.

Q. 7. Alcohols react with halogen acids to form haloalkanes but phenol does not forms halobenzenes.Explain.

Ans. The C—O bond in phenols have some double bond character due to resonance and hence isstronger and cannot be easily cleaved to form halobenzenes. In contrast, the C—O bond inalcohols is a pure single bond and hence can be easily cleaved by X– ions in presence of halogenacids to form haloalkanes.

Q. 8. Unlike phenols, alcohols are easily protonated. Explain.Ans. In phenols, the lone pair of electrons on the oxygen atom is delocalised over the benzene ring due

to resonance and hence is not easily available for protonation. In contrast to alcohols where, thelone pair of electrons on the oxygen atom are localized due to absence of resonance and hence areeasily available for protonation.

Q. 9. Explain why phenols do not undergo substitution of –OH group like alcohols.Ans. C–OH bond in phenols have partial double bond character due to resonance. As a result, this

bond is stronger and hence difficult to cleave. Therefore, phenols do not undergo substitution of–OH group like alcohols. In alcohols C–OH bond is pure single bond and hence can be cleavedrelatively easily.


Q. 10. Explain why phenols do not give protonation reaction readily.Ans. In phenols, the lone pair of oxygen is being shared with benzene ring through resonance (as is

clear from the contributing structures of phenol). Hence, the electron density around oxygen isrelatively less and therefore, phenols do not undergo protonation readily.

Q. 11. Explain why o-nitrophenol is less soluble in water than p-nitrophenol ?

Ans. In o-nitrophenol there is intramolecular hydrogen bonding. This inhibits its hydrogen bondingwith water and reduces its solubility in water.

Q. 12. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomerwhich is steam volatile. Give reason.

Ans. ortho-Nitrophenol is steam volatile because in it there is intramolecular H-bonding.

N

O|

H

O

O

|

← intramolecularhydrogenbonding

Due to intramolecular H-bonding, the intermolecular forces in ortho-nitrophenol are weaker thanin para-nitrophenol (which has intermolecular H-bonding) and hence, it undergoes lessassociation.

Q. 13. How will you distinguish between methyl alcohol and ethyl alcohol ?Ans. Iodoform test : Ethyl alcohol gives a yellow crystalline iodoform when heated with iodine and

sodium hydroxide solution while methyl alcohol does notCH3CH2OH + 4I2 + 6NaOH → CHI3 + HCOONa + 5NaI + 5H2O

Iodoform

Q. 14. Out of ethanol and water which is stronger acid and why ?Ans. Water is stronger acid than ethanol.

(i) C2H5OH + H2O C2H5—O-+ H3O+

Ethoxide ion

(ii) H2O + H2O H—O- + H3O+

Hydroxide ion

Since the OH group in alcohols is less polar than the OH group in water. Thus water is strongeracid than ethanol.

Q. 15. What is the order of reactivity of various types of alcohols in the reactions involving cleavage ofcarbon-oxygen bond ? Explain.

Ans. The order of reactivity of various alcohols towards this type of reaction is :tertiary > secondary > primary

This can be explained in terms of electron releasing inductive effect of alkyl groups. The alkylgroups by their electron releasing effect tends to increase the electron displacement towardsoxygen.

R →⎯

R↓C↑R

→⎯OH > R →⎯

H↓C|R

→⎯OH > R →⎯

H|C|H

→⎯OH

Tertiary alcohol Secondary alcohol Primary alcohol

In other words, the polarity of C—O bond increases and this makes the breaking of the bondbetween carbon and oxygen easier. Therefore, the alcohols with greater number of alkyl groupsattached to the carbon carrying –OH will be more reactive.

Q. 16. How do you explain that phenols are stronger acids than alcohols ?Ans. This can be explained by considering relative stability's of phenol and phenoxide ion.

Phenol as well as phenoxide ion both are resonance stabilized. The various contributingstructures of phenol and phenoxide ion are given below :


: O_H

:

δ-

δ-

δ-

δ+

+ O_H

:

: :

:

+ O_H

:

+ O_H

:

: O_H

:

I II III IV V

O_H

Resonating structures

: O

:

δ- δ-

δ-

δ- O : O :

:

: :

::

O :

:

: O :

:

I II III IV V

O


If we observe these structures, then we find that phenoxide ion is more stabilized by resonancethan phenol. In phenol, three contributing structures (II, III and IV) involves charge separationwhereas in case of phenoxide ion there is no charge separation. Since phenoxide ions are morestabilized by resonance than phenol, therefore, the equilibrium in the dissociation of phenols isvery much in favour of dissociated form and hence phenols furnish a high concentration of H+

ions and behaves as fairly strong acids. On the other hand, in the case of alcohols neither alcoholsnor alkoxide ion is stabilized by resonance and hence they behave as weaker acids than phenols.

R—· ·O· ·

—H R—· ·O· ·

: + H+

Q. 17. Acid catalysed dehydration of tert-butyl alcohol is faster than that of n-butyl alcohol. Givereason.

Ans. Acid-catalysed dehydration of alcohols occurs through formation of carbocation asintermediates.

CH3—

CH3|C—CH3|OH

t-Butanol

H+⎯⎯⎯→

– H2OCH3—

CH3|C+

|CH3

t-Butyl Carbocation(more stable)

CH3CH2CH2CH2—OHn-Butanol

H+⎯⎯⎯→

– H2OCH3CH2CH2C

+H2

n-Butyl carbocation(less stable)

Since tert-butyl carbocation is more stable than n-butyl carbocation, therefore, acid-catalyseddehydration of tert-butyl alcohol occurs faster than that of n-butyl alcohol.

Q. 18. Explain how does the —OH group attached to a carbon of benzene ring activates it towardselectrophilic substitution ?

Ans. The —OH group attached to the benzene ring in phenol activates it towards electrophilicsubstitution. It also directs the incoming group to ortho and para position in the ring as thesepositions become electron rich due to mesomeric effect caused by —OH group as shown below :

Phenol

: O_H

:

δ- δ-

δ-

δ++ O_H

:

:

:

+ O_H

:

+ O_H

:

: O_H

:

I II III IV V

O_H

:



Phenoxide

: O :

:

δ- δ-

δ-

δ- O : O :

:: :

::

O :

:

: O :

:

I II III IV V

O

Resonating structuresQ. 19. Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol ?

Ans. ortho-Nitrophenol is more acidic than ortho-methoxyphenol because nitro group by its electronwithdrawing resonance effect stabilizes the phenoxide ion whereas methoxy group by itselectron releasing effect destabilizes the phenoxide ion. Greater the stability of the phenoxide ion,greater is the dissociation of phenol and greater is its acidic strength.

Q. 20. Why are ethers insoluble in water ?Ans. Ethers are insoluble in water because due to bigger size of alkyl group, the oxygen atom in ethers

fails to form intermolecular H—bonds in water.Q. 21. Diethyl ether does not reacts with sodium. Explain.

Ans. Since diethyl ether does not contain active hydrogen attached to oxygen like alcohols andphenols, it does not reacts with sodium.

Chapter 12. Aldehydes, Ketones and Carboxylic AcidsQ. 1. Formaldehyde undergoes Cannizzaro’s reaction since it has one alpha hydrogen atom.Ans. Formaldehyde undergoes Cannizzaro’s reaction since it has no alpha hydrogen atom.Q. 2. Acetone gives a white precipitate on treatment with sodium chloride.Ans. Acetone gives a white precipitate on treatment with sodium iodide.Q. 3. Acetaldehyde undergoes Cannizzaro’s reaction on treatment with dilute alkali.Ans. Acetaldehyde undergoes Aldol condensation on treatment with dilute alkali.Q. 4. Ketones are less reactive than aldehydes. Why ?Ans. Due to the steric hinderance to the attacking groups, ketones are less reactive than aldehydes.Q. 5. How will you distinguish between formaldehyde and acetaldehyde ?Ans. Acetaldehyde forms yellow precipitate of iodoform with an alkaline solution of iodine (Iodoform

test). Formaldehyde does not gives this test.CH3—CHO + 4 NaOH + 3I2 → HCOONa + CHI3

YellowPpt.

+ 3H2O + 3NaI.

Q. 6. Dipole moments of aldehydes is ketones are different from those of alcohols, though both havepolar C—O bonds, why ?

Ans. The dipole moment of aldehydes and ketones is greater than that of alcohols. Due to the presenceof π bond between carbon and oxygen atom of carbonyl compounds (>C = O) the loosely held πelectron can be readily shifted to oxygen atom. Consequently the magnitude of + ve and – vecharge developed in > C = O bond is higher than C—O bond of alcohols.

Q. 7. Aldehydes are more reactive than ketones in nucleophilic reactions. Why ?Ans. The reactivity of aldehydes is more than that of ketones due to the following reasons :

(a) Inductive effect : The ease with which a nucleophile attacks the carbonyl group dependsupon the electron deficiency. i.e., the magnitude of the positive charge on carbonyl carbon. Sincean alkyl group has electron donating inductive effect (+ I effect), therefore greater the number ofalkyl groups attached to carbonyl carbon greater is the electron density on the carbonyl carbonand hence lower is its reactivity. Aldehydes are more polar than ketones.

Cδ +

= Oδ

H

R

→⎯ _

Aldehyde (more polar)

Cδ +

= Oδ

H

R

→⎯ _

Ketone (less polar)


In ketones we have two alkyl groups whereas in aldehydes only one alkyl group is attached tocarbonyl carbon. Since there is lesser number of alkyl groups in case of aldehydes so theyundergo nucleophilic addition more readily than ketones.(b) Steric effect : The presence of more alkyl groups in case of ketones hinders the attack ofnucleophile on the carbonyl group and so the ketones are less reactive than aldehydes.

Q. 8. During preparation of NH3 derivatives of aldehydes and ketones pH of medium is controlled.Why ?

Ans. Formation of ammonia derivatives (oximes, hydrazone, semi-carbazone etc.) proceeds via theattack of carbonyl carbon with proton to form the conjugate acid.

Cδ +

= Oδ –

+ H+ ⎯→ C +

—OH

Therefore, presence of an acid is must for preparing these derivatives (pH < 7).However, in strongly acid medium, the proton attacks the unshared pair of electrons on nitrogen toform the species RN+H3 which cannot attack the carbonyl carbon.

H++ : NH2R ⎯→ N

+H3R

In basic medium, there is no protonation of carbonyl group and hence no reaction.

C= O basic medium

⎯⎯⎯⎯⎯→ No protonation

Therefore, preparation of ammonia derivatives requires slightly acidic medium (pH = 3·5) and itscareful control is essential.

Q. 9. Arrange the following in the increasing order of acidity and explain your order :Formic acid, acetic acid, chloroacetic acid.

Ans. Increasing order :CH3COOH < HCOOH < ClCH2COOH

pKa value 4·74 3·75 2·86Carboxylic acids are quite strong acids due to the presence of polar OH group. Carboxylic acidsas well as carboxylate ion both are stabilised by resonance. The factors which increases thestability of carboxylate ion more than carboxylic acids, increases the acidic strength.The electron withdrawing groups (– Cl) stabilizes the carboxylate anion, on the other handelectron releasing group (– CH3) destabilizes the carboxylate ion. Further pKa values support theanswer.

Q. 10. For the acetylation reactions, acetic anhydride is preferred to acetyl chloride. Why ?Ans. Both acetyl chloride and acetic anhydride is used for acetylation reactions but acetic anhydride is

preferred to acetyl chloride because the reactions of acetic anhydride are less vigorous and hence,can be controlled easily.

Q. 11. The boiling point of acetamide (494 K) is higher than that of its corresponding acid CH3COOH(391 K). Why ?

Ans. Acetamide has strong intermolecular hydrogen bonding as :

…δ+H —N

|H

—

CH3|C ⎯⎯⎯⎯

δO ……

δ+H —N

|H

—

CH3|C ⎯⎯⎯⎯

δ–O ……

δ+H —N

|H

—

CH3|C ⎯⎯⎯⎯

δ+O …

–

due to this reason acetamide has higher boiling point than acetic acid.Q. 12. Why boiling points of esters are less than that of the carboxylic acids ?

Ans. Because esters do not form hydrogen bonds.Q. 13. Why acid amides are much less basic than amines ?

Ans. In acid amides due to the resonance, the electron deficiency of carbonyl carbon, do not undergothe reactions, hence, acid amides are much less basic.

R—

O||C—

..NH2 ←→ R—

O–

|C =

+NH2

Q. 14. Why carboxylic acids behaves as fairly strong acids ?Ans. Carboxylic acid may be treated as the resonating hybrid of the following two structures.


+ H3O+

In structure (II) a +ve charge is developed on oxygen of the O—H group which weakens theO—H bond because the oxygen atom attracts the electron of this bond and ultimately a proton isreleased and carboxylate ion is formed which is further stabilized by resonance.

≡

Both carboxylic acid as well as carboxylate ion gets stabilized by resonance but the resonance incarboxylate ion is more effective because both the contributing structures (III) and (IV) areidentical, while in carboxylic acid the contributing structures (I) and (II) involves chargeseparation which makes it less stable than carboxylate ion. Because of the more stability ofcarboxylate ion, the carboxylic acid tries to exist in the form of ions i.e., carboxylate ion and aproton. Thus the carboxylic acid behaves as a fairly strong acid.. Hence, greater the stability ofcarboxylate ion, higher will be the ionization and more will be the acidic character.

Q. 15. Explain why the melting point of an aliphatic carboxylic acid containing an even number ofcarbon atom is higher than the next lower and higher member containing odd number of carbonatoms.

Ans. The aliphatic carboxylic acids containing even number of carbon atoms has higher melting pointthan the next higher and just lower member of acid containing odd number of carbon atoms. Thisis because of the alternation effect. This effect is observed in the homologous with less than tencarbon atoms.In the carboxylic acids with even number of carbon atoms, the terminal carboxylic group andmethyl group are present on the opposite sides of the zig-zag carbon chain, due to which themolecules fit better in the crystal lattice, therefore the intermolecular forces becomes strongerwhich results in higher melting point. While in the carboxylic acids with odd number of carbonatoms, the terminal carboxylic group and methyl group are present on the same side of zig-zagcarbon chain. Therefore, they fit poorly in the crystal lattice and shows weaker intermolecularforces which results relatively in lower melting point.

CH3 CH2 COOH CH2 CH2 COOHCH2 CH2 CH3 CH2 CH2

Even number of carbon atoms Odd number of carbon atoms(Terminal groups are on same side) (Terminal groups are on opposite side)

Having lower melting points because Having higher melting points because theythey fit poorly in crystal lattice fit better in crystal lattice

Q. 16. State reason for the following :(i) Monochloro ethanoic acid has a higher pKa value than dichloro ethanoic acid.(ii) Ethanoic acid is a weaker acid than benzoic acid.

Ans. (i) This is because dichloroethanoic acid is a stronger acid than monochloro ethanoic acid.(ii) This is because methyl group due to its positive inductive effect destablises the acetate anionby intensifying the negative charge.

Q. 17. Give plausible explanation of each of the following :(i) Cyclohexanone forms cyanohydrin in good yields but 2,2,6- tri methylcyclo hexanone doesnot.(ii) Hydrazone of aldehydes and ketones are not prepared in highly acidic medium.(iii) During the preparation of ester from a carboxylic acid an alcohol in the presence of an acidcatalyst, the water or the ester formed should be removed as soon as it is formed.


Ans. (i)

HCNHO CNO

Cyclohexanone CyclohexanoneCyanohydrin

⎯⎯→

HCN ⎯⎯→

OCH3 CH3

CH3

2, 2, 6-Trimethylcyclohexanone

HOCH3 CH3

CH3

CN

Due to the presence of three methyl groups at α -position with respect to the C = O, thenucleophilic attack by CN

– ion does not occurs due to steric hindrance. As there is no such steric

hindrance in cyclohexanone hence, nucleophilic attack by the CN– ion occurs readily and hence

cyclohexanone cyanohydrin is obtained in good yield.(ii) In highly acidic medium, the –NH2 group of hydrazine gets protonated.

NH2 —NH2 + H+ ⎯⎯→ NH2—NH3

+

Hydrazine Protonated hydrazineDue to electron withdrawing effect of the NH3

+ group, the lone pair of electrons on the —NH2

group of protonated hydrazine is not available for nucleophilic attack on the C = O and hencehydrazine formation does not occur.(iii) The formation of ester from carboxylic acid and an alcohol in presence of an acid catalyst isa reversible reaction.

RCOOH + R′OH H2SO4

RCOOR′ + H2OCarboxylic acid Alcohol Ester

To shift the equilibrium in the forward direction the water or the ester formed should beremoved as fast as it is formed.

Q. 18. Account for the following :(i) Cl-CH2COOH is a stronger acid than CH3COOH.(ii) Carboxylic acid do not give reaction of carbonyl group.

Ans. (i) Because of –I effect of Cl atom in ClCH2COOH and + I effect of CH3 group in CH3COOH.The electron density in the O—H bond in ClCH2COOH is much lower than CH3COOH. As aresult O–H bond in ClCH2COOH is much weaker than CH3COOH therefore ClCH2COOH losesa proton more easily than CH3COOH. Hence ClCH2COOH is a stronger acid than CH3COOH.(ii) Carboxylic acids are resonance hybrid of the following structure :

R —�CO—H

::

O: :

(I) (II)

←→ R —�CO—H:

O::

:

+

Similarly a carboxyl group of aldehyde and ketones may be regarded as resonance hybrid of thefollowing structures.

(III) (IV)

: :

:

C = O : :C = O+

⎯→

Because of contribution of structure (IV), the carboxyl carbon in aldehydes and ketone iselectrophilic on the other hand electrophillic character of carboxyl carbon is reduced due tocontribution of structure (II).As carboxyl carbon of carboxyl group is less electropositive than carboxyl carbon in aldehydeand ketones therefore carboxylic acid do not give nucleophilic addition reaction of aldehydes andketones.

Q. 19. How would you account for the following :(i) Aldehydes are more reactive than ketones towards nucleophiles.(ii) The boiling points of aldehydes and ketones are lower than that of the corresponding acids.(iii) The aldehydes and ketones undergo a number of addition reactions.


Ans. (i) This is due to steric and electronic reasons sterically. The presence of two similar relativelylarge substituents in Ketones hinders the approach of nucleophile to carboxyl carbon than inaldehyde having only one such substituent. Electronically two alkyl groups reduces the positivityof the carboxyl carbon more effectively in ketone than in aldehydes.(ii) This is due to intermolecular hydrogen bonding in carboxylic acid.(iii) Due to greater electronegativity of oxygen than carbon, the C atom of the C = O groupacquires a partial positive charge in aldehydes and ketones hence readily undergoes nucleophilicaddition reaction.

Chapter 13. Organic Compounds Containing NitrogenQ. 1. Arrange the following compounds in the ascending order of their basic strength and give reasons

for your answer :Methylamine, Aniline, Ethylamine, Diethyl ether

Ans. CH3—··NH2 C2H5··NH2 C2H5—

··O—C2H5

I II III IV+ I effect increases + M effect Bulky C2H5 groupavailability of lp on Nitrogen decreases lp decreases

availability on Nitrogen availability of lp on NBasicity order :Increasing order of basic strengthDiethylether < Aniline < Methylamine < ethylamineReason : Due to + I effect of alkyl groups steric effects of alkyl groups. Aromatic amines areweaker bases than aliphatic amines.

Q. 2. Direct nitration of aniline is not possible. Why ?Ans. Direct nitration of aniline is not possible because the —NH2 group gets oxidized with the

reaction mixture.Q. 3. Why do amines behave as nucleophiles ?Ans. Due to presence of a lone pair of electron on nitrogen atom, amines behaves as nucleophiles.Q. 4. Why is benzene diazonium chloride not stored and used immediately after its preparation ?Ans. Benzene diazonium chloride is very unstable.Q. 5. Why are amines less acidic than alcohols of comparable molecular masses ?Ans. Loss of a proton from an amine gives amide ion while loss of a proton from alcohol gives an

alkoxide ion as shown below :

R — NH2 ⎯⎯→ R — NH– + H+

Amine Amide ion

R — O — H ⎯⎯→ R — O–

+ H+

Alcohol Alkoxide ion

As O is more electronegative than N, RO– can accommodate the negative charge more easily thanthe RNH– ion. Thus RO– is more stable than RNH–. Therefore, amines are less acidic thanalcohols.

Q. 6. Why do primary amines have higher boiling point than tertiary amines ?Ans. In primary amines, two hydrogen atoms of N are present and they undergo extensive

intermolecular hydrogen bonding which results in association of molecules while in tertiaryamines, no hydrogen atom of N is present. Hence there is no H—bonding in tertiary amines. As aresult primary amines have higher boiling point than tertiary amines

Q. 7. Why are aliphatic amines stronger bases than aromatic amines ?Ans. Aliphatic amines are stronger bases than aromatic amines because :

(a) Due to resonance in aromatic amines, the lone pair of electron on the nitrogen atom getsdelocalised over the benzene ring and this is less easily available for protonation.(b) The aryl, amine ions have lower stability than the corresponding alkyl amine i.e. protonationof aromatic amines is not favoured.


Q. 8. Why cyanides are fairly soluble in organic solvents ?Ans. Because with the increase in molecular mass, the bulk of the non-polar portion increases and

hence solubility in water decreases.Q. 9. Why nitroalkanes are called Pseudo acids ?Ans. Nitroalkanes are neutral to litmus, but in strong alkaline medium they behave as acids and forms

salts. Hence they are called pseudo acids. Nitro form is pseudo acid form.

R—CH2—N+

O–

O

R—CH = NO –

OH

Nitroform Acidform

Acid formed is weak acid, and can form salt with strong bases.Q. 10. Why the boiling points of alkyl nitrites are lower than the corresponding nitroalkanes ?

Ans. Because the alkyl nitrites are less polar than that of corresponding nitroalkanes, hence theintermolecular forces are comparatively weak. Due to this reason boiling points of alkyl nitratesare lower than that of corresponding nitroalkanes.

Q. 11. Why the aqueous solution of ethyl amine turns red litmus blue ?Ans. Aqueous solution of ethyl amine is basic in nature because the amino group has a tendency to

accept the proton from water molecule to leave behind the free OH– ions. Due to the presence offree OH– ions, the solution turns red litmus blue.

C2H5

. .NH2 + H—OH ⎯→ [C2H5NH3]+ + OH–

Q. 12. Why the boiling points of alkyl isocyanides are lower than that of corresponding alkyl cyanides ?Ans. Isocyanides are less polar than the cyanides. Due to this reason the intermolecular forces in

isocyanides are weaker than cyanides, thus the boiling points of isocyanides are lower thancorresponding cyanides.

Q. 13. Why isocyanide does not hydrolysis in alkaline medium ?Ans. Because the basic hydrolysis is initiated by the attack of OH– ion on the substrate molecule, But in

alkyl isocyanide, the carbon atom of isocyanide group being negatively charged resists the attackof OH– ions. Hence it readily undergoes the attack of H+ ion in acidic medium.

Q. 14. KCN generally gives cyanides as major product on treatment with alkyl halides, on the otherhand, AgCN gives isocyanides as major product. Why ?

Ans. The formation of cyanides or isocyanides from alkyl halides involves nucleophilic substitutionreaction. The cyanide ion is resonance hybrid of the following structures.

: –C ≡ N : ↔ : C =

. .N – :

⎣⎢⎡

⎦⎥⎤

≡Hybrid structureCδ- ··= Nδ-

From the hybrid structure, it is evident that cyanide is capable of attacking the molecule throughcarbon atom as well as through N atom. Hence CN– is an ambient nucleophile.KCN is predominantly ionic. Therefore both carbon and nitrogen atoms are free to donateelectron pair. Since carbon-carbon bond is relatively stronger than carbon-nitrogen bond, henceattack occurs through the carbon atom of the cyanide group and alkyl cyanides are the majorproduct.While AgCN is predominantly covalent. Thus only N-atom is free to act as donor of electron pairand the attack occurs through the nitrogen atom of cyanide group, giving alkyl isocyanide as themajor product.

Alc. KCN⎯⎯→ R—C ≡ N

(Major) + R — N C

R — XAlkyl halide

Alc. AgCN⎯⎯⎯→ R—N ≡ C

(Major) + R — C N

→

→

Q. 15. Arylamines are weaker bases than alkyl amines ?Ans. Arylamines are weaker bases than alkyl amines because the lone pair of electrons on nitrogen is

withdrawn away from it and partially shared with benzene ring. Therefore, in aniline electrondonating capacity of nitrogen for protonation decreases considerably.


Q. 16. Electron releasing groups like —CH3, —OCH3 —NH2 etc. increases the basic strength whileelectron withdrawing groups like NO2, —CN etc decreases the basic strength of aromatic amines.Why ?

Ans.. Electron Repelling Groups (ERG) increases the basic strength because they tend to stabilise thecation formed by protonation of amines.

On the other hand, the Electron Withdrawing Groups (EWG) decreases the basic strengthbecause they tend to destabilize the protonated amines.

NH2

(ERG)G stabilizes the cation Basic strength decreases

(ERG)G G⎯

→

NH3+

⎯→

H+

NH2

(EWG) (EWG)G′ G′

NH3+

⎯→

⎯→

H+

G′ destabilizes the cation Basic strength increases

Q. 17. Alkyl isocyanides do not undergo basic hydrolysis. Why ?

Ans. The carbon atom of isocyanide (N → C) group in isocyanides bears a negative charge and repels

OH– (from alkalies) which is a nucleophile. But they undergo hydrolysis only in acidic medium.

Q. 18. Alkyl cyanides are water soluble whereas alkyl isocyanides are insoluble in water. Why ?

Ans. The lower alkyl cyanides are soluble in water due to the formation of hydrogen bonding by lonepair of electrons on nitrogen atom

R—C ≡ N ……… H—O

|

H………N ≡ C—R

Because alkyl groups are hydrophobic in nature, the solubility decreases with increase in the sizeof alkyl group.

Q. 19. Why do amines behaves as nucleophiles ?

Ans. Due to presence of lone pair of electrons on nitrogen atom, amines behaves as nucleophiles.

Q. 20. Account for the following :

(i) Ethylamine is soluble in water, whereas aniline is not.

(ii) Although amine group is O-and p-directing in aromatic electrophillic substitution reactions,aniline on nitration gives a substantial amount of m-nitroaniline.

(iii) Aniline does not undergoes Friedel-Craft’s reaction.

Ans. (i) Ethylamine dissolves in water because it forms H—bonds with water molecules as

— H —�N — H H —�N — H — O O

H

H δ– δ+ δ– δ+ δ+

C2H5 H C2H5

H δ–

In aniline due to large hydrocarbon part the extent of H—bond decreases considerably and henceaniline is insoluble is water.

(ii) Nitration is usually carried out with a mixture of conc. HNO3 and conc. H2SO4. In presenceof these acids, most of aniline gets protonated to form anilinium ion. Thus is presence of acid, thereaction mixture consist of aniline and anilinium ion. The NH2 group in aniline is o,p-directing

and activating while the —+NH3 group in anilinium ion is m-directing and deactivating.

Nitration of aniline mainly gives p-nitro aniline. On the other hand, the nitration of anilinium iongives m-nitro aniline.


NH2

– H+ + NO2

+

Aniline

NH2

Aniline

NH2 NH2

Nitroanilin e(2%)

NH2

+

+

NH2

m-Nitroaniline (47%)

p-Nitroaniline (51%)

–H– +H+

⎯⎯⎯→

– H+ + NO2

+ ⎯⎯⎯→ – H+

NH2OH ⎯⎯⎯→

NH2 NH2

+ NH2

NH2

Thus nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of theamino group.(iii) Aniline being a Lewis base, reacts with Lewis acid AlCl3 to form a salt.

C6H5NH2 + AlCl3 ⎯⎯→ C6H5+NH2 AlCl3

–

Lewis base Lewis acidAs a result,N of aniline aquires a positive charge and hence it acts as a strong deactivating groupfor electrophilic substitution reaction. Consequently aniline does not undergoes Friedel-Craft’sreaction.

Q. 21. Accounts for the following :(i) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.(ii) Diazonium salt of aromatic amines are more stable than those of aliphatic amines.(iii) Gabriel phthalimide synthesis is preffered for synthesising primary amines.

Ans. (i) Methylamine being more basic than water, accepts a protron from water, liberating OH–

ions.

CH3 NH2 + OH ⎯→ CH3 — +NH3 + OH

–

These OH– ions combine with Fe3+ ion present in H2O to form brown precipitate of hydrated

ferric oxide.FeCl3 ⎯→ Fe3+ + 3Cl

–

2Fe3+ + 6OH–

⎯→ 2Fe (OH)3 Or Fe2O3·3H2O.(Hydrated Ferric Oxide)(Brown ppt.)

(ii) The diazonium salts of aromatic amine are more stable than those of aliphatic amines due todispersal of the positive charge on the benzene ring as shown below :

+ ::N ≡N +:N = N : +

+

+

:N = N

←→ ←→

: +:N = N

←→

: +:N ≡N

←→

(iii) Gabriel phthalamide reaction gives pure primary amines without any contamination ofsecondary and tertiary amines. Therefore it is preferred for synthesising primary amines.

Q. 22. Account for the following :(i) Aniline gets coloured on standing in air for a long time.(ii) MeNH2 is stronger base than MeOH.

Ans. (i) Due to electron-donating effect (+ R effect) of —NH2 group the electron density on thebenzene ring increases. As a result aniline is easily oxidised on standing in air for long time toform coloured products.(iii) Nitrogen is less electronegative than oxygen therefore lone pair of electrons on nitrogen isreadily available for donation. Hence MeNH2 is more basic than MeOH.


Q. 1. Why is cellulose not digested in human body ?

Ans. It is due to the fact that human beings do not have enzyme to digest cellulose.


Q. 2. Why are carbohydrates generally optically active ?

Ans. Carbohydrates have chiral or asymmetric carbon atom.

Q. 3. Explain why vitamin C cannot be stored in the body ?

Ans. Vitamin C is soluble in water, hence it is readily excreted in urine and thus cannot be stored inthe body.

Q. 4. Why are vitamin A and vitamin C essential to us ? Give their important sources.

Ans. Vitamin A is essential for us because its deficiency can cause Xerophthalmia (hardening of corneaof eye) and night blindness.Sources — Carrot, fish, liver oil, butter and milk.Vitamin C is essential for us because its deficiency causes scurvy (bleeding gums) and pyorrhea(Loosening and bleeding of teeth)Sources — Amla, citrus fruit and green leafy vegetable.

Q. 5. Fructose cannot be oxidized by the action of Br2 water. Why ?Ans. Fructose cannot be oxidized by mild oxidizing agent like Bromine (Br2) water because it contains

ketonic group. Therefore, fructose has no action with bromine water.Q. 6. On electrolysis in acidic solution amino acids migrate towards cathode while in alkaline solution

these migrate towards anode.Ans. In acidic medium amino acids exit as cations (R—

+

CH—|NH3

COOH)

In basic medium amino acids exist as anions (R—CH—|CH2

COO–)

Amino acids migrate towards cathode while in alkaline medium they migrate towards anode.Q. 7. The monoamino, monocarboxylic acids have two pKa values. Why ?

Ans. One pKa value corresponds to dissociation of |+NH3

R—CH—COOH and the second value

corresponds to dissociation of R—CH—|NH2

COOH.

Q. 8. The melting points and solubility in water of amino acids are generally higher than that of thecorresponding haloacids. Explain.

Ans. Amino acids acquire zwitter ion structure due to the presence of both an acidic and a basic groupin the same molecule. As a result they behave like crystalline solids and have high melting pointsand high solubility in water. On the other hand halo acids cannot form zwitter ion structure andhence behave like carboxylic acids.

Q. 9. Amino acid behaves like salt rather than simple amines or carboxylic acid. Explain.Ans. In aqueous solution the carboxylic group loses a proton and amino group accepts a proton to

form a zwitter ion.

R—CH—C—OH

NH2

R—CH—C—O–

|NH3

O||O

||

(Zwitter ion)

| +

Q. 10. Why α-helix named as 3·613 helix ?Ans. The α-helix is also known as 3·613 helix, since each turn of the helix has approximately 3·6 amino

acids residue and a 13-member ring is formed by hydrogen bonding.Q. 11. Give reasons for the following :

(i) On electrolysis in acidic solution amino acids migrate towards cathode, while in alkalinesolution they migrate towards anode.(ii) The mono amino mono carboxylic acids have two pKa values.

Ans. (i) In acidic solution, the carboxylate anion accepts a proton and gets converted into carboxylicacid resulting in the formation of a positive ion.


H3

+N—CH

|R

—COO– H+

acid⎯⎯→ H3N—CH

|R

—COOH

Zwitter ion (Positive ion)

In the presence of a base, the NH3+ ion changes to —NH2 group by losing a proton and this gives

a negative ion.

H3

+N—CH

|R

—COO– H+

base⎯⎯→ H2N—CH

|R

—COOH– + H2O

Zwitter ion (Negative ion)This means that in acidic medium, the amino acid migrates towards the cathode while in alkalinesolution it migrates towards anode on electrolysis.(ii) In aqueous solution monoamino, monocarboxylic acid behaves like salts at isoelectric point.At a pH lower than isoelectric point (i.e. in acidic medium) it shows one pKa value whichcorresponds to structure

|+NH3

R—CH—COOH and at a pH higher than isoelectric point (i.e. basic

medium) it shows another pKa value which corresponds to structure |NH2

R—CH—COO–

Q. 12. Depict the structure of DNA ?Ans. The two strand in DNA molecule are held together through hydrogen bonds between purine

base of one strand and pyrimidine base of another and vice-versa because of different sizes andgeometries of all the base that only possible paring in DNA are G (guanine) and C (cytosine)through three H-bonds (i.e., G ≡ C) and between A (Adenine) and T (Thymine) through two H-bonds (i.e. A = T). Due to this base pairing principle the sequence of base in one strand automatically fixes the sequence of bases in the other strand. Thus the two strands are not identicalbut are complimentary.

Fig. The double strand helix structure of DNAQ. 13. The Ka and Kb value of α-amino acid are very low. Why ?

Ans. In α-amino acid the acidic group is —

+NH3, instead of —COOH group as in carboxylic acids and

basic group is — COO– instead of —NH2 group as in amines,. That is why they have low value ofKa and Kb.

Q. 14. Give reasons for the following :(i) Glucose and fructose gives the same osazone.(ii) Amino acids are amphoteric in nature.

Ans. (i) During osazone formation, the reaction occurs only at C–1 and C–2. As glucose and fructosediffer from each other only in the arrangement of atom at C–1 and C–2, therefore they give thesame osazone.

(ii) Since amino acid have both acidic ( +NH3) as well as basic (COO–) groups. Therefore they are

amphoteric in nature.


Acidic Basic

H3N — CH — COOH H+

←⎯⎯ H3+N CH—COO

–H N—CH—COO

|R|

Rα-Amino acid(Zwitter ion)

+

2 ⎯⎯→ –

|R

Chapter 15. PolymersQ. 1. Why are rubber called as elastomers ?Ans. Rubbers are stretched on application of force and regain original state after the force is removed.

Therefore they are called as elastomers.Q. 2. Why should the monomer used in addition polymerization through free radical pathway be very

pure ?Ans. Pure monomers are required because even the traces of impurities may act like inhibitors which

leads to the formation of polymer with shorter chain length.Q. 3. To have practical application, why are cross links required in rubbers ?Ans. Cross links binds the planar polymer sheets thus increasing its elastomeric properties.Q. 4. Which factor imparts crystalline nature to a polymer like nylon ?Ans. Strong intermolecular forces like hydrogen bonding, leads to close packing of chain that imports

crystalline character.Q. 5. Which type of biomolecules have some structure similarity with synthetic polyamides ? What is

this similarity ?

Ans. Proteins have structural similarity with synthetic polyamide. Polyamide and proteins bothcontain amide linkage.

Q. 6. How does vulcanisation changes the properties of natural rubber ?

Ans. Rubber gets cross linked through —S—S—bond and becomes hard on vulcanisation.

Q. 7. Why are the numbers 6,6 and 6 put in the names of nylon 6, 6 and nylon 6 ?

Ans. In nylon 6,6 two sixes stand for hexamethylene diamine (a monomer with 6 C-atoms) and adipicacid (another monomer with 6 C—atoms)

In nylon-6 six stands for the onlyl monomer caprolactam (6C—atoms)

Q. 8. Why vulcanization is essential ?

Ans. During vulcanization natural rubber converts into vulcanized rubber or the isoprenes chains arecross linked by sulphur. Cross linking helps in preventing the stoppage of the chains onapplication of stress. It also regains the shape after the removal of the stress. Thus vulcanizedrubber is more elastic than natural rubber.

Q. 9. Explain why free radical polymerization of styrene gives a product in which phenyl group are onalternate carbon atom rather than on adjacent carbon atom.

Ans. During free radical polymerization, the addition of free radical to monomer molecule occurs inaccordance with Markovnikov’s rule so as to give more stable benzylic free radical eg.

⎯⎯⎯→

⎯⎯→R + CH2 = CH

R — CH2 — CH

(More stable benzylicradical actually formed)

•

R + CH — CH2•

•

(I)

(II)(Less stable radical

not formed)This process goes on till polystyrene (v) in which the phenyl group are on atterrnate carbon atomis obtained rather than the product (vi) in which the phenyl group are on adjacent carbon atom.


⎯⎯⎯⎯⎯→ ⎯⎯→

⎯⎯→⎯⎯⎯⎯⎯→

•

R + CH2 = CH

+RCH2 — CH — CH2 — CH

•RCH2 — CH — CH — CH2

•

CH = CH2

+

CH = CH2(I)(more stablefree radical)

(III)Benzylic radical

(more stable)

(IV)(Less stable not

formed)

(VI)(Product not

formed)

n

n(V)

Polystyrene(product actually

formed)

CH2 — CH — CH — CH2

CH2 — CH — CH2 — CH

Q. 10. To have practical application, why are cross links require in rubber ?Ans. Cross links bind the planar polymer sheet thus increasing its elastomeric properties.

Q. 11. Can nucleic acids, proteins and starch be considered as step growth polymers ?Ans. Yes, step growth polymers are condensation polymers and they are formed by the loss of simple

molecules like water leading to the formation of high molecular mass polymers.

Chapter 16.Chemistry in Everyday Life

Q. 1. Why is drug metabolism a major consideration of drug design ?Ans. Most drugs and their active metabolities are rendered inactive or less active due to metabolism. A

drug should be so designed that reaches the target without being metabolised.Q. 2. Why is it important to understand the mechanism of drug action and metabolic pathways in the

biological system ?Ans. In order to improve the drug activity and to minimise side effects one should understand the

mechanism of drug action.Q. 3. Why is glycerol added to shaving soap ?Ans. Glycerol is added to shaving soap to prevent rapid drying.Q. 4. Why are detergents preferred over soaps ?Ans. Unlike soaps, detergents can be used even in hard water because like sodium salt, calcium and

magnesium salt of sulphonic acid are soluble in water.Q. 5. Why do we need to classify drugs in different ways ?Ans. Various methods of classification of drug and the usefulness of such classification are as

follows :(i) Classification on the basis of pharamacological effect is useful for doctor because it providesthem the whole range of drugs available for the treatment of a particular type of problem.(ii) Classification on the basis of drug action on a particular biochemical process is useful forchoosing the correct lead compound for designing the synthesis of a desired drug.(iii) Classification on the basis of molecular targets is useful for medicinal chemists so that theycan design a drug which is most effective for a particular receptor site.(iv) Classification on the basis of chemical structure help us to design the synthesis of a numberof structurally similar compound having different substituents and then choosing the drughaving the least toxicity.

Q. 6. Why are certain drugs called enzyme inhibitors ?Ans. Enzyme have active sites that binds the substrate for effective and quick chemical reaction.The

functional group present at the active site of enzymes interacts with functional group of a


substrate via ionic bonding, hydrogen bonding, van der Waal’s interaction, etc. Some drugsinterfere with this interaction by blocking the activity of the enzyme. Therefore, these are calledinhibitors.

Q. 7. Why should medicine not taken without consulting the doctor ?Ans. Side effects are caused when a drug binds to more than one receptor site. So, a doctor must be

consulted to choose the right drug which has the maximum affinity for a particular receptor siteto have the desired effect. The dose of the drug is also crucial because some drugs like opiates inhigher doses acts as poison and may cause death.

Q. 8. Answer the following question :(i) Why is use of aspartame limited to cold food and drink ?(ii) Soaps do not work in hard water.(iii) Why is bithional added to soap ?

Ans. (i) Use of aspartame is limited to cold foods because it is unstable at cooking temperature.(ii) In hard water soap gets precipitated as calcium and magnesium salts which being insolublein water sticks to the clothes as gummy mass.(iii) Bithional acts as an antiseptic agent and reduces the odour produced by bacterialdecomposition of organic matter on the skin.

Q. 9. Pickles have a long shelf-life and do not get spoiled for months. Why ?Ans. Plenty of salt and cover of oil acts as preservative. These do not allow bacteria to act on them.

Q. 10. Between sodium hydrogen carbonate and magnesium hydroxide which is better antacid andwhy

Ans. Magnesium hydroxide is better antacid because being insoluble it does not allows the pH toincrease above neutral. Hydrogen carbonate being soluble in excess can make the stomachalkaline and trigger the production of even more acid.

Q. 11. Why are cimetidine and ranitidine better antacids than sodium bicarbonate or magnesiumaluminium hydroxide ?

Ans. NaHCO3 or Mg(OH)2 or Al(OH)3, if taken in excess makes the stomach alkaline and thustriggers the release of even more HCl which may cause ulcers in the stomach. On the other hand,cimetidine and ranitidine prevents the interaction of histamine with the receptor cells in thestomach wall and thus releases lesser amount of HCl, hence are better antacids.

Q. 12. Why do we require artificial sweetening agents ?Ans. Artificial sweetening agents are used to reduce calorie intake. These also protect teeth from

decaying and acts as a substitute of sugar for diabetics.Q. 13. Why are detergents with straight hydrocarbon groups better than the detergents with branched

chain hydrocarbon groups ?Ans. The detergents with straight chain hydrocarbon groups are more biodegradable than detergents

with branched chain hydrocarbons. Therefore, detergents with straight chain hydrocarbon areeasily decomposed or broken down by micro-organisms like bacteria present in water bodies.Therefore they cause less water pollution and hence are better, than the detergents with branchedchain hydrocarbon groups.

Q. 14. Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it isnot advisable to take its doses without consultation with doctor. Why ?

Ans. Most of the drugs taken in doses higher than recommended may cause harmful effects and actsas poisons leading to death. Therefore a doctor must always be consulted before taking anymedicines, who will advice the patient for proper and safe dose of the drug.

❐

Question

7Set Balance the Chemical Equations

Chapter 7. p-Block ElementsQ. 1. How will you obtain pure potassium permanganate (KMnO4) crystals from its ore, pyrolusite ?Ans. Potassium permanganate (KMnO4) is prepared from mineral pyrosulphite (MnO2)

(a) Conversion of Pyrosulphite ore (MnO2) to Potassium Manganate K2MnO4 :

MnO2 + 4KOH + O2 Heat⎯⎯→ 2K2MnO4 + 2H2O

(b) Oxidation of Potassium Manganate (K2MnO4) to potassium permanganate (KMnO4)2K2MnO4 + Cl2 → 2 KCl + 2KMnO4

Q.2. Give the equation when fluorine is passed through cold, dilute NaOH solution.Ans. 2F2 + 2NaOH ⎯→ OF2 + 2NaF + H2O

Fluorine Sodium Oxygen Sodium Waterhydroxide difluoride fluoride

(cold and dil.)Q. 3. Write balanced chemical equation for the reaction when chlorine is passed through hot concen-

trated NaOH solution.Ans. 6NaOH + 3Cl2 ⎯→ NaClO3 + 5NaCl + 3H2O

Hot conc. Sodiumchlorate

Q. 4. Give the reaction when fluorine is treated with dilute sodium hydroxide solution.Ans. 2F2 + 2NaOH ⎯→ 2NaF + H2O + OF2

Q. 5. Give reactions and the conditions required for preparation of the following compounds :(i) XeF6 (ii) XeOF4

Ans. (i) Xe + 3F2 1 : 20 ratio‚ Ni vessel

300°C‚ 50 atm.⎯⎯⎯⎯⎯⎯⎯⎯→ XeF6

(ii) XeF6 + H2O Partial hydrolysis

with water⎯⎯⎯⎯⎯⎯→ XeOF4 + 2HF

Q. 6. Give the reaction what happens when :(i) PCl5 is heated(ii) White phosphorus is heated with conc. NaOH solution in an inert atmosphere.

Ans. (i) When PCl5 is heated, the less stable axial bond breaks to form PCl3.

PCl5Heat

⎯⎯→ PCl3 + Cl2(ii) White phosphorus reacts with NaOH to form phosphine (PH3)

4PPhosphorus

+ 3NaOH + 3H2O Heat

CO2. atmosphere⎯⎯⎯⎯⎯⎯→

PhosphinePH3 +

Sodium3NaH2PO2

hypophosphiteQ. 7. Write balance chemical equations for the following :

(i) Hydrolysis of PCl3 with heavy water(ii) H3PO3 is heated.

Ans. (i) It reacts with heavy water to form phosphorus oxychloride (POCl3) and deuterium chloride(DCl)

PCl5 + D2O ⎯→ POCl3 + 2DCl(ii) Orthophosphorus acid on heating disproportionates to give orthophosphoric acid andphosphine.

4H3PO3Heat

⎯⎯→ PH3 + 3H3PO4

acidorthophosphorus Phosphine


Q. 8. Complete the following reactions :(i) C2H4 + O2 ⎯→(ii) 4Al + 3O2

Ans. (i) C2H4 + O2 Heat

⎯⎯→ 2CO2 + 2H2O

(ii) 4Al + 3O2 Heat

⎯⎯→ 2Al2O3

Q. 9. Write the reactions of F2 and Cl2 with water.Ans. F2 being a strong oxidising agent oxidises H2O, O2 and O3

2F2 + 2H2O → 4H+ + 4F– + O2

3F2 + 3H2O → 6H+ + 6F– + O3

Cl2 reacts with H2O to form hydrochloric acid and hypochlorous acidCl2 + H2O → HCl + HOCl

Hydro

acidchloric

Hypochlorous

.acid

Q. 10. Write balanced equation for the following :(i) NaCl is heated with sulphuric acid in the presence of MnO2

(ii) Cl2 gas is passed into a solution of NaS in water.Ans. (i) Cl2 is produced

[NaCl + H2SO4 → NaHSO4 + HCl ] × 44HCl + MnO2 → MnCl2 + Cl2 + 2H2O

4NaCl + MnO2 + 4H2SO4 → MnCl2 + 4NaHSO4 + Cl2 + 2H2O

(ii) Cl2 being an oxidising agent oxidises NaI to I2Cl2 + 2NaI → 2 NaCl + I2 (s)

Q. 11. Complete the following equations :(i) P4 + H2O → (ii) XeF4 + O2F2 →

Ans. (i) P4 + H2O → No reaction(ii) XeF4 + O2F2 → XeF6 + O2

Q. 12. Complete the following chemical equations(i) Ca3P2 + H2O →(ii) Cu + H2SO4 (conc.) →

Ans. (i) Ca3P2 + 6H2O → 2PH3 + 3 Ca(OH)2(ii) Cu + 2H2SO4 (conc.) → CuSO4 + SO2 + 2H2O

Q. 13. Complete the following equations :(i) Ag + PCl5 →(ii) CaF2 + H2SO4 →

Ans. (i) 2Ag + PCl5 → 2AgCl + PCl3(ii) CaF2 + H2SO4 → 2HF + CaSO4

Q. 14. Complete the following equations :(i) C + conc. H2SO4 →(ii) XeF4 + H2O →

Ans. (i) C + 2H2SO4 (conc.) → CO2 + 2SO2 + 2H2O(ii) 2XeF4 + 2H2O → 2Xe + 4HF + O2

Q. 15. Write chemical equations for the following when :(i) Orthophosphorus acid is heated(ii) PtF6 and xenon are mixed together

Ans. (i) 4H3PO3 Δ

⎯→ 3H3PO4 + PH3

(ii) PtF6 + Xe → Xe+ [PtF6]–

Q. 16. Complete the following chemical equations :(i) PH3 + HgCl2 →(ii) Br2 + F2 (Excess) →

Balance the Chemical Equations ■ 141

Ans. (i) 2PH3 + 3HgCl2 → Hg3P2 ↓ + 6HClMercuric

phosphide(ii) Br2 + 5F2 (Excess) → 2BrF5

Q. 17. Complete the following chemical equations :(i) P4(s) + NaOH + H2O →(ii) I–(aq) + H2O + O3 →

Ans. (i) P4(s) + 3NaOH + 3H2O → PH3 + 3NaH2PO2Phosphine

(ii) 2I– (aq) + H2O (l) + O3 → 2OH– + I2 + O2Q. 18. Give balanced equation for the following reactions:

(i) Silver nitrate is added to dilute solution of sodium thiosulphate(ii) Potassium dichromate is treated with acidified ferrous sulphate solution(iii) Phosphorus reacts with conc. sulphuric acid.

Ans. (i) 2AgNO3 + Na2S2O3 ⎯→ Ag2S2O3 + 2NaHO3Silver Sodium thionitrate sulphate

(ii) 6FeSO4 + K2Cr2O7 + 7H2O ⎯→ 3Fe2(SO4)3 + K2SO4 + Cr2(SO4)3 + 7H2O acidified errous sulphate(iii) 2P + 3H2SO4 → P2(SO4)3 + 3H2

Phosphorus Sulphuricacid

Q. 19. What happens when Cl2 reacts with(i) Cold and dilute NaOH.(ii) Hot and concentrated NaOH.

Ans. (i) With cold and dilute solution of alkali (NaOH) chlorine reacts to form chloride andhypochlorite.

Cl2 + H2O ⎯→ HCl + HClOHCl + NaOH ⎯→ NaCl + H2O

HClO + NaOH ⎯→ NaClO + H2O

Cl2 + 2NaOH ⎯→ NaCl + NaClO + 2H2OCold and dil. Sodium Sodium

chloride hypochlorite

(ii) With hot and concentrated solution of alkali (NaOH) chlorine reacts to form a mixture ofchloride and chlorate.

Cl2 + H2O ⎯→ HCl + HClO [× 3]HCl + NaOH ⎯→ NaCl + H2O [× 3]

HClO + NaOH ⎯→ NaClO + H2O [× 3]

3 NaClO ⎯→ NaClO3 + 2NaCl

3Cl2 + 6NaOH ⎯→ 5NaCl + NaClO3 + 3H2OHot and Sodium

Conc. chlorate

Q. 20. Give the equation when hydrogen peroxide is treated with acidified KMnO4 solution.

Ans. 2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + 8H2O + 5O2 + K2SO4Potassium Hydrogen Sulphuric Manganesepermanganate Peroxide acid sulphate (Violet) (colourless)

Q. 21. Give the equation when sulphuric acid is treated with hydrogen sulphide.Ans. H2SO4 + H2S ⎯→ 2H2O + SO2↑ + S

Sulphuric Hydrogen Water Sulphur Sulphuracid sulphide dioxide

Q. 22. Write balanced chemical equation for the following reactions ozone and lead sulphide.Ans. PbS + 4O3 ⎯→ PbSO4 + 4O2

Lead Ozone Leadsulphide sulphate


Q. 23. Write balanced chemical equation when sulphuric acid is treated with phosphorous.Ans. P4 + 10H2SO4 ⎯→ 4H3PO4 + 10SO2 + 4H2O

Phosphoricacid

Q. 24. Give the equation when hydrogen sulphide is treated with concentrated sulphuric acid.

Ans. H2S + H2SO4conc.

⎯→ S + SO2 + 2H2O

Q. 25. Give balanced equations when Ozone and Mercury reacts with each other.

Ans. O3 + 2Hg ⎯→ Hg2O + O2

Q. 26. Give the reaction of action of heat on a mixture of sodium chloride and concentrated sulphuricacid.

Ans. NaCl + H2SO4 Δ<200°C

⎯⎯⎯→ NaHSO4 + HCl ↑conc.

Q. 27. Give balanced chemical equation when Sulphur dioxide gas is passed through NaOH solution.

Ans. 2NaOH + SO2 ⎯→ Na2SO3 + H2O

Q. 28. Give balanced equation when Sulphur dioxide and acidified potassium permanganate, reactswith each other

Ans. 2KMnO4 + 5SO2 + 2H2O ⎯→ K2SO4 + 2MnSO4 + 2H2SO4

Q. 29. Write balanced equation when

Ozone and Hydrogen sulphide reacts with each other

Ans. H2S + 4O3 → H2SO4 + 4O2

Q. 30. Give balanced equation for a reaction in which hydrogen peroxide acts as a reducing agent andone in which it acts as an oxidizing agent.

Ans. Hydrogen peroxide as an oxidising agent : It oxidises lead sulphide to lead sulphate.PbS + 4H2O2 → PbSO4 + 4H2O

Hydrogen peroxide as a reducing agent : It decolourises acidified potassium permanganate.2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + SO2

Q. 31. Write balanced equations when dilute hydrochloric acid is added to sodium thiosulphatesolution.

Ans. 2HCl + Na2S2O3 → 2NaCl + SO2 + H2O + S ↓Q. 32. Complete the following chemical equations :

(i) P4 + SO2Cl2 →(ii) XeF2 + H2O →(iii) I2 + HNO3 →

(conc.)Ans. (i) P4 + 2SO2Cl → 4PCl5 + 2SO2

(ii) 2XeF2 + 2H2O → 2Xe + 4HF + O2

(iii) 2HNO3 (Conc.) → H2O + 2NO2 + [O] × 5

I2 + 5 [O] ⎯⎯⎯→ I2O5

I2O5 + H2O ⎯⎯⎯→ 2HIO3 I2 + 10HNO3 (conc.) → 2HIO3 + 10NO2 + 4H2O

Q. 33. Complete the following chemical equations :(i) XeF4 + SbF5 →(ii) Cl2 + F2 (excess) →

Ans. (i) XeF4 + SbF5 ⎯→ [XeF3]+ [SbF6]

–

(ii) Cl2 + 3F2 (excess) 573 K

⎯⎯⎯→ 2ClF3


Q. 34. Complete the following equations :(i) PCl5 + H2O (excess) →(ii) F2 + H2O →(iii) XeF2 + PF5 →(iv) Cl2 (g) + NaOH (aq) →

(hot and

conc.)(v) Cu + HNO3 (dilute) ⎯→

Ans. (i) PCl5 + 4H2O (excess) → H3PO4 + 5HCl

(ii) 2F2(g) + 2H2O(l) → 4H+(aq) + 4F–(aq) + O2(g)

(iii) XeF2 + PF5 → [XeF]+ [PF6]

–

(iv) 3Cl2(g) + 6NaOH(aq) → 5 NaCl + NaClO3 + 3H2O(hot and

conc.)(v) 3Cu + 8HNO3 (dilute) ⎯→ 3Cu [NO3]2 + 2NO + 4H2O

Q. 35. Complete the following equations:(i) C + H2SO4 (conc.) ⎯→(ii) Cl2 + F2 ⎯→

(excess)(iii) AgCl + NH3 (aq) ⎯→

Ans. (i) C + 2H2SO4 (conc.) ⎯→ CO2 + 2SO2 + 2H2O

(ii) Cl2 + 3F2 573K

⎯⎯⎯→ 2ClF3

(excess)

(iii) AgCl + 2NH3 ⎯→ [Ag (NH3)2]+Cl

–

Q. 36. Complete the following reaction :(i) Ca(OCl)2 + HCl →(ii) NaClO3 (aq) + SO2 (g) →

(iii) NaN3 Heat

⎯⎯⎯→Ans. (i) Ca(OCl)2 + 4HCl → CaCl2 + 2H2O + 2Cl2

(ii) 2NaClO3 (aq) + SO2 (g) → 2ClO2 + Na2SO4

chlorine

dioxide

(iii) 2NaNH3 Heat

⎯⎯⎯→ 2Na + 3N2

Chapter 8. d and f-block Elements

Q. 1. Write the balanced equation when potassium iodide is treated with acidified potassiumpermanganate solution.

Ans. 2KMnO4 + 8H2SO4 + 10KI ⎯→ 6K2SO4 + 2MnSO4 + 8H2O + 5I2

Q. 2. Give the equations for the conversion of argentite (Ag2S) to metallic silver.

Ans. Ag2S + 4NaCN 2Na[Ag(CN)2] + Na2S argentite

Na2S + 2O2 Na2SO4

2Na[Ag(CN)2] + Zn→ Na2[Zn(CN)4] + 2Ag ↓Silver metal

Q. 3. Write balanced equation when acidified potassium permanganate and oxalic acid reacts witheach other :

Ans. 2KMnO4 + 3H2SO4 + 5(COOH)2 ⎯→ K2SO4 + 2MnSO4 + 8H2O + 10CO2


Q. 4. Give balanced chemical equation when zinc is added to sodium argentocyanide solution.Ans. Zn + 2Na [Ag(CN)2] ⎯→ Na2[Zn(CN)4] + 2Ag ↓Q. 5. Give the balanced equation for the preparation of silver nitrate in the laboratory.Ans. Equation for the laboratory preparation of AgNO3 :

3Ag + 4HNO3 Δ

⎯→ 3AgNO3 + 2H2O + NO(Silver) (dilute nitric acid) Silver nitrate

Q. 6. Describe the oxidising action of potassium dichromate and write the ionic equation for itsreaction with (i) Iodide (ii) Iron (iii) H2S.

Ans. (i) Cr2O72– + 14H+ + 6I– ⎯→ Cr3+ + 7H2O + 3I2

(ii) Cr2O72– + 14H+ 6Fe2+ ⎯→ 2Cr3+ + 7H2O + 6Fe3+

(iii) Cr2O72– + 8H+ 3H2S ⎯→ 2Cr3+ + 7H2O + 3S

Q. 7. Complete the following equations :(i) 2MnO4– + 5S2– + 16H+ ⎯→(ii) Cr2O7

2– + 2OH– ⎯→Ans. (i) [MnO4

– + 8H+ + 5e– ⎯→ Mn2 + + 4H2O ] × 2S2– ⎯→ S + 2e– ] × 5

2MnO4– + 5e2– 16H+ + 5S2– ⎯→ 2Mn2+ + 5S + 8H2O

(ii) Cr2O72+ + 2OH– ⎯→ 2CrO4

2– + H2OQ. 8. Complete the following equations :

(i) 2MnO4– + 5 NO2

– + 6H+ ⎯→(ii) Cr2O7

2– + 14H+ + 6e– ⎯→Ans. (i) MnO4

– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O ] × 2NO2

– + H2O ⎯→ NO32– + 2H+ × 2e–] × 5

2MnO4– + 5NO2

– + 6H+ ⎯→ 2Mn2+ + 5 NO32– + 2H2O

(ii) Cr2O72– + 14H+ + 6e– ⎯→ 2Cr3+ + 7H2O

Q. 9. Complete the following equations :(i) 2MnO4

2– + 5SO32– + 6H+ ⎯→

(ii) 2CrO42– + 2H+ ⎯→

Ans. (i) MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O ] × 2[SO3

2– + H2O ⎯→ SO42– + 2H+ + 2e–] × 5

5SO32– + 2MnO4

– + 6H+ ⎯→ 2Mn2+ + 5SO42– + 3H2O

(ii) 2CrO42– + 2H+ ⎯→ Cr2O7

2– + H2OQ. 10. Complete the following chemical equations :

(i) Fe3+ + I– ⎯→(ii) CrO4

2– + H+ ⎯→Ans. (i) 2Fe3+ + 2I ⎯→ 2Fe2+ + I2

(ii) 2CrO42– + 2H+ ⎯→ Cr2O7

2– + H2OQ. 11. Complete the following equations :

(i) Cr2O72– + I– + H+ ⎯→

(ii) MnO4 + SO32– + H+ ⎯→

Ans. (i) Cr2O72– + 14H+ + 6e– ⎯→ 2Cr3+ + 7H2O

[2I– ⎯→ I2 + 2e–] × 3

Cr2O72– + 6I– + 14H+ ⎯→ 2Cr3+ + 3I2 + 7H2O

(ii) [MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O] × 2

[SO32– + H2O ⎯→ SO4

2– + 2H+ + 2e–] × 5

2MnO4– + 5SO3

2– + 6H+ ⎯→ 2Mn2+ + 5SO42– + 2H2O

Q. 12. Complete the following equations :(i) Cr2O7

2– + 2OH– ⎯→(ii) MnO4

– + 4H+ + 3e– ⎯→


Ans. (i) Cr2O72– + 2OH– ⎯→ 2CrO4

2– + H2O

(ii) MnO4– + 4H+ + 3e–

Heat⎯⎯→ MnO2 + 2H2O

Q. 13. Write the balanced chemical equations for reactions involved during extraction of silver.Ans. (i) Treatment with sodium cyanide.

4Ag + 8NaCN + O2 + 2H2O 4Na[Ag(CN2] + 4NaOH

Ag2S + 4NaCN 2Na [Ag(CN)2] + Na2S

AgCl + 2NaCN Na[Ag(CN)2] + NaCl.(ii) Conversion of sodium sulphide to sodium sulphate.

4Na2S + 5O2 + 2H2O ⎯⎯→ 2Na2SO4 + 4NaOH + 2S.(iii) Precipitation of silver.

2Na [Ag(CN)2] + Zn ⎯⎯→ 2Ag + Na2 [Zn(CN)4]Q. 14. Write the balanced equation. How do you prepare

(i) K2MnO4 from MnO2 ?

(iii) Na2Cr2O7 from Na2CrO4 ?

Ans. (i) Pyrolusite is fused with KOH in the presence of atmospheric oxygen to give K2MnO4

2MnO2 + 4KOH + O2 Heat

⎯⎯⎯⎯→ 2K2MnO4 + 2H2O

Pyrolusite Potassium manganate

(ii) Na2CrO4 is extracted with water and acidified with dil. H2SO4 to give Na2Cr2O7

2Na2CrO4 + H2SO4 ⎯⎯⎯→ Na2Cr2O7 + Na2SO4 + H2OSodium Sodiumchromate dichromate


Q. 1. Methyl magnesium iodide is treated with carbon dioxide and the product is hydrolyzed in acidicmedium.

Ans.

O||C

=O + CH3 MgI ⎯→

O ||C— |CH3

OMgI + HOH⎯⎯→acid

CH3COOH + Mg (OH) I

Q. 2. Complete the following reaction and name the reaction :

……………+ 3I2 + 4KOH ⎯→ CHI3 + CH3COOK + 3KI + 3H2O

Ans. Iodoform reaction

CH3COCH3 + 3I2 + 4KOH ⎯→ CHI3 + CH3COOK + 3KI + 3H2O

Q. 3. Write balanced equations for the preparation of DDT.

Ans. → CCl3CHO + 2 Cl CCl3CH

DDT

Cl

Cl

Q. 4. Complete the following reactions :

(i)

+ HBr ⎯⎯→

H

HH

(ii) CH3CH2CH = CH2 + HBr Peroxide

⎯⎯⎯→


Ans. (i)+ HBr ⎯⎯→

H

H H H

H

H

Br

H

1-Bromo-1-phenyl ethane

(ii) CH3CH2CH = CH2 + HBr Peroxide⎯⎯⎯→

1 bromobutaneCH3—CH2—CH2—CH2 —Br

Q. 5. Write equation for the preparation of 1-iodobutane from :

(i) 1-butanol, (ii) 1-chlorobutane (iii) but-1-ene.

Ans. (i) 3CH3CH2CH2CH2OH + PI3 ⎯→ 3CH3CH2CH2CH2I + H3PO3

1-butanol 1-Iodobutane

(ii)1-chlorobutane

CH3CH2CH2CH2Cl + KI Acetone‚ heat

Finklelsteinreaction

⎯⎯⎯⎯⎯→ 1-Iodobutane

CH3CH2CH2CH2I + KCl

(iii) CH3CH2CH = CH2 + HBr Peroxide

Anti- Markonikoff’saddition

⎯⎯⎯⎯⎯⎯⎯→ CH3CH2CH2CH2Br KI/Acetone

Δ⎯⎯⎯⎯→

(Finkelstein reaction)

1-IodobutaneCH3CH2CH2CH2I

Q. 6. Write the complete equation and the structure of the major organic product in each of thefollowing reactions :

(i) CH3CH2CH2Cl + NaI Acetone

Heat⎯⎯⎯→

(ii) (CH3)3 CBr + KOH Ethanol‚ heat

⎯⎯⎯⎯⎯→

Ans. (i) CH3CH2CH2Cl + NaI1-chloropropane

Acetone‚ heat

⎯⎯⎯⎯⎯→(Finkelstein reaction)

CH3CH2CH2I 1-Iodopropane

+ NaCl

(ii) (CH3)3CBr

2-methyl propane2-Bromo-

+ KOHEthanol‚ heat

(Dehydrohalogenation)⎯⎯⎯⎯⎯⎯⎯⎯⎯→

2-methyl propeneCH3—

CH3|C = CH2 + KBr + H2O

Q. 7. Write the structure of the major organic product in each of the following reactions :

(i) CH3CH (Br) CH2CH3 + KOH Water

⎯⎯⎯⎯→

(ii) CH3CH2Br + KCN Aq. Ethanol

⎯⎯⎯⎯⎯⎯→

(iii) C6H5ONa + C2H5Cl ⎯⎯→

Ans. (i) CH3CH|Br

—CH2CH3 + KOH water

hydrolysis⎯⎯⎯⎯→ CH3—CH

|OH

—CH2—CH3 + KBr

2-Bromo butane Butan-2-ol

(ii) CH2CH2Br + KCNBromo ethane

Aq. Ethanol

⎯⎯⎯⎯⎯⎯⎯⎯⎯→(Nucleophilic Substitution reaction)

CH3CH2CN + KBrPropene nitrile

(iii) C6H5ONa + C2H5Cl Williamson’s

Synthesis⎯⎯⎯⎯⎯→ C6H5—O—C2H5 + NaCl

Sod. phenoxide Ethyl Phenetolchloride

Q. 8. Write the structure of the major organic product in each of the following :

(i) CH3CH2CH2OH + SOCl2 ⎯→


(ii) CH3CH2CH = CH2 + HBr Peroxide

⎯⎯⎯→

(iii) CH3—CH = C(CH3)2 + HBr ⎯→

Ans. (i)Propanol

CH3CH2CH2OH + SOCl2 Nucleophilic

Substitution⎯⎯⎯⎯⎯→

1-Chloro propaneCH3CH2CH2Cl + HCl + SO2

(ii)But-1 -ene

CH3CH2CH = CH2 + HBr Peroxide

(Anti-mark .add.)⎯⎯⎯⎯⎯⎯→

1-bromo butaneCH3CH2CH2CH2Br

(iii)2-Methyl but-2-ene

CH3—CH =

CH3|C— CH3 + HBr

Mark. add.⎯⎯⎯⎯→

2-Bromo-2-methyl butane

CH3—CH2—

CH3|C—|Br

CH3


(i) C6H5N2Cl + KI ⎯→

(ii)H H

C = C H H

+ Br2 CCl4⎯⎯→

Ans. (i) C6 H5N2Cl + KI → Iodobenzene

C6H5I + KCl + N2

(ii)H H

C = C H H

+ Br2 CCl4⎯⎯→ H—C—

|Br

|H

C—|Br

|H

H

1, 2-Dibromoethane


(i) OH + SOCl2 ⎯→ (ii)

CH2 — OH+ HCl ⎯→

HO

Ans. (i) OH + SOCl2 ⎯→ Cl + SO2 + HCl

(ii)

CH2 — OH+ HCl ⎯→

HO

CH2 Cl+ H2O

HO

Q. 11. What happens when (Give chemical equations only) :

(i) Ethanol is warmed with iodine and an aqueous solution of Na2CO3.

(ii) Chloral is treated with chlorobenzene in the presence of Conc. H2SO4.Ans. (i) CH3CH2OH + I2 + aq. 3Na2CO3 ⎯⎯⎯→

Iodoform CHI3 + HCOONa + 5NaI + 5H2O + 3CO2

(ii) CCl3CHO + Cl2Conc. H2SO4

⎯⎯⎯⎯⎯⎯→

Cl

Cl

Cl

C CH

Cl

Cl

+ H2O

DDT


Q. 12. How is iodoform prepared from propanone ? Give the reaction sequence also.

Ans. CH3COCH3 + 4NaOH + 3I2 Δ

⎯⎯→ CHI3 + CH3COONa + 3H2O + 3NaIPropanone Iodoform

The reaction sequence is :

2NaOH + I2 ⎯⎯→ NaOI + NaI + H2OSod.hypoiodite

CH3—COCH3 + 3NaOI ⎯⎯→ CI3COCH3 + 3NaOH

I3CCOCH3 + NaOH ⎯⎯→ CHI3 + CH3COONaIodoform

Q. 13. Complete the following giving the structure :

(i) CH3

HBr

⎯⎯⎯→Peroxide

(ii) CH3

+ HBr

⎯⎯→

Ans. (i) CH3

HBr

⎯⎯⎯⎯→Peroxide

CH3Br

(ii)CH3

+ HBr

⎯⎯⎯⎯→ CH3 Br

Q. 14. Complete the following giving the structure of major organic products :

(i) (CH3)3 — C—Br Mg

⎯⎯→ ……… H2O

⎯⎯⎯→ ………

(ii) (CH3)3—C—CH2—Br C2H5ONa/C2H5OH

⎯⎯⎯⎯⎯⎯⎯⎯→ …………

(iii) + (CH3)3—C—CH2—Br anhyd. AlCl3

⎯⎯⎯⎯⎯→ …………

Ans. (i)

CH3

CH3C—Br

CH3

+ Mg

dry ether

CH3

CH3

C—MgBrCH3H2 O

⎯⎯→

CH3

CH3—CH—CH3

2-Methyl propane

— ⎯⎯⎯⎯⎯→ —

2-Bromo-2-methyl tert-butylpropane magnesium

bromide

(ii)

neo-pentyl bromide

CH3—C—CH2—Br

CH3

CH3

C2H5ONa/C2H5OH

⎯⎯⎯⎯⎯⎯⎯⎯→Δ

CH3—C—CH2—O—CH2—CH3

CH3

CH3Ethyl neo-pentyl ether

(iii)

CH3

CH3

+ CH3—C—CH2—Br anhyd. AlCl3⎯⎯⎯⎯⎯⎯→ (1‚2-methyl shift)

1,1-Dimethyl propyl benzene

—C— —CH3 CH2

CH3

CH3

Benzene


Chapter 11. Alcohols, Phenols and EtherQ. 1. Give balanced equations for the following name reaction Reimer-Tiemann reaction

Ans.

Q. 2. Write the equation when sodium ethoxide is treated with ethyl bromide.

Ans. C2H5ONa + C2H5Br ⎯→ C2H5OC2H5 + NaBrSodium ethyl diethyl etherethoxide bromide

Q. 3. Give balanced equation for the following reaction between 1-butanol and hydrogen chloride.Ans. CH3CH2CH2CH2OH + HCl ⎯→ CH3CH2CH2CH2Cl + H2OQ. 4. Give balanced equation for the following :

Glycerol is heated with oxalic acid at 110°C (383 K). Ans. CH2 OH

|CHOH|CH2OH

+ COOH|COOH

110°C

CH2 O.OC.COOH|CHOH|CH2 OH

+ H2O

110°C

CH2 OH|CHOH|CH2 OH

+ Formic acidHCOOH + CO2

Glycerol Glycerol

⎯⎯→ ⎯⎯→

Oxalic acid

Q. 5. Give balanced equation for the following reaction :Phenol heated with chloroform and sodium hydroxide.

Ans.

OH

+ CHCl3 + NaOH Heated⎯⎯→

ONa

CHCl2+ H2O + HCl

Phenol Chloroform


(i) CH3—CH = CH2 H2O/H+

⎯⎯⎯→

(ii) CH3—CH2—CH—|CH3

CHO NaBH4⎯⎯⎯→

Ans. (i)Propene

CH3—CH = CH2 H+/H2O

⎯⎯⎯→

Propan-2-ol

CH3—CH—|OH

CH3

(ii)

2-Methyl butanal

CH3—CH2—CH—|CH3

CHO NaBH4⎯⎯⎯→

2-Methyl butan-1-ol

CH3—CH2—CH—|CH3

CH2OH

Q. 7. Write the balanced equation of the following reactions :(i) Oxidation of propan-1-ol with alkaline KMnO4 solution(ii) Bromine in CS2 with phenol.

Ans. (i)Propan-1-ol

CH3CH2CH2OH + 2[O] Alk·KMnO4⎯⎯⎯⎯⎯→

Propanoic acid

CH2CH2COOH +H2O

(ii) ⎯⎯→Br2

CS2

OHOH OHBr

Br

+

o-Bromophenol(Minor Product)

p-Bromophenol(Major Product)


Q. 8. Complete and balance the following equations :

(i) ⎯⎯⎯⎯⎯→

OH

Phenol

Dil HNO3

(ii) ⎯⎯⎯⎯⎯→

OH

343 KCHCl ,NaOH3

Ans. (i) ⎯⎯⎯⎯→Dil. HNO3

NO2

NO2

OHOH OH

+

o-Nitrophenol

p-Nitrophenol

Phenol

(ii) ⎯⎯⎯⎯→⎯⎯⎯⎯⎯⎯⎯⎯→CHCl3 , NaOH, 343 K H+, H2OCHO

ONaCHO

OHOH

Phenol(Riemer - Tiemann

Reaction) Salicylaldehyde(Major product)

Q. 9. Complete the following reaction :

⎯⎯⎯→

OH……………………

Ans.⎯⎯⎯→H2O/H+

OH

4-Methylhept-3-ene 4-Methylheptan-4-ol

Q. 10. Write chemical equation for the preparation of phenol from chlorobenzene.

Ans.

Cl|

+ NaOH

623 K300 atm

⎯⎯⎯⎯→– HCl

ONa|

H2O+

⎯⎯→

OH|

+ Na+ + H2O

Q. 11. Give the equations for the preparation of phenol from cumene.Ans. Cumene is oxidised into cumene hydroperoxide by air at high pressure which on decomposition

by aqueous acid gives phenol and acetone.

|

Cumene

CH3—CH—CH3

+ O2 Air Oxidation

⎯⎯⎯⎯⎯⎯→High pressure

|

Cumenehydroperoxide

CH3—C—CH3

|OOH

Dil. H2SO4⎯⎯⎯⎯→

OH|

Phenol

+ CH3—C—CH3

||O

Acetone

Q. 12. Give balanced equations for the reaction of ethanol with :(a) Sodium (b) Acetyl chloride(c) PCl5 (d) SOCl2 (e) Acid anhydride

Ans. (a) 2C2H5OH + 2Na ⎯→ 2 C2H5ONa + H2

(b) C2H5OH + CH3COCl ⎯→ C2H5O.COCH3 + HCl(c) C2H5OH + PCl5 ⎯→ C2H5Cl + POCl3 + HCl(d) C2H5OH + SOCl2 ⎯⎯→ C2H5Cl + SO2 + HCl(e) C2H5OH + (CH3CO)2O ⎯→ C2H5O.COCH3 + CH3COOH


Q. 13. Complete the following equations :1. CH3CH2OH + SOCl2 ⎯→ ……… + ……… + ………

2. RCH2OH + ……… ⎯→ RCH2Br + ………

RCH2Br + ……… ⎯→ RCH2CN + ………

RCH2CN + ……… ⎯→ RCH2COOH + ………

3.

ONa|

+ ………… 400 K

⎯⎯⎯→4-7 atm

OH| COONa

4. CH3CH2OH Cu

⎯⎯→573 K

……… + ………

5.OH

+ HNO3 conc.

⎯⎯⎯→excess

……… + ………

Ans. 1. CH3CH2OH + SOCl2 ⎯⎯→ CH3CH2Cl + SO2 + HCl

2. RCH2OH + HBr ⎯⎯→ RCH2Br + H2O

RCH2Br + KCN ⎯⎯→ RCH2CN + KBr

RCH2CN + 2H2O HCl

⎯→ RCH2COOH + NH3

3.

ONa|

+ CO2 400 K

⎯⎯⎯→4-7 atm

OH| COONa

4. CH3CH2OH Cu

⎯⎯⎯→573 K

CH3CHO

⎯⎯⎯→NO2

+ H2

5.OH

+ 3 HNO3 ⎯⎯⎯→

NO2

O2N

OH

NO2

+ 3H2O

Q. 14. Give balanced equation for the following reaction :Chlorine is passed through diethyl ether.

Ans. CH3CH2OCH2CH3 + Cl2diethyl ether

– HCl⎯⎯→ CH3CHCl– O–CH2CH3

α-chloro di ethyl ether

+ Cl2

– HCl⎯⎯→ CH3CHCl·O·CHCl·CH3

α-α' di chloro diethyl ether

Q. 15. Give balanced equation for the following reaction :Diethyl ether with phosphorus pentachloride

Ans. C2H5OC2H5 + PCl5 ⎯→ 2C2H5Cl + POCl3

ethylchloride

Phosphorylchloride

Q. 16. State the product of the following reactions :

(i) CH3CH2CH2—O—CH3 + HBr ⎯→

(ii)OC2H5

+ HBr ⎯⎯⎯→

(iii) (CH3)3 C—O—C2H5 HI

⎯→

Ans. (i) CH3—CH2—CH2—O—CH3 + H-Br ⎯→ CH3—Br + CH3CH2CH2OH

(ii)OC2H5

OH

+ HBr ⎯⎯⎯→ C2H5 Br +

(iii) (CH3)3 C—O—C2H5 + HI ⎯→ (CH3)3 C—I + C2H5 —OH.


Q. 17. Give the major product that are formed by heating each of the following ethers with HI :

(i) CH3—CH2—

CH3|CH—CH2—O—CH2—CH3

(ii) CH3—CH2—CH2—O—C—|CH3

|CH3

CH2—CH3

(iii) — CH2 — O —

Ans. (i) CH3—CH2—CH—|CH3

CH2—OH + CH3CH2I

(ii) CH3—CH2CH2OH + CH3CH2—C—|CH3

|CH3

I

(iii) — CH2 I + — OH

Chapter 12. Aldehyde, Ketones and Carboxylic acidQ. 1. Give the chemical reaction when A is treated with NaOH and name the reaction.Ans. Cannizzaro Reaction :

2C6H5CHO + NaOH 50% NaOH

⎯⎯⎯⎯⎯→ C6H5CH2OH + C6H5COONaBenzaldehyde Benzyl Alcohol Sodium benzoate

Q. 2. Give balanced equation for the Rosenmund reaction.

Ans.

Q. 3. Write the reaction when formaldehyde is treated with ammonia.Ans. 6HCHO + 4NH3 ⎯→ (CH2)6 N4 + 6H2O

Formaldehyde Urotropine

Q. 4. Write the reaction when acetaldehyde is heated with hydroiodic acid in the presence of redphosphorous.

Ans. CH3 CHO + 4HIred P

ΔCH3·CH3 + H2 O + 2I2

Acetaldehyde Ethane

⎯⎯→

Q. 5. Write the reaction when calcium acetate is subjected to dry distillation.

Ans. (CH3COO)2CaΔ

⎯→ CH3COCH3 + CaCO3Calcium acetate Acetone

Q. 6. Write the reaction when benzaldehyde is treated with sodium bisulphite.

Ans.Benzaldehyde

C6H5 CHO + Sodium bisulphite

NaHSO 3 C6H5 —

H|C—OH|SO3Na

⎯→

Benzaldehyde sodium bisulphite

Q. 7. Give balanced equation for the Clemmensen’s reduction.Ans. Clemmensen’s reduction : It is a reaction in which reduction of aldehyde to alkane takes place

using Zn/Hg and HCl.

CH3CHO + 4H Zn/Hg

+ HCl⎯⎯→ CH3·CH3 + H2O

Acetaldehyde Ethane

Q. 8. Give the reaction when benzaldehyde is treated with 50% sodium hydroxide solution.Ans. 2C6H5CHO + NaOH ⎯→ C6H5COONa + C6H5CH2OH

benzaldehyde 50% sodium Benzylalcoholhydroxide

This is Cannizzaro’s reaction.


Q. 9. Give balanced equation for the following reaction between benzaldehyde and hydroxylamine.

Ans. C6H5CHO + H2NOH ⎯→ C6H5CH = NOH + H2O

Q. 10. Give balanced equations when :Acetone reacts with hydrogen in the presence of heated copper.

Ans.Acetone

CH3COCH3 + H2 CuΔ

⎯⎯→ Propanol-2

CH3CHOHCH3

Q. 11. Give balanced equations when :Benzaldehyde is treated with hydrogen cyanide.

Ans.Benzaldehyde

C6H5CHO + hydrogen cyanide

HCN pH 9 – 10

⎯⎯⎯⎯→ C6H5—

OH|C—H|CN

Q. 12. Give balanced equation for the following reaction :Acetaldehyde reacts with phenyl hydrazine.

Ans. CH3CHO + NH2NHC6H5 H+⎯→ CH3—CH = N—

C6H5|NH + H2O

Acetaldehyde Phenyl hydrazine Acetaldehyde phenylhydrazine

Q. 13. Write balanced equation of the reaction when

Acetone and Phosphorus pentachloride.

Ans. CH3 —

Cl|

C—|

Cl

CH3 + POCl3

C = O + PCl5

CH3

CH3

⎯→

Acetone 2, 2 Dichloropropane

Q. 14. Predict the product of the following reactions :

(i)——

O + HO—NH2

H+⎯→

(ii) ⎯⎯→+ NH2 — NH —

—

—NO2

O2N——

O

(iii) R—CH = CH—CHO + NH2—C||O

—NH—NH2 H+

⎯→

Ans. (i)——

O(ii) NNH —NO2

—O2N

(iii) R—CH = CH—CH = N—NH—C||O

—NH2.Q. 15. Predict the product of the following reactions :

(i) + CH3CH2NH2 ⎯⎯→H+

——

C

O——

CH3

(ii) ⎯⎯⎯⎯→

C2H5

AlCl3 , CS2

— —

+ ——

AnhydC

O

Cl


Ans. (i)

—

—C = N — CH2 — CH2

H3C

(ii)

—

Propiophenone

— —C— C2H5

O

Q. 16. Complete the following synthesis

(i) C6H5CHO + CH3CH2CHO dil. NaOH

⎯⎯⎯⎯→

(ii) CH3COCH2COOC2H5 (i) NaBH4

H+⎯⎯⎯⎯→

(iii) ⎯⎯→— OHCrO3

Ans. (i)Benzaldehyde

C6H5CH O + H2 C—|CH3

CHO dil·NaOH

Claisen-Schmidt

condensation

⎯⎯⎯⎯⎯→

2-methyl-3-phenyl-prop-2-enal

C6H5CH = C—|CH3

CHO

(ii) Only keto group reduced by NaBH4

Ethyl-3-oxobutanoateCH3—C

||O

—CH2COOC2H5 (i) NaBH4

(ii) H+⎯⎯⎯⎯→

Ethyl-3-hydroxybutanoate

CH3—CH—|OH

CH2COOC2H5

(iii) ——O⎯⎯→— OHCrO3

Q. 17. Complete the following synthesis by giving missing starting material, reagent or products :

(i) ——CH ⎯→2 — CHO

(ii) ⎯⎯⎯⎯⎯→ 2 ——O(i) O3

(ii) Zn—H2O

Ans. (i) ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→B2H6 H2O2

3

—CH2OH——CH2

——CH2 — CHOB PCC

NaOHTHFMethylene

CyclohexaneCyclohexane

Carbaldehyde

(ii)

——

⎯⎯⎯⎯⎯→ 2 ——O(i) O3

(ii) Zn—H2OCyclohexylidene

CyclohexaneCyclohexane

Q. 18. Complete the following synthesis by giving starting material, reagent or products :

(i) ⎯⎯⎯⎯→—CH2CH3 KMnO4

KOH, Heat(ii) C6H5CHO

NH2CONHNH2⎯⎯⎯⎯⎯→


(iii) ⎯⎯→

——

— —

C

O

Ans. (i) —CH2CH3

—COOK

Ethyl benzene

⎯⎯⎯⎯→KMnO4

KOH, HeatPot. benzoate

(ii)Benzaldehyde

C6H5CHO NH2CO NHMH2⎯⎯⎯⎯⎯⎯⎯→

Benzaldehyde semi carbazone

C6H5CH = NNHCONH2 + H2O

(iii) ⎯⎯⎯⎯⎯⎯→+—

— —

C

O

— —COCl

+ HClFridel-Craftsacylation

Anhyd . AlCl3

Benzene Benzoylchloride

Q. 19. Write the products of the following reactions :

(i) CH3—C||O

—CH3 Zn – Hg

Conc. HCl⎯⎯⎯⎯→ ? (ii) CH3—C

||O

—Cl + H2 Pd – BaSO4⎯⎯⎯⎯⎯→ ?

Ans. (i) CH3—C||O

—CH3 Zn – Hg

Conc. HCl⎯⎯⎯⎯→ CH3—CH2—CH3 + H2O

(ii) CH3—C||O

—Cl + H2 Pd – BaSO4⎯⎯⎯⎯⎯→ CH3CHO + HCl

Q. 20. Write the balance equation when Acetyl chloride is treated with ethyl alcohol.

Ans.Acetyl chlorideCH3COCl +

EthanolCH3CH2OH ⎯⎯→

Ethyl acetate

CH3—C||O

—C2H5 + HCl

Q. 21. Give balanced equation for Kolbe’s electrolytic reaction.Ans. Kolbe’s electrolytic reaction : In this reaction aqueous solution of sodium salt of carboxylic acid

is electrolyzed and alkane is formed at anode.

CH3COONa(aq.) CH3COO– + Na+

At anode : CH3COO– – e– ⎯⎯⎯→ CH3CO·O

– CO2⎯⎯→ CH3 +

·CH3⎯⎯→ CH3·CH3

(Ethane)At cathode : 2H2O + 2e– ⎯→ 2OH– + H2

Q. 22. Give reaction when oxalic acid is treated with acidified potassium permanganate solution.Ans. 5 (COOH)2 + 2KMnO4 + 3H2SO4 ⎯→ K2SO4 + 2MnSO4 + 10CO2 + 8H2O

Q. 23. Give reaction when benzoic acid is treated with a mixture of concentrated nitric acid andconcentrated sulphuric acid.

Ans.

COOH COOH

NO2Benzoic acidconc.

Meta nitrobenzoic acid

conc H2SO4+ HNO3

Q. 24. Give reaction when ethylacetate is treated with ammonia.

Ans. CH3COOC2H5 + NH3 ⎯→ CH3CONH2 + C2H5OH

Q. 25. Give balanced equations for the reaction when benzoic acid and phosphorus pentachloridereacts with each other.

Ans. C6H5COOH + PCl5 ⎯→ C6H5COCl + POCl3 + HCl


Q. 26. Give balanced equation for the reactionBenzoic acid solution is treated with sodium carbonate.

Ans. 2C6H5COOH + Na2CO3 ⎯⎯→ 2C6H5COONa + H2O + CO2. Benzoic acid Sodium carbonate Sodium benzoate

Q. 27. Give balanced equation when formic acid is heated with Tollen’s reagent.Ans. HCOOH

Formic acid + Ag2O

Tollen’s reagent ⎯→ 2Ag + H2O + CO2

Q. 28. Complete the following synthesis :

(i) ⎯⎯→COOH

COOH Heat

SOCl2

(ii) ⎯⎯⎯⎯→CHO

COOH

NaCN/HCl

(iii)

COOHBr2/FeBr3

⎯⎯⎯⎯→

Ans. (i)

Phthalic acid

⎯⎯→COOH

COOH

COCl

COClHeat

SOCl2

Phthaloyl chloride

(ii)

2-Formyl benzoicacid

CHO⎯⎯→

COOH

CH—CN

OH

COOH

NaCN

2 - (1-Hydroxy, cyanomethylbenzoic acid)

HCl

(iii)

COOHBr2/FeBr3

⎯⎯⎯⎯→

COOH

Br+ HBr

Q. 29. Give the balanced equation and name the products when acetic acid is treated with PCl5.Ans. CH3COOH + PCl5 ⎯→ CH3COCl + POCl3 + HCl

Acetic acid Phosphorus Acetylchloride Phosphorus Hydrogenpentachloride (Ethanoylchloride) oxychloride chloride

Q. 30. Write the chemical equations of the carboxylic acid with :

(i) Zn metal (ii) NaHCO3

(iii) P2O5 (iv) HI/P4 at 473°K

(v) Na metal

Ans. (i) Carboxylic acid with Zn metal forms zinc salts with the liberation of H2 gas.

2RCOOH + Zn ⎯→ (RCOO)2Zn + H2 ↑2CH3COOH + Zn ⎯→ (CH3COO)2Zn + H2 ↑

Ethanoic acid Zinc ethanoate

(ii) Carbon dioxide is liberated.CH3COOH + NaHCO3 ⎯→ CH3COONa + H2O + CO2 ↑

Acetic acid Sodium acetate

(iii) Acid anhydride is obtained.CH3COOH

+CH3COOH

+ P2O5Δ

⎯→ CH3CO CH3CO

O + H2O

Ethanoic acid Ethanoic anhydride


(iv) Carboxylic acids are reduced to alkane by HI and red P at 473 K.

CH3COOH + 6HIEthanoic acid

Red P4⎯⎯⎯→473°K Ethane

CH3—CH3 + 2H2O + 3I2

(v) H2 gas is liberated.2CH3COOH

Acetic acid + 2Na ⎯→ 2CH3COONa

Sodium acetate + H2 ↑

Q. 31. Give a chemical test to distinguish between oxalic acid and formic acid.Ans. Formic acid forms silver mirror with Tollen’s reagent while oxalic acid does not forms silver

mirror.HCOOH + Ag2O ⎯→ 2Ag ↓ + CO2 + H2O

Formic acid Tollen’s reagent Silver mirror

Chapter 13. Organic Compounds Containing NitrogenQ. 1. Give balanced equation for Hoffmann’s degradation reaction

Ans. C2H5CONH2 + Br2 + 4KOH ⎯→ C2H5NH2 + 2KBr + 2H2O + K2CO3Propanamide Ethylamine

Q. 2. Write the balance equation how will you convert ethyl amine to methyl amine ?

Ans. C2H5NH2 NaNO2,HCl

⎯⎯⎯⎯→ C2H5OH K2Cr2O7

Conc.H 2SO4‚Δ⎯⎯⎯⎯⎯⎯→ CH3COOH

NH 3⎯⎯→ CH3CONH2

Br2 + conc. KOH←⎯⎯⎯⎯⎯⎯⎯CH3NH2

Methyl amine(1° amine)

Ethylamine

Δ

ΔThen

Q. 3. Aniline is treated with nitrous acid and HCl at low temperature. Give the equation.

Ans.

NH2

HNO2 + conc. HCl

0 – 5°C⎯⎯⎯⎯⎯⎯→

N2+Cl–

+ 2H2O

Aniline Benzene

diazonium

chloride

Q. 4. Give balanced equation reaction Balz-Schiemann’s reaction.Ans. Balz-Schiemann reaction : Reaction in which benzene diazonium salt is converted to

fluorobenzene via diazonium fluoroborate.

C6H5N2Cl + HBF4 – HCl

⎯⎯→ C6H5N2BF4 Δ

⎯→ C6H5F + N2 + BF3

Benzene Benzene diazonium fluoro-

diazonium fluoroborate benzenechloride

Q. 5. Give the balanced equation when acetamide is heated with bromine and sodium hydroxidesolution.

Ans. CH3CONH2acetamide

+ Br2 + 4NaOH Δ

⎯→ CH3NH2 + Na2CO3 + 2NaBr + 2H2O

This is Hoffmann’s bromamide reaction or Hoffmann’s degradation reaction.Q. 6. Complete the following reaction and name the reaction :

C3H7NH2 + CHCl3 + 3KOH (alc.) Δ→ ⎯⎯⎯⎯ + 3KCl + 3H2O

Ans. Carbylamine reaction

C3H7NH2 + CHCl3 + 3KOH (alc.) Δ

⎯→ C3H7NC + 3KCl + 3H2O

Q. 7. Give balanced equation when aniline is treated with benzoyl chloride.

Ans.Aniline

C6H5NH2 + Benzoyl chloride

C6H5COCl ⎯→ N-Phenylbenzamide‚ (Benzanilide)

C6H5NH—C||O

—C6H5 + HCl


Q. 8. Write balanced equation when aniline and bromine water reacts with each other.

Ans. C6H5NH2

Aniline

+ 3Br2 ⎯→

Br

Br

BrNH2

+ 3HBr

2, 4, 6 Tribromoaniline

Q. 9. Write balanced equation when ethyl amine and nitrous acid reacts with each other.Ans. C2H5NH2 + HNO2 ⎯→ C2H5OH + N2 + H2O

Ethylamine Ethyl alcohol

Q. 10. Give balanced equation for the reaction when methyl isocyanide is warmed with dilutehydrochloric acid.

Ans. CH3NC + 2H2O HCl

⎯⎯→ CH3NH2 + HCOOHMethylamine Formic acid

Q. 11. Give balanced equation when aniline is treated with a mixture of NaNO2 and excess of HClat low temperature.

Ans. NaNO2 + HCl ⎯⎯→ HNO2 + NaCl

C6H5NH2 + HNO2 + HCl 0–5°C

reacts⎯⎯⎯→ C6H5N2Cl + 2H2O.

Q. 12. Give balanced equation whenaniline reacts with acetyl chlorideAns. Aniline with acetyl chloride

— —

(Acetanilide)

+ CH3COCl ⎯⎯⎯→ + HCl

NH2 NH—C—CH3

O

(Aniline)

Q. 13. Complete the following reactions :(i) C6H5NH2 + CHCl3 + alc. KOH ⎯→(ii) C6H5NH2 + H2SO4 (conc.) ⎯→(iii) C6H5NH2 + Br2 (aq) ⎯→

Ans. (i)Aniline

C6H5NH2 + CHCl3 + (alc.)

3KOH Heat

Carbylaminereaction

⎯⎯⎯→ Phenyl isocyanide

C6H5N →= C + 3KCl + 3H2O

(ii) C6H5NH2 + H2SO4 (conc.) ⎯→ C6H5N+H3HSO4–

Aniline Anilinium hydrogen sulphate

(iii)

Br

BrBr

2, 4, 6 - Tribromo aniline

+ 3 Br2 (aq) ⎯⎯⎯→ + 3 HBr

NH2 NH2

Aniline


(i) C6H5N2Cl + H3PO2 + H2O Reduction

⎯⎯⎯⎯→

(ii) C6H5N2Cl + C2H5OH Reduction

⎯⎯⎯⎯→

Ans. (i) C6H5N2Cl + H3PO2 + H2O Reduction

⎯⎯⎯⎯→ C6H6 + N2 + H3 PO3 + HCl

(ii) C6H5N2Cl + C2H5OH Reduction

⎯⎯⎯⎯→ C6H6 + CH3CHO + N2 + HClQ. 15. Complete the following reactions :

(i) C6H5NH2 + (CH3CO)2O ⎯→ (ii) C6H5N2Cl (i) HBF4

(ii) NaNO2/Cu‚ Heat⎯⎯⎯⎯⎯⎯⎯⎯→


Ans. (i)Aniline

C6H5NH2 + (CH3CO)2O ⎯→ Acetanilide

C6H5NHCOCH3 + CH3COOH

(ii) C6H5N2Cl HBF4⎯⎯→ C6H5N+

2BF–4

NaNO2/Cu⎯⎯⎯⎯⎯→ C6H5NO2 + BF3 + NaF

BenzendiazoniumChloride

Benzene diazoniumtetra fluoroborate

Nitro benzene


(i) C6H5NH2 conc.H2SO4

Δ⎯⎯⎯⎯⎯→ (ii) CH3CH2NH2

HCl⎯⎯→

(iii) RNH2 CHCl3/KOH

⎯⎯⎯⎯⎯→

Ans. (i)Conc. H2 SO4C6H5NH2 ⎯⎯⎯⎯⎯→ H2N — — SO3H + H2O

(ii) CH3CH2NH2 + HCl ⎯→ CH3CH2N+H3Cl–

(iii) RNH2 + CHCl3 + 3KOH ⎯→ RN →= C + 3KCl + 3H2O.

Q. 17. Write the main products of the following reactions :

(i) C6H5N2+Cl–

H3PO2 + H2O⎯⎯⎯⎯⎯→ ?

(ii) ?Br2 (aq)

⎯⎯⎯⎯→

NH2

(iii) CH3—C||O

—NH2 Br2 + NaOH

⎯⎯⎯⎯⎯→ ?

Ans. (i) C6H5N2+Cl–

H3PO2 + H2O⎯⎯⎯⎯⎯→ C6H6 + N2 + H3PO3 + HCl

(ii)Br2 (aq)

⎯⎯⎯⎯→

NH2

Br

BrBrNH2

(iii) CH3—C||O

—NH2 Br2 + 4NaOH

⎯⎯⎯⎯⎯→ CH3—NH2 + Na2CO3 + 2NaBr + 2H2O.

Q. 18. Complete and balance the following reactions.

(i) C2H5NH2 + CH3COCl ⎯→

(ii) R—NH2 + HNO2 ⎯→

(iii) R—NH2 + CH3MgI ⎯→

Ans. (i) C2H5NH2 + CH3COCl ⎯→ C2H5NHCOCH3 + HClEthanamine Ethanoylchloride N-methylethanamide

(ii) R—NH2 + HNO2 ⎯→ ROH + N2↑ + H2OPrimary amine Alcohol

(iii) R—NH 2

CH3

I Mg

R.NH

I Mg CH

4

Methyl magnesium iodide Alkylammoniummagnesium

iodide

⎯→ CH+ +

Q. 19. Complete the following by identifying A, B, C and D.

(i) CH3NH2 HNO2⎯⎯→ A

PCl5⎯→ B KCN⎯→ C

Na + C2H5OH⎯⎯⎯⎯→ D

(ii) C2H5NH2 HNO2⎯⎯→ A

[O]⎯→ B

[O]⎯→ C

NH3 + heat⎯⎯⎯⎯→ D

(iii) CH3CH2COOH NH3⎯⎯→ A

Δ⎯→ B

Br2/KOH⎯⎯⎯→ C

HNO2⎯⎯⎯⎯→ D


(iv) CH3 —C

O

||—NH2

Δ

P2O5⎯⎯⎯→ A

LiAlH4⎯⎯⎯→

(v) C2H5OH P/I2

–H3PO3⎯⎯⎯→ A

AgNO2

–AgI⎯⎯⎯→ B

ZnNH4Cl

2H⎯⎯⎯→ C

(vi) CH2 = CH2 HI

⎯→ A KCN

⎯⎯⎯⎯→C2H5OH B H2 /Ni⎯→ C

(vii) C6H5NH2 Br2⎯⎯→aq. A

NaNO2/HCl⎯⎯⎯⎯⎯→ B

H3PO2⎯⎯→H2O C

Ans. (i) CH3NH2Methanamine

HNO2⎯→ CH3OH

Methanol PCl5⎯→ CH3Cl

Chloromethane KCN⎯→ CH3CN

Ethanenitrile Na + C2H5OH

⎯⎯⎯→ CH3CH2NH2Ethanamine

(A) (B) (C) (D)

(ii) C2H5NH2Ethaneamine

HNO2⎯→ C2H5OH

Ethanol

[O]⎯→ CH3CHO

Ethanal

[O]⎯→ CH3COOH

Ethanoic acid

NH3/heat⎯⎯⎯⎯→ CH3CONH2

ethanamide

(A) (B) (C) (D)

(iii) CH3CH2COOHPropanoic acid

NH3⎯→ CH3CH2COONH4

Ammoniumpropanoate

Δ⎯⎯→ –H2O

CH3CH2CONH2Propanamide

Br2/KOH

⎯→ CH3CH2NH2Ethanamine

(A) (B) (C)

HNO2⎯→ CH3CH2OH

Ethanol

(D)

(iv)Acetamide

CH3 —C

O||

—NH2 Δ

P2O5⎯⎯⎯→

Methyl

CH3 —C≡N LiAlH4

⎯⎯⎯→ Ethylamine

CH3—CH2—NH2

isocyanide

(v) C2H5OH P/I2⎯⎯→–H3PO3

C2H5IEthyliodide

AgNO2⎯⎯→–AgI C2H5NO2

Nitroethane ZnNH4Cl

⎯⎯→2 [H]

CH3CH2

H|N—OH

Ethyl hydroxyl amine

(A) (B) (C)

(vi) CH2 = CH2 HI

⎯→ CH3CH2IEthyl Iodide

KCN

⎯⎯→C2H5OH CH3CH2CNPropanenitrile

H2/Ni⎯→ CH3CH2CH2.NH2

Propane 1-amine

(A) (B) (C)

(vii)

NH2|

Aniline

Br2 (aq.)⎯⎯→

–HBr

BrNH2| Br

|Br

2, 4, 6–tribromoaniline(A)

NaNO2/HCl

⎯⎯→278 K

N2+Cl–

| Br

|Br

2, 4, 6–tribromobenzenediazoniumchloride

(B)

H3PO2/H2O⎯⎯⎯⎯→

|Br

1,3,5–tribromobenzene

Br

(C)

Br Br

Q. 20. Complete and balance the following equations and name the organic compound formed:(i) C2H5NH2 + CH3COCl ⎯→ (ii) CH3CONH2 + Br2 + 4KOH ⎯→

(iii) CH3CN + H2O2 NaOH⎯→ (iv) CH3NH2 + HNO2 ⎯→

Ans. (i)Ethanamine

C2H5NH2 + EthanoylchlorideCH3COCl ⎯→ C2H5NH C

||O

— CH3 + HCl

N-Ethylethanamide

(ii) CH3CONH2 + Br2 + 4KOH ⎯→ CH3NH2 + 2KBr + K2CO3 + 2H2OEthanamide Methanamide


(iii)Ethane nitrileCH3—CN + H2O2

NaOH⎯→ CH3—C

||O

.NH2 + 12O2

Acetamide

(iv) CH3NH2 + HNO2 ⎯→ CH3OH + N2 + H2OMethanamine Methanol

Q. 21. Complete the following giving appropriate reagents :CH2Cl

?⎯→

CH2 NO2

Sn/HCl⎯⎯⎯→ ?

KOH (aq.) Br2←⎯⎯⎯⎯⎯⎯ ?

CH2NC

? HCl/H2O

Warm←⎯⎯⎯⎯

⎯⎯

⎯→

Ans. The complete sequence is as follows :

CH2Cl|

⎯⎯

⎯→

AgNO2

(Alc.)

CH2NO2

Sn/HCl

6 [H]

CH2NH2

KOH (aq.)/Br 2

CH2CO.NH2

Benzyl chloride Phenylnitro methane

Benzyl amine Phenylacetamide

CHCl2/KOH

CH2NCCH2NH2

Benzylamine

HCOOH +Formic acid

HCl/H2O

Warm

| | |

| |

⎯⎯⎯→⎯⎯⎯→ ←⎯⎯⎯⎯⎯

←⎯⎯⎯⎯


Q. 1. Write down the balance chemical equations for :

(i)

CHO

(CHOH)4

CH2OHGlucose

+ HCN ⎯⎯⎯→

(ii)

CHO

(CHOH)4

CH2OH

+ HI ⎯⎯⎯→Δ

(iii)

CHO

(CHOH)4

CH2O

⎯⎯⎯⎯→+ Br2 water


(iv)

CHO

(CHOH)4

CH2OH

⎯⎯⎯→+ HNO3

(v)

CHO

(CHOH)4

CH2OH

⎯⎯⎯→NH2OH+

Ans. (i)

CHO

(CHOH)4

CH2OH

+ HCN ⎯⎯⎯→

CH CN

OH(CHOH)4

CH2OHGlucose cyanohydrin

(ii)

CHO

(CHOH)4

CH2OH

+ HI ⎯⎯⎯→Δ

CH3—CH2—CH2—CH2—CH2—CH3

n-Hexane

(iii)

CHO

(CHOH)4

CH2O

⎯⎯⎯⎯→+ Br2 water

COOH

(CHOH)4

CH2OHGluconic acid

(iv)

CHO

(CHOH)4

CH2OH

⎯⎯⎯→+ HNO3

COOH

(CHOH)4

COOHSaccharic acid

(v)

CHO

(CHOH)4

CH2OH

⎯⎯⎯→NH2OH+

CH = N—OH

(CHOH)4

CH2OHGlucose oxime

Chapter 15. PolymersQ. 1. Complete the following reactions :

(i) ⎯⎯→

CH = CH2

1, 3 Butadiene Styrene

nCH2 = CH —�CH = CH2 +

(ii)+ HCHO ⎯→

Formal-dehyde

MelamineNH2

NH2NH2

N

N

N


(iii) CH3—CH|OH

—CH2—COOH + CH3—CH2—CH|OH

—CH2—COOH ⎯⎯→

3-Hydroxy butanoic acid 3-Hydroxy pentanoic acid

(iv)

Caprolactum

H2C⎯⎯⎯⎯→533–543K

H2OH2C

CH2—CH2

CH2

C = O

H

N

(v) ⎯⎯⎯⎯⎯→(2n – 1)H2OnHO — CH2 — CH2 — OH + nHO —

O

C

O

Ethylene glycolC—OH

Ans. (i) ⎯⎯→

CH = CH2

1, 3 Butadiene Styrene

nCH2 = CH — CH = CH2 +

Butadiene-styreneCopolymer

CH2CH = CH — CH2 — CH — CH2

n

––

(ii)+ HCHO ⎯→

Formal-dehyde

MelamineNH2

NH2NH2

N

N

N

Melamine polymerNH

NH—CH2NH

N

N

N

––

(iii) CH3—CH|OH

—CH2—COOH + CH3—CH2—CH|OH

—CH2—COOH ⎯⎯→

3-Hydroxy butanoic acid 3-Hydroxy pentanoic acid

⎣⎢⎢⎡

⎦⎥⎥⎤

O—CH—|CH3

CH2—C

| |O

—O—CH—CH2—|CH2—CH3

C

| |O

––

(iv)

Caprolactum

H2C⎯⎯⎯⎯→533–543K

H2OH2C

CH2—CH2

CH2

C = O

H

N

—

⎣⎢⎢⎢⎡

C | |O

—(CH2)5—N|H

Nylon–6

—

⎦⎥⎥⎥⎤

(v)

n

––⎯⎯⎯⎯⎯→nHO — CH2 — OH + nHO —

O

Ethyleneglycol

Glyptal

C

O

C—OHO — CH2—CH2 — O —

O

C

O

C

❐

Question

8Set Formula/Structure based Questions

Chapter 1. Solid StateQ. 1. A compound formed by elements A and B has a cubic structure in which A atoms are at the

corners of cube and B atoms are at the face centres. Derive the formula of compound.

Ans. No. of A atoms per unit cell = 8 (at corner) × 18 = 1

Number of B atoms per unit cell = 6 (at the face centre) × 12 = 3

A : B = 1 : 3

The formula of compound is AB3.

Q. 2. A solid is made up of two elements X and Y. Atoms of Y are in ccp arrangement while atoms of Xoccupy all the tetrahedral sites. What is the formula of the compound ?

Ans. Let number of atoms of Y in ccp arrangement = N

So, number of tetrahedral sites = 2N

therefore number of X atoms = 2N

X : Y = 2N : N

Hence, formula of compound = X2Y


Q. 1. Draw the structure of the ClF3 molecule and state its geometry.Ans. ClF3 molecule structure.

It is a ‘T’ shaped molecule.

Q. 2. Draw the structure of XeF2 molecule indicating the

(i) Lone pairs.

(ii) State the hybridisation of the central atom.

(iii) State the geometry of the molecule.

Ans. (i) XeF2

Lone pair of electron (lp)}

(ii) Central atom Xe shows sp3d hybridisation.

(iii) Molecule is linear.

Formula/Structure based Questions ■ 165

Q. 3. To which class of compounds does IF7 belong ? What is the structure of the molecule ?

Ans. IF7 belongs to class of interhalogen compounds.

It has pentagonal bipyramidal structure.

I

Q. 4. Draw the structure of the following compounds:

(i) BrF3

(ii) (HPO3)3

(iii) XeF4

Ans.

Br ——

—

—

—

——

—

—

—

—

—— ——

——

—

O

OO

OO

O

OH

OHHO

F

FF

FF

FF

P

P

P

BrF3

(HPO3)3

Xe

XeF4

(i) (ii) (iii)

Q. 5. Draw the structure of the following :

(i) H2S2O8 (ii) HClO4

Ans.

——— — —

—

—— ————

————

Peroxo sulphuric acid (H2S2O8)

H—

—OO

O

O

O

O

O O

S

— —

O

O

S Cl

OH

—

OHPerchloric acid

(HClO4)

(ii)(i)


(i) H2S2O7 (ii) HClO3

Ans. (i) H2S2O7 (ii) HClO3

—

— ——

——

————

Pyrosulphuric acid (Oleum)(H2S2O7)

O

O

OHO

S

— —

O

O

S

OH

—

——

—

——

H

O

O

O

Cl

Chloric acid(HClO3)



(a) Hypochlorous acid (HOCl)

(b) Chlorous acid (HO2Cl)

(c) Chloric acid (HO3Cl)

(d) Perchloric acid (HO4Cl)

Ans. (a) H

O

ClHypochlorus acid

(b) H

O

Cl

O

Chlorous acid

(c)O

Cl

O

O—H

Chloric acid

(d) O OO

Cl

O—H

Perchloric acid

Q. 8. Draw the structure of the following:

(i) NF3 (ii) H3PO3

Ans.

F

F

FOrthophosphorus acid

(H3PO3)

— ——

——H

O

OH

P

OHN

NF3

(i) (ii)


(i) XeF6 (ii) HClO4 (iii) S8

Ans.

— ———

———

———

OO

S SSS S

SS S

FF

Fe

F

F

F

Xe

—

—

——

H

O

O

Cl

HClO4XeF6

(i) (ii) (iii)

S8

Q. 10. Draw the structure of the following :(i) XeO3 (ii) XeF2 (iii) XeOF4


Ans.

——

———

—

F

F

F

FF

FXe XeXe

— ———

OO— — —

O—

O

XeOF4XeF2XeO3

(i) (ii) (iii)

Q. 11. Draw the structure of red phosphorus and white phosphorus.

Ans.

P P

P

P

Whitephosphorus

Redphosphorus

P P

P

P

Q. 12. For the molecule IF7 :(i) Draw the structure of the molecule.(ii) State the hybridisation of the central atom.(iii) State the geometry of the molecule.

Ans. (i)

(ii) Hybridisation of I is sp3d3.(iii) Geometry is pentagonal bipyramidal.

Q. 13. Draw the resonating structure of SO2.

Ans.

S S S

O[I] [II] [III]

O O O O O

Sp2 orbital

143pm

Q. 14. Give resonating structure of O3.Ans.

OOO O

O O

O116.8°

127 .8 pm

O

O

Resonating Structure of O3

Q. 15. Draw the structure of H2SO4 .

Ans. HO S

O

O

OH


Q. 16. Write the structure of an ozonide.Ans. O

CH2 CH2

O O ozonide

Chapter 8. d and f Block ElementsQ. 1. Draw the structure of xenon tetrafluoride molecule and state the hybridization of the central

atom and the geometry of the molecule.Ans. Structure of Xenon tetrafluoride :

Hybridization of the central atom : sp3d2

Geometry of XeF4 molecule : Square planarQ. 2. Draw the structure of chromate and dichromate ion.

Ans.

——

——

—

—— 126°—

OOO

O OO

O

OOO

O

Cr Cr —Cr

—

Chromate ion

2 – 2 –

179 p

m

163 pm

Dichromate ion

Q. 3. Draw the structure of MnO42– and MnO4

–.

Ans.

— —

—

— ——— ——O

O

OO O

Mn

— —

— —

O

OO

Mn

Tetrahedralmanganate ion

(green)

Tetrahedralpermanganate ion

(Purple)

––

MnO4–2 MnO4

–

–


Q. 1. Write the structures of optical isomers of the complex ion [Co(en)2Cl2]+.

Ans. Optical Isomers of [Co(en)2Cl2] +

Cl

Co

N

N

N

ClN

N

enen

en

+

enen

Co

Cl

N

N

NCl

N

en

cis d-isomer l-isomer trans-mesoisomer

Co

N

Cl

Cl

NN

+ +


Q. 2. Write the formula of Triamminetriaquachromium (III) chloride.

Ans. [Cr(NH3)3(H2O)3]Cl3

Q. 3. Write the formula of Potassium hexacyanoferrate (III).

Ans. K3[Fe(CN)6]

Q. 4. Write the formulae of the following coordination compounds :

(i) Potassium tetracyanonickel(0).

(ii) Triamminetrinitrocobalt(III).

Ans. (i) K4 [Ni(CN)4] (ii) [Co(NH3)3(NO2)3]

Q. 5. Draw the structural isomer of [Co(NH3)5NO2)Cl2 and name the type of isomerism.Ans. Structural isomer → [Co(NH3)5Cl] Cl.NO2.

Type of isomerism → Ionisation isomerism.

Q. 6. What is the hybridization of the chlorine atom in ClF3 molecule ?

Ans. The hybridization of Cl atom in ClF3 is sp3d.

Cl atom ↑↓ ↑↓ ↑↓ ↑

3s 3p 3d

Cl atom in ClF3 with sp3d ↑↓ ↑↓ ↑ ↑ ↑

hybridization sp3d hybrid 3d

Q. 7. Draw the geometrical isomers exhibited by the compound [PtCl2(NH3)2].Ans. Geometrical isomers of [PtCl2(NH3)2]

cis isomer trans isomer

Q. 8. Write the formulae of the following co-ordination compounds :

(i) tetracarbonyl nickel (0)

(ii) potassium dicyanoargentate(1)

Ans. (i) [Ni(CO)4] → tetracarbonyl nickel (0).

(ii) K[Ag(CN)2] → potassium dicyanoargentate (I).Q. 9. Draw all the isomers (geometrical and optical) of the following :

(i) [Cr (C2O4)3]3– (ii) [PtCl2(en)2]2+ (iii) [Cr(NH3)2Cl2 (en)]+

Ans. (i) [Cr (C2O4)3]3–

Cr

Mirror

3–ox ox

ox ox

ox oxCr

3–


(ii) [PtCl2(en)2]2+

Pt Pt en

enen

en

Cl

ClCl

Cl

Mirror

2+ 2+

(iii) [Cr(NH3)2Cl2 (en)]+

(a) Cis –[Cr (NH3)2Cl2 (en)]+

enen

ClCl

ClClH3N

NH3

NH3NH3

Cr Cr

Mirror

+ +

(b) Trans–[Cr (NH3]2 Cl2 (en)]+

H3N

enen

Cl

Cl

ClCl H3N

H3N NH3

Cr Cr

Mirror

+ +

Q. 10. Draw all the isomers (geometrical and optical) of the following :

(i) [CoCl2(en)2]+

(ii) [Co(NH3)Cl (en)2]2+

(iii) [Co(NH3)2Cl2 (en)]+

Ans. (i) [CoCl2(en)2]+

enen

enen

Cl

Cl Cl

Cl ClCl

CoCoCo

MirrorTrans-

+ ++

Cis-

Geometrical isomersOptical isomers

enen


(ii) [Co(NH3)Cl (en)2]2+

en en

en enenen

Cl Cl Cl

CoCoCo

H3N

NH3

NH3

Trans-

2+

Mirror

2+ 2+

Cis-


(iii) [Co(NH3)2Cl2 (en)]+

enen en

Cl

Cl

Cl Cl

Cl

Co Co Co

H3NNH3

NH3NH3

NH3

+

NH3

Cl

Mirror

+ +

Trans- Cis-


Q. 11. Write all the geomertical isomers of [Pt(NH3) (Br) (Cl) (Py)].Ans. Three isomers are possible as follows :

Pt Pt Pt

Cl

Cl

Cl

Br

Br

Br

H3NNH3NH3

PyPyPy

Q. 12. Write the structure and names of all the stereoisomers of the following compound:

(i) [Co(en)3]3+ (ii) [PfCl2(NH3)2] (iii) [Fe(NH3)4Cl2]+

Ans.

d-Tris-(ethane-1, 2,diamine cobalt III)

d-Tris-(ethane-1, 2,diamine cobalt III)

en

en

en en

en

enCo Co

Mirror

3+ 3+(i)

Cis-Diamine dichloridoMirrorplatinum (II)

Pt Pt

Cl

ClCl NH3

NH3

NH3

NH3Cl

Trans-Diaminedichlorido platinum (II)

(ii)


Cis-Tetra amine dichloridoiron (III)

Cl

Cl

Cl

Cl

NH3NH3

NH3 NH3

NH3

NH3

NH3

NH3

Mirror

+ +

Fe Fe

Trans-Tetra aminedichloridoiron (III)

(iii)

Q. 13. Draw the geometrical isomers exhibited by the compound [PtCl2(NH3)2].

Ans. Structures of geometrical isomers of [PtCl2(NH3)2]

Cis isomer Trans isomerQ. 14. Draw structures of geometrical isomers of [Fe (NH3)2 (CN)4]–.

Ans.NC

NC CN

CN

NH3

NH3

Fe

–

NC

NC CN

CN

NH3

NH3

Fe

transcis

–

Q. 15. Write the formula of the following species :

(a) 2, 2′-Bipyridyl

(b) Acetylacetonate.Ans. (a) 2, 2′-Bipyridyl

� �

(b) Acetylacetonate

Q. 16. For the complex ion [Fe(CN)6]3 – state :

(i) The geometry of the ion.

(ii) The magnetic property of the ion.

Ans. (i) Octahedral geometry

(ii) Ion is paramagnetic.

Chapter 10. Haloalkanes and HaloarensQ. 1. Give the structure of the following compounds.

(i) BHC (ii) D. D. T.


Ans. (i) BHC (ii) D. D. T.

Cl

ClCl

ClCl

Cl

BHC

H

— ——

Cl C — C — Cl

Cl

Cl

Cl

[2, 3 - Bis - (4-chlorophenyl) –1, –1, 1, trichloro methane]

Q. 2. Write structure of the compound having formula C4H9Br.Ans. It has four isomers.

(i) CH3CH2CH2CH2Br 1-bromobutane

CH3

|(ii) CH3—CH—CH2—Br 1-Bromo-2-methyl propane(iii) CH3CH2—CH—CH3

|Br

2-Bromo butane

(iv) CH3—C—Br|CH3

|CH3

2-Bromo-2-methyl propane

Q. 3. Write the structure of the following compounds :(i) 2—Chloro-3-methyl pentane(ii) p-Bromochloro benzene(iii) 1-Chloro-4-ethyl cyclohexane

Ans. (i) CH3—CH—CH—CH2—CH3

Cl CH3

| |

(ii) ClBr

(iii)

Cl

C2H5


(i) 2-(2-Chlorophenyl)-1-iodooctane.

(ii) Perfluoro benzene

(iii) 4-tert-Butyl-3-iodoheptane.

Ans. (i)Cl

ICH2—CH—(CH2)5—CH3


(ii) F

FF

FF

F

(iii) CH3—CH2—CH—CH—CH2CH2CH3|I

|C(CH3)3

Q. 5. Write the structure of the following compounds :(i) 1-Bromo-4-sec-butyl-2-methyl benzene.(ii) 1-Chloro-4-ethyl cyclo hexane.(iii) 1-Bromo-4-sec-butyl-2-methyl benzene.

Ans. (i)

—— —

—

CH3 — CH2 — CH3

CH3

CH3

Br

(ii) Cl

C2H5

(iii) Br

—

—

——

—CH3

CH3

CH

C2H5

Q. 6. Write the structure of the alkyl halide synthesised in the following reactions :

(i) (CH3)3 COH Conc. HCl

⎯⎯⎯⎯→ room temp.

(ii) CH3 CH—CH2CH3 PBr3

⎯⎯⎯→

|

OH

(iii) (CH3)2 CHCH2OH SOCl2

⎯⎯⎯⎯→

Ans. CH3

CH3

CH3

— C—Cl(i)

CH3

Br

—CH—CH —CH (ii) 2 3

CH3

CH3

— CH—CH Cl (iii) 2


Chapter 11. Alcohols, Phenols and Ether

Q. 1. Draw the structure of Methanol.

Ans.

σ σσ

σσ

H

H

H

CH

—

—— O

142 pm

108.5°

Structure of Methanol

Lone pairs

Q. 2. Write the structure of the compound whose IUPAC names are as follows :

(i) 3-cyclohexyl pentan-3-ol

(ii) Cyclopent-3-en-1-ol

(iii) 3-Chloromethyl pentan-1-ol

Ans. (i)

——

OH

CH3 — CH2 — C — CH2CH3

(ii) OH

(iii) HO—CH2—CH2—CH—CH2—CH3|CH2Cl

Q. 3. Write the structure of the following compounds:

(i) 1-phenyl propan-2-ol

(ii) 3,5-Dimethyl hexane 1,3,5-triol

(iii) 2,3-Diethyl phenol

Ans. (i)

——

OH

H

— CH2 — C — CH3

(ii)

— —

—

OH

—

OH

—

OH

CH3 CH3

CH2 — CH2 — C — CH2 — C — CH3

(iii)

—

OHC2H5

C2H5

——


Q. 4. Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC

names. Classify them as primary, secondary and tertiary alcohols.

Ans. Eight isomeric alcohols are possible for the molecular formula C5H12O.1. CH3—CH2—CH2—CH2—CH2OH

Pentan-1-ol

(Primary)

2. CH3—CH2—CH2—CH—|OH

CH3

Pentan-2-ol (Secondary)

3. CH3—CH2—CH—CH2—|OH

CH3

Pentan-3-ol (Secondary)

4. CH3—CH—|CH3

CH2—CH2OH

3-Methylbutan-1-ol (Primary)

5. CH3—CH—|CH3

CH—|OH

CH3

3-Methylbutan-2-ol (Secondary)

6. CH3—

OH|C—|CH3

CH2—CH3

2-Methylbutan-2-ol (Tertiary)

7. CH3—CH2—CH—|CH3

CH2OH

2-Methylbutan-1-ol (Primary)

8. CH3—

CH3|C—|CH3

CH2OH

2, 2-Dimethylpropan-1-ol (Primary)

Q. 5. Give the structure of ethereal group.

Ans. In ethers, the oxygen atom is sp3 hybridized. The twoC—O σ bonds are formed by sp3—sp3 overlap. The twounshared pairs of electrons on oxygen occupy sp3

hybrid orbitals. The C—O—C angle is very close to 110°.

Unshared Electron Pairs in sp3 orbitals

O O

R

R

110

°

R

σ

σ

R

sp3_

sp3

sp 3_sp 3

Chapter 12. Aldehydes, Ketones and Carboxylic acid

Q. 1. Write the structure of three ethers with molecular formula C4H10O.

Ans. Structural formula of three ethers with molecular formula → [C4H10O]

(1) H5C2—O—C2H5 (2) H3C—O—C3H7Dimethyl ether Methyl propyl ether

(3) H3C—C—CH3

CH3

|OH

sec Butyl alcohol

|

Q. 2. Draw the structure of the following compounds:(i) p—p′—Dihydroxy benzophenone(ii) p—Nitropropiophenone

(iii) p—Methyl benzaldehyde

Ans. (i)

— —

O

CHO —— OH——

(ii) O2N —— COCH2CH3

(iii) CH3 —— CHO


Q. 3. Draw the structure of the following derivatives:(i) The 2,4-dinitrophenyl hydrozone of benzaldehyde(ii) Cyclo propene oxime(iii) Acetaldehyde dimethyl acetal(iv) The semicarbazone of cyclobutanone(v) The ethyl ketal of hexan-3-one(vi) The methyl hemiacetal of formaldehyde

Ans. (i)—

— CH = NNH — — NO2

NO2

(ii) N — OH

— —

(iii)

———

—C

CH3 OCH3

OCH3 H

(iv)

— —

O

= NNH — C — NH2

(v)

——

H2C CH2

OO

H3C —�CH2 —�C —�CH2 —�CH2 —�CH3

(vi)

———

—CH OH

OCH3 H

Q. 4. Draw the structures of the following compounds :(a) 3-methyl butanal (b) p-methylbenzaldehyde(c) 4-chloropentan-2-one (d) p, p-dihydroxybenzophenone(e) p-nitropropiophenone (f) 4-methylpent-3-en-2-one(g) 3-bromo-4-phenyl pentanoic acid (h) Hex-2-en-4-noic acid(i) 2, 4-dimethylpentanoyl chloride (j) Formic acetic anhydride

Ans. (a) CH3—CHC|CH3

H2CHO (b) H3C— —CHO

(c) CH3COCH2CH(Cl)CH3 (d) HO— —

O||C— —OH

(e) O2N— —COCH2CH3 (f) CH3COCH =

CH3|C— CH3

(g) CH3—CH—CH|C6H5

(Br) CH2COOH (h) CH3C ≡ C—CH = CH—COOH

(i) CH3—

CH3|CH—CH2

CH3|CH—COCl (j) H—

O||C—O—

O||C—CH3


Q. 5. Write the structural formula of :(i) 1, 2-Dimethoxy ethane (ii) Phenylacetic acid(iii) Iso-Valeric acid (iv) Adipic acid(v) Succinic acid (vi) Benzoyl chloride(vii) Ethyl benzoate

Ans. (i) CH3OCH2—CH2OCH3 (ii) C6H5CH2COOH(iii) (CH3)2CHCH2COOH (iv) HOOC (CH2)4COOH

(v)CH2COOH|CH2COOH

(vi) C6H5COCl

(vii) C6H5COOC2H5


Q. 1. Write the structure of 4-ethoxypentanenitrile.

Ans. CH3—CHCH2CH2CN|OC2H5

Q. 2. Give the structure of vinyl cyanide and write its IUPAC name.Ans. CH2 = CHCN, Prop-2–enenitrileQ. 3. Draw the structures of the tautomers of nitroethane.

Ans. CH3CH2—+ +N

O–

–OCH3CH = N

OH O

Nitroform Isonitro form(Pseudo acid) (aciform or nitrolic acid)

Q. 4. Write structures and names of all the nitro compounds having formula C7H7NO2 and C2H5NO2.Ans. The structures of nitro compounds with formula C7H7NO2 are as follows :

CH3| |

CH3

—NO2

CH3|

NO2

CH2NO2

|—NO2

|

m-Nitro Toluene Phenyl Nitro Methaneo-Nitro Toluene

p-Nitro Toluene

The structures of nitro compounds with formula C2H5NO2 are as follows :

CH3CH2—

O↑N = O, CH3CH2—O—N = O

Q. 5. Write isomeric structures of nitriles and isonitriles having molecular formula C4H7N.Ans. Isomeric structures of molecular formula C4H7N are :

(i) CH3CH2CH2CN and CH3CH2CH2NCButanenitrile n-propylcarbylamine

(ii) CH3

CH3|

CH —CN and CH3

CH3|

CH —NC2-methylpropane nitrile Isopropylcarbylamine

Q. 6. Give structures and names of various amines represented by the formula C3H9N.

Ans. (i) CH3CH2CH2NH2 (ii) CH3CHCH3

|NH2

Propan-1-amine (1°) Propan-2-amine (2°)

(iii) CH3CH2NH|CH3

(iv) CH3—N—CH3

|CH3

N-Methyl ethanamine (2°) N-N-Dimethylmethanamine (3°)


Q. 7. Give the structures of :(i) 3-(N-methylamino) butanal (ii) 4-Bromo-N ethyl aniline(iii) 2, 6-Diaminohexanoic acid (iv) 2- (N-ethyl-N methylamino propane)

Ans. (i) CH3—CH—CH2—CHO|NH—CH3

(ii) Br — —NH—C2H5

(iii) H2N—CH2—CH2—CH2—CH2—CH—COOH|NH2

(iv) CH3—N—CH—CH3

|CH3

|C2H5

Chapter 14. BiomoleculesQ. 1. Draw the structure of Nucleoside and Nucleotide.

Ans.

H

HH

H

BaseOHOH2C

OH OHNucleoside

O — P — O — H2C

H

HH

H

BaseO

OH OHNucleotide

Q. 2. Draw the structure of Pyran and Furan ring.

Ans. O O

Pyran Furan

Q. 3. What is meant by pyranose structure of glucose ?Ans. The six membered cyclic structure of glucose is called pyranose (α-or β-) in analogy with pyran.

The cyclic structure of glucose is more correctly represented by Haworth structure as given.

Pyranα-D (+)

Glucopyranoseβ-D (+)

Glucopyranose

OCH2OH

OH

OH3 2

6

14

5

HO OHH

H

H

H

HO

CH2OH

OH

HH

3 2

6

14

5

HO OHH

H

H

OHO

Q. 4. Give the Fisher projection of L-Glucose.Ans.

CHO

HO H

H OH

HO H

HO H

H2OHL-Glucose



Q. 1. Write the structure of monomers used for getting the following polymer :

(i) PVC (ii) Teflon (iii) BakeliteAns. (i)

———

—

H H

C = C

Vinyl chlorideCl Cl

(ii)

———

—

F F

C = C

F F Tetrafluoroethylene

(iii)

— —

O

OH and H — C — H—

Phenol Formaldehyde

Q. 2. Give the structure of Bakelite.Ans. OH

Bakelite

CH2

CH2

OH

CH2

CH2

OH

CH2

CH2CH2

OH

CH2

OH

CH2

OH

CH2CH2

Q. 3. Draw the structure of the following polymers :

(i) Polystyrene

(ii) PAN

(iii) PMMA

Ans. (i)

n

Polystyrene

— CH — CH2 —

(ii)

—

CN

n

— CH2 — CH —

PAN

(iii)

——

CH3

COOCH3 n

— CH2 — C —————

PMMA


Q. 4. Draw the structures of linear, branched and cross lineked polymers.

Ans.

Linear Chain

Branched Chain Cross-Linked polymer

Q. 5. Draw the structure of monomer units of the following polymers :

(a) Buna–S

(b) Nylon 6

(c) Teflon

(d) PMMA

(e) Polystyrene.

Ans. (a)1‚3 Butadiene

CH2 = CH = CH2 and

CH = CH2

Styrene

(b)

Caprolactum

CH2

CH2

CH2–CH2

CH2

C = O

H

N

(c) CF2 = CF2

Tetrafluoroethene

(d) CH2 = C|CH3

|COOCH3

Methyl methacrylate

(e) CH = CH2

Styrene

❐

Question

9Set Short Answer Questions / Definitions

Chapter 1. Solid StateQ. 1. For a crystal of sodium chloride, state :

(i) The type of lattice in which it crystallizes.(ii) The coordination number of each sodium ion and chloride ion in the crystal lattice.(iii) The number of sodium ions and chloride ions present in a unit cell of sodium chloride.(iv) The structural arrangement of the sodium chloride crystal.

Ans. (i) For a crystal of sodium, the type of lattice in which it crystallises is Face Centered Cubic(fcc).(ii) The coordination number of each sodium ion and chloride ion in the crystal lattice is 6 : 6.(iii) The number of sodium ions and chloride ions present in a unit cell is 4 each.(iv) The structural arrangement of the sodium chloride crystal is cubic close packing with Cl–

ions at corners and face centres and Na+ ions at edge centres and body centre. With respect toNa+ ion it is octahedral.

Q. 2. An ionic compound is made up of A cations and B anions. If A cations are present at thealternate corners and B anion is present on the body of the diagonal, what is the formula of theionic compound ?

Ans. Formula of ionic compound is A+ B– or AB.Q. 3. Define Frenkel defect in solid crystal.Ans. Frenkel defect : This defect arises in a crystal when an ion, generally cation, leaves its lattice

site and occupies an interstitial site so that electrical neutrality as well as stoichiometry ismaintained.

Q. 4. In a crystal of diamond :

(i) How many carbon atoms are present per unit cell ?

(ii) What type of lattice does diamond crystallize in ?

(iii) How many carbon atoms surround each carbon atom ?

(iv) How are they arranged ?

(v) What is the coordination number of each carbon atom ?

Ans. (i) Four carbon atoms are present per unit cell.

(ii) Diamond crystallizes in a network type of lattice with fcc arrangement of carbon atoms.

(iii) Four carbon atoms surround each carbon atom.

Short Answer Questions /Definitions ■ 183

(iv) Four carbon atoms are tetrahedrally arranged around each central carbon atom

C

CC

CC

Tetrahedral arrangement(v) Coordination number of each carbon atom is 4.

Q. 5. What is Schottky defect in a solid ?Ans. Schottky defect : It is a stoichiometric defect in which some of the lattice points in a crystal are

vacant such that cation-anion ratio remains the same and electrical neutrality is maintained.

Q. 6. What are F-centres in an ionic crystal ?Ans. F-centres : In ionic compounds having non-stoichiometric defects anionic vacancies are

occupied by electrons. These are called F-centres. These electrons remain delocalised and areresponsible for imparting colours to crystals.

Ionic Crystal with F-centre

Q. 7. Why are crystals having F-centres paramagnetic ?Ans. Free, unpaired electrons occupying anionic vacancies undergo spin in a particular direction.

This produces a magnetic moment and ionic crystals with F-centres, therefore, showsparamagnetic behaviour.

Q. 8. Define piezoelectricity and give one use of piezoelectric crystals.Ans. Piezoelectricity : When electricity is produced in insulators due to displacement of ion by

mechanical stress, it is called piezoelectricity. Piezoelectric crystals are used as pick ups inrecord players.

Q. 9. What are semiconductors ? What is the effect of increasing temperature on the conductivity of asemiconductor ?

Ans. Semiconductors : These are the crystals which allow only partial conduction of currentthrough them. Their conductivities are intermediate between those of conductors andinsulators. Their conductivity increases with increase in temperature.

Q. 10. Compare the crystals of copper and diamond giving one similarity and one difference.Ans. Similarity : Crystals of both, copper and diamond shows fcc arrangement of constituent

particles.Difference : In copper coordination number is 12 and lattice sites are occupied by Cu2+ ions.In diamond coordination number is 4 and lattice sites are occupied by C atoms.

Q. 11. In a body centred and face centred arrangement of atoms of an element, what will be thenumber of atoms present in respective unit cells ?

Ans. Body centred cubic unit cell : This type of unit cell has eight atoms at corners and one at thebody centre. Each corner atom makes 1/8 contribution and the atom at the body centre belongsonly to the particular unit cell. Hence, a body centred unit cell has

8 (at the corner) × 18 + 1 (at body centre) × 1 = 2 atoms.


Face centred cubic unit cell : A face centred cubic unit cell has one atom at each corner (thereare eight corners of a cube) and one atom at each face centre (there are six faces of a cube). Anatom at the face centre is being shared by two unit cells and makes a contribution of only 1/2 toa particular unit cell. Hence a face centred cubic unit cell has

8 (at corners) × 18 + 6 (at face centre) ×

12 = 4 atoms.

Q. 12. For diamond, state the element present at the lattice sites, the number of nearest neighbours foreach atom and the type of cell. State the hybridization of the carbon atom in diamond.

Ans. Carbon is present in lattice sites, one carbon atom is linked with other four carbon atoms i.e.,tetrahedral units. Diamond has covalent network crystals. It has sp3 hybridization.

Q. 13. How many sodium ions and chloride ions are present in a unit cell of sodium chloride ?Ans. 4 sodium ions and 4 chloride ions are present in a unit cell of NaCl.

Q. 14. What is the coordination number of sodium and chloride ions in sodium chloride crystals ?Ans. The coordination number of sodium as well as chloride ions is 6.

Q. 15. Define face-centred cubic structure.Ans. A face-centred cubic structure has one atom at each corner and one atom at each face centre.

Q. 16. Define void.Ans. The empty spaces present between the atoms or the ions when they are packed within the

crystal are called voids.Q. 17. What are interstitials in crystal ?

Ans. Atoms or ions that fill the normal vacant interstitial void in a crystal are called interstitials.Q. 18. Define the unit cell, or what do you understand by the term unit cell with reference to a crystal.

Ans. The unit cell is the smallest portion of a space lattice which by repeating own dimensions invarious directions, can generate the complete crystal i.e., the repetition of unit cells in three.dimensions makes the space lattice.

Q. 19. Define space lattice.

Ans. The three-dimensional arrangement of structural units or constituent particles (atoms,molecules or ions) in a crystal is called space lattice.

Q. 20. What do you mean by coordination number ?

Ans. The number of nearest neighbours (atoms or ions) of any atom or ion is known as itscoordination number.

Q. 21. What is the non-stoichiometry defect in crystals ?

Ans. These defects occur when the ratio of cation and anion in resulting compound is different fromthat as indicated by the laws of chemical combination. It arises when a neutral atom occupiessome of the interstitial sites in a crystal or when ions of one valency state are substituted bythose of another valency state.

Q. 22. State the type of structure possessed by a unit cell of CsCl.Ans. Body-centered cubic lattice.

Q. 23. What is the effect of Schottky defect and Frenkel defects on electrical conductivity ? In whichcompound both are present together.

Ans. Both the defects increases the electrical conductivity of the crystal. In silver bromide (AgBr)both defects are present together.

Q. 24. Name different types of crystalline solids.Ans. The types of crystalline solids are :

(i) Molecular solids (ii) Ionic solids(iii) Covalent solids (iv) Metallic solids

Q. 25. What happens when CsCl structure is heated at 760 K ?Ans. CsCl on heating at 760 K transforms to NaCl structure, i.e., its coordination number decreases

from 8 to 6.Q. 26. What happens when NaCl structure is under high pressure ?

Ans. NaCl changes to CsCl structure i.e., its coordination number increases from 6 to 8.


Q. 27. What is the relation of edge length and the radius of the atom in a face centered unit cell ?Ans. Edge length of the face centered unit cell,

a = 2√⎯ 2 . r or a = √⎯ 2 dwhere, r is the radius and d is the distance between two neighbouring atoms.

Q. 28. What is doping ?Ans. The process of incorporating impurities in a crystal lattice is called doping.

Q. 29. Account for the statement that graphite is anisotropic with respect to conduction of electricity.Ans. In graphite delocalised π-electrons are free to move within the layer, thus graphite conducts the

electricity only in the plane of layer and not in the direction perpendicular to its laminalcrystals. Hence in graphite, the values of electric conduction are different in differentdirections, therefore, it is anisotropic with respect to electric conduction.

Q. 30. What is Pyroelectric effect ?Ans. The electricity produced when a dielectric solid with net electric dipole moment is subjected to

heat.Q. 31. What is meant by ferro-magnetism ?

Ans. Magnetic property of solids that arises due to spontaneous alignment of magnetic moments inthe same direction.

Q. 32. What is meant by anti-ferromagnetism ?Ans. Magnetic property due to alignment of same number of parallel and anti-parallel magnetic

dipoles.Q. 33. For sodium chloride crystal, state :

(i) The type of unit cell, (ii) The nature of forces holding the particles together, (iii) The numberof nearest neighbours around each sodium ion, (iv) The geometry of the sodium ions which arearranged around a chloride ion.

Ans. (i) Face centered cubic unit cell.(ii) Electrostatic forces.(iii) The number of nearest neighbours around each sodium ion is 6 (coordination number).(iv) Octahedral.

Q. 34. What type of bonds are present between the particles of crystals of NaCl, Cu, diamond andgraphite ?

Ans. Name of the solid Type of the bond present

(i) NaCl Ionic bond (Electrostatic forces)(ii) Cu Metallic bond

(iii) Diamond Covalent bond(iv) Graphite Covalent bond

Q. 35. The figures given belows show the location of atoms in three crystallographic planes in a fcclattice. Draw the unit cell for the corresponding structure and identify these planes in yourdiagram.

(a) (b) (c)Ans. (a) Face plane (b) Face Diagonal plane (c) Diagonal plane


Chapter 2. SolutionsQ. 1. Two liquids A and B forms type II non-ideal solution which shows a minimum in its

temperature-mole fraction plot (T-χ diagram). Can the two liquids be completely separated byfractional distillation ?

Ans. Two liquids A and B forms type II non-ideal solution which shows a minimum in itstemperature-mole fraction plot. They cannot be completely separated by fractional distillation.Upon fractional distillation either A or B is obtained pure along with composite composition.

Q. 2. Define Raoult’s law for the elevation of boiling point of a solution.Ans. According to Raoults’ law Elevation of boiling point of a solution is directly proportional to

the lowering in vapour pressure caused by the number of particles of solute present in thesolution.

Q. 3. Define cryoscopic constant.Ans. Cryoscopic constant : It is the depression in freezing point of the solvent produced on

dissolving one mole of a substance in 1000 g of it. It is also called molal depression constant.Q. 4. A solution X is prepared by dissolving three moles of glucose in one litre of water and a

solution Y is prepared by dissolving 1·5 mole of sodium chloride in one litre of water. Will theosmotic pressure of X be higher, lower or equal to that of Y ? Give reason for your answer.

Ans. Osmotic pressure of X and Y will be equal.

Reason : This is because osmotic pressure is a colligative property. Its value depends upon thenumber of particles of solute present in the solution. In the given case 3 mole of glucose gives3 mole of molecules (particles) and 1·5 mole of NaCl, due to complete dissociation it provides1·5 moles of Na+ ions and 1·5 mole of Cl– ions → Total 3 moles of ions (particles).

NaClwater

Na+ + Cl–

1·5 mole 1·5 mole + 1·5 mole = 3 mole.Q. 5. 0·1 M urea solution shows less depression in freezing point than 0·1 M MgCl2 solution. Explain.Ans. Depression in freezing point is a colligative property which depends upon number of particles

present in the solution.MgCl2 when dissolved in water gives three particles per molecules (MgCl2 Mg2+ + 2 Cl–)while urea remains undissociated.

Q. 6. The elevation of boiling point produced by dilute equimolal solutions of three substances are inthe order A > glucose > B. Suggest a reason for this observation.

Ans. This observation suggests that substance A dissociates while substance B associates in thesolution because elevation in boiling point depends upon number of particles of the solutepresent in the solution and not on the molal concentration.

Q. 7. Define mole fraction.

Ans. Mole fraction of a component in a solution may be defined as the ratio of moles of thatcomponent to the total number of moles of all the components present in the solution.

Q. 8. Define molal elevation constant or ebullioscopic constant.

Ans. Molal elevation constant is defined as the elevation in boiling point when one mole of non-volatile solute is added to one kilogram of solvent.

Q. 9. Define molal depression constant or cryoscopic constant.

Ans. It is the depression in freezing point when 1 mole of non-volatile solute is dissolved in 1000 g ofsolvent.

Q. 10. What is an antifreeze ?Ans. A substance such as ethylene glycol which is added to water to lower its freezing point is called

an antifreeze. It is named so as it delays freezing.Q. 11. Define ebullioscopic constant.

Ans. It is defined as the elevation in boiling point that takes place when molality of the solution isunity. It is also called molal elevation constant.


Q. 12. What is an azeotropic mixture ?Ans. Azeotropic mixture is a mixture of two liquids having a definite composition, which boils at a

constant temperature like a pure liquid. For example 95·57% of ethanol and 4·43% of waterforms an azeotropic mixture which boils at 351·15 K.

Q. 13. What is the value of van’t Hoff factor for KCl in dilute aqueous solutions ?Ans. 2, because it dissociates completely in dilute solutions.

Q. 14. The weights of solutes present in two isotonic solutions A and B are in the ratio 2 : 3. If thesolutes are non-electrolytes, how are their molecular weights related ?

Ans. For an isotonic solution the relation between weights and molecular weights of two solutes Aand B will be :

WA

MA =

WB

MB

According to the question the ratio is 2 : 3 i.e., the molecular weights of A and B will be in theratio of 2 : 3.

Q. 15. What is reverse osmosis ?

Ans. On applying pressure (higher than osmotic pressure) over solution, solvent particles movesfrom solution towards solvent through semi-permeable membrane and is named as reverseosmosis.

Q. 16. Name two ways of measuring the concentration of a solution which are not dependent ontemperature .

Ans. Molality and mole fraction are the two methods to measure the concentration which are notdependent on temperature.

Q. 17. State Konowaloff’s rule.Ans. The mole fraction of more volatile component of the solution is always greater in vapour phase

than in the solution phase. It is known as Konowaloff’s rule.Q. 18. If the density of solution is dg/cm3, molar mass of the solute is MB, what would be the relation

between molarity (M) and molality (m) of the solution ?

Ans. m = M

d – MMB/1000 or M = md

1 + mMB/1000Q. 19. What is a semi-permeable membrane ? Give examples.

Ans. A semi-permeable membrane is a membrane which allows solvent molecules to pass through itbut does not allows the solute particles to pass through it.For example : Animal bladder, Parchment paper, cellophane, etc.

Q. 20. What are ideal solutions ? Give two examples.Ans. The solutions which completely obey’s Raoult’s law are ideal solutions.

For example : (i) Benzene + Toluene, (ii) Carbon tetrachloride + Silicon tetrachloride.Q. 21. What is meant by centi-normal solution and semi-normal solution ?

Ans. A centi-normal solution contains 1/100 gm equivalents per litre of solution and a semi-normalsolution contains 1/2 gm equivalents per litre of solution.

Q. 22. Write down the expression relating elevation of boiling point of a solvent, when a solute isdissolved in it to form a dilute solution and the concentration of the solution, defining theterms.

Ans. If non-volatile solute is added to the pure solvent, its boiling point is elevated. If the boilingpoint of the solvent is T Kelvin and boiling point of the solution as Tsol , the elevation of boilingpoint ΔTb will be

ΔTb = Tsol – TThe elevation of boiling point ΔTb is related to the concentration of the solution by :

ΔTb = Kbmwhere, Kb = Molal elevation constant

m = Molality of the solution.


Q. 23. The molecular weights of sodium chloride and glucose are determined by the depression offreezing point method. Compared to their theoretical molecular weights, what will be theirobserved molecular weights when determined by the above method ? Justify your answer.

Ans. Sodium chloride is a strong electrolyte therefore, dissociates in solution to give twice the actualnumber of particles, therefore, the molecular weight determined by colligative property will behalf the actual weight. The molecular weight of the non-electrolyte glucose will be same as thetheoretical weight.

Q. 24. What is a colligative property ? Give two examples.Ans. The properties of dilute solutions which depends upon the number of solute particles

(molecules or ions) but not upon their nature are called colligative properties.For example : relative lowering of vapour pressure, elevation of boiling point.

Chapter 3. ElectrochemistryQ. 1. What happens when a nickel rod is dipped into a copper sulphate solution ? Justify your

answer.

[E°Ni+2/Ni

= – 0·25 V and E°Cu+2/Cu

= + 0·34 V]

Ans. Copper metal starts precipitating.

This is because E°Ni+2/Ni

is less than E°Cu+2/Cu

. So Ni displaces Cu+2 ions from the solution.

Ni + Cu+2(aq.) ⎯→ Ni+2

(aq.) + Cumetal metal

Q. 2. What is standard hydrogen electrode ?

Ans. Standard hydrogen electrode : It is a reference electrode against which the electrode potentialsof all electrodes are measured.

When hydrogen gas at 1 atm-pressure is adsorbed over platinum electrode dipped in 1 M HClat 25°C, it is standard hydrogen electrode and its potential is, E° = 0 volt.

Q. 3. Two metallic elements A and B have the following standard oxidation potentials :

A = 0·40 V, B = – 0·80 V. What would you expect if element A was added to an aqueous saltsolution of element B ? Give a reason for your answer.

Ans. Metal A will displace metal B from its salt solution because oxidation potential of A is higherthan that of B.

Q. 4. Arrange Ag, Cr and Hg metals in the increasing order of reducing power. Given :

E°Ag+/Ag

= + 0·80 V

E°Cr+3/cr = – 0·74 V and E°

Hg+2/Hg = + 0·79 V

Ans.Ag < Hg < Cr

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→Increasing reducing power

Q. 5. Define molar conductance of a solution. State its unit. How is it related to the specificconductance of a solution ?

Ans. Molar conductance : It is the conductance of all ions furnished by one mole of an electrolytepresent in a definite volume of the solution.

Its unit is ohm– 1 cm2 mol– 1. Molar conductance ∧m is related to specific conductance k by thefollowing equation

∧m =k × 1000

Mwhere k is ∧sp and M is molar concentration of the solution.

Q. 6. State Faraday’s First Law of Electrolysis.Ans. Faraday’s first law of electrolysis : The law states that the amount of a substance decomposed

as a result of electrolysis is directly proportional to the quantity of electricity passed.


Q. 7. The reduction potential of a metal X is – 0.76 volt while that of Y is – 2.38 volt. Which of the twometals is a stronger reducing agent ? Give a reason for your answer.

Ans. Reduction potential of metal Y is lesser than that of metal X. So metal Y is a stronger reducingagent out of the two.

Q. 8. Mention any two factors affecting the electrode potential of a metal.Ans. Factors affecting the electrode potential of a metal.

(i) Nature of metal : Extremely active metals have low electrode potential, while less activehave high electrode potential.

(ii) Temperature : The change in the temperature of the solution also changes the electrodepotential.

Q. 9. Define specific conductance or conductivity.Ans. Specific conductance is defined as conductance of electrolyte when distance between electrodes

is 1 cm and area of cross-section is 1 cm2.Q. 10. Define electro-chemical series.

Ans. The arrangement of elements in the increasing or decreasing order of their standard reductionpotentials is called electro-chemical series.

Q. 11. What are secondary cells ?Ans. Secondary cells are those cells which are rechargeable, the products can be changed back to

reactants.Q. 12. What is galvanization ?

Ans. The process of coating zinc over iron is called galvanization.Q. 13. What are conductors ? Give the unit of electric conductance.

Ans. The substance through which an electric current can pass is known as conductor. The unit ofelectric conductance is ohm–1.

Q. 14. What do you mean by insulator ?Ans. A substance which does not allows the electricity to pass through it, is known as insulator or

non-conductor.Q. 15. Name two sulphides which are used as conductors.

Ans. PbS and CdS.Q. 16. Define electrolysis.

Ans. The process of decomposition of an electrolyte by the passage of electricity through itsdissolved or molten state is known as electrolysis.

Q. 17. Define degree of dissociation of an electrolyte.Ans. The fraction of total molecules of an electrolyte which dissociates into the constituent ions in

the solution is known as degree of dissociation.Q. 18. Define Faraday constant.

Ans. It is defined as the charge present on one mole of electrons. It is equal to 96,500 coulomb.1 F = 96,500 coulomb

Q. 19. If the Faraday were to be 60230 coulomb instead of 96500 coulomb what will be the charge ofan electron ?

Ans. The charge on the electron is given by :

e =FN =

602306·023 × 1023

where F is value of 1 Faraday in coulomb and N is Avogadro’s number

e =60230

60230 × 1019

∴ Charge on an electron = 1 × 10–19 coulomb.Q. 20. What are semiconductors ?

Ans. Semiconductors are the crystals which allow only partial conduction of current through them.Q. 21. What is the effect of increasing temperature on the conductivity of a semiconductor ?

Ans. Conductivities of semiconductors are intermediate between those of conductors and insulators.Their conductivity increases with increase in temperature.


Q. 22. For the cell Cu(s) |Cu2+ (aq.) || Ag+ (aq.)|Ag(s). Write the overall reaction.Ans. Cu(s) + 2Ag+(aq.) ⎯→ 2Ag(s) + Cu2+(aq.)

Q. 23. Write the Nernst equation for the following cell at 298K Pt, Cl2 (1 atom)|2Cl– (0·1 M).

Ans. E = E° + 0·059

2 log [Cl2][Cl–] = E° +

0·0592 log

1(0·1)2

Q. 24. What is standard electrode potential ?Ans. It is the reduction electrode potential of that half cell in which the concentration of ionic species

is taken 1M and the concentration of gases are taken 1 atm., and the reduction reaction takesplace at 25°C, and electrolyte and electrode are in their standard states.

Q. 25. Define the E.M.F. of a cell.Ans. The potential difference of two electrodes of a cell when no current is drawn from the cell is

known as its E.M.F.Q. 26. Define Ohm’s law.

Ans. At constant temperature the ratio of the potential difference across the terminals of a conductorto the current flowing through it is constant.

Q. 27. Define equivalent conductance. In what units is it expressed ?Ans. It is defined as the conductance of a solution containing 1 gram equivalent of electrolyte and

the solution contained in between two electrodes which are 1 cm apart.It’s unit is expressed as ohm– 1 cm2 eq– 1.

Q. 28. What is the relationship between equivalent conductivity and molar conductivity ? Ans. Relationship between equivalent conductivity (λeq) and molar conductivity (λm) is

λeq =λm

Z

where, Z = charge carried by the ions of electrolyte.Q. 29. List out three factors which affects the electrolytic conductivity.

Ans. The three factors are :(i) The inter-ionic attraction (depends on solute - solute interaction)(ii) The solvation of ions (depends on solute-solvent interaction)(iii) The viscosity of the solvent (depends on solvent-solvent interaction)

Q. 30. What is fuel cell ?Ans. Cells that can convert chemical energy of the fuel directly into electrical energy are called fuel

cells.Q. 31. What is over voltage ?

Ans. The additional voltage required for electrolysis is called over voltage.Q. 32. What is the composition of salt bridge ?

Ans. Salt bridge is made from the mixture of KNO3 or NaNO3 or NH4NO3 and agar-agar gel.Q. 33. What are the functions of salt-bridge ?

Ans. (i) It allows the flow of current by completing the circuit.(ii) It maintains electrical neutrality.

Q. 34. Electrolysis of KBr (aq.) gives Br2 at anode but KF (aq.) does not give F2 at anode. Why ?Ans. Br– ions have higher oxidation electrode potential than water, hence KBr gets oxidized to give

Br2 while the oxidation electrode potential of F– ions is lower than water hence KF would notoxidize in comparison of water.

Q. 35. Why an aqueous solution of NaCl on electrolysis gives H2 gas at cathode and not theNa metal ?

Ans. Because the standard electrode potential of hydrogen is higher than that of Na, therefore, watergets reduced and gives H2 gas at cathode.

Q. 36. State and explain Kohlrausch’s law.Ans. “At infinite dilution when the electrolyte is completely dissociated each ion makes a definite

contribution towards the molar conductivity of electrolyte irrespective of the nature of theother ion with which it is associated.”


If λ+∞ and λ–

∞ are molar conductivities of cation and anion at infinite dilution, the molarconductivity of the electrolyte at infinite dilution λ∞

m is given by :λm

∞ = n1λ+∞ + n2λ–

∞

where n1 and n2 are the number of +ve and –ve ions furnished by one formula unit of theelectrolyte.

Q. 37. What happens when a Zn rod is dipped in CuSO4 solution ? Give the reason for thisobservations.

Ans. The Cu from copper sulphate starts depositing on Zn rod and Zn rod starts dissolving. It is dueto the following redox reaction.

Zn(s) + Cu+2(aq.) ⎯→ Zn+2(aq.) + Cu(s) Zinc rod From CuSO4Because the standard oxidation potential of Zn is higher than that of Cu, therefore, zinc getsoxidized and Cu ion gets reduced.

Q. 38. What is rust ?Ans. Chemically rust is hydrated iron(III) oxide, Fe2O3. xH2O. It is generally caused by moisture,

CO2, O2 of air. Rust is a non-sticking brown-coloured material which can be easily removed byscratching.

Q. 39. Define galvanization with example.Ans. Zinc metal is generally used for protecting iron from corrosion and the process is called

galvanization. Galvanized iron sheets maintain their shine due to the formation of a thinprotective layer of basic zinc carbonate, (ZnCO3), Zn(OH)2 due to the reaction between zinc,oxygen, CO2 and moisture in air.

Q. 40. When 96500 coulomb of electricity is passed through a solution of CuSO4, 3·18 gm of Cu isdeposited on cathode. Name the law or principle to which this observation confirms.

Ans. Faraday’s first law of electrolysis.Q. 41. Define the term transport number.

Ans. The fraction of current carried out by an ion is known as its transport number. Hence,

Transport number of anion (na) or cation (nc) = Current carried by cation or anion

Total current carriedThe sum of the transport number of cation and transport number of anion is equal to one i.e.na + nc = 1.

Q. 42. Represent electrochemical cell corresponding to following reactions :(i) H2(g) + Cu+2

(aq.) ⎯→ 2H+ + Cu(ii) 2Cr(s) + 3Cu+2

(aq.) ⎯→ 2Cr+3(aq.) + 3Cu(s)

Ans. Electrochemical cell of these reactions can be represented as :(i) Pt–

(s), H2(g) | H+(aq.) || Cu+2

(aq.) | Cu(s)(ii) Cr(s) | Cr+3

(aq.) || Cu+2(aq.) | Cu(s)

Q. 43. The standard reduction potentials for Cu and Zn electrodes are + 0·350 V and –0·763 Vrespectively for the cell : Zn|Zn+2 (1·0M) || Cu+2 (1·0 M) | Cu(i) Write the cell reaction.(ii) Calculate the emf of the cell.(iii) Is the cell reaction spontaneous ? Why ?

Ans. (i) Cell reaction will beZn(s) + Cu+2

(aq.) ⎯→ Zn+2(aq.) + Cu(s)

(ii) E.M.F. of the cellECell = ERight – ELeft

= + 0·350 – (– 0·763) = +1·113 V(iii) As the value of Ecell is +ve, the cell reaction will be spontaneous.

Q. 44. What are the concentration cells ?Ans. These are the Galvanic cells in which the electrodes are made up of same metal but they have

different concentration of electrolytes. The difference in the concentration of electrolytes createspotential difference across the two electrodes. The electrode with smaller concentration ofelectrolyte acts as anode or –ve terminal, whereas the electrode with larger concentration ofelectrolyte acts as cathode or +ve terminal. For example,

Cu(s)|Cu+2 (0·01 M) || Cu+2 (0·1 M) | Cu


If C2 is the concentration of electrolyte in cathodic half and C1 is the concentration of electrolytein anodic half cell, then the net cell voltage at 298 K is given by—

ECell = 0·059

n log

C2

C1

Q. 45. Explain the terms :(i) Conductivity (ii) Molar conductivity (iii) Equivalent conductivity

Ans. (i) Conductivity : The term conductance is used as the ease with which the current flowsthrough a conductor. Therefore, the conductivity is defined as the reciprocal of resistancei.e., C = 1/R where C represents the conductivity and R represents the resistance. Theunit of conductivity is ohm–1 or mhos.The electrical conductance or conductivity is also expressed in Siemens, S

1 S = 1 Ω–1

(ii) Molar conductivity : Molar conductivity is defined as the conductance of the solutioncontaining one gram mole of the electrolyte such that the entire solution is placedbetween two parallel electrodes one centimetre apart. It is denoted by Λm. Molar

conductivity is related to conductivity (K) by the relation Λm = K × 1000

M where M is the

molarity of the solution.The unit of molar conductivity is ohm–1 cm2 mol–1 or Ω–1 cm2 mol–1 or S cm2 mol–1.

(iii) Equivalent Conductivity : Equivalent conductivity is defined as the conductance of asolution, containing 1 gm equivalent of electrolyte such that the entire solution is placedbetween two electrodes one centimetre apart. It is denoted by the symbol Λeq.

Λeq = Volume × Λsp

= (cm3 × equiv–1) × (ohm–1 × cm–1)= ohm–1 cm2 equiv–1

= S cm2 equiv–1

Q. 46. Explain the factors on which the conductance of an electrolytic solution depends.Ans. The factors on which electrolytic conductance depends are :

(i) Nature of electrolyte : All the electrolytes do not ionize in their aqueous solution to thesame extent, they can be divided into two categories depending upon their extent ofionization.

(a) Strong electrolytes : These are completely ionised in aqueous solution and have veryhigh values of Λm.

(b) Weak electrolytes : These are not completely ionised in aqueous solution and possess lowvalue of Λm.

(ii) Concentration of the solution : Molar conductivity of electrolytes generally increases withdilution. In case of strong electrolytes molar conductivity approaches a certain limitingvalue when concentration approaches zero and this position is known as molarconductivity at zero concentration or at infinite dilution.Whereas in case of weak electrolytes, the value of Λm at infinite dilution cannot beobtained by extrapolating the graph to the zero concentration.

(iii) Temperature : Conductivity is independent of temperature.Q. 47. Explain the influence of solvent on conductivity.

Ans. The properties of an electrolyte solution depends upon the dielectric constant of the mediumand on the size and space distribution of ions. In water there is not much detectable differencesin the ionization of KCl, KI or KBr. But in the case of liquid ammonia there is a markeddifference of ionization of each electrolyte. In the solvents other than water the conductanceviscosity product of a given ions are approximately constant.

Q. 48. Answer the following questions on hydrogen electrode :(i) Of which metal is the electrode made ?(ii) Why is the electrode coated with a layer of Pt black ?(iii) What is the concentration of HCl and pressure of the hydrogen gas taken in NHE ?(iv) What reaction takes place at this electrode ?


Ans. (i) Hydrogen electrode is made of Pt metal.(ii) Electrode is coated with a layer of Pt black which absorbs H2 and thus, helps to establish

a rapid equilibrium between H2 and H+ ions.(iii) The concentration of HCl is 1 M and pressure of hydrogen gas is 1 atm. in NHE.(iv) When this electrode acts as the anode i.e., oxidation takes place, the following reaction

occurs :H2 (g) ⎯→ 2H+ (aq.) + 2e –

When this electrode acts as the cathode i.e., reduction takes place, the following reaction occurs :2H+ (aq) + 2e– ⎯→ H2 (g)

The pH of the solution using the NHE electrode will be Ecell = 0·059.

Chapter 4. Chemical KineticsQ. 1. What is the order of reaction whose rate constant has the same unit as the rate of reaction ?Ans. Zero order reactionQ. 2. Write the mathematical expression relating the variation of rate constant of a reaction with

temperature.Ans. Mathematical expression for Arrhenius equation :

k = Ae–Ea/RT

where k is rate constant of the given reactionA is another constant called frequency factor.Ea is activation energyR is gas constant andT is temperature in Kelvin.

Q. 3. How can you graphically find the activation energy of the reaction from the above expression ?Ans. Finding activation energy graphically :

Log form of Arrhenius equation is

loge k = loge A – Ea

RT

or log10 k = log10 A – Ea

2·303 RTIt is a straight line equation (y = mx + c) when different values ofrate constant k are calculated at different temperatures for a given

reaction and if a graph is plotted between log10 k against 1T then

slope of the graph line should be equal to – Ea

2·303 R

Using dydx

= – Ea

2·303 R ,

value of Ea can be calculatedQ. 4. Give one example (equation) of a homogeneously catalysed reaction and name the catalyst.Ans. Hydrolysis of ethyl acetate to give ethanol and acetic acid is catalysed by dil. H2SO4.

ethyl acetateCH3COO·C2H5 +

waterHOH

Catalystdil. H2SO4⎯⎯⎯→

acetic acidCH3COOH +

ethanolC2H5OH

Q. 5. What is meant by promoter ? Give an example.Ans. Promoters : These are substances which when added to substrate in very small amount along

with the catalyst, enhances the effectiveness of the catalyst. These are not catalysts themselves.Example : In Haber’s process iron powder acts as catalyst while Al2O3 is a promoter.

Q. 6. Define molecularity of a reaction. Give one difference between the order of a reaction and itsmolecularity.

Ans. The minimum number of reacting particles (molecules, atoms, ions) that comes together orcollide in the rate determining step, is called molecularity of the reaction.Molecularity is always a whole number while order of reaction can be zero, whole number oreven fractional.


Q. 7. Mention any two factors that influence the rate of a chemical reaction.Ans. Rate of a chemical reaction can be affected by :

(i) Concentration (ii) Temperature.Q. 8. Define specific reaction rate or rate constant.Ans. Specific reaction rate is the rate of reaction when the molal concentration of each of the

reactants is unity.Q. 9. Define threshold energy of a reaction.Ans. Threshold energy is the minimum energy which must be possessed by reacting molecules in

order to undergo effective collisions which leads to formation of product molecules.Q. 10. Define the order of reaction.

Ans. Order of reaction is the sum of the exponents of the molar concentration of the reactants in therate equation for a general reaction

xA + yB ⎯→ ProductIf the rate law is, Rate = k [A]x [B]y

then, Order = x + yQ. 11. Name the three factors that usually modify the rate of a reaction.

Ans. (i) Increase in the concentration of reactants.(ii) Increase in the temperature.(iii) Use of catalysts.

Q. 12. Give the units of a zero-order rate constant.Ans. The unit of zero order rate constant is mol l–1sec–1.

Q. 13. Give two examples of zero order reaction.Ans. (i) Photochemical reaction of H2 and Cl2 over water surface.

(ii) Bromination of acetone.Q. 14. For a reaction, the rate constant k is independent of the initial concentration of the reactant.

What is the order of reaction ?Ans. The order of the reaction would be first because for the first order reaction t1/2 ∝ (a)°.

Q. 15. What is the unit in which the rate constant of a first order is expressed ?Ans. Rate law for first order is, rate = K [A]

Hence, the rate constant, k =Rate[A]

Units of K =Units of rateUnits of [A]

=Mol L–1 sec–1

Mol L–1 = sec–1

Hence, the units of k is time–1, which may sec–1, min–1 or hr–1.Q. 16. What is an effective or fruitful collision ?

Ans. A collision in which colliding molecules possess energy equal to or more than threshold energyand the collision has proper orientation of molecules is called an effective collision. It leads tothe formation of products.

Q. 17. Define pseudo-unimolecular reactions.Ans. Any first order reaction whose molecularity is higher than 1 is known as pseudo-unimolecular

reaction.(i) Hydrolysis of ester in acidic medium

CH3COOC2H5 + H2O H+

⎯→CH3COOH + C2H5OH(order = 1, molecularity = 2)

(ii) Hydrolysis of cane sugar.

C12H22O11(aq) + H2O H+

⎯→ C6H12O6 + C6H12O6(aq)

(order = 1, molecularity = 2)Q. 18. A single step reaction involves the simultaneous collision of three particles. What is its

molecularity ?Ans. Molecularity is three.


Q. 19. Give one example of the reaction where order and molecularity are same.Ans. Both order as well as molecularity is one for the decomposition of hydrogen peroxide.

Q. 20. Define the half-life period of a reaction.

Ans. Half-life period is time in which the concentration of a reactant reduces to half of its originalvalue.

Q. 21. What is the relation between the rate constant and half-life period for a first order reaction ?

Ans. t1/2 = 0·693

k where, t1/2 = half-life period, k = rate constant.

Q. 22. A reaction is found to be zero order reaction, will its molecularity be zero ?Ans. No, because molecularity cannot be zero.

Q. 23. What will be the shape of the curve if concentration of the reactant is plotted against time for azero order reaction ?

Ans. A straight line with a ‘–ve’ (Negative) slope is obtained.Q. 24. A reaction between A and B is of second order. Write three different rate law expressions

which might possible apply to the reaction.Ans. (i) Rate = k [A] [B]

(ii) Rate = k [A]2

(iii) Rate = k [B]2

Q. 25. Draw a graph which can be used to calculate the activation energy of a reaction.Ans.

1/T

Slope

Value of log A

log

k

Fig. A plot between log k vs 1/ TQ. 26. Give two important characteristics of a catalyst.

Ans. Two important characteristics of a catalyst are :(i) Remains unchanged at the end of the reaction(ii) Selectivity.

Q. 27. What is promoter ? How does it works ?Ans. Any substance which enhances the activity of a catalyst is called promoter. It encourages the

formation of lattice defects.Q. 28. What is a catalytic poison ? How does it works ?

Ans. It reduces the efficiency of the catalyst by blocking the active sites of catalyst.Q. 29. What are shape selective catalysts ?

Ans. Certain reactions are made to proceed in a specific manner by using certain zeolites whosecages and tunnels have sizes compared to the size of the reactant and product molecule.

Q. 30. Give the example of a shape selective catalyst.Ans. Zeolites are shape selective catalysts. Example. ZSM-5.

Q. 31. How are alcohols converted to gasoline ?Ans. Alcohols are converted to gasoline by using ZSM-5.

Q. 32. What is sorption ?Ans. A process in which both adsorption and absorption takes place simultaneously is called

sorption.Q. 33. What is adsorption ?

Ans. The existence of higher concentration of any molecular species at the surface than in the bulk ofa solid (or liquid) is known as adsorption.


Q. 34. What is desorption ?Ans. The process of removing an adsorbed substance from a surface on which it is adsorbed is called

desorption.Q. 35. What is chemisorption ?

Ans. Adsorption in which adsorbent and adsorbate are held by chemical bonds, and these are highlyspecific.

Q. 36. Define enthalpy of adsorption.Ans. The enthalpy change when one mole of the adsorbate is adsorbed on the surface of the

adsorbent is called enthalpy of adsorption.Q. 37. Write general formula of zeolites.

Ans. Mx/n [(AlO2)x (SiO2)y]. mH2O.Q. 38. Give any four examples of heterogeneous catalytic reactions.

Ans. Heterogeneous catalytic reactions :

(i) N2(g) + 3H2(g)Fe

⎯⎯⎯→ 2NH3(g) Haber’s process

(ii) 4NH3(g) + 5O2(g)Pt

⎯⎯⎯→ 4NO(g) + 6H2O(g) Ostwald’s process

(iii) 2SO2(g) + O2(g)V2O5⎯⎯⎯→ 2SO3(g) Contact process

(iv) CO(g) + 2 H2(g)ZnO

⎯⎯⎯→Cr2O3

CH3OH(g) Synthesis of methanol

Q. 39. Explain two aspects for choosing a heterogeneous catalyst.Ans. The main two aspects to be considered are :

(i) Activity : The ability of a catalyst to accelerate the rate of chemical reaction is calledactivity. For example, Pt or Ni or Pd accelerates the hydrogenation reactions.

(ii) Selectivity : The ability of a catalyst to direct a reaction to yield a particular product iscalled its selectivity. For example hydrogenation of acetylene with Pt gives ethane, whilewith Lindlar’s catalyst it gives alkene.

Q. 40. Define homogeneous and heterogeneous reactions.Ans. Homogeneous reaction : In this reaction takes place in one phase only. For example,

2CO(g) + O2(g) → 2CO2(g)Heterogeneous reaction : In this reaction involves two or more phases. For example,

Zn(s) + H2O(l) → ZnO(s) + H2(g) ↑Q. 41. What is the role of chemisorption in a catalytic reaction ?

Ans. The process of chemisorption on a catalyst can alter the structure and the activity of a reactantin the same way as the formation of an activated complex by weakening some bonds andallowing the formation of others, there by enhancing the rate of the reaction.

Q. 42. If at a given temperature a catalyst is added to the reactants, what will happen to the nature ofthe equilibrium constant ?

Ans. Equilibrium constant will remain unchanged, as the catalyst only accelerates or retards theapproach of equilibrium, with the same ratio of the velocities of forward and backwardreaction.

Chapter 5. Surface ChemistryQ. 1. What is physical adsorption ?Ans. If the adsorbate is held on an adsorbent surface by weak van der Waals forces, the adsorption is

called physical adsorption.Q. 2. What type of forces are responsible for the occurrence of physiosorption ?Ans. van der Waals forces.Q. 3. What is meant by chemical adsorption ?Ans. If the adsorbate is held on the surface of the adsorbent as a result of chemical reaction forming

surface compound it is called chemical adsorption.


Q. 4. What is sorption ?Ans. Sorption is the process in which adsorption and absorption takes place simultaneously e.g.

dying of cotton fibres by azo dyes.Q. 5. Define desorption ?Ans. The process of removal of an adsorbed substance from a surface on which it is adsorbed is

called desorption.Q. 6. What is occlusion ?Ans. The adsorption of gases on the surface of metals is called occlusion.Q. 7. What is the effect of temperature on chemisorption ?Ans. Chemisorption initially increases then decreases with rise in temperature. The initial increase is

due to the fact that heat supplied acts as activation energy. The decrease afterward is due tothe exothemic nature of adsorption equilibrium.

Q. 8. What is common in aquasols and solid aerosols ? How they differ ?Ans. Aquasols and solid aerosol both have solid as the dispersed phase. They differ in dispersion

medium. Aquasol have water as the dispersed medium while aerosols have gas as thedispersion medium.

Q. 9. What is the role of desorption in the process of catalysis ?Ans. The reaction products formed on the catalyst surface gets detached from the surface as a result

of desorption thereby making the surface available again for more reaction.Q. 10. What is the role of diffusion in heterogeneous catalysis ?

Ans. The gaseous molecule diffuses on to the surface of the solid catalyst and gets absorbed. Afterthe required chemical changes the product diffuses away from the surface of the catalystleaving the surface free for more reactant molecules to get adsorbed and undergo a reaction.

Q. 11. What is collodion ?Ans. It is a 4% solution of nitrocellulose in a mixture of alcohol and ether.

Q. 12. What is the principle of dialysis ?Ans. Dialysis is based on the principle that ions can pass through semi-permeable membrane

whereas colloidal particle cannot pass through it.Q. 13. What happens when dialysis is prolonged ?

Ans. Due to excessive dialysis, traces of electrolyte which stabilises the colloids are removedcompletely, making the colloid unstable. As a result, coagulation takes place.

Q. 14. What happens when an electric field is applied to a colloidal dispersion ?Ans. The colloidal particle moves towards the oppositely charged electrode and gets neutralised and

coagulated there.Q. 15. What is electrodialysis ?

Ans. It is process by which colloidal solution containing ionic impurities are purified. The colloidalsolution containing ionic impurities is placed in bag of parchment paper in distilled water in anelectric field. The ions comes out through parchment paper and the salt is purified.

Q. 16. What do you mean by (i) Ultrafiltration (ii) Tyndall effect.Ans. (i) Ultrafiltration : In this process colloidal solution are purified by carrying out filtration

through special type of graded filters called ultra-filters. Filter paper allows the passage ofelectrolyte but does not allows the passage of colloidal particles.(ii) Tyndall effect : The scattering of light by colloidal particles is known as Tyndall effect.

Q. 17. What causes Brownian movement in a colloidal solution ?Ans. Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion

medium causes Brownian motion. This stabilizes the sols.Q. 18. What is the main cause of charge on a colloidal solution ?

Ans. The charge on the colloidal particles is due to adsorption of common ions of the electrolyte onthe surface of the colloidal particles e.g. Fe3+ from FeCl3 on the surface of Fe(OH)3 particles.

Q. 19. What causes electrophoresis ?Ans. Electrophoresis is due to charge on colloidal particles the charged particle moves towards one

of the electrode in electric field.Q. 20. What is ‘coagulation’ process ?

Ans. The process of settling of colloidal particles is called coagulation.


Q. 21. What is an emulsion ?Ans. Emulsion is a colloidal solution in which both the dispersed phase and dispersion medium are

liquids e.g. milk, cod-liver oil etc.Q. 22. How is adsorption of gas related to its critical temperature ?

Ans. Higher the critical temperature of a gas greater is the ease of liquification i.e. greater are thevan der Waals forces of attraction and hence greater is adsorption.

Q. 23. Give the expression of Freundlich isotherm.

Ans.xm

= KP1/n or log xm

= log K + 1n

log P

where m is the mass of adsorbent, x is the mass of adsorbate, P is the pressure of the gas and nis an integer.

Q. 24. Indicate a chemical reaction involving a homogeneous catalyst.

Ans. 2SO2 (g) + O2(g) NO(g)

⎯⎯⎯→ 2SO3(g).Q. 25. How does a catalyst works ?

Ans. Catalyst provides an alternative path involving lower activation energy for the reactants.Q. 26. Define colloidal solution.

Ans. A colloidal solution is a state in which the particle size lies between 1 nm and 1000 nm. Itappears to be homogeneous but actually it is heterogeneous.

Q. 27. Give one example of sol and gel.Ans. Sol : Paint, cell fluids.

Gel : Butter, cheese.Q. 28. Define peptization.

Ans. The process of converting a precipitate into colloidal sol by shaking it with dispersion mediumin the presence of small amount of electrolyte is called peptization.

Q. 29. Write two differences between sols and emulsions.Ans. (i) Sols are dispersion of solids in liquids while emulsion are dispersion of liquid in liquid.

(ii) Sols are quite stable whereas emulsions are less stable.Q. 30. How do emulsifying agents stabilise the emulsion ?

Ans. The emulsifying agent forms a interfacial layer between suspended particle and the dispersionmedium thereby stabilising the emulsion.

Q. 31. Define emulsification.Ans. The process of making an emulsion from an oil is termed as emulsification.

Q. 32. Give one example each of ‘oil in water’ and ‘water in oil’ emulsion.Ans. Oil in water emulsion—milk, varnishing cream.

Water in oil emulsion — Butter, cold cream.Q. 33. Define ultra-filtration.

Ans. In this process colloidal solutions are purified by carrying out filtration through special types ofgraded filter called ultra-filters. Filter paper allows the passsage of electrolyte but does notallows the passage of colloidal particles.

Q. 34. Give one example of positively charged sol and one of negatively charged sols.Ans. Fe(OH)3 is positively charged sol whereas As2S3 is a negatively charged sol.

Q. 35. How can a lyophillic sol be coagulated ?Ans. This can be done (i) by adding an electrolyte (ii) by adding suitable solvent.

Q. 36. Give one example each of lyophobic sol and lyophilic sol.Ans. Lyophobic sol : Gold sol, As2S3 sol

Lyophilic sol : Sol of starch, sol of gum.Q. 37. Based on the type of dispersed phase, what type of colloid is micelle ?

Ans. Associated colloids.Q. 38. Name the temperature above which the formation of micelle takes place.

Ans. Kraft temperature.


Q. 39. Describe some features of zeolites.Ans. Features of zeolites are :

(i) Zeolites are hydrated alumino silicates which have a three-dimensional network structurecontaining water molecules in their pores.(ii) To use them as catalyst, they are heated so that water of hydration present in the pores islost and the pores become vacant.

Q. 40. (i) How does BF3 acts as a catalyst in industrial process ?(ii) Give an example of shape-selective catalyst.

Ans. (i) It is because BF3 is an electron deficient compound and helps to generate electrophile.(ii) ZSM-5 (Zeolite sieve with molecular porosity (5).

Chapter 6. General Principles and Processes of Isolation of ElementsQ. 1. What is gangue ?Ans. The earthy and silicious impurities associated with the ores are called gangue.Q. 2. What is the function of collectors in the froth floatation process for concentration of ores ?Ans. Collectors (e.g. pine oil, xanthates etc.) enhances non-wettability of the ore particles.Q. 3. What is the significance of leaching in the extraction of aluminium ?Ans. Leaching is significant as it helps in removing the impurities like SiO2, Fe2O3, TiO2 etc. from the

bauxite ore.Q. 4. What is meant by beneficiation process ?Ans. The process of removal of unwanted earthy and silicious impurities (gangue) from the ore is

known as beneficiation process.Q. 5. What is liquation ?Ans. Liquation is a method of refining of metals and is used when the impurities are not miscible

with the metal and the melting temperature of the metal is lower than that of impurities.Q. 6. What name is given to carbon reduction process for extracting the metals ?Ans. Smelting.Q. 7. What is the role of cryolite in the metallurgy of aluminium ?Ans. The role of cryolite is as follows :

(i) It increases the electrical conductivity of the mixture.(ii) It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Q. 8. What is the role of flux in metallurgical process ?Ans. Flux is used for making the molten mass more conducting.Q. 9. What are fluxes ? How are they useful ?Ans. Flux is a substance that combines with gangue which may still be present in the roasted or

calcinated ore to form an easily fusible substance called the slag.Q. 10. What is slag ?

Ans. A slag is an easily fusible material which is formed when gangue still present in the roasted orthe calcinated ore combines with the flux. For example in the metallurgy of iron, CaO (flux)combines with silica gangue to form easily fusible calcium silicate (CaSiO3) slag.

CaCO3 ⎯→ CaO + CO2

CaO + SiO2 ⎯→ CaSiO3 (slag)Q. 11. What is the role of depressant in the froth floatation process ?

Ans. In froth floatation process, the role of the depressant is to prevent certain type of particles fromforming the froth with the air bubbles. For example NaCN is used as a depressant to separatelead sulphide ore (PbS) from zinc sulphide (ZnS), NaCN form a zinc complex Na2 [Zn(CN)4] onthe surface of ZnS thereby preventing it from the formation of froth.

Q. 12. What should be the considerations during the extraction of metals by electrochemicalmethod ?

Ans. Generally two things are considered so that proper precautions can be taken :(i) Reactivity of metal produced.(ii) Suitablity of electrodes.


Q. 13. What is meant by the term chromatography ?Ans. The term chromatography was originally derived from Greek word ‘chroma’ colour and

graphy meaning ‘writing’ because it was first used for the separation of coloured substances(plant pigments) into individual components. Presently the term chromatography has lost itsoriginal meaning and the method is widely used for separation, purification andcharacterisation of the components of a mixture whether coloured or colourless.

Q. 14. What criterion is followed for selection of the stationary phase in chromatography ?Ans. The stationary phase is selected in such a way that the impurities are more strongly adsorbed

or are more soluble in the stationary phase than elements to be purified. Under theseconditions, when the mixture is extracted impurities will be retained by the stationary phaseand the pure component is easily eluted.

Q. 15. What do you mean by activity or selectivity of a catalyst ?Ans. Activity of catalyst : The ability of a catalyst to increase the rate of reaction is called its activity

For example a mixture of hydrogen and oxygen does not reacts at all, however in the presenceof platinum as a catalyst the mixture reacts with explosive voilence to form water.

2H2 + O2 Pt

⎯⎯→ 2H2OSelectivity of catalyst : It is the ability of a catalyst to direct a reaction to yield a particularproduct for example both dehydrogenation and dehydration of ethanol are possible. But inpresence of suitable catalyst only one reaction is made to occur.

CH3CH2OHCu

⎯⎯→573K

CH3CHO + H2O (dehydrogenation)

CH3CH2OHAl2O3

⎯⎯⎯⎯→ CH2 = CH2 + H2O (dehydration)Q. 16. What is shape-selective catalysis ?

Ans. Shape-selective catalysis is a chemical reaction in which the rate depends on the pore size of thecatalyst, and also on the shape and size of the reactant and product molecule. A shape selectivecatalyst has a variety of active sites of different shape and size. Zeolites acts as a shape selectivecatalyst.

Q. 17. What happens when gelatin is added to gold sol ?Ans. Gold sol which is lyophobic starts behaving like a lyophilic colloid when gelatin is added to it.

Q. 18. What are the factors, which influence the adsorption of a gas on a solid ?Ans. The extent of adsorption of a gas on a solid surface depends on following factors :

(i) The nature of gas.(ii) Surface area of adsorbent.(iii) Pressure .(iv) Temperature.(v) Activation of adsorbent.

Q. 19. What role does adsorption play in heterogeneous catalysis ?Ans. In heterogeneous catalysis generally the reactants are gaseous whereas the catalyst is a solid.

The reactant molecules are adsorbed on the surface of the solid catalyst by physical adsorptionor chemical adsorption. As a result the concentration of the reactant molecules on the surfaceincreases and hence the rate of reaction increases. Alternatively one of the reactant moleculeundergoes fragmentation on the surface of the solid catalyst producing active species whichreacts faster. The product molecules in either case have no affinity for the solid catalyst and aredesorbed making the surface free for fresh adsorption. This theory is called theory ofadsorption.

Q. 20. An ore sample of galena (PbS) is contaminated with zinc blende (ZnS). Name one chemicalwhich an be used to concentrate galena selectively by froth floatation method.

Ans. NaCN (sodium cyanide) used as a depressant.Q. 21. When is electrolytic reduction applied for getting a metal ?

Ans. When chemical reaction is not feasible e.g. for the oxide of highly reactive metal, alkali, metals,alkaline earth metals and other elements like Al, Zn, etc. electrolytic reduction is applied forgetting a metal.

Q. 22. Name the method that is used for refining of nickel.Ans. Mond process (vapour phase refining).


Q. 23. Name the method used for refining of zirconium.Ans. Van Arkel method.

Q. 24. Which reducing agent is employed to get copper from the leached low grade copper ore ?Ans. Scrap ion, Cu2+ + 2Fe ⎯⎯⎯→ Cu + 2Fe2+

or H2 gas Cu2+ + H2 ⎯⎯⎯→ Cu + 2H+

Q. 25. Indicate the temperature at which carbon can be used as a reducing agent for FeO.Ans. Above 1123 K carbon can reduce FeO to Fe.

Q. 26. Name one of (a) acidic flux (b) basic flux.Ans. (a) Silica (b) Lime

Q. 27. Name the principle ore of aluminum. Explain the significance of leaching in extraction ofaluminum.

Ans. Bauxite (Al2O3. 2H2O).Leaching is significant as it helps in removing the impurities like SiO2, Fe2O3, TiO2 etc. from thebauxite ore and thus alumina is obtained.

Q. 28. How metals used as semiconductors refined ? What is the principle of the method used ?Ans. Semiconductor metal is produced by zone-refining method which is based on the principle that

the impurities are more soluble in melt than in the solid state of metal.Q. 29. Give two requirements for vapour phase refining.

Ans. (i) The metal should form volatile compound with a suitable reagent.(ii) The volatile compound should be easily decomposable so that the recovery of metal is easy.

Q. 30. Outline the principles behind the refining of metal by following methods :(i) Zone refining method.(ii) Chromatographic method.

Ans. (i) Zone refining method is based on the principle that the impurities are more soluble in themelt than in the solid state of the metal.(ii) Chromatographic method is based on the principle that different components of a mixtureare adsorbed to different extent on an adsorbent. The adsorbed components are removed byusing suitable solvents.

Q. 31. In the extraction of zinc from zinc blende :(i) Give an equation to show how zinc oxide is converted to zinc.(ii) How is impure zinc finally electro-refined ?

Ans. (i) ZnOzinc oxide

+ Ccoke

heated

⎯⎯⎯→ Znzinc

+ CO ↑

(ii) Electro refining of zinc :Electrolyte used : ZnSO4 solution containing little H2SO4

Anode : Block of impure zincCathode : Thin sheet of pure zinc.On passing current pure zinc is obtained at cathode.Anode reaction : Zn

impure – 2e– ⎯→ Zn

2+(aq.)

Cathode reaction : Zn2+(aq.) + 2e– ⎯→ Zn

pure metalQ. 32. What is the significance of leaching in the extraction of aluminium ?

Ans. Leaching of bauxite with NaOH is significant as it helps in removing the impurities like SiO2,Fe2O3, NaOH removes SiO2 as sodium silicate.

Q. 33. What is the role of graphite rod in the electrometallurgy of aluminium ?Ans. Graphite acts as a reducing agent and reduces Al2O3 to Al.

2Al2O3 + 3C ⎯⎯→ 4Al + 3CO2.Q. 34. State the role of silica in the metallurgy of copper.

Ans. The role of silica is to remove FeO present as an impurity in the form of a slag, FeSiO3.FeO + SiO2 ⎯⎯→ FeSiO3GangueFlux Slag


Q. 35. Explain the electrolytic refining for the purification of copper.Ans. Impure copper can be purified by electrolytic refining. The anode is of impure copper while

cathode is made up of pure copper. Electrolyte is the solution of metal salt on passingelectricity the metal from the anode goes into solution as Cu2+ ions while pure metal getsdeposited at the cathode. The impurities settle down below anode as anode mud. The reactionsinvolved are :Anode :

ImpureCu ⎯⎯→ Cu2+ + 2e– + impurities

Cathode : Cu2+ + 2e– ⎯⎯→PureCu

Chapter 7. p-Block ElementsQ. 1. Which gas is filled in sodium vapour lamps ?Ans. Neon gas.Q. 2. Obtaining light of different colours in neon signs is due to presence of which gas.Ans. Argon mixed with neon gas.Q. 3. Give the outer electronic configuration of group sixteen element.Ans. ns2np4, the outer p-electrons of the outermost shell are arranged as px2py1pz1.Q. 4. Name two compounds in which oxygen has oxidation state different from – 2. Give oxidation

states also.Ans. (i) OF2, oxidation state of O = + 2

(ii) H2O2, oxidation state of O = – 1.Q. 5. Metallic character increases or decreases as we move down the group 16. Why ?Ans. As we move down the group, the ionisation energy decreases and therefore, the metallic

character increases.Q. 6. Which is the most stable allotropic form of sulphur.Ans. Orthorhombic S8.Q. 7. What are the oxidation states of S in the following compounds ?

(a) PbS (b) SO2 (c) SF6 (d) Na2S2O3 (e) H2SO3.Ans. (a) – 2 (b) + 4 (c) + 6 (d) + 2 (e) + 4Q. 8. Define catenation.Ans. The property of element due to which it forms covalent bonds with other atoms of same

element to form chain of atoms is called catenation.Q. 9. Why H2O is a liquid and H2S is a gas ?Ans. There is H-bonding in water molecule due to high electronegativity and small size of oxygen

atom but there is no H-bonding in H2S.Q. 10. What is the state of hybridisation of sulphur in :

(i) SF6 (ii) SF4 (iii) SO3 ?Ans. (i) sp3d2 (ii) sp3d (iii) sp2.

Q. 11. Which one among the following is the strongest oxidising agent :

ClO–4, BrO–

4, IO–4

Ans. The strongest oxidising agent among the following compounds is BrO–4.

Q. 12. Among the hydrides of halogens, HX :

(i) Which is most stable ?

(ii) Which is most acidic ?

(iii) Which has low boiling point ?

Ans. (i) HF (ii) HI (iii) HCl.

Q. 13. Write two uses of ClO2.

Ans. Chlorine dioxide (ClO2) is used as :

(i) Oxidising agent (ii) Bleaching agent.


Q. 14. With which neutral molecule is ClO– isoelectronic ? Is that molecule a Lewis base ?Ans. ClO– species has 26 electrons (17 (Cl) + 8(O) + 1 e– (charge). The neutral molecule which is

isoelectronic with it is ClF (17 + 9 = 26 e–).ClF can act as a Lewis base due to presence of lone pair of electrons.

Q. 15. Which one of the following does not exist why ?(a) XeOF4 (b) NeF2

(c) XeF2 (d) XeF6

Ans. (b) NeF2 cannot exist because Neon (Z = 10) do not contain vacant d-orbital in its valence shelland hence cannot extend its valency.

Q. 16. Write the formula of noble gas compound that is iso-structural with :

(a) ICl4– (b) IBr2

– (c) BrO3–

Ans. (a) XeF4 (b) XeF2 (c) XeO3.

Q. 17. Does the hydrolysis of XeF6 leads to a redox reaction ?

Ans. No, the products of hydrolysis are XeOF4 and XeO2F2 where oxidation states of all the elementsremains the same as it was in the reacting state.

Q. 18. Give the outer electronic configuration of group 17 elements.

Ans. ns2 np5.

Q. 19. Give the outer electronic configuration of noble gases.

Ans. ns2 np6.

Q. 20. Give an example of oxidation of one halide by another halogen.

Ans. Cl2

(0) + KI

(– 1) ⎯⎯→ KCl

(– 1) + I

(0)

Q. 21. Give two elements of halogen family and two of chalcogen family.

Ans. Fluorine and chlorine are from halogen family and oxygen and sulphur are from chalcogenfamily.

Q. 22. Which property is involved in the separation of noble gases by Dewar’s method ?

Ans. Adsorption.

Q. 23. Which elements are mostly involved in the compounds of Xenon ?

Ans. Oxygen and fluorine.

Q. 24. Oxygen generally exhibits an oxidation state of – 2 only whereas other members of its familyshow oxidation states of + 2, + 4 and + 6. Why ?

Ans. The electronic configuration of oxygen is 1s2 2s2 2px2 2py

1 2pz1 i.e., it has two half filled orbitals

and there is no d-orbital available for excitation of electrons while other members of this grouphas d-orbitals, therefore, they show +2, +4 and + 6 oxidation states.

Q. 25. Arrange the following in the order of property mentioned for elements :(a) F2, Cl2, Br2, I2 – increasing bond energy.

(b) HF, HCl, HBr, HI – increasing acid strength.

(c) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.

(d) H2O,H2S, H2Se, H2Te – increasing acid strength.Ans. (a) In the order of increasing bond energy.

Cl—Cl bond energy is highest while I—I bond energy is least.I—I < F—F < Br—Br < Cl—Cl

Bond energyleast

increasing bond⎯⎯⎯⎯⎯⎯⎯→energy

Bond energyhighest

(b) In the order of increasing acid strength in water (i.e., aqueous solution) HI is strongest acidwhile HF is the weakest acid.

HF < HCl < HBr < HI


Weakestacid

acid strength⎯⎯⎯⎯⎯⎯⎯→increases

Strongestacid

(c) In order of increasing base strengthBiH3

least basic < SbH3 < AsH3 < PH3 < NH3

most basic

(d) In order of increasing acid strength.H2O

least acidic < H2Te < H2Se < H2S

most acidic

Q. 26. Give five uses of noble gases.Ans. The five uses of noble gases are :

(i) Helium is used in gas-cooled atomic reactors as a heat transfer gas.(ii) Neon is used widely in preparing neon signals in many electronic appliances.(iii) Argon is mainly used in filling electrical lamps.(iv) Krypton is used in filling luminous sign tubes and filament lamps.(v) Radon is used in the preparation of ointments for the treatment of cancer.

Q. 27. How is hydrogen peroxide prepared in the laboratory ?Ans. Lab preparation of hydrogen peroxide : A thin paste of hydrated barium peroxide made with

ice cold water is slowly added to an ice cold dilute sulphuric acid solution.BaO2 · 8H2O + H2SO4 → BaSO4 ↓ + H2O2 + 8H2O

Q. 28. What is tailing of mercury ?Ans. Due to the dissolution of Hg2O in mercury, the mercury loses its meniscus and starts sticking to

the sides of the glass. This is called tailing of mercury.Q. 29. Give two uses of O3.

Ans. (i) As disinfectant and germicide in the purification of drinking water.(ii) As an oxidising agent.

Q. 30. Which substances are responsible for depletion of ozone layer in the atmosphere ?Ans. Chemicals responsible for destruction or depletion of O3 layer are NOX released by ultrasound

aircrafts or lightning and CFC used as aerosol propellants and in refrigerators.Q. 31. What happens when hydrogen peroxide is added to acidified potassium permanganate

solution ?Ans. When hydrogen peroxide is added to acidified potassium permanganate solution, the pink

colour of the acid solution is reduced to colourless solution, i.e.,2KMnO4 + 3H2SO4 + 5H2O2 ⎯→ K2SO4 + 2MnSO4 + 8H2O + 5O2

Q. 32. Which form of sulphur shows paramagnetic behaviour ?Ans. In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the

antibonding (II) orbitals like O2 and hence, exhibits paramagnetism.Q. 33. How is the presence of SO2 detected ?

Ans. When SO2 is passed through acidified potassium permanganate it discharges its pink colour2KMnO4 + 5SO2 + 2H2O⎯⎯→ K2SO4 + 2MnSO4 + 2H 2SO4

Q. 34. Give two uses of sulphuric acid.Ans. (i) In storage batteries and lead accumulators.

(ii) In chemical industries for the preparation of hydrochloric acid, nitric acid, sulphates,ether, dyes and paints.

Q. 35. What is aqua regia ?Ans. When three parts of concentrated HCl and one part of concentrated HNO3 are mixed, aqua

regia is formed.

Q. 36. A green coloured solution turns pink when O3 is bubbled through it. The pink colour furtherdecolourises when zinc and dilute H2SO4 is added to it. Identify the green and pink colouredcompounds giving sequence of reactions.

Ans. K2MnO4 (green)KMnO4 (Pink)


Q. 37. Give two uses of SO2.Ans. (i) It is used in the manufacturing of H2SO4.

(ii) It is used as bleaching agent for delicate fibres.Q. 38. Give the bleaching action of SO2, why bleaching is temporary ?

Ans. SO2 in presence of moisture acts as a bleaching agent. This is due to the reducing nature of SO2SO2 + 2H2O ⎯⎯→ H2SO4 + 2H

Coloured matter + H ⎯⎯→ Colourless matter.The bleaching is temporary. The bleached matter when exposed to air regains its colour due tooxidation.

Q. 39. How is SO2 an air pollutant ? Explain.Ans. SO2 released into atmosphere through volcanic eruption is about 67% and rest from human

activity. SO2 is oxidised to SO3 and then converted to H2SO4 in the presence of moisture. Thiscomes down in the form of acid rain. SO2 is strongly irritating to respiratory tract of humansand animals. It causes irritation to eyes. It causes necrotic blotching.

Q. 40. Briefly explain the structure of H2O2 in gaseous phase.Ans. The structure of hydrogen peroxide in gaseous phase is similar to partly open book. It is a non-

planar structure the O—O bond is along the binding of the book and the two O—H bonds onthe planes of two separate pages. The O—O bond distance is 1·48 Å and the O—H bonddistance is 0·97 Å. The planes in which the two O—H bonds lie makes an angle of 94° while theangle H—O—O is 97°.

Chapter 8. f-Block ElementsQ. 1. What is meant by Lanthanide contraction ? Write the general electronic configuration of inner

transition elements.Ans. The steady decrease in the atomic and ionic radii of lanthanide elements with increasing atomic

number is called Lanthanide contraction.General electronic configuration of inner transition elements :

5f1–14 6d1 7s2

Q. 2. What is the electronic configuration of chromium atom (Z = 24) ? Give a reason for youranswer.

Ans. Electronic configuration of Cr atom is :1s2, 2s2p6, 3s2 p6d5, 4s1

This is because an electronic configuration with half-filled and completely filled orbitals ismore stable. 24Cr has 3d orbitals half-filled which is more stable configuration than normallyexpected i.e., 3d4 4s2.

Q. 3. State the common oxidation state of :(i) Lanthanides (ii) Actinides

Ans. (i) Oxidation state of lanthanides is + 3. Some lanthanides show + 2, + 4 as well.

(ii) Oxidation state of actinides is + 3 but many actinides also show + 2, + 4 and some + 5,+ 6, + 7 also.

Q. 4. Describe the process of conversion of blister copper to copper.Ans. The blister copper is converted into pure copper by electrolytic refining.

Impure copper — anode, Pure copper plate — cathode, 15% CuSO4 Soln + 5% H2SO4electrolyte.

Cell reaction CuSO4Δ→ Cu2+ + SO4

2 –

At anode At cathode Cu (s) → Cu2+ + 2e– Cu2+ + 2e– → Cu

Pure copper gets deposited at cathode.Q. 5. What is the shape of chromate ion ?Ans. Tetrahedral.Q. 6. Which three elements belonging to d-block are usually not considered as transition elements ?Ans. Zn, Cd, Hg.


Q. 7. Which metal is used for galvanisation of Fe ?Ans. Zinc.Q. 8. What is the most characteristic oxidation state of lanthanoids ?Ans. + 3.Q. 9. What happens when K2Cr2O7 is heated ?Ans. When potassium dichromate is heated it gives potassium chromate and chromic oxide.

4K2Cr2O7 Δ⎯⎯→ 4K2CrO4 + 2Cr2O3 + 3O 2

Q. 10. Which element of the first transition series exhibits the largest oxidation states ?Ans. Manganese + 2 to + 7.

Q. 11. On what ground can you say that Sc (Z = 21) is a transition element while Zn (Z = 30) is not ?Ans. Sc : [Ar] 3d1 4s2 Scandium has incompletely filled 3d - orbitals whereas zinc has completely

filled 3d orbitals in its ground state as well as in common oxidation state.Q. 12. Which of the following ions would form white complexes ?

Cu2+, Zn2+, Ti3+, Ti4+, Cd2+.Ans. Zn2+, Ti3+, Ti4+, Cd2+.

Q. 13. What are the different oxidation states exhibited by the lanthanoids ?Ans. The principal oxidation state of lanthanoids is +3. However, some lanthanoids also exhibit

oxidation states of +2 and + 4. For example, Eu exhibit oxidation state of +2 and Ce exhibitsoxidation state of + 4.

Q. 14. Which metal in the first series of transition metals exhibit + 1 oxidation state most frequentlyand why ?

Ans. In the first transition series copper exhibit +1 oxidation state most frequently. This is due to thereason that Cu(I) has stable electronic configuration.

Cu(I) : [At] 3d10.Q. 15. Give two uses of K2Cr2O7

Ans. (i) For the volumetric estimation of ferrous salts, iodides and sulphites.(ii) It is used in chrome tanning in leather industry.

Q. 16. Give two uses of KMnO4.Ans. (i) As disinfectant for water.

(ii) In dry cells.Q. 17. Give first and last atomic number of all the three transition series.

Ans. Ist —21 to 30.IInd —39 to 48.IIIrd —57, 72 to 80 (1 + 9).

Q. 18. Write first element of Ist , IInd and IIIrd transition series.Ans. Scandium, Yttrium, Lanthanum.

Q. 19. In what way is the electronic configuration of the transition elements different from that of thenon-transition elements ?

Ans. Transition elements have partially filled d-subshell belonging to penultimate energy levelwhereas non-transition elements do not have any partially filled d-subshell. In non-transitionelements, the last electron enters the s or p-subshell whereas in transition elements the lastelectron enters the d-subshell of penultimate energy level.

Q. 20. Write down the electronic configuration of the following ions :(a) Cr3+ (b) Cu+ (c) Co2+ (d) Mn2+

(Atomic no. Cr = 24, Cu = 29, Co = 27, Mn = 25)Ans. (a) Cr3+ : 1s22s22p63s23p63d3 (b) Cu+ : 1s22s22p63s23p63d10

(c) Co2+ : 1s22s22p63s23p63d7 (d) Mn2+: 1s22s22p63s23p63d5

Q. 21. What is the reason for the greater tendency of transition metals to form complex compounds ?Ans. The transition metals have great tendency to form complex compounds and there are large

number of such compounds due to :


(i) Small size of metal ions.(ii) High ionic charge on them.(iii) Availability of d-orbitals for bond formation.

Q. 22. What is Lanthanoid contraction ? Discuss its causes.Ans. As the atomic number increases in lanthanoid series, for every proton in the nucleus the extra

electron goes to fill 4ƒ-orbitals. The 4ƒ-electrons constitute inner shells and are ratherineffective in screening the nuclear charge. Thus, there is a gradual increase in the effectivenuclear charge experienced by the outer electrons. Consequently, the attraction of the nucleusfor the electrons in the outermost shell increases as the atomic number of lanthanoids increasesand the electron cloud shrinks. This results in gradual decrease in size of lanthanoids withincrease in atomic number.

Q. 23. What are alloys ? Name an important alloy which contains some of the lanthanoid metals.Mention its uses.

Ans. An alloy is a homogeneous mixture of a metal with other metals or non-metals. A large numberof alloys of transition metals are known and extensively used in modern industries. Animportant alloy which contains some of lanthanoid metals is ‘misch-metal’.Uses of Misch-metal :

(i) Addition of about 3% misch metal to magnesium increases its strength and thus is used inmaking jet engine parts.

(ii) It is pyrophoric and is used in cigarette and gas lighters, tracer bullets, shells, etc.

Q. 24. What are inner transition elements ? Describe which of the following atomic numbers are theatomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

Ans. Inner transition elements are the elements which have partly filled ƒ-orbitals. These are alsocalled ƒ-block elements. There are two series of inner transition elements.

(i) Lanthanoids (4ƒ-series). These are the 14 elements from atomic numbers 58-71.(ii) Actinoids (5f-series). These are the 14 elements from atomic numbers 90-103.Among the given atomic numbers, only 59, 95 and 102 are the atomic numbers of innertransition elements.

Q. 25. What happens when ?(a) KMnO4 is heated.(b) K2Cr2O7 is dissolved in H2O2 Solution (acidified)(c) K2Cr2O7 reacts with sodium chloride in the presence of conc. H2SO4.

Ans. (a) 2KMnO4 ⎯⎯→Δ

K2MnO4 + MnO2 + O2

It gives potassium manganate and oxygen is evolved.(b) K2Cr2O7 + 4H2O2 + 2H+ ⎯⎯→ 2K+ + 2CrO5 + 5H2O.

It forms deep blue colour due to the formation of [CrO(O2)2]. The deep blue colour fadesaway gradually due to decomposition of CrO5 into Cr3+ ion and oxygen.(c) K2Cr2O7 +4NaCl + 6H2SO4 ⎯⎯→ 2KHSO4 + 4NaHSO4+ 2CrO2Cl2 + 3H2O

Orange colour chromyl chloride fumes are formed

Chapter 9. Coordination CompoundsQ. 1. What type of isomerism is exhibited by the following pairs of compounds :

(i) [PtCl2 (NH3)4] Br2 and [PtBr2 (NH3)4] Cl2

(ii) [Cr (SCN) (H2O)5]2+ and [Cr (NCS ) (H2O)5]2+

Ans. (i) Ionization isomerism(ii) Linkage isomerism

Q. 2. State the hybridization and magnetic property of [Fe(CN)6]3– ion according to the valence bondtheory.

Ans. Hybridization of [Fe (CN)6]3– : d2sp3

Magnetic property : Due to presence of one unpaired electron in 3d shell, [Fe(CN)6]3– ion isparamagnetic in nature.


Q. 3. Name the type of isomerism shown by the following pairs of compounds :(i) [CoCl(H2O)(NH3)4]Cl2 and [CoCl2(NH3)4]Cl.H2O(ii) [Pt(NH3)4][PtCl6] and [Pt (NH3)4Cl2][PtCl4]

Ans. (i) Hydrate isomerism(ii) Coordination isomerism.

Q. 4. For the complex ion of [Co(NH3)6]3+ :(i) State the hybridisation of the complex.(ii) State the magnetic nature of the complex.

Ans. (i) Complex has d2sp3 hybridisation.(ii) Complex is diamagnetic in nature.

Q. 5. For the complex ion [Fe(CN)6]3 – state :(i) The geometry of the ion.(ii) The magnetic property of the ion.

Ans. (i) Octahedral geometry(ii) Ion is paramagnetic.

Q. 6. What type of structural isomers are [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br ? Give a chemicaltest to distinguish the isomers.

Ans. [Co (NH3)5 Br] SO4 and [Co (NH3)5SO4] Br are ionisation isomers.Aqueous solution of [Co (NH3)5 Br] SO4 will give a white ppt with BaCl2 solution. The ppt isinsoluble in conc. HNO3 confirming presence of free SO4

2– ion. Aqueous solution of[Co (NH3)5 SO4] Br will not give this test.

[Co (NH3)5 Br] SO4 [Co(NH3)5 Br]2+

SO42–

BaCl2 + SO2 –4 ⎯→ BaSO4 ↓

white ppt + 2Cl–

[Co (NH3)5 SO4] Br [Co (NH3)5 SO4]+ + Br–

BaCl2 + Br– ⎯→ No pptQ. 7. [CoF6]3– is a coordination complex ion.

(i) What is the oxidation number of cobalt in the complex ?(ii) How many unpaired electrons are there in the complex ?(iii) State the magnetic behaviour of the complex.(iv) Give the I.U.P.A.C. name of the complex.

Ans. (i) [CoF6]3–

x + (6 × – 1) = – 3∴ Oxidation number of Co, ‘x’ = + 3

(ii) 4 unpaired electrons.(iii) Complex ion is paramagnetic.(iv) Hexafluorocobaltate(III) ion.

Q. 8. Name the type of isomerism shown by the following pair of compounds :[PtCl2(NH3)4] Br2 and [PtBr2(NH3)4]Cl2.

Give a chemical test to distinguish between the given pair of isomers.Ans. Ionization isomerism is exhibited by the given pair of coordination compounds.

[PtCl2(NH3)4] Br2

H2O [PtCl2(NH3)4]2+ + 2Br–

[PtBr2(NH3)4]Cl2

H2O [PtBr2(NH3)4]2+ + 2Cl–

Test to distinguish : Aqueous solution of [PtCl2(NH3)4]Br2 when treated with AgNO3 solutiongives a pale yellow ppt of AgBr which dissolves in excess of NH4OH with difficulty.On the other hand, aqueous solution of [Pt Br2 (NH3)4]Cl2, when treated with AgNO3 solutiongives a white ppt of AgCl which readily dissolves in NH4OH.


Q. 9. Name the type of isomerism shown by the following pairs of coordination compounds :(i) [Co (NH3)5 NO2]Cl2 and [Co (NH3)5 ONO]Cl2

(ii) [Cr (H2O)5 Cl] Cl2·H2O and [Cr (H2O)4 Cl2] Cl·2H2OAns. (i) Linkage isomerism

(ii) Hydrate isomerismQ. 10. What are crystal fields ?

Ans. The ligands especially anionic (or polar ligands) has negatively charged field around thembecause of which they are called crystal fields.

Q. 11. What is meant by chelate effect ?Ans. When a bidentate or a polydentate ligand donor atoms are positioned in such a way that when

they coordinate with the central metal ion, a five or a six-membered ring is formed this effect iscalled chelate effect.

Q. 12. Define the following terms :(a) Coordination number (b) Counter ions (c) Ligands.

Ans. (a) The number of sigma (σ) bonds formed by the central metal atom with the ligands iscalled the coordination number of central metal atom.

(b) The ion present outside the coordination sphere to neutralise the charge present is calledcounter ion.

(c) Ligands are those atoms or groups of atoms which have a tendency to donate a lone pairof electrons to the central metal atom or ion.

Q. 13. What are donor sites ?Ans. A donor atom may contain one or more unshared pair of electrons which are known as donor

sites of the ligand.Q. 14. The ligands are attached to the central metal ion by means of which bond ?

Ans. Coordinate or dative bond.Q. 15. Define anionic complex and cationic complex.

Ans. The complex ion carrying a negative charge is called anionic complex.The complex ion carrying a positive charge is called cationic complex.

Q. 16. Give two examples of anionic complexes.Ans. (a) [PtCl6]2– (b) [Ag(CN)2]–.

Q. 17. Name the complex ion carrying neither a positive charge nor a negative charge.Ans. Neutral complex e.g., [Ni (CO)4].

Q. 18. Define multi-dentate ligands.Ans. Ligands with more than one donor sites are collectively called multi-dentate ligands.

Q. 19. Give three examples of bidentate ligands.

Ans. (a) Ethylenediammine, ··NH2CH2CH2 ··NH2.

(b) Oxalate ion, COO–

|COO–

(c) Dimethyl glyoxime ion, C N OH

CH3—C

··

N—OH|

CH3—

··

==

==

Q. 20. What are ambidentate ligands ?Ans. These are unidentate ligands having more than one donor atoms through which they can

coordinate to the central atom.Q. 21. Give two examples of ambidentate ligands.

Ans. (a) M ← N�OO : Nitro group M ←O—N=O : Nitrito group

(N donor atom) (O donor atom)

(b) M ← SCN — S–Thiocyanate M ← NCS — N-Thiocyanate(S donor atom) (N donor atom)


Q. 22. Write two examples of positive unidentate ligands :Ans. (a) Nitrosonium, NO+. (b) Nitronium, NO2

+.Q. 23. Identify the ligands from the following coordinate compounds:

(a) [Co(en)2 Cl(NO2)2] (b) K [Co(CN) (CO)2 (NO)](c) K3 [Al (OH)6] (d) [Co(H2O)2 (NH3)4] (OH)3

Ans. (a) Ethylenediamine, Cl– and NO–2 (b) CN–, CO and NO

(c) OH– only (d) H2O and NH3.Q. 24. What are bridging ligands ?

Ans. These are the monodentate ligands which can simultaneously coordinate with two or moremetal atoms.

eg. ⎣⎡

⎦⎤(NH3)4 Co

NH2

NO2

Co (NH 3)4] 2+

Here NH–2 and NO+

2 are bridging ligands.

Q. 25. Define oxidation number.Ans. It is the residual charge that appears on the central atom when all other atoms or ions are

removed from it.Q. 26. What is the oxidation number of the central metal atom (underlined) in each of the following

complex species :(a) K4[Fe (CN)6] (b) [Fe (H2O)6] Cl3

(c) [Pt (NH3)2Cl2] Cl2 (d) [Co(NH3)4 (H2O) Br] (NO3)2

Ans. (a) + 2 (b) + 3(c) + 4 (d) + 3

Q. 27. Write the abbreviations for the given complicated molecules :(a) Ethylenediammine (b) Pyridine (c) Acetylacetonate

Ans. (a) en, (b) Py, (c) acac.Q. 28. The spin only magnetic moment of [MnBr4]2– is 5.9 BM. Predict the geometry of the complex

ion.Ans. Since the coordination number of Mn2+ ion in the complex ion is 4, it will be either tetrahedral

(sp3) or square planar (dsp2). But the fact that the magnetic moment of the complex ion is 5.9BM, it should be tetrahedral in shape rather than square planar because of the presence of fiveunpaired electrons in the d-orbitals.

Q. 29. Define coordination compounds.Ans. The compounds in which ions or neutral molecules are bonded to a central metal atom by

coordinate bonds are called coordination compounds e.g., [Ni(CO)4], [COCl3(NH3)3] etc. Theyare also called complex compounds.

Q. 30. Explain giving examples :(a) Coordination entity (b) Coordination polyhedron(c) Homoleptic and heteroleptic (d) Ligand(e) Coordination number (f) Complex ion

Ans. (a) Coordination entity : It usually constitutes a central metal atom or ion, to which a fixednumber of other atoms or ions or groups are attached by coordinate bonds.

For example, [Ni(CO)4], [Co(NH3)6]+, [PtCl4]2– etc.

(b) Coordination polyhedron : The spatial arrangement of ligands that are directly attachedto the central atom or ion defines a coordination polyhedron about the central atom orion. For example, [Co(NH3)6]3+ is octahedral, [PtCl4]2– is square planar etc.

(c) Homoleptic and Heteroleptic : Complexes in which a metal is bound to only one kind ofdonor group are known as homoleptic. e.g., [Co(NH3)6]3+.

Complexes in which a metal is bound to more than one kind of donor groups are knownas heteroleptic. For example, [Co(NH3)4Cl2]+.


(d) Ligands : The atoms or molecules or ions which donote pair of electrons to the centralmetal atom and thus forms coordinate bond with the central metal atom are calledligands. For example, NH3, CO, CN, Cl–, Br–, H2O.

(e) Coordination Number : It is the total number of ligands attached to the central metalatom in the coordination sphere. For example, in the complex [Ni(NH3)4]Cl2, thecoordination number of Ni is 4.

(f) Complex ion : When the species formed by the linking of ions or neutral molecules to thecentral metal atom through coordinate bond carries a +ve or –ve, charge it is called acomplex ion. For example, [Cu (NH3)4]2 +, [Fe (CN)6]3 – etc.

Q. 31. What is spectrochemical series ? What is the difference between a weak field ligand and astrong field ligand ?

Ans. The arrangement of the ligands in order of increasing field strength is known asspectrochemical series.

I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < EDTA4– < NH3 < en < CN– < CO

Ligands have different field strengths and as a result the crystal field splitting Δo or Δt dependsupon the field produced by the ligand and charge on metal ions. Some ligands are able toproduce strong fields in this case the splitting will be large whereas others produce weak fieldsand consequently results in small splitting of d-orbitals.

Chapter 10. Haloalkanes and HaloarenesQ. 1. What will be the product formed when chlorobenzene is heated with sodium metal in the

presence of dry ether ?

Ans. The product formed will be biphenyl ( ) and sodium chloride (NaCl) Wurtz

Fittig reaction

2 + 2 Na + 2 NaClDry ether

Cl

Q. 2. What is plane polarised light ?Ans. A beam of light which has vibration in only one plane is called plane polarised light.Q. 3. What is known as racemic mixture ?Ans. An equimolar mixture of a pair of enantiomers is called racemic mixture.Q. 4. What are alkyl halides ? What is their general formula ?Ans. Halogen derivatives of alkanes are called alkyl halides. A general formula of monohalo-alkanes

is CnH2n+1 X.Q. 5. Classify each of the following alkyl halides as 1°, 2° or 3°.

(i) CH

Cl(ii) (CH3CH2)3CCl(iii) (CH3)3C CH3 Br

(iv)

BrAns. (i) 2°, (ii) 3°, (iii) 1°, (iv) 2°Q. 6. Discuss the ability of alkyl halides to dissolve (i) in H2O, (ii) in organic solvents.Ans. (i) They are insoluble in H2O, probably because there is little (in case of RF) to no H–bonding

with H2O.

(ii) They are soluble in organic solvents such as alcohol, ether, and benzene.


Q. 7. List the densities of chloromethane, fluoromethane, iodomethane, bromomethane, water andmethane in decreasing order.

Ans. The decreasing order is CH3I > CH3Br > H2O > CH3Cl > CH3F > CH4

Q. 8. Define nucleophilicity as used for nucleophiles.

Ans. Nucleophilicity is defined by a reaction of a :Nu with an electrophilic C.

Nu:– + C L Nu→ C + :L–

Q. 9. Arrange the following halides in order of increasing SN2 reactivity :

CH3Cl, CH3Br, CH3CH2Cl, (CH3)2 CHCl

Ans. (CH3)2CHCl < CH3CH2Cl < CH3Cl < CH3Br. As the size of R group increases, the reactivitytowards SN2 reaction decreases. Further, C—Br bond is weaker than C—Cl bond and it isdifficult to cleave C—Cl bond.

Q. 10. Arrange the following in order of increasing property as indicated :

(i) CH3F, CH3Cl, CH3Br, CH3I → reactivity towards nucleophilic substitution reaction.

(ii) (CH3)3CCl, (CH3)2CHCl, CH3CH2Cl → reactivity towards SN1 reaction.

(iii) CH3Cl, CH3CH2Br, (CH3)2CHBr → boiling point.

Ans. (i) CH3F < CH3Cl < CH3Br < CH3I (C—X bond strength decreases as size of X increases)

(ii) CH3CH2Cl < (CH3)2CHCl < (CH3)3 CCl (as size of R group increases reactivity towards SN1increases).

(iii) CH3Cl < CH3CH 2Br < (CH3)2CHBr.Q. 11. Explain Fittig reaction.

Ans. Haloarenes on reaction with sodium in the presence of dry ether forms diphenyl (or biphenyl)(a compound in which two benzene rings unite with each other). This reaction is known asFittig reaction.

⎯⎯⎯⎯→ether

+ 2NaBrBr + 2Na + Br

BiphenylQ. 12. Illustrate Wurtz-Fittig reaction.

Ans. A mixture of an alkyl halide and an aryl halide in the presence of sodium gives in alkyl phenyl.Alkane and biphenyls are also formed as side products in the Wurtz-Fittig reaction.

Cl + 2Na + Cl—CH 3 ether

⎯⎯⎯→

side productsmajor

CH3 + CH3—CH3 + ⎩⎨⎧

Q. 13. How are following compounds prepared by halogenation of ethane :(i) Chloroethane (ii) Bromoethane.

Ans. (i) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorinein the presence of diffused sunlight or U.V. light or at high temperature chloroethane isobtained.

C2H6 + Cl2 diffused sunlight

⎯⎯⎯⎯⎯⎯⎯⎯→ C2H5—Cl + HCl Ethane

(excess)Ethane Chloroethane

(ii) Bromination of ethane : When ethane is heated with Br2 in the presence of anhydrousAlBr3, bromoethane is obtained.

C2H6 + Br2 Anhydrous AlBr3

⎯⎯⎯⎯⎯⎯⎯⎯→ C2H5—Br + HBrEthane Bromoethane

Q. 14. How are following compounds obtained from alkenes :(i) C2H5Cl (ii) C2H5Br (iii) CH3—CH—CH3

|I


Ans. (i) Ethene on reaction with hydrogen chloride forms C2H5Cl.H2C = CH2 + HCl ⎯→ CH3—CH2—Cl

Ethene Chloroethane

(ii) Ethene on reaction with hydrogen bromide forms C2H5Br.H2C=CH2 + HBr ⎯→ CH3—CH2—Br

Ethene Bromoethane

(iii) Propene on reaction with hydrogen iodide forms CH3—CHI—CH3CH3—

Propene

CH =CH2 + HI ⎯→

2-iodopropane

CH3—CH—CH3 |I

Q. 15. How will you prepare ethyl chloride (chloroethane) from ethyl alcohol using (i) PCl3 , (ii) PCl5and (iii) SOCl2 ?

Ans. (i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

CH3CH2OH + PCl3 Δ⎯⎯→ 3CH3CH2Cl + H3PO3

ethyl alcohol ethyl chloride phosphorus acid

(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

CH3CH2OH + PCl5 Δ⎯⎯→ CH3CH2Cl + POCl3 + HCl

Ethyl alcohol Ethyl chloride Phosphoryl chloride`

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethylchloride is formed.

CH3CH2OHEthyl alcohol

+ SOCl2 pyridine

ethyl chloride⎯⎯⎯⎯⎯⎯⎯→ CH3CH2Cl

Ethylchloride + SO2↑ + HCl ↑

Q. 16. How chloroform is prepared from ethanol ?Ans. Chloroform is prepared from ethyl alcohol and bleaching powder. This is a three step reaction

process and is called as haloform reaction. The steps involved in the reaction are as follows :Step 1. Oxidation :

CH3—CH2OH + Cl2 ⎯⎯→ CH3—CHO + 2HCl Ethyl alcohol Acetaldehyde

Step 2. Chlorination :CH3—CHO + 3Cl2 ⎯⎯→ CCl3—CHO + 3HCl

Chloral(Trichloroacetaldehyde)

Step 3. Hydrolysis :2CCl3—CHO + Ca(OH)2 ⎯⎯→ 2CHCl3 + (HCOO)2Ca

Chloroform Calciumformate

Q. 17. C–F bond in CH3F is the strongest bond and C–I bond in CH3I is the weakest bond. Explain.Ans. Methyl fluoride (CH3F) is highly polar molecule and has the shortest C–F bond length and the

strongest C–F bond due to greater overlap of orbitals of the same principal quantum numberi.e., overlap of 2sp3 orbital of carbon with 2pz orbital of fluorine.On the other hand, methyl iodide (CH3I) is much less polar and has the longest (C–I) bondlength and the weakest C–I bond due to poor overlap of 2sp3 orbital carbon with 5pz orbital ofiodine i.e., 2sp3 orbital of carbon cannot penetrate into larger orbitals.

Q. 18. What is the action of moist silver oxide (Ag2O) and aqueous alcoholic potassium cyanide onbromoethane ?

Ans. (i) Action of Moist Ag2O : Bromoethane (C2H5Br) when boiled with moist Ag2O undergoeshydrolysis and forms C2H5OH.

Ag2O + H2O ⎯⎯→ 2AgOH

Bromoethane

C2H5—Br + AgOH moist Ag2O

Δ⎯⎯⎯⎯⎯→

Ethanol

C2H5—OH + AgBr

(ii) Action of aqueous alc. KCN : When ethyl bromide (bromoethane) is boiled with alcoholicsolution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

CH3—CH2—Br + KCN boil

⎯⎯→ CH3—CH2—CN + KBr Ethyl bromide (alc.) Ethyl cyanide

Q. 19. Explain reaction of (i) ethyl bromide , (ii) methyl chloride with AgCN alc.Ans. (i) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl

isocyanide is formed.


CH3—CH2—Br + AgCN Δ

⎯⎯⎯→ CH3—CH2—NC + AgBrEthyl chloride (alc.) Ethyl isocyanide

(ii) Methyl chloride : When methyl chloride is heated with alcoholic silver cyanide, methylisocyanide is formed.

CH3—Cl + AgCN Δ

⎯⎯⎯→ CH3—NC + AgCl Methyl chloride (alc.) Methyl isocyanide

The above reactions are called isocyanide reaction.

Q. 20. Explain the following substitution reactions of chlorobenzene :

(i) Halogenation, (ii) Nitration, (iii) Sulphonation.

Ans. (i) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrousferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.

2

Cl

+ 2Cl2

Cl

Cl1, 4-dichlorobenzene

(major product)

+ 2HCl

ClCl

+

1, 2-dichlorobenzene

AnhydrousFeCl 3⎯⎯⎯⎯⎯⎯→

(ii) Nitration : When chlorobenzene is heated with nitrating mixture (conc. nitric acid + conc.sulphuric acid) a mixture of 1-chloro-4-nitro benzene (major product) and 1-chloro-2-nitrobenzene is formed.

2

Cl

+ Conc.2HNO3

Conc. H2SO 4

Δ⎯⎯⎯⎯⎯⎯→

ClNO

+

2

1-chloro-

Cl

+ 2H O

1-chloro-4-nitrobenzene(major product)

2

2NO

(iii) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixtureof 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid isformed.

+ 2H 2SO 4 Δ

⎯⎯→

SO 3H

SO 3H

+ 2H 2O

2-chlorobenzenesulphonic acid

4-chlorobenzenesulphonic acid(major product)

Cl Cl Cl

2 +Conc.

Q. 21. Explain Fittig reaction.Ans. Haloarenes on reaction with sodium in the presence of dry ether forms diphenyl (or biphenyl)

a compound in which two benzene rings unite with each other. This reaction is known as Fittigreaction.

Br + 2Na + Brether

⎯⎯⎯→

Biphenyl

+ 2NaBr

In this reaction, two aryl groups undergo coupling reaction to form biphenyl.


Q. 22. State the harmful effects of the following compounds :(i) Trichloromethane or Chloroform (CHCl3).(ii) Tetrachloromethane or Carbontetrachloride (CCl4).(iii) Freons (CCl2F2, CCl3F, CHClF2),(iv) DDT

Ans. (i) Chloroform (CHCl3) vapour when inhaled for a short time causes dizziness, headacheand fatigue and if inhaled for a long time affects central nervous system.(ii) Carbon tetrachloride (CCl4) causes eye irritation, damages nerve cells, vomitingsensation, dizziness, unconciousness or death.(iii) Ferons causes ozone depletion and are toxic to nature.(iv) DDT is not readily metabolised by animals and gets deposited and stored in fatty tissues.

Q. 23. List the uses of following :(i) Chloropicrin (ii) Chloretone(iii) p-dichlorobenzene (iv) BHC(v) DDT (vi) Carbon tetrachloride(vii) Iodoform.

Ans. (i) Chloropicrin :(a) It is used as an insecticide.(b) It is used as a war gas.

(ii) Chloretone :It is used as a hypnotic drug.

(iii) p-dichlorobenzene :(a) It is used as a general insecticide, germicide and soil fumigant.(b) It is used as deodorant and moth repellent

(iv) BHC :It is used as a pesticide in agriculture.

(v) DDT :It is used as a powerful insecticide.

(vi) Carbon tetrachloride :(a) It is used as an industrial solvent for oils, fats, resins etc.(b) It is used as a fire-extinguisher.

(vii) Iodoform :(a) It is used as an antiseptic due to liberation of free iodine.(b) It is used in pharmaceutical preparation like iodex.

Chapter 11. Alcohols, Phenols and EthersQ. 1. Give balanced equation for the preparation of salicylaldehyde from phenol.

Ans. OH

+ CHCl3 CHCl2

HCl+ NaOH

H2OPhenol

Salicylaldehyde

ONa

2HCl+ NaOH

NaCl+ dil HCl H2O

CHO ONa

CHO OH

CH(OH)2 ONa

Q. 2. What is absolute alcohol ?Ans. 100% ethyl alcohol is called absolute alcohol.Q. 3. What is denatured alcohol ?

Ans. Alcohol is made unfit for drinking by mixing some copper sulphate and pyridine in it. This iscalled denatured alcohol.

Q. 4. Name the three systems for naming alcohols.Ans. (i) Common system (ii) Carbinol system (iii) IUPAC system


Q. 5. Branching increases or decreases the boiling point of alcohols ?Ans. Decreases.Q. 6. O—H bond of alcohol is polar or not ?Ans. It is highly polar.Q. 7. Name the product which is obtained on oxidation of :

(i) Primary alcohol (ii) Secondary alcohol (iii) Tertiary alcoholAns. (i) Carboxylic acid (ii) A Ketone (iii) No-reactionQ. 8. Alcohols 1°, 2°, and 3° reacts with hot copper. Give the products obtained.Ans. 1° Alcohols gives aldehyde, 2° Alcohols gives acetone and 3° Alcohols gives alkene.Q. 9. What is esterification ?Ans. When alcohol reacts with a carboxylic acid it gives an ester and water the reaction is called

esterification.Q. 10. State the basic difference between alcohol and phenol.

Ans. The alcohols contain –OH group attached to alkyl group whereas in phenols, the – OH group isattached to aromatic rings.

Q. 11. What is fermentation ?Ans. Fermentation is a process carried out under anaerobic conditions. It is slow decomposition of

big organic molecules into simpler ones under the influence of non-living complex subtancescalled ferments.

Q. 12. Can we use sodium metal to remove last traces of water from (i) benzene (ii) ethanol ?Ans. (i) Sodium metal can be used to remove last traces of water from benzene because it does not

reacts with benzene.(ii) Sodium metal cannot be used to remove traces of water from ethanol because it reacts withethanol.

2C2H5OH + 2Na ⎯→ 2C2H5ONa + H2

Q. 13. Arrange the following alcohols in order of increasing reactivity towards Lucas reagent :2-butanol, 1-butanol, 2-methyl-2-propanol

Ans. During the reaction of alcohols with Lucas reagent C—O bond is cleaved. Therefore, the orderof reactivity of alcohols towards Lucas reagent follows the order : 3° > 2° > 1°. Thus, the correctorder of reactivity is :

2-methyl-2-propanol > (3°)

2-butanol > (2°)

1-butanol(1°)

Q. 14. What is the main product formed when phenol is subjected to Kolbe’s reaction ?Ans. Salicylic acid.

Q. 15. Arrange the following compounds in increasing order of their acidic strength :(a) propan-1-ol (b) 2, 4, 6-trinitrophenol

(c) 3 nitrophenol (d) 3, 5-dinitrophenol

(e) phenol (f) 4-methyl phenol

Ans. Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitro-phenol.

Q. 16. Give uses of alcohols.Ans. (i) In the manufacturing of dyes, synthetic drugs and detergents.

(ii) As antifreeze.

(iii) As a solvent for resins, fats, oils, fatty acids and hydrocarbons.

Q. 17. What is meant by hydroboration-oxidation reaction ? Illustrate it with an example.

Ans. It is indirect hydration of an alkene using diborane.

For example : Propene is converted to propan-1-ol when treated with diborane followed byoxidation with alkaline peroxide, propane gets converted to propan-1-ol

CH3—CH = CH2 (i) B2H6

⎯⎯⎯⎯⎯→H2O2/OH–

CH3CH2—CH2OH.propan-1-ol


Q. 18. Give the effect of substituents on acidic strength of phenols.

Ans. The presence of electron withdrawing groups (NO2, CHO, etc.) and electron releasing groups(R, NH2, —OR, etc.) also affects the acidic strength. In general,

(i) The electron withdrawing groups (X) increase the acidic strength by stabilizing thephenoxide ion by dispersal of negative charge.

(ii) The electron releasing groups (Y) decrease the acidic strength by destabilizing thephenoxide ion by concentrating the negative charge.

OH|

Acidic strength Xis more

– H+

O|

X stabilises Xthe anion

–

OH|

Acidic strength Yis less

– H+

Y destabilises the anion

–O|

Y

(iii) The electron releasing or electron attracting effect is generally more pronounced if thesubstituent is present at the ortho or para position.

OH|

|OH

>

OH| NO2

>

OH|

NO2

>

OH|

Acid strength decreases

(iv) The increase in the number of substituents further increases or decreases the acidicstrength.

OHNO2NO2

NO2

>

OHNO2

NO2

>

OH

NO2

>

OH|

Acid strength decreases

Q. 19. Classify the following as primary, secondary and tertiary alcohols :

(i) CH3—

CH3|C—CH2OH|CH3

(ii) H2C = CH—CH2OH

(iii) CH3—CH2—CH2—OH (iv)

OH

CH—CH3

(v) OH

CH2—CH—CH3(vi) CH2

CH3

CH = CH—C—OH

Ans. (i), (ii) and (iii) are primary alcohols, (iv) and (v) are secondary alcohols and (vi) is tertiaryalcohol.


Q. 20. Describe Williamson’s ether synthesis.Ans. Williamson’s ether synthesis : This involves the treatment of a sodium alkoxide with an alkyl

halide. Sodium alkoxides are obtained by treating alcohols with sodium metal. Bothsymmetrical and unsymmetrical ethers can be made by this method.

(ROH + Na → RONa + H2).

R—–O

+Na + R′—X ⎯⎯→ R—O—R′ + NaX

Sodium alkoxide Alkyl halide Ether

CH3—–O

+Na + C2H5—Br ⎯⎯→ CH3—O—C2H5 + NaBr

Sodium methoxide Bromoethane Ethyl methyl ether

C2H5—–O

+Na + C2H5—Br ⎯⎯→ C2H5—O—C2H5 + NaBr

Sodium ethoxide Bromoethane Diethyl ether

Chapter 12. Aldehydes, Ketones and Carboxylic acidsQ. 1. Which compound gives odour of bitter almonds ?Ans. Benzaldehyde.Q. 2. Name the force that explains the boiling points of aldehydes and ketones.Ans. Dipole-dipole attraction.Q. 3. What type of addition reaction aldehydes and ketones undergoes ?Ans. Nucleophilic addition reaction.Q. 4. Which carbonyl compound does not gives the silver mirror test ?

Ans. Ketones.

Q. 5. What happen when calcium benzoate is distilled with calcium formate ?

Ans. When calcium benzoate is distilled with calcium formate it forms benzaldehyde.

Q. 6. Name the hybridisation which the carbon atom in carbonyl group involves.Ans. sp2 hybridisation.Q. 7. Which group gives Cannizzaro’s reaction ?Ans. Aldehyde with α-hydrogen.Q. 8. What happens when acetaldehyde is heated with Fehling’s solution ?Ans. It gives a precipitate of Cu2O.Q. 9. Acetaldehyde cyanohydrin on hydrolysis gives which acid ?Ans. α-hydroxy propionic acid.

Q. 10. When two molecules of acetaldehyde condenses in the presence of a dilute alkali it forms :Ans. Aldol.

Q. 11. The reaction 2C6H5CHO + NaOH → C6H5COONa + C6H5CH2OH is known as :Ans. Cannizzaro’s reaction.

Q. 12. Aldehydes gives silver mirror test with ammonical silver nitrate solution due to the formationof :

Ans. Ag (silver)Q. 13. Trimer of formaldehyde is called as ?

Ans. Meta formaldehyde.Q. 14. Reduction of aldehydes and ketones into hydrocarbons using Zn/Hg and conc. HCl is called :

Ans. Clemmensen’s reduction.Q. 15. Is Cannizzaro reaction given by all the aldehydes ?

Ans. No, only by those aldehydes which do not have α-hydrogen atom.Q. 16. What is a Cannizzaro reaction ?

Ans. A chemical reaction that involves the base induced disproportionation of an aldehyde lacking ahydrogen atom in the alpha position is called Cannizzaro reaction

2C6H5CHO + NaOH ⎯⎯→ C6H5COONa + C6H5CH2OHQ. 17. What is the name of 40% solution of formaldehyde ?

Ans. Formalin.


Q. 18. Name the synthetic plastic formed by fomaldehyde.Ans. Bakelite is the synthetic plastic formed by formaldehyde.

Q. 19. Name two dyes prepared by formaldehyde.Ans. Pararosaniline and Indigo.

Q. 20. Explain acidity of α-hydrogen in carbonyl compounds.Ans. Acidity of αααα-hydrogens : A carbon atom next to the carbonyl group is called an α-carbon. A

hydrogen attached to an α-carbon is referred to as an α -hydrogen. The α-hydrogens ofaldehydes and ketones are acidic in nature. The acidity is due to the fact that the anion, whichresults from the removal of an α -hydrogen by a base B

–, is stabilized by resonance. The

resonance-stabilized anion is called Enolate Ion.B

Removal of H by Base

Resonance-StabilizedEnolate ion

_C_C_H|

O||

|BH

: :

_C_C_O||

|

:..

..

_C=C_O|

|

: :..

The α-carbon of the enolate ion is negatively charged. It can act as a nucleophile. The formationof the enolate ion followed by its addition to a carbonyl group is the process involved in all thecondensation reactions of aldehydes and ketones.

Q. 21. Properties of aldehydes and ketones are due to carbonyl group, explain.Ans. The reactivity of aldehydes and ketones towards nucleophilic addition reaction is :

H H

C = O

CH3

H

C = O >

CH3

CH3

C = O

C2H5

CH3

C = O

C2H5

C2H5

C = O> > >

This order is due to :(i) Electron releasing nature of alkyl group, the electron density on the carbon atom of CO

group increases. As a result, the attacking tendency of the nucleophile decreases.(ii) The bulky alkyl group causes steric hindrance for the incoming nucleophile.

Q. 22. Write a note on aldol condensation and cross aldol condensation.Ans. (i) Aldol condensation : Two molecules of an aldehyde having at least one α-hydrogen atom,

condenses in the presence of a dilute alkali to give a β-hydroxyaldehyde.

CH3 — C

O| |

|H

+ HCH2 CHC dil. NaOH

⎯⎯⎯⎯⎯→ CH3 — C —

OH|

|H

CH2CHO

Aldol(3-hydroxybutanal)

When aldol heated with dilute acids it undergoes dehydration to form, α , β -unsaturatedaldehydes.

Aldol

CH3 — CH|OH

— CHCHO|OH

H+‚ Δ

⎯⎯⎯→–H2O

CH3

αCH =

αCHCHO

But-2-enal

(ii) Crossed Aldol condensation : Aldol condensation is not confined to similar molecules ofaldehydes but can also take place between different aldehydes. This type of condensationis called crossed or mixed aldol condensation.

CH3—C|H

= O + HCH2COCH3Acetone

KCN⎯→ CH3 — C —

OH|

H|

CH2COCH3

Acetaldehyde 4-hydroxy pentan-2-one


Q. 23. How benzaldehyde is prepared in lab ?Ans. Benzaldehyde is prepared by the oxidation of toluene in laboratory.

Toluene

CH3

CrO2Cl2CCl4

⎯⎯⎯→ ⎣⎢⎢⎡

⎦⎥⎥⎤CH(OCrCl2OH)2

Complex

H2O⎯⎯→

CHO

Benzaldehyde

Q. 24. Aldehydes are readily oxidised with the following reagents (give reaction) :(a) K2Cr2O7 (b) Tollen’s reagent (c) Fehling’s reagent

Ans. Aldehydes are readily oxidised to carboxylic acids with same number of carbon atoms withmild oxidising agents.

R—C|H

= O ⎯→

K2Cr2O7⎯⎯⎯⎯⎯⎯⎯→

Ag(NH3)2OH

Tollen’s reagent⎯⎯⎯⎯⎯⎯⎯→

RCOOH

RCOOH + Ag (Silver mirror)

CU2+‚ OH–

Fehlingsolution

⎯⎯⎯⎯⎯⎯⎯→ RCOOH + Cu2O (red ppt.)

Q. 25. What is obtained on oxidation of acetone ?Ans. Oxidation of acetone (Ketones) : Acetone (Ketones) are oxidised only under vigorous

conditions using powerful oxidising agents such as conc. HNO3 , KMnO4/H2SO4,K2Cr2O7/H2SO4, etc. Oxidation of ketones involves cleavage of bond between carbonyl carbonatom and α-carbon on either side of ketone group giving a mixture of carboxylic acids.

CH3—C||O

—CH3Acetone

[O]⎯⎯⎯⎯→Conc. HNO3

HCOOHFormic

acid

+ CH3COOHAcetic acid

Q. 26. Arrange the following compounds in increasing order of their property as indicated :Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone.(reactivity towards HCN)

Ans. The reactivity of aldehydes and ketones towards nucleophilic addition depends upon,(i) + I effect (ii) steric hindrance. Hence, the order isDi-tert-butyl ketone < methyl-tert-butyl ketone < Acetone < Acetaldehyde.

Q. 27. How would you obtain the following from named sources :(i) Tertiary butyl alcohol from acetone.(ii) Acetone from acetic acid. Mention conditions for reactions involved.

Ans. (i) Acetone on reaction with methyl magnesium bromide in presence of dry ether givesadditional compound and further on hydrolysis gives tertiary butyl alcohol.

CH3

CH3C

Acetone

= O +CH3MgBrMethyl

magnesiumbromide

etherdry

⎯⎯→

⎣⎢⎢⎢⎢⎡

⎦⎥⎥⎥⎥⎤CH3

CH3C

CH3

OMgBrAddition compound

H+/H2O

⎯⎯⎯→ Mg OH Br

+ CH3

CH3C

CH3

OHtert-butyl alcohol

(ii) Acetic acid is first converted to its calcium salt which is then subjected to dry distillation.

2CH3COOHAcetic acid

+ Ca(OH)2Slacked lime

(or lime water)

⎯⎯→ CH3CO

CH3COCa

Calcium acetate

heat

⎯⎯⎯→dry distill.

CH3

CH3C

Acetone

= O +CaCO3

Q. 28. Write the reaction to prepare acetaldehyde from hydrogen gas and an acid chloride.Ans.

CH3—||O

C—Cl 2 CH3—||O

C—H + HCl Pd/BaSO4

Acetyl chloride Hydrogengas

⎯⎯⎯⎯⎯⎯→Acetaldehyde(S)

+ H


Q. 29. Which of the following is a stronger acid in each of the following pairs ?(i) Monochloro, dichloro and trichloroacetic acid(ii) α-chloropropionic acid and β-fluropropionic acid(iii) Propionic acid, acetic acid and formic acid(iv) m-hydroxy benzoic acid and p-hydroxybenzoic acid(v) Acetic acid and benzoic acid

Ans. (i) Trichloroacetic acid (ii) α-chloropropionic acid(iii) Formic acid (iv) m-hydroxybenzoic acid(v) Benzoic acid

Q. 30. What is pyruvic acid ? Which optically active acid is obtained from it ?Ans. Pyruvic acid is a α-ketoacid with carboxylic acid and a ketone functional group. It can be made

from glucose through glycolysis CH3—

O||C—COOH (2-oxopropanoic acid). The optically active

acid obtained from it is lactic acid (2-hydroxypropanoic acid).Q. 31. Write the name of the esters which gives the flavour of pineapple and jasmine.

Ans. Pineapple—Ethyl butanoateJasmine—Benzyl ethanoate

Q. 32. During the formation of acid chloride, which bond C—OH or CO—H of carboxylic acid isbroken ?

Ans. C—OH bond of carboxylic acid is broken during the formation of acid chloride.Q. 33. What is H.V.Z. reaction ?

Ans. Hell Volhard Zelinsky reaction is known as HVZ reaction.

CH3COOHAcetic acid

Cl2⎯⎯→P CH2ClCOOH

Chloroacetic acid

Cl2⎯⎯→P CHCl2COOHDichloroacetic acid

Cl2⎯⎯→P CCl3COOH

Trichloroacetic acidQ. 34. What is vinegar and how could you obtain it ?

Ans. Vinegar is the dilute form of acetic acid. It is obtained by the fermentation of ethyl alcohol withbacteria acetobacter in the presence of air.

EthanolCH3CH2OH + O2

Acetobacter⎯⎯⎯⎯→

Acetic acidCH3COOH + H2O

Chapter 13. Organic Compounds Containing NitrogenQ. 1. What are cyanides ?Ans. Cyanides are the derivatives of hydrogen cyanide, in which H atom is replaced by alkyl or aryl

group.

Q. 2. Which out of C2H5CN and C2H5NC has repelling smell ?Ans. C2H5NC.Q. 3. What is Hinsberg reagent ? What is its use ?Ans. Benzesulphonyl chloride, C6H5SO2Cl2 is known as Hinsberg reagent. It is used to distinguish

primary, secondary and tertiary amines.Q. 4. What is vapour phase nitration ? Give one example.Ans. When nitration is carried out at higher temperature, so that the reactant and product molecules

exist in vapour phase only, it is known as vapour phase nitration.Vapours of methane and nitric acid gives the vapours of nitromethane at 623K.

CH4 + HNO3 623K

⎯⎯⎯→ CH3—NO2 + H2O

Q. 5. Which reagent can be used to prepare chloropicrin ?Ans. CH3NO2 and Cl2/NaOH or CHCl3 and HNO3.Q. 6. Which organic compound is formed when 2-nitropropane is hydrolyzed with boiling dilute

HCl ?Ans. Acetone.


Q. 7. Arrange the following in the decreasing order of basicity :NH3, CH3NH2, (CH3)2 NH, (CH)3 N

Ans. (CH3)2 NH > CH3 NH2 > (CH3) 3 N > NH3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→Decreasing order of basicityQ. 8. What do you mean by exhaustive methylation of amine ?Ans. The process of converting primary, secondary or tertiary amine into its quarternary ammonium

salt by using methyl iodide, is known as exhaustive methylation.

CH3NH21°-amine

CH3I

⎯⎯→–HI (CH3)2 NH2°-amine

CH3I

⎯⎯→–HI (CH3)3 N3°-amine

CH3I

⎯⎯→ [(CH3)4 N]+ I–

Quaternary ammonium salt

Q. 9. How would you prepare alkyl cyanides from :(i) Primary amines (ii) Alkyl halides (iii) Acid amides

Ans. (i) Alkyl cyanides are prepared by the dehydrogenation of primary amines. When primaryamines are passed over copper or nickel at 773 K alkyl isocyanides are formed. This isalso a commercial method.

RCH2NH2Cu or Ni⎯⎯→773°K RCN + 2H2

CH3CH2NH2Cu or Ni⎯⎯→773°K CH3CH2CN + 2H2

(ii) When alkyl halides are treated with ethanol aqueous solution of potassium cyanide,alkylcyanides are formed.

C2H5—Cl + KCN ⎯→ C2H5CN + KCl Ethylchloride Ethylcyanide

(iii) When acid amides are distilled with P2O5 they get dehydrated and forms alkyl cyanides.

CH3CONH2P2O5⎯⎯→distilled CH3CN + H2O

Ethanamide Ethanenitrile

Q. 10. Explain alkyl amines are stronger bases than ammonia.

Ans. Because of the electron releasing inductive effect of alkyl groups, it concentrates electrondensity on nitrogen and hence makes the lone pair of nitrogen more easily available for sharingwith acids. Thus, electron releasing effect of alkyl groups stabilizes the alkyl ammonium ionformed and hence shifts the equilibrium in forward direction making the alkyl amines strongerbases than ammonia.

R→ . .N + H2O R →

⊕NH3 + H

–

Q. 11. What is the relation between the basic character of amines, dissociation constant and pKb ?Ans. When any amine is dissolved in water it produces free OH– ions. Basic characters can be shown

by the following equations.

R. .NH2 + H—OH ⎯→ RNH3

+ + OH–

Amine (base) Water (acid) Alkylammonium ion Hydroxyl ion

Thus, the value of Kb is directly proportional to the basic strength of amine.

But pKb = – log Kb. Hence lower the value of pKb stronger will be the basic strength.Q. 12. Describe mild reduction in cyanides.

Ans. In cyanides when reduction is carried out with acidified stannous chloride (SnCl2/HCl) atroom temperature imine hydrochloride is obtained which on subsequent hydrolysis withboiling water gives aldehyde. This specific type of reduction of cyanides is called Stephen’sreduction.

CH3—C ≡ N SnCl2/HCl

2[H]⎯⎯⎯⎯⎯⎯⎯⎯→ CH3 CH = NH·HCl

Acetaldiminehydrochloride

H2O/Boil⎯⎯⎯⎯⎯→

Acetaldehyde

CH3CHO + NH4Cl


Q. 13. What happens when :(a) Ammonium acetate is heated in presence of P2O5 ?(b) Propane nitrile is heated in presence of LiAlH4 ?(c) Alkyl halides, on heating with ethanolic solution of AgCN ?(d) Acetamide is distilled with phosphorus pentaoxide ?(e) Ethyl iodide reacts with AgNO2 and product is reduced ?(f) Ethyl bromide is treated with AgCN and product is hydrolyzed ?(g) Ethyl magnesium bromide is treated with cyanogen chloride ?(h) Ethyl amine is treated with acetic anhydride ?(i) Ethyl magnesium bromide is treated with chloramine ?

Ans. (a) Ethanenitrile is formed.

CH3COONH4 P2O5⎯⎯→

Δ EthanenitrileCH3CN + 2H2O

(b) Propanamine obtained.

CH3CH2CN LiAlH4⎯⎯→ CH3CH2CH2NH2

Propanamine(c) Ethyl isocyanide obtained.

C2H5Br + AgCN C2H5OH⎯⎯→

ΔC2H5—N ⎯→⎯⎯⎯⎯ C + AgBr

Ethylisocyanide(d) Ethanenitrile (methyl cyanide) is formed.

CH3CONH2P2O5⎯⎯→distill. Ethanenitrile

CH3CN + H2O

(e) Ethyl amine is formed.

C2H5I + AgNO2Ethyl iodide

–AgI⎯⎯→ C2H5NO2

Sn/HCl ⎯⎯⎯→

[H] C2H5NH2

Ethylamine (f) Ethyl amine is formed.

C2H5Br + AgCN–AgBr⎯⎯→ C2H5NC

H2O/H+⎯→ C2H5NH2 + HCOOH

Ethylamine(g) Ethyl cyanide is formed.

C2H5MgBrEthyl magnesium

bromide

+ Cl—CNCyanogen

chloride

⎯⎯→ C2H5CN Ethyl cyanide

+ Mg

Cl Br

(h) N-Ethyl acetamide is formed.

CH3—CH2—NH|

H

+ O

O||C C||O

—CH3

—CH3

–CH3COOH⎯⎯⎯⎯→ CH3—CH2—N

|H

—

O||C—CH3

EthanamineAcetic anhydride

N-Ethylacetamide

(i) Ethanamine is formed.

C2H5MgBrEthyl magnesium

bromide

+ Cl.NH2Chloramine

⎯⎯⎯→ C2H5NH2Ethanamine

+ MgCl Br

Q. 14. Comparing the basic strength arrange the following in order of decreasing basic strength

· ·NH

N··

NH

O

· ·NH

I II III IV


Ans. (a) Among I and II, I is more basic. The lone pair in (I) is in sp3 hybrid orbital while, (II) it is insp2 hybrid orbital. Since sp2 hybrid orbital is relatively more electronegative than sp3 hybridorbital. Hence electron donating capacity of I is more than that of II.(b) Among I and III, electron donating capacity of III is less due to electron withdrawing effectof oxygen. However III is more basic than II.(c) Among II and IV, IV is less basic because IV is aromatic and the lone pair of N atom is usedup in cyclic II cloud of 6π electrons (According to Huckel’s rule). Hence availability of lone pairof N in IV is less.Conclusion : The basic strength decreases as

I > III > II > IVQ. 15. How the ethylamine is prepared from :

(i) Propanamide (ii) Ethanamide(iii) Nitroethane (iv) Acetaldehyde(v) Ethanol

Ans. (i) When propanamide is treated with Br2 and KOH, ethanamine is obtained.CH3CH2CONH2 + Br2 + 4KOH⎯→ CH3CH2NH2 + K2CO3 + 2KBr + 2H2O Propanamide Ethylamine

(ii) When ethanamide is reduced with LiAlH4, it gives ethylamine.

CH3CONH2 + 4 [H] ⎯→ CH3CH2NH2 + 2H2O Ethanamide Ethylamine

(iii) Nitroethane on reduction with Sn/HCl gives ethylamine.

CH3CH2.NO2 + 6 [H]Sn/HCl⎯⎯→ CH3CH2NH2 + 2H2O

Nitroethane Ethylamine

(iv) Acetaldehyde is first converted into amine by treating it with ammonia and amine onreduction form ethanamine.

AcetaldehydeCH3—

O||C—H + NH3

–H2O⎯→ CH3—C|H

= NH H2/Ni⎯⎯→[H] CH3CH2NH2

Ethylamine

Acetaldiamine

(v) When the vapours of ethanol and ammonia are passed over alumina at 723K and highpressure, ethylamine is formed.

CH3CH2OHEthanol

+ NH3 Al2O3⎯⎯→723 K Ethylamine

CH3CH2NH2 + H2O

Chapter 14. BiomoleculesQ. 1. The deficiency of which vitamin will cause the following disease :

(i) Scurvy (ii) HaemorrhagesAns. (i) Vitamin C (ii) Vitamin K.Q. 2. What do you observe when glucose is treated with bromine water ?Ans. When glucose is treated with bromine water, bromine water gets decolourised.Q. 3. What is isoelectric point ?Ans. Isoelectric point : It is the pH at which an amino acid has an equal tendency to migrate

towards oppositely charged electrodes during electrolysis.Q. 4. Deficiency of which vitamin will cause the following disease :

(i) Night blindness. (ii) Scurvy.Ans. (i) Vitamin A (ii) Vitamin CQ. 5. What is mutarotation ?Ans. The spontaneous change of specific rotation of an optically active substance with time is called

mutarotation.


Q. 6. What are the constituents of starch ?Ans. Amylose and amylopectin.Q. 7. What are zwitter ions ?Ans. A zwitter ion is a dipolar ion formed by neutralisation of acidic and basic centres present

within the molecule.Q. 8. Define native state with reference to protein.Ans. The sequence in which amino acids are linked together with the help of peptide bonds forms

native state of protein.Q. 9. What are the monosaccharides ? Name the functional group common to both glucose and

fructose.Ans. Monosaccharides are simple carbohydrates which cannot be hydrolysed to still simpler

carbohydrates. Alcoholic group is common to both glucose and fructose.Q. 10. What are the reducing sugars ?

Ans. The carbohydrates which are capable of reducing Tollen’s reagent and Fehling’s solution arecalled reducing sugar e.g., all monosaccharide and diasaccharide except sucrose is reducingsugar.

Q. 11. What do you understand by the term glycosidic linkage ?Ans. In oligosaccharides and polysaccharides, the two monosaccharide units are linked together by

an oxide or other linkage formed by the loss of a water molecule. Such a linkage between twomonosaccharide units through oxygen atom is called glycosidic linkage.

Q. 12. Classify the following into monosaccharides and disaccharides.Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Ans. Ribose, 2-deoxyribose, galactose and fructose are monosaccharides.Maltose and lactose are disaccharides.

Q. 13. How would you obtain glucose and fructose from sucrose ?Ans. By hydrolysis of sucrose in presence of mineral acid, glucose and fructose are obtained.

SucroseC12H22O11 + H2O

dil. HCl⎯⎯⎯⎯→

GlucoseC6H12O6 +

FructoseC6H12O6

Q. 14. What are the hydrolysis products of (a) sucrose and (b) lactose ?Ans. Hydrolysis products of sucrose are α-D-glucose and β-D-fructose whereas hydrolysis products

of lactose are β-D-galactose and β-D-glucose.Q. 15. Define proteins.

Ans. Proteins are polymers of amino acids, their molecular mass is more than 10000.Q. 16. Define amino acids.Ans. Amino substituted carboxylic acids are called amino acids.

Q. 17. What are α-amino acids ? Write their general formula.Ans. In α-amino acids, the amino group is present on the α-carbon atom, i.e., carbon atom next to

COOH— group.R—CH—

|NH2

COOH

Q. 18. What are essential and non-essential amino acids ?Ans. The amino acids which the body can synthesise are called non-essential amino acids. Those

amino acids which our body is not able to synthesise are called essential amino acids.Q. 19. What is zwitter ion ? Represent the zwitter ion of glycine.Ans. The doubly charged ion is known as zwitter ion. Glycine exists in aqueous solution as zwitter

ion.

H2NCH2COOH + H2O H3+NCH2CO

–O + H2O

Glycine Zwitter ionQ. 20. What is isoelectric point ? Do all amino acids have same isoelectric points ?Ans. The pH at which zwitter ion exists is known as isoelectric point of the α-amino acid. No,

different amino acids have different isoelectric points.Q. 21. Give one example of a fibrous protein. Name the final product of hydrolysis of proteins. What

is denaturation of proteins ?Ans. Myosin in muscles is a fibrous protein.

On hydrolysis of proteins, amino acid is obtained as a final product.


Denaturation of proteins : When proteins are heated or subjected to the action of alkalis etc.their physical and biological properties changes drastically without change in their chemicalnature. This process is called denaturation of proteins.

Q. 22. Define native proteins.Ans. A protein found in a living system with definite configuration and biological activity is called

native protein.Q. 23. What type of bonding occurs in globular proteins ?

Ans. Globular proteins have :(i) Cross-linked bonding (ii) Weak intermolecular hydrogen bonding.

Q. 24. Give two characteristics of enzymes.Ans. (a) They are highly specific in their action, each enzyme can catalyse only a specific type of a

reaction.(b) They can speed up an uncatalysed reaction up to the extent of ten million times.

Q. 25. Name the monomeric units of nucleic acids. Write three parts of nucleotide.Ans. Monomeric unit of nucleic acids are nucleotides.

The three parts of nucleotide are :(a) A phosphate group(b) A five carbon sugar(c) A nitrogen containing heterocyclic base.

Q. 26. Define gene and genetic code.Ans. The DNA sequences that acts as a code for the specific protein is called gene. The relation

between the nucleotide triplets and the amino acids is called genetic code.Q. 27. Which forces are responsible for the stability of α-helix ? Why is it named as 3·613 helix ?

Ans. Hydrogen bonds between —CO—and—NH—groups of peptide chains gives stability to theα-helix.The α-helix is also known as 3·613 helix because each turn of the helix has approximately 3·6aminoacids and a 13 member ring is formed by hydrogen bonding between —CO— and—NH— groups in different parts of the helix.

Q. 28. Name the following :(i) An α-amino acid which is not optically active.(ii) Pyrimidine bases present in RNA.(iii) Disease caused by deficiency of enzyme phenylalanine hydroxylase.(iv) The secondary structure of proteins.(v) The sweetest carbohydrate.(vi) A reducing sugar and a non-reducing sugar.

Ans. (i) Glycine(ii) Cytosine and uracil(iii) Phenylketone urea(iv) α-helix and β-sheet structure(v) Fructose(vi) Glucose is a reducing sugar while sucrose is a non-reducing sugar.

Q. 29. Explain the following terms :(i) Native protein (ii) Denaturation (iii) Renaturation

Ans. (i) Native protein : It is the protein found in a biological system with definite configurationand biological activity.(ii) Denaturation : When a protein is subjected to some physical or chemical treatment whichdisrupts its higher structure without affecting its primary structure, the process is calleddenaturation. The denatured protein loses its biological activity. For example, boiling of an egg.The denaturation may be reversible or irreversible.(iii)Renaturation : In many cases a denatured protein recovers its physical and chemicalproperties and biological activity when the disruptive agent is removed. This process, which isreverse of denaturation, is known as renaturation.

Q. 30. What are enzymes ? How do enzymes differ from ordinary chemical catalysts ?Ans. Enzymes are naturally occurring simple or conjugate proteins which acts as catalysts in the

biochemical reactions in living organisms.


Ordinary chemical catalysts were generally metals or inorganic compounds whereas enzymesare simple or conjugate proteins. Ordinary chemical catalysts are effective even at hightemperatures whereas enzymes are effective at body temperature. Enzymes are highly specificin nature.

Q. 31. Name five enzymes along with their functions.Ans. (i) Amylase converts Starch to Glucose.

(ii) Urease converts Urea to CO2 + NH3.(iii) Pepsin converts Proteins to Amino acids.(iv) Nuclease converts DNA, RNA to Nucleotides.(v) DNA Polymerase converts Deoxynucleotide triphosphates to DNA.

Q. 32. Write four characteristic features of enzymes. Name a disease which is caused by the deficiencyof a particular enzyme.

Ans. The four characteristic features of enzymes are :(i) They work at moderate temperature. (ii) They work at specific pH.(iii) They are specific in their action. (iv) Their activity is very high.

Deficiency of Enzyme Disease1. Pheylalanine hydroxylase 1. Phenyl ketone Urea2. Tyrosinase 2. Albinism

Q. 33. What are the peptides ? How are they classified ? What is peptide bond ?Ans. Peptides are the products formed by the condensation of two or more amino acids through

their amino and carboxylic groups involving elimination of water molecules. They may beclassified as dipeptides, tripeptides and polypeptides, depending upon whether the number ofamino acid molecules taking part in condensation is two, three or more than three respectively.The linkage (—CO—NH—) which unites various amino acid units in a peptide molecule is calledpeptide linkage or peptide bond.

H2N—CH|R

—C | |O

— OH + H N|H

Amino acid

—C|R

HCOOH ⎯⎯⎯⎯→Condense

H2NC|R

H —C—

O||C N

|H

—C|R

H—COOH

Q. 34. What is a nucleoside ? Write down the structure of a nucleoside which is present only in RNA.Ans. The N-glycosides of purine or pyrimidine bases with pentose sugars are

known as nucleosides.Base + sugar Nucleoside

The pentose sugar present in RNA is ribose while in DNA is 2-deoxyribose.The nucleoside formed by combination of uracil and ribose is presentonly in RNA.

C H2

HOHOH||

O

H

O

NH

ON

OH

H| |

H

Q. 35. What type of linkages are responsible for the formation of(i) Primary structure of proteins (ii) Cross-linking of polypeptide chains(iii) α-helix formation (iv) β-sheet structure.

Ans. (i) Primary structure of proteins arises due to peptide bonds (CO—NH—bond) betweenvarious constituent amino acids.

(ii) Hydrogen bonds or disulphide bonds.(iii) Hydrogen bonds between —C—

||O

and N|H

— groups of the same polypeptide chain.

(iv) Hydrogen bonds between —C— ||O

and N|H

— groups belonging to different polypeptide

chains.Q. 36. Describe what do you understand by primary structure and secondary structure of proteins.

Ans. Primary structure : The specific sequence in which the various α-amino acids are present in aprotein and are linked to one another is called its primary structure. Any change in primarystructure creates a different proteins.


Secondary structure : The conformation in which the polypeptide chain assembles as a result ofhydrogen bonding is known as secondary structure. The two types of secondary structure are α-helix and β-pleated sheet structure.In α-helix structure, the polypeptide chain form all the possible hydrogen bonds by twisting intoright handed screw (helix) with NH group of each amino acids residue hydrogen bonded to the

C = O group of an adjacent turn of the helix. In β-pleated structure, all peptide chains are

streched out to nearly maximum extention then laid side by side and are held together byhydrogen bonds.

Chapter 15. PolymersQ. 1. Name the monomers and the type of polymerisation in each of the following polymers :

(1) Terylene (2) Polyvinyl chloride

Ans. Polymer Monomer Type of Polymerisation

(1) Terylene Ethyl Glycol and Terephthalicacid

Condensation polymerisation

(2) Polyvinyl chloride Vinyl chloride Addition Polymerisation

Q. 2. Name the type of polymerisation (addition or condensation) and name the monomers in eachof the following polymers :

(i) Protein.

(ii) Polyethylene.

Ans. (i) Condensation polymerisation.

Monomer → Amino acids.(ii) Addition polymerization

Monomer → EthyleneQ. 3. What type of isomers are glucose and fructose ?Ans. Functional isomers.Q. 4. Name the functional group common to both glucose and fructose.Ans. Alcoholic group.Q. 5. What are polyamides ? Give one example of a polyamide and name its monomers.Ans. Polyamides are condensation polymers formed from acids and amines.

Nylon-6, 6 is a polyamide. Monomer units present in it are hexamethylene diamine and adipicacid.

Q. 6. What type of polymerization takes place when a polyester is formed ? Give one example of apolyester and name the monomers from which it is formed ?

Ans. Condensation polymerization, having ester linkage. For example : Terylene.It is an example of polyester, monomer units from which it is formed are ethylene glycol andterephthalic acid.

nHO—CH2—CH2—OH + n HO—

O||C— —

O||C—OH

Ethylene glycol Terepthalic Acid

425 – 475 K ↓– 2 n H2O

—CH2—CH2—O—

O||C— —

O||C n

Terylene

(O— —)—

Q. 7. Give two examples of natural polymers.Ans. Natural polymers

(a) Polysaccharides like starch(b) Proteins


Q. 8. Give one example each of addition and condensation polymer. Name the monomers in eachcase.

Ans. Addition polymer : For example , —[ CF2—CF2—] n Monomer : Tetrafluoroethylene CF2 = CF2Teflon

Condensation polymer : For example ,

O CH2 CH2 O C

O

C

O

nDacron(Terylene)

Monomer : Ethylene glycol HO—CH2—CH2—OH, Ethane–1, 2–diol and Terephthalic

acid CO

HO C OHO

Q. 9. What are polyolefins ? Give the reaction for the preparation of polythene, a polyolefin.Ans. Polymers obtained from unsaturated hydrocarbons (olefins) are called polyolefins.

Polythene is prepared by the addition polymerization of ethylene by two methods :(i) Polymerization at high pressure :

nCH2 = CH2 1000 – 5000 atm

⎯⎯⎯⎯⎯⎯→520 K‚ O2

( CH 2—CH2 )nLow density polythene

(ii) Polymerization at low pressure :

nCH2 = CH2 6-7 atm and 333 – 343 K⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Ziegler-Natta catalyst ( CH 2—CH2 )n

High density polythene

Q. 10. What is a plasticiser ?Ans. Organic compounds which when added to plastic makes them soft and workable are called

plasticiser.Q. 11. Define the term monomer.

Ans. The smallest unit which combines to form macromolecule (polymer) are called monomer.Q. 12. What is polymerization ?

Ans. The process in which large number of small molecules combines to form a giant molecule as amacromolecule.

Q. 13. Give one example each for chain growth and step growth polymerization.Ans. Polyvinyl chloride—chain growth polymerization.

Nylon-66—step growth polymerization.Q. 14. Give any one example of a cross-linked synthetic polymer.

Ans. Urea formaldehyde resin is a cross-linked synthetic polymer.Q. 15. Give one example of thermosetting and thermoplastic polymer.

Ans. Bakelite (thermosetting) and polyethene (thermoplastic).Q. 16. Give the example of linear polymer and a polyester polymer.

Ans. Linear polymer—PolyethylenePolyester polymer—Terylene.

Q. 17. To which category does each of these polymers belong, natural or synthetic ?(i) Starch (ii) Nylon-6,6(iii) Cellulose (iv) PVC

Ans. Natural polymer—Starch, Cellulose

Synthetic polymer—Nylon-6, 6, PVC

Q. 18. Give the name of one natural elastomer.

Ans. Natural rubber (linear chains).Q. 19. Which polymer is known as gutta-percha ?

Ans. A kind of rubber obtained from the bark of some trees. Here the carbon chains are trans withrespect to C = C bond.


Q. 20. How would you define plasticity ?Ans. When the shape of any substance is changed under pressure especially when hot and on

cooling the shape is gained due to sufficient viscous nature, this property is termed asplasticity. e.g., sealing wax shows plasticity.

Q. 21. Name the monomer which on polymerization can give benzene. What type of polymerizationis this– Addition or Condensation ?

Ans. Acetylene (monomer) on polymerization gives benzene. This is an addition type ofpolymerization.

Acetylene (monomer)

CH

CH+

CH

CH

CH|||CH

400° C⎯→

|||

|||+

Benzene (Polymer)

Q. 22. What is the monomer unit in nylon-6 ?Ans. Monomer unit of nylon-6 is caprolactum.

H2C

H2C

H2C—CH2

C = O

CH2

N

H|

Q. 23. What type of polymer is :(i) Protein (ii) Polyethylene

Ans. Protein—Natural condensation polymer of α-amino acids.Polyethylene—Synthetic addition polymer of ethylene.

Q. 24. Write the monomer of Terylene, Bakelite, Teflon, Kevlar or Nomex, Thiocol and Orlon.Ans. Terylene—terephthalic acid and ethylene glycol.

Bakelite—phenol and formaldehyde.Teflon—tetrafluoroethylene (F2C = CF2).Kevlar or Nomex—p-phenylene diamine + terephthaloyl chloride .Thiocol—Ethylene dichloride + sodium polysulphide.Orlon—Acrylonitrile.

Q. 25. Give two examples each of natural polymers, semi-synthetic polymer and polydienes.Ans. Natural Polymers : Starch (polymer of α–D glucose) and Natural rubber (polymer of isoprene).

Semi-synthetic polymers : Nitrocellulose and cellulose acetate.Polydienes : Buna rubber and Neoprene.

Q. 26. What is the difference between Nylon-6 and Nylon-6, 6 ?Ans. Nylon-6 is the polymer of caprolactum while nylon-6, 6 is the polymer of adipic acid

(1, 6-hexaminedioic acid) and hexamethylene diamine (1, 6-diaminohexane).Q. 27. Define the term co-polymer.

Ans. When a polymer is formed due to polymerization of two or more different monomer units, it isknown as copolymer. For example, buna-S rubber is the copolymer of 1, 3-butadiene andstyrene.

Q. 28. What do you mean by resins ?Ans. Amorphous, translucent or transparent organic solids or semisolids which usually have a

typical lustre are known as resins. For example, shellac secreted by the lac insects.Q. 29. What is Zeigler Natta catalyst ?

Ans. Zeigler Natta catalyst is a mixture of triethyl aluminium and titanium chloride.[Al (C2H5)3 + TiCl4]


Q. 30. Answer the following :(i) Which polymer is used in coating the utensils ?(ii) Which polymer is used in asbestos and cement ?(iii) Under which name polystyrene is sold.(iv) Which substance is used in the manufacture of magnetic recording tapes ?(v) What structural changes are produced during vulcanization of natural rubber ?

Ans. (i) Teflon(ii) Glyptal (poly-ethylene glycol phthalate)(iii) Styrofoan or styron(iv) Terylene(v) Sulphur cross-links are introduced between the chains.

Q. 31. Give abbreviation and chemical name of teflon.Ans. PTFE (Polytetra fluoroethylene).

Q. 32. (i) Give one example of a polyester used as a synthetic fibre.(ii) Name the compounds from which this polymer is prepared.

Ans. (i) Dacron or terylene is the polyester polymer which is used as a synthetic fibre.(ii) It is prepared by the condensation of ethylene glycol and terephthalic acid.

Q. 33. (i) Give any one example of a cross-linked synthetic polymer.(ii) With reference to the polymer named by you :

(1) Write the compounds from which it is prepared.(2) Give one physical property of the cross-linked synthetic polymer.

Ans. (i) Urea formaldehyde resin is a cross-linked synthetic polymer.(ii) (1) It is prepared from urea and formaldehyde.

(2) It is hard, rigid and thermosetting type of plastic.Q. 34. What is Buna-S and how it is synthesized ?

Ans. Buna-S is a copolymer of 1, 3-butadiene and styrene. It is obtained by the polymerization ofbutadiene and styrene in the ratio of 3 : 1 in the presence of Na. It is also known as StyreneButadiene Rubber (SBR).

n(CH2 = Butadiene CH—CH = CH

2) +

Styrenen(C6H5CH2) = CH2

Δ⎯⎯→Na (—CH2—CH = CH—CH2 —

|C6

(SBR)CH

H5

— CH2)n

Q. 35. Give the starting material of each of the following :

(i) PMMA (ii) PCTEF (iii) PAN (iv) PVC.

Ans. (i) PMMA—Poly methylmethacrylate

Starting material is—methylmethacrylate

nCH2 = C|CH3

|COOCH3

(ii) PCTFE—Polymonochlorotrifluoro ethylene.Starting material is—ChlorotrifluoroeEthylene.

nClFC = CF2

(iii) PAN—Polyacrylo nitrileStarting material is—Acrylonitrile

nCH3 = CHCN(iv) PVC—Polylinyl chloride

Starting material is—Vinyl chloridenCH2 = CHCl

Q. 36. Write the names and structures of the monomers of each of the following. Give one use of eachof the polymers.(i) Terylene (ii) Neoprene(iii) Natural rubber (iv) Polyvinyl chloride


Ans. (i) Terylene : Monomer is an ester, obtained by the condensation between ethylene glycol andterephthalic acid.

HO—C

O

+HO—CH2—CH2—OH

O

Terephthalicacid

Ethylene glycol

C—OH

Uses : Used as fabrics.

(ii) Neoprene : Monomer is chloroprene (2 chloro-1, 3-butadiene)

CH2 = CH— C = CH2|Cl

Chloroprene

Uses : Used for making shoes, shoeheels, stoppers.

(iii) Natural rubber : Monomer is isoprene (2-methyl, 1, 3-butadiene)CH2 = CH—C = CH2

|CH3

Isoprene

Uses : Used for making shoes, gloves, car tyres, conveyor.(iv) Polyvinyl chloride : Monomer is vinyl chloride.

CH2 = CH |Cl

Uses : Used in water pipes, PVC sheets, hoses, hand bags etc.

Q. 37. Give examples and the structure of polymers belonging to following categories.(i) Polyhalolefins (ii) Polyesters (iii) Polyacrylates

Ans. (i) Polyhalolefins : Examples—Teflon, PVC

Teflon(—CF2—CF2—)n ,

⎝⎜⎛

⎠⎟⎞—CH2—CH—

|Cl n

PVC(ii) Polyesters : Example—Terylene, Glyptal, Nylon 6, 6, Nylon 6, 10, Nylon 6

⎣⎢⎢⎡

⎦⎥⎥⎤

–

H|N—(CH2)6—

H|N—C

||O

—(CH2)4—

O||C–

n

(Nylon-6, 6)

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

—HN—(CH2)6—

H|N—

O||C—(CH2)8—

O||C— n

(Nylon-6, 10)

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

– HN—(CH2)5—

O||C – n

Nylon-6(iii) Polyacrylates : Examples—(a) Plexiglass or acrylate or perspex or lucite or PMMA.

(b) Polyethyl acrylate, (c) Orlon or Acrilon or PAN.

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

—CH2—

CH3|C|COOCH3

n

Lucite

,

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤—CH2—CH—

|COOC2H5

n Polyethyl acrylate

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤—CH2—CH

|CN

—

n

Orlon


Chapter 16. Chemistry in Everyday LifeQ. 1. What are competitive inhibitors ?Ans. The drugs which compete with the natural substrate or co-enzyme for active site and thus

inhibit the enzyme activity are called competitive inhibitors.Q. 2. What are receptors ?Ans. Proteins which are crucial to communication system in the body are called receptors.Q. 3. What are agonists ?Ans. Drugs that minic the natural messenger by switching on the receptor are called agonist.Q. 4. What is antiseptic ? Give an example.Ans. Chemical substance which prevents the growth of micro-organism or kill them but are not

harmful to living tissue are called antiseptic. Example : Dettol, Savlon etc.Q. 5. What are the main constituents of dettol ?Ans. Chloroxylenol and α-terpineol in suitable solvent.Q. 6. What is tincture of Iodine ?Ans. A 2-3 percent solution of iodine in alcohol water mixture is known as tincture of Iodine. It is

used as antiseptic.Q. 7. What is the mode of action of antimicrobial drugs ?Ans. Antimicrobial drugs can kill the micro-organisms such as bacteria, virus, fungi and other

parasites they can alternatively, inhibit the pathogen action of microbes.Q. 8. What is bacteriostatic drug ? Give example.Ans. A drug which inhibits or arrests the growth of the disease-causing organism is called a

bacterio-static drug, e.g. erthronycin, tetracycline.Q. 9. What is meant by narrow spectrum antibiotics ?Ans. Antibiotics which are mainly effective against gram positive or gram negative bacteria are

known as narrow spectrum antibiotics. e.g. penicillin.Q. 10. What is an antipyretic ? Give an example.

Ans. Chemicals which are used to bring down the body temperature during high fever are calledantipyretics. For example : Paracetamol, aspirin etc.

Q. 11. What is the nature of an antacid ?Ans. Substance which reduces and release excess HCl by preventing the interaction of histamine

with the receptors present in the stomach wall are called antacids. The most commonly usedantacids are cimctidine and reanitidine.

Q. 12. What are harmful effects of hyperacidity ?Ans. Ulcer development in stomach.

Q. 13. What are food preservatives ?Ans. Chemical substance which are used to protect food against bacteria, yeast and moulds are

called preservatives. For example : Sodium meta bisulphite, sodium benzoate. etc.Q. 14. What are artificial sweetening agents ? Give two examples.

Ans. Artificial sweetening agents are chemical substances which are sweet in taste but do not addany calories to our body for example : saccharin, aspartame etc.

Q. 15. What is aspartame and what is its use ?Ans. It is methyl ester of the dipeptide derived from phenyl alanine and aspartic acid. It is used as

an artificial sweetener.Q. 16. What are antioxidants ?

Ans. Antioxidants are important and necessary food additives that helps in food preservation byretarding the action of oxygen on food. These are more reactive towards oxygen than the foodmaterial they are protecting. For example: Butylated hydroxyl toluene (BHT) and butylatedhydroxyl anisole (B.H.A.)

Q. 17. What are soaps ?Ans. Soaps are sodium or potassium salts of fatty acids e.g. sodium palmitate.

Q. 18. What is the composition of cationic detergents ?Ans. Mostly acetates or chlorides of quarternary amine.


Q. 19. What is meant by the term broad spectrum antibiotics ? Explain.Ans. Broad spectrum antibiotics are effective against several different types of harmful bacteria.

Examples are tetracycline, toxacin, chloramphenicol etc. chloramphenicol can be used in case oftyphoid, acute fever, dysentry, urinary infections, meningitis and pneumonia.

Q. 20. What are biodegradable and non-biodegradable detergents ? Give one example of each.Ans. Detergents having straight hydrocarbon chains are easily degraded by micro-organisms and

hence are called biodegradable detergents, whereas detergents containing branchedhydrocarbon chains are not easily degraded by the micro-organisms and hence are called non-biodegradable detergents. As, a result, non-biodegradable detergents accumulate in river andwater ways thereby causing severe water pollution.

Q. 21. What is medicinal use of narcotic drugs ?Ans. Since narcotic drugs releive pain and produce sleep, these are chiefly used for the relief of post-

operative pain, cardiac pain and pain of terminal cancer and in child birth.Q. 22. What are scientific explanation for the feeling of depression ?

Ans. A person suffers from depression when he has low levels of noradrenaline. Noradrenaline is aneuro-transmitter that plays a role in mood changing. It lowers the signal sending activity andmakes the person suffer from depression.

Q. 23. What are the functions performed by histamine in the body ?Ans. Histamine is a potent vasodilator. It contracts muscles in the gut and bronchi. It relaxes some

other muscles, e.g. in the walls of blood vessels. Histamine is also responsible for congestion inthe nose associated with common cold and allergies. Also histamine stimulates the release ofpepsin and hydrochloric acid in the stomach.

Q. 24. What problem arises in using alitame as artificial sweetener ?Ans. Alitame is a high potency artificial sweetener. Therefore, the control of sweetness of food is

difficult while using it.Q. 25. What is common between the antibiotic arsphenamine and azodye ?

Ans. Arsphenamine possesses—As = As—linkage that resembles —N = N—linkage in azodyes.Q. 26. What is the basic difference between antiseptics and disinfectants ?

Ans. Antiseptics are applied to living tissues whereas disinfectants are applied to non-living objects.Q. 27. What is the advantage of using histamines over antacids in the treatment of acidity ?

Ans. Antacids control only the symptoms and not the cause. They work by neutralising one acidproduced in the stomach. They do not control and cause the production of more acid.Antihistamines are the drugs that suppress the action of histamines which is the chemicalresponsible for stimulation of secretion of pepsin and HCl in the stomach. Antihistaminesinfluences and prevents the binding of histamine with the receptors present in the stomachwall resulting in lower acid production and therefore better treatment.

Q. 28. Define the term chemotherapy.Ans. The branch of chemistry which deals with the treatment of disease using chemicals is called

chemotherapy.Q. 29. Give two examples of macromolecules that are chosen as a drug target.

Ans. Biological macromolecules such as carbohydrates, proteins, lipids and nucleic acids are chosenas drug targets.

Q. 30. Which forces are involved in holding the drug to the active site of enzymes ?Ans. Ionic bonding, hydrogen bonding, dipole-dipole interaction or van der Waals interaction.

Q. 31. Which site of an enzyme is called allosteric site ?Ans. Sites different from active sites of enzyme where a molecule can bind that affects the active site

are called allosteric sites. Some drugs may also bind at these sites.Q. 32. Name two types of chemical messengers.

Ans. Neurotransmitters and hormones.Q. 33. Where are receptor located ?

Ans. Receptors are embedded in cell membrane.Q. 34. Name two narcotics which are used as analgesics.

Ans. Morphine and Codeine are used as analgesics.


Q. 35. Name a drug used in mental depression.

Ans. Equanil or barbituric acid derivatives such as seconal, veronal, luminal etc.

Q. 36. Name a substance which can be used as an antiseptic as well as disinfectant.

Ans. 0·2% solution of phenol acts as antiseptic while 1% of its solution acts as disinfectant.

Q. 37. Name the chemical responsible for the antiseptic properties of dettol.

Ans. Chloroxylenol and terpineol.

Q. 38. Which type of drug comes under antimicrobial drugs ?

Ans. Antiseptic, antibiotics and disinfectants.

Q. 39. Mention the pharmacological effect of most sulphonamides.

Ans. Sulphonamides are antibacterial.

Q. 40. Name the antibiotic used specifically for treatment of typhoid fever.

Ans. Chloramphenicol.

Q. 41. Give the name of one broad spectrum antibiotic.

Ans. Tetracycline is a broad spectrum antibiotic.

Q. 42. Give one example of a (i) Bactericidal antibiotic (ii) Bacteriostatic antibiotic.

Ans. (i) Penicillin (ii) Tetracycline.

Q. 43. Name an antacid which prevents the formation of acid in the stomach.

Ans. Cimetidine or Ranitidine.

Q. 44. Write the name of antacid which is often used as a medicine.

Ans. Ranitidine (Zantac).

Q. 45. Name the sweetening agent used in the preparation of sweet for a diabetic patient.

Ans. Saccharin.

Q. 46. Name an artificial sweetener which is derivative of sucrose.

Ans. Sucralose.

Q. 47. Name two α-amino acids which forms a dipeptide which is 100 times more sweet than canesugar.

Ans. Asparatic acid and Phenylalanine.

Q. 48. How transparent soaps are manufactured ?

Ans. Dissolving soap in ethanol followed by evaporating excess of solvent.

Q. 49. Give an example of an alkyl benzene sulphonate detergent.

Ans. Sodium 4-(2-dedecyl) benzene sulphonate.❐

Question

10Set Numerical Problems

Chapter 1. Solid StateQ. 1. A compound AB has a simple cubic structure and has molecular mass 99. Its density is 3·4 g cm–3.

What will be the edge length of the unit cell ?Sol. Simple cubic structure molecular mass = 99

ρ = 3·4 g/cm3

Edge length = ?

ρ =Z × Ma3 × NA

g/cm3

3·4 =1 × 99

a3 × 6·023 × 1023

a3 =99

3·4 × 6·023 × 1023

a = 3·6 × 10–8 cmQ. 2. Chromium metal crystallizes with a body centered cubic lattice. The edge length of the unit cell is

found to be 287 pm. Calculate the atomic radius. What would be the density of chromium ing/cm3 ? (atomic mass of Cr = 52·99)

Sol. For body centred cubic cell atomic radius,

r = √⎯ 3 × Edge length4

= √⎯ 3 × 2874 = 124·27 pm.

Thus, atomic radius of chromium metal is 124·27 pm.

Density of the unit cell of the solid = Mass of unit cell

Volume of unit cellMass of unit cell = No. of atoms in unit cell (Z) × Mass of each atom (m)For body centered cubic lattice, the number of atoms in unit cell (Z) is

= 1 + ⎝⎜⎛

⎠⎟⎞1

8 × 8 = 1 + 1 = 2

Mass of each atom (m) =Atomic mass

Avogadro’s no. = 52·99

6·023 × 1023

The volume of the cubic lattice = (Edge length)3

= (287 × 10– 10)3

∴ Density of the unit cell, ρ =2 × 52·99

6·023 × 1023 × (287)3 × 10– 30 g/cm3

= 7·44 g cm–3

Thus, the density of chromium is 7·44 g/cm3.

Q. 3. A bcc element (atomic mass 65) has a cell edge of 420 pm. Calculate its density in g/cm3.

Sol. Density, ρ =Z × Ma3 × NA

Given : Z = 2 (for bcc arrangement),

M = 65, a = 420 pm = 420 × 10– 10 cmNA = 6·023 × 1023

Numerical Problems ■ 237

∴ ρ =2 × 65

(420 × 10– 10)3 × 6·023 × 1023

=2 × 65 × 104

(42)3 × 6·023= 2·91 g cm– 3

Density of given element is 2·91 g cm– 3.Q. 4. Lead(II) sulphide has FCC crystal structure. The edge length of the unit cell of PbS crystal is 500

pm. What is its density ? [Pb = 207.2, S = 32]Sol. For a cubic crystal, density ρ is given by

ρ =Z × Ma3 × NA

Given : Z = 4 formula units for FCC arrangement,M = (207·2 + 32) for PbS, a = 500 pm = 500 × 10–10 cm and NA = 6·02 × 1023

∴ Density of PbS, ρ = 4 × 239·2

(500 × 10–10)3 × 6·02 × 1023 = 12·715 g cm–3

Q. 5. Calculate the value of Avogadro’s number from the data : Density of NaCl = 2·165 g cm–3.Distance between Na+ and Cl– in NaCl crystal = 281 pm.

Sol. Edge length, a = 281 pmVolume of unit cell = (2·81)3 × 10–24 cm3

Density = 2·165 g cm–3

Weight of each unit cell = (2·81)3 × 10–24 × 2·165 g

Theoretical mass of unit cell =Atomic mass

Avogadro’s No.

∴ Avogadro’s No. =58·5/2

(2·81)3 × 10–24 × 2·165= 6·09 × 1023 mol–1.

Q. 6. A metal (atomic mass = 50) has a body centered cubic crystal lattice. The density of the metal is5·96 g cm–3. Find out the volume of the unit cell. (N = 6·023 × 1023).

Sol. Density of the metal = Density of the unit cell= 5·96 gm cm–3

No. of atoms in unit cell = 2 (BCC)Molar mass of metal = 50 g mol–1

Mass of one atom =50 g mol–1

6·023 × 1023 atoms mol–1

Mass of unit cell =2 × 50

6·023 × 1023 g

Volume of unit cell =Mass

Density

=2 × 50 g

6·023 × 1023 × 5·96 g cm–3

=100

6·023 × 5·96 × 10–23 cm3

= 2·785 × 10–23 cm3.Q. 7. An element A crystallizes in FCC structure 208 g of it has 4·2832 × 1024 atoms. Calculate the edge

length of the unit cell if density of A is 7·2 g cm–3.Sol. Let the edge length of the unit cell = a

∴ Volume of the unit cell = a3

Mass of an atom of A =208

4·2832 × 1024 = 4·856 × 10–23 g

Since in FCC structure, there are 4 atoms per unit cell


Density of unit cell =Mass of unit cell

Volume of unit cell

=4 × Mass of an atom of A

a3

7·2 g cm–3 =4 (4·856 × 10–23 g)

a3

a3 =4 × (4·856 × 10–23 g)

7·2 g cm–3

= 2·7 × 10–23 cm3 = 27 × 10–24 cm3

∴ a = 3 × 10–8 cm.The edge length of the unit cell = 3 × 10–8 cm.

Q. 8. Show by simple calculation that the percentage of space occupied by spheres in a simple cubicunit cell is 52·4%.

Sol. In a simple cubic unit cell there is only one atom per unit cell.Let a be the edge length of the unit cell and r be the radius of sphere.

Volume of the sphere =43 πr3

As the spheres at corners are touching each other, the edge length a is equal to 2ra = 2r

∴ Volume of the cube = a3 = (2r)3 = 8r3

% of the space occupied by the spheres =Volume of sphereVolume of cube × 100 =

4/3 πr3

8r3 × 100

=43 ×

227 ×

18 × 100 = 52·4%.

Q. 9. An element crystallizes in FCC lattice having edge length 400 pm. Calculate the maximumdiameter of atom which can be placed in interstitial site without disjoining the structure.

Sol. In FCC, interstitial sites will be octahedral and tetrahedral voids.

For tetrahedral voids, r1

r2= 0·225

For octahedral voids,r1

r2= 0·414

where r1 = radius of atom in interstitial sites, r2 = radius of atom arranged in FCC

So, 4r2 = √⎯ 2aFor maximum diameter of atom in interstitial site, octahedral voids will be considered

Diameter = 2r1 = 2 (0·414 r2)

= 2 × 0·414 × 400

2√⎯ 2 = 117·1 pm.

Q. 10. The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 Å whichcrystallizes in cubic lattice. If the density is 2 g/cc then find the radius of metal atom,(NA = 6·02 × 1023). Give the answer in pm.

Sol. For cubic lattice, ρ =Z × A

NA × V

or Z =ρ × NA × V

AGiven : A = 75 g/mol, V = (5Å)3 = (5 × 10–8)3 cm3 = 125 × 10–24 cm3 and NA = 6·02 × 1023,ρ = 2 g/cc

∴ Z =2 × 6·02 × 1023 × 125 × 10–24

75 ≈ 2

The value of Z represents that element must have body centered cubic (BCC) structure.

For BCC, atomic radius, r = √⎯ 34 a


or r = √⎯ 34 × 5 = 2·165 Å = 216·6 pm.

Hence, the radius of metal atom is 216·6 pm.Q. 11. The length of the unit cell edge of a body centered cubic metal crystal is 352 pm. Calculate the

radius of the metal atom.Sol. Let the radius of the metal atom is r.

∴ Body diagonal = r + 2r + r = 4r [In a BCC unit cell, the atoms touch eachother along the body diagonal]

Body diagonal = √⎯ 3 × Edge length = √⎯ 3 a

∴ 4 r = √⎯ 3 a

or r = √⎯ 34 a

or r = √⎯ 34 × 352 = 152.4 pm.

Hence, the radius of metal atom is 152·4 pm.Q. 12. The edge length of unit cell of a body centered cubic (BCC) crystal is 352 pm. Calculate the radius

of the atom.Sol. BCC crystal a = 352 pm, r = ?

∴√⎯ 34 a = r

r = 152·42 pmQ. 13. In a body centered and face centered arrangement of atoms of an element, what will be the

number of atoms present in respective unit cells ? Justify your answer with calculation.

Sol. Body centered cubic : Number of atoms at the corners per unit cell = 18 × 8 = 1

Number of atoms present at the centre of the cube = 1∴ Total number of atoms in BCC arrangement = 1 + 1 = 2

Face centered cubic : Number of atoms present at the corners per unit cell = 18 × 8 = 1

Number of atoms present at the faces per unit cell = 12 × 6 = 3

∴ Total number of atoms in FCC arrangement = 1 + 3 = 4

Chapter 2. SolutionsQ. 1. A 10% aqueous solution of cane sugar (mol. wt. 342) is isotonic with 1·754% aqueous solution of

urea. Find the molecular mass of urea.Sol. van’t Hoff factor (i) = 1

πcane sugar = πurea

C1RT = C2RTC1 = C2

⎣⎢⎢⎡

⎦⎥⎥⎤10

342

=

⎣⎢⎢⎡

⎦⎥⎥⎤1·754

x

x =1·754 × 342

10x = 59·9868 gm

Molecular mass of urea is 59·9868 gm.Q. 2. The molecular weight of an organic compound is 58 g mol–1. What will be the boiling point of a

solution containing 48 g of the solute in 1200 gram of water ?[Kb for water = 0·513°C kg mole–1; Boiling point of water = 100°C.]


Sol. Given : Molecular weight w = 58 g/mol, M = 48 g, W = 1200 g H2O, Kb for water = 0·513°C kgmole–1 and Boiling point of water = 100°C (Tb)Boiling point of solution = ?

ΔTb =1000 × Kb × w

M × W

ΔTb =1000 × 0·513 × 58

48 × 1200ΔTb = 0·516°CΔTb = T + Tb

T = 100·516°CQ. 3. What will be the value of van’t Hoff factor (i) of benzoic acid if it dimerises in aqueous solution ?

Sol. i =Particles after association

Particles before association

=12 = 0·5

Q. 4. A solution containing 0·5 g of KCl dissolved in 100 g of water and freezes at – 0·24°C. Calculatethe degree of dissociation of the salt. [Kf for water = 1·86°C] Atomic weights of K = 39 andCl = 35·5]

Sol. Given : Wt. of KCl (w) = 0·5 gWt. of water (W) = 100 g

Kf for water = 1·86°CMolecular wt. (m) of KCl = 39 + 35·5 = 74·5

Apply the relation of freezing point depression (ΔTf) with molecular weight of the solute :

ΔTfcal =Kf × 1000 × w

W × m = 1·86 × 1000 × 0·5

100 × 74·5= 0·124°C

ΔTfobs = 0°C + 0·24°C = 0·24°C

van’t Hoff factor (i) =ΔTfobs

ΔTfcal =

0·240·124 = 1·93

The degree of dissociation (α) of KCl is :

α =i – 1n – 1 where, n = number of ions.

Here n = 2 for K+ and Cl–

=1·93 – 1

2 – 1 = 0·93

1∴ α = 0·93 or 93%Thus, the degree of dissociation of the salt is 93% or 0·93.

Q. 5. If 1·71 g of sugar (molar mass = 342) is dissolved in 500 ml of an aqueous solution at 300 K, whatwill be its osmotic pressure ?

Sol. Given : Wt. of solute (w) = 1·71gMolar mass of solute (m) = 342

Volume of solution (V) = 500 ml = 0·5 LTemperature (T) = 300 KGas constant (R) = 0·0821 lit atm K–1 mol– 1

Osmotic pressure (π) =wRTmV

=1·71 × 0·0821 × 300

342 × 0·5 = 0·25 atm

Thus, osmotic pressure of the solution is 0·25 atm.


Q. 6. 0·70 g of an organic compound when dissolved in 32 g of acetone produces an elevation of 0·25°Cin the boiling point. Calculate the molecular mass of organic compound (Kb for acetone = 1·72Kmol–1).

Sol. Given : Wt. of the solute (w) = 0·70 gWt. of the solvent (W) = 32 g

Elevation of boiling point (ΔTb) = 0·25°CKb = 1·72 K kg mol– 1

The relation of elevation of boiling point and molecular mass is :

m =1000 × Kb × w

W × ΔTb

=1000 × 1·72 × 0·70

32 × 0·25= 150·5 g

Thus, the molecular mass of the organic compound is 150·5 g.Q. 7. What will be the vapour pressure of a solution containing 5 mole of sucrose (C12H22O11) in 1 kg of

water, if the vapour pressure of pure water is 4·57 mm of Hg ? [C = 12, H = 1, O = 16]Sol. According to Raoult’s law :

Psolution = Posolvent × xsolvent

= 4·57 ×

100018

100018 + 5

= 4·57 × 55·55

55·55 + 5= 4·192 mm of Hg.

Q. 8. A 2 molal solution of sodium chloride in water causes an elevation in the boiling point of waterby 1·88 K. What is the value of van’t Hoff factor ? What does it signify ? [Kb = 0·52 K kg mol–1]

Sol. Elevation in boiling point ΔTb is a colligative property and

ΔTb = i × Kb × Molality

Given : ΔTb = 1·88 K.

Kb = 0·52 K kg mol–1, Molality = 2 and i is van’t Hoff factor.

∴ i =1·88

0·52 × 2

i = 1·8076

Value of ‘i’ in this case signifies that when 1 mole of NaCl is dissolved in water it dissociates andfurnishes 1·8076 mole of particles in the solution.

Q. 9. Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol to beadded to 4 kg of water to prevent it from freezing at – 6°C. (Kf for H2O = 1·85 K mole– 1 kg)

Sol. Depression in freezing point ΔTf = 1000 × Kf × w

W × MIn the given case w = 4 kg = 4000 g, Kf = 1·85, ΔTf = 0 – (– 6°) = 6°C andMass of ethylene glycol (CH2OH—CH2OH ) = 62∴ Amount of ethylene glycol required,

w =M × w × ΔTf

1000 × Kf

=62 × 4000 × 61000 × 1·85 = 804·32 g.

Q. 10. The freezing point of a solution containing 0·3 gm of acetic acid in 30 gm of benzene is loweredby 0·45 K. Calculate the van't Hoff factor.(At. wt. of C = 12, H = 1, O = 16, Kf for benzene = 5·12 K kg mole– 1).


Sol. If i is the van’t Hoff factor then depression in freezing point

ΔTf =i × 1000 × Kf × w

W × MGiven : ΔTf = 0·45 K, w = 0·3 g, W = 30 gM of acetic acid, CH3COOH = 60 and Kf = 5·12

∴ i =ΔTf × W × M1000 × Kf × w =

0·45 × 30 × 601000 × 5·12 × 0·3

i = 0·5273van’t Hoff factor is 0·5273.

Q. 11. A solution of urea in water has a boiling point of 100·18°C. Calculate the freezing point of thesolution. (Kf for water is 1·86 K kg mol–1 and Kb for water is 0·512 K kg mol–1).

Sol. Given : Boiling point of solution = 100·18°C, Kb = 0·512 K kg mol–1, Kf = 1·86 K kg mol–1,ΔT Boiling = 100·18° – 100° = 0·18, ΔT = Kb × Molality∴ Molality = 0·18/0·512Now ΔTfreezing = Kf × Molality

= 1·86 × 0·180·512 = 0·6539

∴ Freezing point of solution = 0 – 0·6539°C= – 0·6539°C.

Q. 12. A solution of lactose containing 8·45 g of lactose in 100 g of water has a vapour pressure of 4·559mm of Hg at 0°C. If the vapour pressure of pure water is 4·579 mm of Hg, calculate the molecularweight of lactose.

Sol. Given : Weight of lactose, w = 8·45 g,Weight of water, W = 100 g,

Vapour pressure of pure water = 4·579 mm of Hgand Vapour pressure of solution = 4·559 mm of Hg.Now, according to Raoult’s law,

ΔPP =

w × Mm × W

4·579 – 4·5594·579 =

8·45 × 18m × 100

m =8·45 × 18 × 4·579

0·02 × 100 = 348·23 ≈ 348

Molecular mass of lactose is 348 g.Q. 13. 46 gm of ethyl alcohol is dissolved in 18 gm of water. Calculate the mole fraction of ethyl alcohol.

(At. wt. of C = 12, O = 16, H = 1).

Sol. Xalc =nalc

nalc + nwater

=46/46

46/46 + 18/18 = 12 = 0·5.

Q. 14. The osmotic pressure of 0·01 molar solution of an electrolyte is found to be 0·65 atm at 27°C.Calculate the van’t Hoff factor. What conclusion can you draw about the molecular state of thesolute in the solution ?

Sol. Given : π = 0·65 atm; c = 0·01 M, R = 0·0821 L atm K–1 mol–1 and T = 273 + 27 = 300 KNow, π = i × CRT

0·65 = i × 0·01 × 300 × 0·0821

i =0·65

0·01 × 300 × 0·0821 = 2·639.

van’t Hoff factor i = 2·639. It indicates that the electrolyte undergoes dissociation in givensolution.


Q. 15. What is the mass of a non-volatile solute (molar mass 60) that needs to be dissolved in 100 g ofwater in order to decrease the vapour pressure of water by 25% ? What will be the molality of thesolution ?

Sol. Given : ΔP = 25 (atm), P = 100 (atm).m = 60, W = 100 g and M = 18 (H2O)

According to Raoult’s law,ΔPP = xsolute =

nn + N

If w is the mass of non-volatile solute then on entering given values in the equation25100 =

w/60w/60 + 100/18

4 = 1 + 60 × 100w × 18

3 =600018 w ⇒ w = 111·11 g.

Molality of the solution =No. of mole of soluteMass of solvent in kg

=w/60

100/1000

=111·11/60100/1000 = 18·518

∴ Mass of solute required is 111·11 g and molality of the solution is 18·518.

Q. 16. The boiling point of pure water is 373 K. Calculate the boiling point of an aqueous solutioncontaining 18 gm of glucose (MW = 180) in 100 gm of water. Molal elevation constant of water is0·52 K kg mol– 1.

Sol. Mass of solvent (WA) = 100 g, mass of solute (WB) = 18 g

molecular mass of solute (MB) = 180

Molality =WB × 1000MB × WA

=18 × 1000180 × 100 = 1 mol kg– 1

Δ Tb = Kb × m = 0·52 × 1 = 0·52 K

Boiling point of solution = ⎣⎢⎢⎡

⎦⎥⎥⎤Boiling point of

pure solvent

+ ⎣⎢

⎢⎡

⎦⎥⎥⎤Elevation in

boiling point

= 373 + 0·52 = 373·52 KQ. 17. Equal weights of two substances X and Y were dissolved in equal volumes of water. The osmotic

pressure of the solution containing Y is five times the osmotic pressure of the solution containingX. What is the molecular weight of X if that of Y is 60 ?

Sol. From van’t Hoff equation π =nV · RT

For solution X, πx =nX

V ·RT …(i)

For solution Y, πY =nY

V ·RT …(ii)

From equation (i) and (ii),

πX

πY=

nX

nY…(iii)

but number of mole, n =Weight of the substance‚ WMol wt. of the substance‚ M


Hence nX =WX

MX and nY =

WY

MY

but WX = WY

and πY = 5πX

Therefore, from equation (iii),πX

5πX=

WX/MX

WY/MY[˙.˙ WX = WY and πY = 5πX]

15 =

MY

MX

but molecular weight of Y, MY = 60

Hence, molecular wt. of X, MX = 5 × 60 = 300

Q. 18. Which of the following solutions will have a lower vapour pressure and why ?(1) A 5% solution of cane sugar (C12 H22 O11)(2) A 5% solution of urea (NH2CONH2)

(Relative atomic masses : H = 1, C = 12, O = 16, N = 14)

Sol. No. of mole of cane sugar =Wt. of cane sugar

Mol. wt. of cane sugar

=5

342 = 0·014

No. of mole of urea =Wt. of urea

Mol. wt. of urea

=560 = 0·083

5% urea solution will have lower vapour pressure because according to Raoult's law larger thenumber of mole of solute in the solution greater is the lowering of vapour pressure.

Q. 19. If the molality of an aqueous solution of cane sugar is 0.4445, what is the mole fraction of canesugar ?

Sol. The mole fraction of cane sugar solution

=0·4445

0·4445 + 100018

= 0·4445

0·4445 + 55·55

=0·4445

56Mole fraction = 0·0079.

Q. 20. Albumins are the most abundant proteins in blood. At 25°C, 3.5 g of albumin in 100 ml of waterproduces an osmotic pressure of 0.014 atm. What is the molecular weight of albumin ?

Sol. From m =wRTπV

Given : w = 3·5 g, R = 0·0821 atm mol– K– 1, T = 273 + 25 = 298 K, π = 0·014 atm, V = 100 ml= 100 × 10– 3 litre.

m =3·5 × 0·0821 × 2980·014 × 100 × 10– 3

=35 × 821 × 298

140= 61164·5 g mol– 1

Q. 21. An aqueous solution containing 1·70 g of cane sugar in 100 ml water begins to freeze at – 0·093°C.The cryoscopic constant (molal depression constant) of water is 1·86 K kg mole– 1. Calculate themolecular weight of cane sugar.


Sol. ΔTf =Kf × w × 1000

m × W

0 – (– 0·093) = 1·86 × 1·70m

× 1,000100

∴ m =1·86 × 1·70 × 1,000

0·093 × 100 = 340 g mol– 1.

Q. 22. A decinormal solution of sodium chloride exerts an osmotic pressure of 4·82 atmosphere at 27°C.Calculate the degree of dissociation of sodium chloride.

Sol. From van’t Hoff equationπ = i CRT …(i)

NaCl dissociates in solution to give Na+ and Cl–. Suppose we start with 1 mole of NaCl and α isthe degree of dissociation.

NaClΔ

⎯→ Na+ + Cl–

Initially : 1 0 0At equilibrium : 1 – α mole α mole α moleTotal moles 1 – α + α + α = 1 + α

van’t Hoff factor i =1 + α

1Substituting the values in the equation (i)

4·82 = i × 0·1 × 0·0821 × 300

i =4·82

0·1 × 0·0821 × 3001 + α

1 = 48201 0 0821 300

⋅⋅ × ⋅ ×( )

α = 48201 0 0821 300

⋅⋅ × ⋅ ×( ) – 1

α = 0·96Hence degree of dissociation = 96%.

Q. 23. Calculate the vapour pressure of a solution at 100°C containing 3 g of cane sugar in 33 g of water.Sol. We know that,

PoA – PA

PoA

= XB = WB/MB

WA/MA…(1)

Given : PoA =760 mm of Hg, WB = 3 g, MB = 342, WA = 33 g and MA = 18, PA = ?

Putting the values in equation (1)

760 – PA

760 =

33423318

760 – PA

760 = 0·00478.

∴ PA = 756·36 mm of Hg

∴ Vapour pressure of the solution = 756·36 mm of Hg.

Q. 24. A solution to be used in a hand lotion is prepared by mixing 90 gm of water, 9·2 gm ethyl alcoholand 18·4 gm of glycerol (C3H8O3). Calculate the mole fraction of glycerol present in it.

Sol. Mole of water, nH2O =Wt. of water

Mol. wt. of water = 9018 = 5 mole

Moles of ethyl alcohol, nEA =Wt. of E.A.

Mol. wt. of E.A. = 9·246 = 0·2 mole


Mole of glycerol, nG =Wt. of glycerol

Mol. wt. of glycerol = 18·492 = 0·2 mole

Mole fraction of glycerol =nG

nH2O + nEA + nG =

0·25 + 0·2 + 0·2 = 0·037.

Q. 25. How many gram of carbon dioxide gas is dissolved in a 1 L bottle of carbonated water if themanufacturer uses a pressure of 2·4 atm in the bottling process at 25°C ?

Sol. According to Henry’s law,P = KHC

where, P = Partial pressure of gas, KH = Conc. of dissolved gas in solution

C =P

KH

C = 2·4 atm/29·76 atm/C mol/LC = 0·08 mol/L

Since we only have 1 L of water, we have 0·08 mole of CO2

Converting mole to gramMass of 1 mol of CO2 = 12 + (16 × 2) = 12 + 32 = 44 g

g of CO2 = mol CO2 × (44 g/mol)g of CO2 = 8·06 × 10– 2 mol × 44 g/mol.g of CO2 = 3·55 g.

Thus, 3·52 g of CO2 is dissolved in a 1 L bottle of carbonated water from the manufacturer.Q. 26. The partial pressure of O2 in air is 0·20 bar. The solubility of O2 in water in presence of air at 25°C

is 3·2 × 10– 4 M. Calculate Henry’s constant. (KH = Henry’s constant)Sol. P = KHm

KH =Pm

=0·20 bar

3·2 × 10– 4 M= 625 bar/M.

Q. 27. The solubility of oxygen in water is 0·06945 g per litre at STP. Calculate the pressure at which1 litre of water dissolves 2·0 g of oxygen at 0°C.

Sol.P1

P2=

C1

C2

P2 = P1 × C2

C1 = 1 atm ×

2·0 g0·06945 g = 28·8 atm.

Q. 28. If 27 g of acetylene, C2H2, dissolves in 1·00 L of acetone at 1·0 atm pressure, (a) what is theHenry’s law constant and (b) what is the solubility in acetone if the partial pressure of C2H2 is12·5 atom ?

Sol. Begin with Henry’s law m = KH P(a) Substituting, 27·0 g C2H2/1·00 L = 27·0 g/L

KH = 1 atm.

So, KH =(27·0 g/L)

1·0 atm = 27·0 g/L atm– 1

(b) m = KH P = (27·0 g/L atm– 1) (12·5 atm)m = 337·5 g/L

The mass or solubility of gas soluble in 1L of acetone is 337·5 g/L.Q. 29. The Henry’s law constant of H2(g) in water is 5·34 × 107 torr. Calculate the solubility of this gas in

water at 29 K if its partial pressure over the solution is 760 torr. Assume that the density of thesolution is same as the density of solvent.

Sol. X2 =n2

n1 + n2 ≈

n2

n1(assuming that the number of mole of H2, ( n2) is negligible in

comparison with the number of moles, (n1) of the solvent)


n1 =1000 g

18 mol–1 = 55·5 mol, so that

X2 =n2

55·5P2 = KHX2 (Henry’s law)

KH = p2

X2

5·34 × 107 =760

n2/55·5 = 760 torr × 55·5 mol

n2

∴ n2 =760 torr × 55·5 mol

5·34 × 107 torr= 8 × 10– 4

The solubility of H2 gas in water is 8 × 10– 4 mol/dm3

Q. 30. The osmotic pressure of a dilute aqueous solution of a compound X containing 0·12 g/L is twicethe osmotic pressure of a dilute aqueous solution of another compound Y containing 0·18 g/L.What is the ratio of the molecular weight of X to that of Y ? Both X and Y remain in molecularform in solution.

Sol. We know that, πV =WM . RT

For the solution containing X, π1V1 =W1

M1 . RT …(i)

For the solution containing Y, π2V2 =W2

M2 . RT …(ii)

But V1 = V2 = 1 litre, W1 = 0·12 g, W 2 = 0·18 g and π1 = 2π2

After putting these values in equations (i) and (ii), we get

2π2 =0·12M1

RT …(iii)

and π2 =0·18M2

RT …(iv)

From equations (iii) and (iv),2π2

π2=

0·120·18 ×

M2

M1

or2 × 0·18

0·12 =M2

M1

3 =M2

M1 or 3M1 = M2

Hence, the ratio of the molecular weight of X to Y will be 3 : 1.Q. 31. 5·0 gm of a substance with molecular weight 200 is dissolved in 50 gm of a solvent with

molecular weight 60 and vapour pressure 40 cm Hg at 27°C. Calculate the vapour pressure of thesolution at this temperature.

Sol. Number of mole of solute nB =Wt. of solute

Mol. wt. of solute

=5

200 = 0·025

Mole of solvent, nA =5060 = 0·833

Mole fraction of solute, XB =nB

nA + nB =

0·0250·025 + 0·833

=0·0250·858 = 0·029


According to Raoult’s law, P°A – PS

P°A= XB P°A and Ps = V.P. of pure solvent and solution.

Given, P°A = 40 cm Hg

Hence,40 – PS

40 = 0·029

PS = 38·84 cm Hg.Q. 32. (i) The constant for the depression of freezing point of water is 1·86 K per mole per kg. An

aqueous solution of sugarcane freezes at – 0·135 °C, what is the molality of the solution ?(ii) If the molality of an aqueous solution of sugarcane is 0·4445, what is the mole fraction of

sugarcane ? Sol. (i) We know that the depression of freezing point,

ΔTf = Kf.m ⇒ m = ΔTf

Kf

Given, Kf = 1·86 K, ΔTf = 0° – (– 0·135°C) = 0·135°C

Hence , m =0·1351·86 = 0·073

(ii) Given the molality of aqueous solution of sugarcane = 0·4445, i.e., this quantity present in1 kg or 1000 gm of water

Hence, the number of mole of water (nA) =Wt. of water

Mol. wt. of water

=100018 = 55·5555

Mole fraction of sugarcane XB =nB

nA + nB

=0·4445

55·5555 + 0·4445

=0·4445

56 = 0·0079.

Q. 33. At 37°C the osmotic pressure of blood is 7·65 atm. How much glucose (Molecular wt. = 180)should be used per litre for an intravenous injection that is isotonic with blood ?[Hint : Isotonic solution have the same osmotic pressure, thus, the intravenous injection shouldhave 7·65 atm osmotic pressure.]

Sol. Given π = 7·65 atm, MB = 180, V = 1 L, T = 273 + 37 = 310 K, R = 0·082 L atom K–1mol–1, WB = ?

We know, π =WB × RTMB × V

WB =π × MB × V

RT

=7·65 × 180 × 10·082 × 310

= 54·169 gm.Q. 34. The freezing point of a solution containing 0·2 gm of acetic acid in 20·0 g of benzene is lowered by

0·45°C. Calculate degree of association of acetic acid in benzene. Kf for benzene is = 5·12 K gmmole–1.

Sol. ΔTf (Theoretical) =1000 × Kf × WA

WB × MB

=1000 × 5·12 × 0·2

20 × 60= 0·853°C

Δ Tf (obs) = 0·45°

i =0·450·853

= 0·528


2CH3COOH → (CH3COOH)2

∴ Association = i – 1

1/m – 1 =0·528 – 11/2 – 1 =

– 0·472– 0·5

= 0·944 or 94·4%.Q. 35. The freezing point of ether was lowered by 5·4°C on dissolving 20 g of phenol in 100 g of ether,

Calculate the molar mass of phenol [Kf (ether) = 5·12 K mol– 1].Sol. We know that,

Given : ΔTf = Depression of freezing pointΔTf = 273 – (5·4 + 273) = 5·4 K, WB = 20 g,WA = 100 g, and Kf = 5·12 K mol– 1.MB = Molecular weight of solute.

= Kf × WB × 1000WA × MB

…(1)

Putting the values in equestion (1)

5·4 = 5·12 × 20 × 1000100 × MB

MB = 189·6∴ Observed molecular mass of phenol is 189·6 g/mol.

Q. 36. The depression in the freezing point of a sugar solution was found to be 0·402°C. Calculate theosmotic pressure of the sugar solution at 27°C (Kf = 1·86 K kg mol–1)

Sol. For sugar solution,

Δ T = m × Kf

m =Kf

ΔT = 0·4021.86 = 0·216 m.

Now, for the same solution, osmotic pressureπ = CRT

=nV × R × T

⎝⎜⎛

⎠⎟⎞When water is the solvent

molality ~– molarity

= 0.216 × 0·0821 × 300= 5·32 atm.

Q. 37. 0·750 gm of a compound A dissolved in 25 gm of the solvent lowered the freezing point of thesolvent by 0·502°C. Calculate the molecular weight of the substance. The molecular depressionconstant of the solvent is 50·2°C per 100 gm of the solvent.

Sol. Given : w = 0·750 gm, Kf = 50·2°C/100 gm, ΔTf = 0·502°C, W = 25 gm and m = ?

m =100 × Kf × w

ΔTf × W

=100 × 50·2 × 0·750

0·502 × 25 = 300.

Q. 38. Calculate the amount of ice that will separate out on cooling the solution containing 50 gm ofethylene glycol (CH2OH)2 in 200 gm of water at – 9·3°C. Kf for water = 1·86°C kg mol–1.

Sol. ΔTf = Kf × WB × 1000MB × WA

WA = Kf × WB × 1000MB × ΔTf

=1·86 × 50 × 1000

62 × 9·3

= 161·29

∴ Amount of ice that will separate = 200 – 161·29

= 38·71 g.


Q. 39. To 500 cm3 of water, 3·0 × 10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated, whatwill be the depression in freezing point ? Kf and density of water are 1·86 K kg–1 mol–1 and 0·997gcm–3, respectively.

Sol. Density of water = 0·997 g/cm3

Weight of water (W) = 500 × 0·997 = 498·5 gWeight of acetic acid (w) = 3·0 × 10–3 kg = 3·0 gm

mol. wt. of CH3COOH(n) = 60 ΔTf =1000 × Kf × w

m × W

(ΔTf )cal =1000 × 1·86 × 3·0

60 × 498·5 = 0·186

CH3COOH CH3COO– + H+

At t = 0 1 mole 0 0At equilibrium (1 – α) α gm ion α-gm ionNo. of particles after dissociation = 1 – α + α + α = 1 + α

α of CH3COOH =23100 = 0·23 (on 23 of dissociated)

So, no. of particles after dissociation = 1 + 0·23 = 1·23.By van’t Hoff factor,

(ΔTf )ob

(ΔTf )cal=

No. of particles after dissociationNo. of particles before dissociation =

1·231

(ΔTf )ob = 1·23 × 0·186 = 0·228 KHence, depression in freezing point (ΔTf ) = 0·228 K.

Q. 40. 75·2 gm of C6H5OH (Phenol) is dissolved in a solvent of Kf = 14. If the depression in freezingpoint is 7 K, then find the % of phenol that dimerizes.

Sol. Phenol dimerizes as follows :2C6H5OH (C6H5OH)2

At t = 0 1 0At equilibrium, (1 – α) α/2

Total mole after dimerization = 1 – α + α/2 = 1 – α/2

van’t Hoff factor (i) =Normal mol. wt. of phenol

Abnormal mol. wt. of phenol = 94mab

mab =1000 × Kf × w

W × ΔTfGiven that, Kf = 14, w = 72·5 gm, W = 1000 gm

∴ mab =1000 × 14 × 72·5

1000 × 7 = 145

∴ van’t Hoff factor (i) =94145 …(1)

Also, van’t Hoff factor (i) =No. of particles after dimerization

No. of particles before dimerization …(2)

From equations (1) and (2) 1 – α

2 =94145 = 0·65

α

2 = 1 – 0·65 = 0·35

α = 2 × 0·35 = 0·70Hence, 70% phenol is present in dimeric form.

Q. 41. The osmotic pressure of 0·25 M urea solution is 2·67 atm. What will be the osmotic pressure of0·25 M solution of potassium sulphate ?

Sol. For urea, πurea = iCRTFor Potassium sulphate solution, πK2SO4 = iCRTBecause molar concentrations of both the solutions is same.


∴πurea

iRT =πK2SO4

iRTπurea

iurea=

πK2SO4

iK2SO4

For urea, i = 1, for K2SO4, i = 3∴ πK2SO4 = 2·67 × 3 = 8·01 atm.

Chapter 3. ElectrochemistryQ. 1. Consider the following cell reaction at 298 K :

2Ag+ + Cd → 2 Ag + Cd2+

The standard reduction potential (E°) for Ag+/Ag and Cd2+/Cd are 0·80 V and – 0·40 Vrespectively :(1) Write the cell representation.(2) What will be the emf of the cell if the concentration of Cd2+ is 0·1 M and that of Ag+ is 0·2 M

?(3) Will the cell work spontaneously for the condition given in (2) above ?

Sol. (1) Cell Representation Cd Cd2+

(0·1M) 2Ag2+

(0·2M) 2Ag

(2) E°(red) cell = E°(red) cathode – E°(red) anode

= 1·2[n = 2]

Ecell = E°(red)cell – 0·059

n log

⎣⎢⎢⎡

⎦⎥⎥⎤M

Mn+

= 1·2 – 0·059

2 log ⎣⎢⎢⎡

⎦⎥⎥⎤0·1

0·04

Ecell = 1·1882177 volt.(3) Since Ecell is +ve therefore, cell reaction will be spontaneous.

ΔG° = – nFEcell = – 229·32 kJSince ΔG° is (–ve) it will also favour spontaneous reaction.

Q. 2. If a current of 0·5 ampere flows through metallic wire for two hours, then how many electronsflow through the wires ?

Sol. Q (coulomb) = 1 (ampere) × t (s)= 0·5 × 2 × 60 × 60= 3600 C

A flow of 96500 c is equivalent to the flow of 1 mole of electrons= 6·02 × 1023 electrons

∴ 3600 C is equivalent to flow of electrons

=6·02 × 1023

96500 × 3600

= 2·246 × 1022 electron.Q. 3. The conductivity of 0·20 M solution of KCl at 298 K is 0·0248 S cm–1. Calculate its molar

conductivity.

Sol. λm = K × 1000Molarity =

0·0248 S cm–1 × 1000 cm3 L–1

0·20 mol L–1

= 124 S cm2 mol–1

Q. 4. Consider the reaction

Cr2O72–

+ 14H+ + 6e

– ⎯⎯⎯→ 2Cr3+ + 7H2O

What is the quantity of electricity in Coulomb’s needed to reduce 1 mol of Cr2O72–

?

Sol. From the given reaction 1 mole of Cr2O72–

ion require6F = 6 × 96500 C = 579000 C of electricity for reduction of Cr

3+


Q. 5. How much charge is required for the following reduction :(i) 1 mole of Al3+ to Al(ii) 1 mol of Cu2+ to Cu(iii) 1 mol of MnO4

– to Mn2+

Sol. (i) Al3+ + 3e– Al

∴ Quantity of charge required for reduction of 1 mole ofAl3+ = 3F

= 3 × 96500C = 289500 C

(ii) Cu2+ + 2e– ⎯⎯→ Cu∴ Quantity of charge required for reduction of 1 mole of Cu2+ = 2F

= 2 × 96500 CC = 193000 C(iii) MnO4

– ⎯⎯→ Mn2+

Oxidation number of Mn changes from + 7 to + 2∴ Quantity of charge required = 5F

= 5 × 96500 C= 4825000 C

Q. 6. The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1500 Ω . What isthe cell constant if conductivity of 0·001 M KCl solution at 298 K is 0·146 × 10–3 S cm–1 ?

Sol. Conductivity (K) = 1

Resistance (R) × Cell constant

K = 0·146 × 10–3 S cm–1, R = 1500 Ω

0·146 × 10–3 S cm–1 = 1

1500 Ω × Cell constant

Cell constant = 0·146 × 10–3 S cm–1 × 1500 Ω= 219 × 10–3 cm–1

= 0·219 cm–1

Q. 7. A 0·05 M NH4OH solution offers the resistance of 50 ohm to a conductivity cell at 298 K. If thecell constant is 0·50 cm–1 and molar conductance of NH4OH at infinite dilution is 471·4 ohm–1 cm2

mol–1, calculate :(i) Specific conductance(ii) Molar conductance(iii) Degree of dissociation

Sol. (i) Specific conductance : =Cell constant

Resistance

=0·50 cm–1

50 ohmλsp = 0·01 ohm–1 cm–1

(ii) Molar conductance : =1000 λsp

c

=1000 × 0·01 ohm–1 cm–1

0·05 m

=1000 × 0·01

0·05λm = 200 m–1 ohm–1 cm–1

(iii) Degree of dissociation (α)

α =∧c

m

∧∞m

=200 m–1 ohm–1 cm–1

471·4 ohm–1 cm2 m–1

α = 0·424


Q. 8. The specific conductance of a 0·01 M solution of acetic acid at 298K is 1·65 × 10– 4 ohm– 1 cm– 1.The molar conductance at infinite dilution for H+ ion and CH3COO– ion are 349·1 ohm– 1 cm2

mol– 1 and 40·9 ohm– 1 cm2 mol– 1 respectively. Calculate :(i) Molar conductance of the solution.(ii) Degree of dissociation of CH3COOH.(iii) Dissociation constant for acetic acid.

Sol. (i) Molar conductance ∧m =K × 1000

C

=1·65 × 10–4 × 1000

0·01= 16·5 ohm– 1 cm2 mol– 1

(ii) Degree of dissociation α =Λm

Λ∞m

∧m = 16·5 ohm– 1 cm2 mol– 1

Λ∞m (CH3COOH) = λ∞[H+] + λ∞[CH3COO–]

= 349·1 + 40·9= 390 ohm– 1 cm2 mol– 1

α =16·5390 = 0·0423

(iii) CH3COOH CH3COO– + H+

Initial 0·01 M 0 0 At equilibrium 0·01 (1 – α) 0·01 α 0·01 α

Dissociation constant Kd =[CH3COO–] [H+]

[CH3COO–] = 0·01 α × 0·01 α

0·01 (1 – α)

=0·01 α2

1 – α = 0·01α2

[Taking 1 – α = 1, as α is negligible as compared to 1]

= 0·01 × (0·0423)2

Kd = 1·86 × 10– 5

Hence, the dissociation constant (Kd) for acetic acid is 1·86 × 10–5.Q. 9. (i) Calculate the e.m.f. of the following cell reaction at 298K :

Mg(s) + Cu2+ (0.0001 M) → Mg2+ (0.001M) + Cu(s)

The standard potential (Eo) of the cell is 2.71 V.(ii) The solubility product (Ksp) of BaSO4 is 1·5 × 10– 9. Calculate the solubility of barium

sulphate in pure water and in 0.1 M BaCl2.Sol. Mg(s) + Cu2+ (0·0001M) → Mg2+ (0·001 M) + Cu

(i) By Nernst equation the EMF of the cell at 298K is :

E = E° – 0·0591

2 log [Mg2+][Cu2+]

= 2·71 – 0·0591

2 log 0·0010·0001

= 2·71 – 0·02955

= 2·68V

Thus, the EMF of the cell is 2·68V

(ii) BaSO4 ⎯→ Ba2+ + SO42–

∴ Ksp = [Ba2+] [SO42–]

Let solubility of Ba2+ = SO42– = x


Then,1·5 × 10– 9 = x × x or x2 = 15 × 10– 10 or x = 3·87 × 10– 5

Thus, solubility of BaSO4 in pure water is 3·87 × 10–5.

Let solubility of BaSO4 in 0·1 M BaCl2 be ‘s’.

BaSO4(s) ⎯⎯→ Ba2+(aq) + SO42– (aq)

Hence, 1·5 × 10– 9 = (s + 0·1) × s = s × 0·1 (Since s << 1)

s = 1·5 × 10– 8

Thus, the solubility of BaSO4 in presence of 0·1M BaCl2 is 1·5 × 10– 8.

Q. 10. The electrical resistance of a column of 0·05 M NaOH solution of diameter 1 cm and length 50 cmis 5·55 × 103 ohm. Calculate its resistivity, conductivity, and molar conductivity.

Sol. A = πr2 = 3·14 × ⎝⎜⎛

⎠⎟⎞1

2 cm 2

= 0·785 cm2, l = 50 cm

Resistivity (ρ) =R × A

l =

5·55 × 103 ohm × 0·785 cm2

50 cm= 0·087135 × 103 ohm cm

Conductivity (K) =1ρ =

187·135 ohm cm = 0·01148 S cm–1

Molar conductivity (λm) =K × 1000

M

=0·01148 S cm–1 × 1000 cm3 L–1

0·05 mol L–1

= 229·6 S cm2 mol–1

Q. 11. λm for NaCl, HCl and NaAc are 126·4, 425·9 and 91·0 S cm2 mol–1

respectivity. Calculate Λ°m for

HAc.

Sol. λ°m (HAc) = λ°

H+ + λ°AC = λ°

H+ + λ°Cl– + λ°

AC + λ°Na+ – λ°

Cl+ – λ°Na+

= Λ°m (HCl) + Λ°

m (NaAc) – Λ°m (NaCl)

= (425·9 + 91·0 – 126·4) S cm2 mol–1

= 390·5 S cm2 mo–1

Q. 12. How much electricity in terms of Faraday is required to produce(i) 20·0 g of Ca from molten CaCl2 ?(ii) 40·0 g of Al from molten Al2O3 ?

Sol. (i) Ca2+ + 2e– ⎯⎯⎯→ CaAs , 40 g of Ca require electricity 2F

∴ 20 g of Ca will require electricity = 2F40 × 2 = 0·1 F

(ii) Al3+ + 3e– ⎯⎯⎯→ AlAs, 1 mol of Al i.e. 27 g of Al require electricity 3F

∴ 40 g of Al will require electricity = 3F27 × 40 = 4·44 F

Q. 13. 0·05 M NaOH solution offered a resistance of 31·6 ohm in a conductivity cell at 298 K. If the cellconstant of the cell is 0·367 cm–1, calculate the molar conductivity of the NaOH solution.

Sol. Given : Molarity of NaOH solution = 0·05 M

Resistance = 31·6 ohm

Cell constant = 0·367 cm–1

Now, Molar conductivity =Specific conductivity × 1000

Molarity


=Cell constant

Resistance × 1000

Molarity

=0·367 × 100031·6 × 0·05

Λm = 232·3 ohm–1 cm2 mol–1

Q. 14. Calculate the value of Ecell at 298K for the following cell :Al | Al3+ (0·01M) || Sn2+ (0·015M) | Sn

EoAl3+/Al = – 1·66 volt and E

oSn2+/Sn = – 0·14 volt

Sol. Given cell is : Al /Al3 + (0·01 M) || Sn2 + (0·015 M)/Sn

Given : EoAl3+//Al = – 1·66V, Eo

Sn2+/Sn = – 0·14V

∴ Eocell = Eo

R – EoL

= – 0·14 – (– 1·66)= – 0·14 + 1·66= 1·52V

and net cell reaction is2Al + 3Sn2+ ⎯→ 2Al3+ + 3Sn

so net cell reaction involves transfer of 6 mole of electrons∴ n = 6According to Nernst equation, at 298K

Ecell = Eocell –

0·059n

log [Al3+]2

[Sn2+]3

= 1·52 – 0·059

6 log [0·01]2

[0·015]3

= 1·52 – 0·01447Ecell = 1·50553V.

Q. 15. The conductivity of 0·2M KCl solution is 3 × 10–2 ohm–1 cm–1. Calculate its molar conductance.Sol. Given : Conductivity = 3 × 10–2 ohm–1 cm–1 and Molarity = 0·2 M

Now, Λmolar =1000 × Conductivity

Molarity

=1000 × 3 × 10–2

0·2 = 150 ohm cm2 mol–1

Q. 16. How many electrons will flow when a current of 5 ampere is passed through a solution for 200second ?

Sol. 1 F = charge on 1 mole of e–

or 96500 coulomb = charge on 6·023 × 1023 electrons.

∴ 5 × 200 coulomb =6·023 × 1023 × 1000

96500 electrons

= 6·24 × 1021 electrons.= 6·24 × 1021 electrons will flow.

Q. 17. A solution of 0·1(N) KCl offers a resistance of 245 ohm. Calculate the specific conductance andthe equivalent conductance of the solution if the cell constant is 0.571 cm-1.

Sol. Given : R = 245 ohm, cell constant = 0·571 cm–1, Normality = 0·1

Now, specific conductance, κ =1R × cell constant

=1

245 × 0·571 = 2·33 × 10–3 ohm–1 cm–1

Equivalent conductance, Λeq =1000 × κ

Normality = 1000 × 2·33 × 10–3

0·1= 23·3 ohm–1 cm2 eq–1

For the given solution, specific conductance is 2·33 × 10– 3 ohm– 1 cm– 1 and Λe q is23·3 ohm–1 cm2 eq–1.


Q. 18. From the following conductivities at infinite dilution, calculate Λ°m for NH4OH.

Λ°m for Ba(OH)2 = 457·6 Ω–1 mol–1

Λ°m for BaCl2 = 240·6 Ω–1 cm2 mol–1

Λ°m for NH4Cl = 129·8 Ω–1 mol–1

Sol. Λ°m (NH4OH) = λ°

NH4+ + λ°OH–

= ( )λ°NH4+ + λ°

Cl– + 12 ( )λ°

Ba2+ + λ°OH– –

12 [ ]λ°

Ba2+ + 2λ°Cl–

= λ°m(NH4Cl) +

12 [ ]λ°

m(Ba(OH)2) – 12 [λ°

m(BaCl2)]

= 129·8 + 12 457·6 –

12 × 240·6

= 238·3 ohm–1 cm2 mol–1

Q. 19. A solution of Ni(NO3)2 is electrolysed between platinum electrode using a current of 5·0 amperefor 20 minutes. What mass of nickel will be deposited at the cathode ? [At. mass of Ni = 58·74]

Sol. Quantity of electricity passedI × t = I × t = 5 × 20 × 60 = 6000 C

The electrode reaction is Ni2+ + 2e– ⎯⎯→ Ni2 × 96500 C deposit Ni = 58·74 g

∴ 6000 C will deposit Ni = 58·74

2 × 96500 × 6000 g = 1·825 g

Q. 20. Chromium metal can be plated out from an acidic solution containing CrO3 according to thefollowing equation :

CrO3 + 6H+ + 6e – ⎯⎯→ Cr + 3H2O

Calculate (i) how many gram of chromium will be plated out by 24,000 coulomb and (ii) howlong will it take to plate out 1·5 g chromium by using 12·5 A current ? [At. mass of Cr = 52]

Sol. (i) 6 × 96,500 coulomb deposit Cr = 1 mole = 52 g

∴ 24,000 coulomb deposit Cr =52 × 240006 × 96500 g

= 2·1554 g(ii) 52 g of Cr is deposited Cr = 6 × 96500 C

∴ 1·5 g require electricity =6 × 96500

52 × 1·5 C

=16701·9 C

12·5 A = 1336·15

Q. 21. Three electrolytic cell A, B and C containing electrolytes ZnSO4, AgNO3 and CuSO4 respectivelywere connected in series. A steady current of 7·5 ampere were passed through them until 1·45g ofAg were deposited at cathode of cell B. How long did the current flow ? What mass of copperand zinc were deposited ? (At. wt.) of Cu = 63·5, Zn = 65·3, Ag = 108)

Sol. Ag+ + e– ⎯⎯→ Ag108 g of Ag is deposited 96500 C

1·45 g of Ag will be deposited by 96500108 × 1·45 C = 1295·6 C

t = QI =

1295·67·50 = 863·7 sec.

Cu2+ + 2e– ⎯⎯→ Cu2 × 96500 C deposit Cu = 63·5 g

∴ Cu deposited by 1295·6 C =63·5

2 × 96500 × 1295·6

= 0·426 g


Zn2+ + 2e– ⎯⎯→ Zn

2 × 96500 C deposit Zn = 65·3 g

∴ Zn deposited by 1295·6 C =65·3

2 × 96500 × 1295·6

= 0·438 gQ. 22. How many coulombs are required for the following oxidation :

(i) 1 mol of H2O to O2 (ii) 1 mol of FeO to Fe2O3 ?Sol. (i) The electrode reaction for 1 mol of H2O is given as :

H2O ⎯⎯→ 2H+ + 12 O2 + 2e

–

∴ The quantity of electricity required= 2F= 2 × 96500 C = 193000 C

(ii) The electrode reaction is given as

2FeO + 12 O2 ⎯⎯→ Fe2O3

i.e. 2Fe ⎯⎯→ 2Fe3+ + 3e–

For the oxidation of 2 mole of FeO required charge = 3F∴ For the oxidation of 1 mol of FeO

required charge = 1 F = 96500 c.Q. 23. Calculate the emf of the cell in which the following reaction takes place :

Ni(s) + Ag+ (0·002M) ⎯⎯→ Ni2+ (0·160M) + 2Ag(s)Given that E°Cell = 1·05 V.

Sol. From Nernst equation to the given cell reaction

Ecell = E°cell – 0·0591

n log

[Ni2+][Ag+]2

= 1·05 V – 0·0591

2 log 0·160

(0·002)2

= 1·05 – 0·0591

2 log 4× 104

= 1·05 – 0·0591

2 (4·6021) = 1·05 – 0·14 V = 0·91V.

Q. 24. A 0·05 M NaOH solution offered a resistance of 31·6 ohm in a conductivity cell. If the cellconstant of the conductivity cell is 0·378 cm– 1, determine the molar conductivity of sodiumhydroxide solution at this temperature.

Sol. Given, Molarity = 0·05 M, R = 31·6 ohmCell constant = 0·378 cm– 1, Molar conductivity, ∧m = ?

Specific conductivity, κ =1R × cell constant

=1

31·6 × 0·378 = 0·0119 ohm– 1 cm– 1

∧m =κ × 1000Molarity

=0·0119 × 1000

0·05 = 238 ohm– 1 mol– 1

Q. 25. A current of 10 A is passed for 80 min and 27 seconds through a cell containing dilutesulphuric acid.(1) How many moles of oxygen gas will be liberated at the anode ?(2) Calculate the amount of zinc deposited at the cathode when another cell

containing ZnSO4 solution is connected in series (Zn = 65).


Sol. On passing electric current through dilute H2SO4 hydrogen gas is evolved at cathode andoxygen gas at anode.

Calculation of charge, Q = i × t = 10 × (80 × 60 + 27)

Q = 48270 coulomb

(1) The number of moles of O2 liberated will be :

96500 C deposits one gm equivalent of element = 8 g of O2 = 14 mole

96500 C =14 mole O2

∴ 48270 C = 0·125 mole of O2(2) Calculation of mass of zinc :

96500 coulomb = 65/2 g zinc∴ 48270 coulomb = 16·25 g zinc.

Q. 26. Calculate : Ecell at 25°C for the reaction :Zn + Cu2+ (0.20 M) → Zn2+ (0.50 M) + Cu

Given : E°(Zn2+/Zn) = – 0.76 voltE°(Cu2+/Cu) = 0.34 volt.

Sol. Ecell = E°cell – 0·0591

n log

[Product][Reactant]

Ecell = 0·34 – (– 0·76) – 0·0591

2 log [Zn2+][Cu2+]

Ecell = 1·10 – 0·0591

2 log 0 500 20⋅⋅[ ]

Ecell = 1·10 – 0·02955 × 0·3979

Ecell = 1·088 V.

Q. 27. The standard electrode potentials of Pb|Pb+2 and Pt|I–|I2 are – 0·126 and 0·536 volt

respectively. When a galvanic cell is constructed using 0·1 molar concentrations of therespective ions Pt is found to be cathode. What is the voltage generated in the cell ?

Sol. Because Pt works as cathode, the cell reduction reaction will take place on it thus, the cellreaction will be

Pb + I2 ⎯→ Pb+2 + 2I–

E°Cell = E°Cathode – E°Anode

= 0·536 – (– 0·126) = 0·662 V

Voltage generated i.e., Ecell of the 0·1 M concentration of respective ions can be calculated by

ECell = E°Cell + 0·059

n log

1[Pb+2] [I–]2

ECell = 0·662 + 0·059

2 log 1

[0·1] [0·1]2

= 0·662 + 0·059

2 log 103 = 0·7505 V.

Q. 28. Calculate the e.m.f., of 298 K of the cell Zn|ZnSO4 (aq. 0·01 M)|KCl (aq. saturated) Hg2Cl2(s)|Hg.Given the potential of calomel electrode is 0·242 V and the standard potential of the zincelectrode is – 0·763 V.

Sol. We know that Ecell = E°cell – 0·059

n log

CAnode

CCathode

On substituting the various values, we get,

E°cell = 0·242 – (– 0·763) = 1·005 V


Ecell = 1·005 – 0·059

2 log 0·01(0·1)2

= 1·005 – 0·059

2 log (1)

= 1·005 V.

∴ E.M.F. of the cell = 1·005 V.

Q. 29. The resistivity of a 0·8 M , P = 5 × 10–3 Ω cm. Calculate molar conductivity of the solution.

Sol. The resistivity of (P) = 5 × 10–3 Ω cm.

The conductivity of the solution = 1P =

15 × 10–3

15 × 10–3 =

15 × 103

Now the molar conductivity = Λm = 1000 × K

M

= 103 × 103

5 × 1

0·8= 2·5 × 105 Ω–1 cm2 mol–1.

Q. 30. A zinc rod is dipped in 0·1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298K. Calculate the electrode potential (E°Zn2+/Zn = – 0·76V)

Sol. The electrode reaction written as reduction reaction is

Zn2+ + 2e– ⎯⎯→ Zn (n= 2)

Applying Nernst equation, we get

EZn2+/Zn = E°Zn2+/Zn – 0·0591

2 log 1

[Zn2+]As 0·1 M ZnSO4 solution is 95% dissociated, this means that in the solution

[Zn2+] =95100 × 0·1 M = 0·095 M

EZn2+/Zn = – 0·76 – 0·0591

2 log 1

0·095= – 0·76 – 0·02955 (log 1000 – log 95)= – 0·76 – 0·02955 (3 – 1·9777)= – 0·76 – 0·03021= – 0·79021 V

Q. 31. Write the Nernst equation and emf of the following cell at 298 K.Mg (s) / Mg2+ (0·001 M) || Cu2+ (0·0001 M) /Cu(s)Given : E°Mg2+/Mg = – 2·37 V,

E°Cu2+/Cu = + 0·34 VSol. Cell equation : Mg + Cu2+ ⎯⎯→ Mg2+ + Cu n = 2

Nernst equation

Ecell = E°cell – 0·0591

2 log [Mg2+][Cu2+]

Ecell = 0·34 – (– 2·37) – 0·0591

2 log 10–3

10–4

= 2·71 – 0·02955 = 2·68 V.Q. 32. The cell in which the following reaction occurs :

2Fe3+ (aq) + 2I– (aq) ⎯⎯→ 2Fe2+ (aq) + I2(s)has E°cell = 0·236 V at 298 V. Calculate the standard Gibbs energy and the equilibrium constant ofthe cell.


Sol. 2Fe3+ + 2e–⎯⎯→ 2Fe2+

or 2I–

⎯⎯→ I2 + 2e–

Thus, for the given cell reaction, n = 2ΔrG° = – n FE°cell

= – 2 × 96500 × 0·236 J mol–1

= – 45548 J mol–1

= – 45·55 kJ mol–1

ΔrG° = – 2·303 RT log KC

log KC =– Δr G°

2·303 kRT = 45548 × 103 kJ mol–

2·303 × 8·314 × 298 kJ K– 1 mol– 1

= 7·983∴ KC = Antilog (7·983)

= 9·616 × 107

Q. 33. For the cell :

Zn ⏐ Zn 2+ (a = 1) ⏐⏐ Cu2+ (a = 1) ⏐ Cu

Given that EZn , Zn2+ = + 0.761 volt; ECu2+

, Cu = + 0.339 volt

(i) Write the cell reaction.

(ii) Calculate the emf and free energy change at 298 K involved in the cell.[Faraday’s constant F = 96500 coulomb equivalent–1].

Sol. (i) Cell reaction Zn + Cu2+ ⎯→ Zn2+ + Cu(ii) E°cell = E°anode + E°cathode

= 0·761 + 0·339= 1·100 V.

Ecell = E° – 2·303 RT

nF log [Products][Reactants]

= 1·10 – 2·303 × 8·31 × 298

2 × 96500 log 11

= 1·10 – 0 = 1·10 VFree energy change at 298 K,

ΔG° = – nFE°cell

= – 2 × 96500 × 1·10= – 212300 joule= – 212·3 kJ= 3·172 × 1034

Q. 34. Calculate the standard cell potential of galvanic cell in which the following reaction takesplace :

2 Cr (s) + 3Cd2+(aq) ⎯⎯→ 2Cr3+

(aq) + 3Cd(s)

Given : E°Cr3+

/Cr = – 0·74 V, E°Cd2+/Cd = – 0·4VAlso calculate ΔrG° and Equilibrium constant of the reaction.

Sol. E°Cell = E°cathode – E°anode

= – 0·40 V – (– 0·74 V)= + 0·34V

Δr G° = – n F E°Cell

= – 6 mol × 96500 C mol–1

× 0·34V

= – 196860 C V mol–1

= – 196860 J

= – 196·86 kJ mol–1


Δr G° = – 2·303 RT log KC

– 196860 = – 2·303 × 8·314 × 298 log KC

or KC = Antilog 34·5014Q. 35. Calculate the standard cell potentials of galvanic cell in which the following reaction takes place :

Fe2+(aq) + Ag+

(aq) ⎯→ Fe3+ (aq) + Ag(s)

Given : E°Ag+/Ag = 0·80V, E°Fe3+/Fe2+= 0·77VAlso calculate ΔrG° and equilibrium constant of the reaction.

Sol. E°Cell = + 0·80 V – 0·77V = + 0·03VΔnG° = –– nFE°

Cell

= – (1 mol) × (96500 C mol–1) × (0·03V)= – 2895 C V mol–1

= – 2895 J mol–1

= – 2·895 kJ mol–1

ΔrG° = – 2·303 RT log KC

– 2895 = – 2·303 × 8·314 × 298 × log KC

or log KC = 0·5074KC = Antilog (0·5074)

= 3·22Q. 36. Conductivity of 2·5 × 10–4 M methanoic acid is 5·25 × 10–5 S cm–1. Calculate its molar conductivity

and degree of dissociation.Given : λ°

H+ = 349·5 S cm2 mol–1 and λ°(HCOO–) = 50·5 S cm2 mol–1

Sol. Λcm =

K × 1000M

Λcm =

5·25 × 10–5 S cm–1 × 1000 cm3 L–1

2·5 × 10–4 mol L–1

Λcm = 210 S cm2 mol

–1

Λcm (HCOOH) = Λ°

HCOO– + Λ°

H+

= 50·5 S cm2 mol–1 + 349·5 S cm2 mol–1

= 400 S cm2 mol–1

Degree of dissociation, α =Λc

m

Λ°m

=

210 S cm2 mol–1

400 S cm2 mol–1

= 0·525α = 52·5%

Q. 37. The value of Λ∞ of Al2 (SO4)3 is 858 S cm2 mol–1 while λ∞ of CO42– is 160 S cm2 mol–1. Calculate

the value of ionic conductivity of Al+3.Sol. According to Kohlrausch’s law,

Λ∞m (Al2(SO4)3) = 2 λ∞ (Al3+) + 3λ∞ (SO4

2–)858 = 2 λ∞ (Al3+) + 3 × 160

or 2 λ∞ (Al3+) = 858 – 3 × 160 = 378 S cm2 mol–1

or λ∞ (Al3+) =3782 = 189 S cm2 mol–1

Q. 38. What is the electrode potential of Mg+2 | Mg electrode in which conc., of Mg+2 is 0·01 M ?E°Mg2+ | Mg = – 2·36 V.

Sol. Mg2+ (aq.) + 2e– = Mg(s)

EMg2+|Mg = E°Mg2+/Mg + 0·059

2 log [Mg+2][Mg]

[Mg2+] = 0·01 M, [Mg] = 1

EMg2+/Mg = –2·36 + 0·059

2 log (0·01) = – 2·42 V.


Q. 39. The resistance of 0·01 M NaCl solution in a conductivity cell was found to be 210 Ω. Theequivalent conductance of this solution is 4·5 × 10–3 Ω–1 cm2 eq–1. What is the cell constant of thecell ?

Sol. Equivalent conductance =1000 × Conductivity

N

Conductivity =Equivalent conductance × N

1000= 4·5 × 10–3 Ω–1 cm2 eq–1 × 0·01 M/1000= 4·5 × 10–8 Ω–1 cm–1

Cell constant = Conductivity × Resistance

= 4·5 × 10–8 Ω–1 cm–1 × 210 Ω

= 945 × 10–8 cm–1

= 9·45 × 10–10 cm–1.Q. 40. A cell is constructed by dipping a zinc rod in 0·1 M zinc nitrate solution and a lead rod in 0·2 M

lead nitrate solution.

EoPb2+/Pb = – 0·13 V and E

oZn2+/Zn = – 0·76 V.

(i) Write the spontaneous cell reaction.(ii) Calculate standard emf and emf of the cell.

Sol. (i) Spontaneous cell reactionAt anode Zn – 2e– ⎯→ Zn2+ E = + 0·76 volt.At cathode Pb2+ + 2e– ⎯→ Pb E = – 0·13 voltCell reaction : Zn + Pb2+ ⎯→ Zn2+ + Pbas cell representation Zn/Zn2+ || Pb2+/Pb

(ii) Eocell = Eanode + Ecathode

= 0·76 + (– 0·13) = 0·76 – 0·13

Eocell = + 0·63 volt

⇒ Ecell = Eocell –

0·05912n

log [Products][Reactants]

= 0·63 – 0·05912

2 log [Zn2+][Pb2+]

= 0·63 – 0·0295 log [0·1][0·2]

Ecell = 0·63 – 0·0295 log ⎝⎜⎛

⎠⎟⎞1

2

volt.

= 0·63 – 0·0295 log (·5) volt. = 0·638 V

Q. 41. Calculate the e.m.f., of the cell

Cr(s) | Cr3+ (0·1 M) || Fe2+ (0·01 M) | Fe (s)

E°Cr+3/Cr = – 0·75, E° Fe2+/Fe = – 0·45 V

Sol. E°Cell = E°Right – E°Left

= – 0·45 – (– 0·75)

= + 0·30 V

2Cr + 3Fe2+ ⎯⎯→ 2Cr3+ + 3Fe

ECell = E°Cell – 0·059

n log

[Cr3+]2

[Fe2+]3

= +0·30 – 0·059

6 log [0·1]2

[0·01]3


= 0·30 – 0·059

6 log [10–2][10–6]

= 0·30 – 0·059

6 log 104

= 0·30 – 0·059

6 × 4

= 0·30 – 0·0393 V= 0·2607 V.

Q. 42. For the following cell, calculate the emf :Al/Al3+ (0·01 M) // Fe2+ (0·02M) /Fe

Given : E°Al3+ / Al = – 1·66 V

E°Fe2+ / Fel = – 0·44 V

Sol. For the cellAl | Al3+ (0·01 M) || Fe2+ (0·02M) | FeAnode Cathode

E°Al3+ / Al = – 1·66 V

E°Fe2+ / Fel = – 0·44 V

Eocell = Eo

Fe – EoAl = – 0·44 – (– 1·66) = + 1·22 V

Nernst equation for the cell reaction

2Al + 3Fe2+ ⎯→ 2Al3+ + 3Fe

Ecell = Eocell +

0·05916 log

[Fe2+]3

[Al3+]2

= 1·22 + 0·0591

6 log (0·02)3

(0·01)2

= 1·22 + 0·0591

6 log 8 × 10– 2

= 1·22 + 0·00985 [log 8 – 2 log 10]= 1·22 + 0·00985 [0·8451 – 2 × 1]

= 1·22 + 0·00985 [– 1·1549]

= 1·220 – 0·011 = 1·209 V

Q. 43. Calculate the e.m.f., of the following cell.

Mg(s) | Mg2+ (0·001 M) || Cu2+ (0·0001 M)| Cu(s)

E°Cu2+/Cu = 0·337 V, E°Mg2+/Mg = –2·37 V

Sol. The cell reaction is

Mg + Cu2+(aq) ⎯→ Mg2+

(aq) + Cu

E°Cell = E°Right – E°Left

= + 0·337 – (–2·37)= 2·707 V

ECell = E°Cell – 0·059

n log

[Mg+2][Cu+2]

= 2·707 – 0‚059

2 log 10–3

10–4

= 2·707 – ·0295 log 10

= 2·707 – 0·0295

= 2·6775 V


Q. 44. Calculate the maximum work that can be obtained from the given electro-chemical cellconstructed with two metals M and N.

⎣⎢⎢⎡

⎦⎥⎥⎤

EoM2+/M = – 0·76 V‚ E

o

N2+/N = + 0·34 V

The cell reaction is M + N+2 → M+2 + N

Sol. According to given cell reaction :

Eoanode (M2+/M) = – 0·76 V

EoCathode (N2+/N) = + 0·34 V

n = 2

Now, Welectrical = – n FEocell

= – 2 × 96500 × [0·34 – (– 0·76)]

= – 2 × 96500 × 1·1

= – 212·3 kJQ. 45. How many electrons will flow when a current of 5 ampere is passed through a solution for 200

second ?

Sol. 1 F = charge on 1 mole of e–

or 96500 coulomb = charge on 6·023 × 1023 electrons.

∴ 5 × 200 coulomb =6·023 × 1023 × 1000

96500 electrons charge

= 6·24 × 1021 electrons.

= 6·24 × 1021 electrons will flow.

Q. 46. Predict if the reaction between the following is feasible :

(i) Fe3+ (aq) and I–(aq)

(ii) Ag+ (aq) and Cu (s)

(iii) Fe3+ (aq) and Br– (aq)

(iv) Ag (s) and Fe3+ (aq)(v) Br2 (aq) and Fe2+ (aq)Given standard electrode potentials

E°12

I2/I–

= 0·54 V, E°Cu2+/Cu = + 0·34V

E°12

Br2/Br–

= + 1·09 V, E°Ag+/Ag = + 0·80 V

E° Fe3+/Fe2+ = + 0·77V

Sol. A reaction is feasible if emf of the cell reaction is positive :

(i) Fe3+ (aq) + I– (aq) ⎯→ Fe2+ (aq) +

12 I2

∴ E°Cell = E °Fe3+/ Fe2 – E°1

2

I2/I–

0·77V – 0·54 V = 0·23 V (feasible)(ii) 2Ag+ (aq) + Cu ⎯⎯→ 2Ag(s) + Cu2+ (aq)

E°Cell = E°Ag+/Ag – E°Cu2+/Cu = 0·80V – 0·34 V

= 0·46 V (feasible)

(iii) Fe3+ (aq) + Br– (aq) ⎯→ Fe3+ (aq) + 12 Br2

E°Cell = 0·77V – 1·09 V = – 0·32 V (not feasible)


(iv) Ag(s) + Fe3+ (aq) ⎯⎯→ Ag+ (aq) + Fe2+(aq)E°

Cell = 0·77V – 0·80V

= – 0·03 V (not feasible)

(v)12 Br2 (aq) + Fe2+(aq) ⎯⎯→ Br– + Fe3+

E°Cell = 1·09 V – 0·77 V

= 0·32 V (feasible)Q. 47. Conductivity of 0·00241 M acetic acid is 7·896 × 10–5 S cm–1. Calculate its molar conductivity and

if Δ°m for acetic acid is 390·5 S cm2 mol–1, what is its dissociation constant ?

Sol. Δ°m =

K × 1000m

=7·896 × 10–5 S cm–1 × 1000 cm3 L–1

0·00241 mol L–1

= 32·76 S cm2 mol–1

α =Δ°

m

Δ°m

=32·76390·5 = 8·39 × 10–2

Kα =Cα2

1 – α = 0·00241 × 1(8·39 × 10–2)2

(1 – 8·39 × 10–2)= 1·86 × 10–5

Q. 48. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.Sol. pH = 10

[H+] = 10–10

E = E° – 0·059

1 log 1

[H+]

E = 0 – 0·059

1 log 1

10–10

E = 0 – 0·059 × 10 = – 0·59 VQ. 49. A voltaic cell is set up at 25°C with the following half cells.

Al3+ (0·001 M) and Ni2+ (0·50 M)write an equation for the reaction that occurs when the cell generates an electric current.Determine the cell potential.

(Given : E°Ni2+/Ni = – 0·25 V, E°

Al3+

/Al = – 1·66 V)Sol. The cell reaction is

2Al + 3Ni2+ (aq) ⎯⎯→ 2Al3+ (aq) + 3 Ni (s)

ECell = E°Cell –

0·059n

log [Al3+]2

[Ni2+]3

E°Cell = E°

(Ni2+/Ni) – E°(Al3+/Al)

= – 0·25V – (– 1·66 V) = 1·41 Vn = 6

Substituting the values

ECell = 1·41 – 0·059

6 log (0·001)2

(0·50)3

= 1·41 – 0·059

6 log 8 × 10–6


= 1·41 – 0·059

6 × (– 5·097)

= 1·41 – 0·05 = 1·46 VQ. 50. Silver is electrodeposited on a metallic vessel of total surface area 500 cm2 by passing a current of

0·5 ampere for 2 hours. Calculate the thickness of silver deposited.

(Given density of silver = 10·5 g cm–3. Atomic mass of silver = 108 amu, F = 96500 C mol–1)Sol. m = ZIt

=108

96500 × 0·5 × 2 × 3600 = 4·029g

d =mV ⇒ V =

md

V =4·029 g

10·5 g cm–3 = 0·3837 cm3

Let the thickness of silver deposited be x cmV = A × x

or x =VA

x =0·3837

500 = 7·67 × 10–4 cm.

Q. 51. A copper-silver cell is set up. The copper ion concentration is 0·10 M. The concentration of silverion is not known. The cell potential was found to be 0·422V. Determine the concentraion of silverions in the cell. (Given : E°Ag+/Ag = + 0·80V, E°Cu2+/Cu = + 0·34V)

Sol. Given : E°Ag+/Ag = + 0·80V, E°Cu2+/Cu = + 0·34V[Cu2+] = 0·10M, [Ag+] = ?

E°Cell = E°Ag+/Ag – E°Cu2+/Cu

= 0·80 – 0·34 V= 0·46 V

From Nernst equation

ECell = E°Cell –

0·0591n

log [Cu2+][Ag+]2

0·422 = 0·46 – 0·0591

2 log [0·10][Ag+]2

log 0·10

[Ag+]2 = 1·2881

[Ag+]2 = 0·0051[Ag+] = 7·1 × 10–2 M

Q. 52. Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. The free energychange for the decomposition reaction.

23 Al2O3 ⎯⎯→

43 Al + O2 is 960 kJ (F = 96500 C mol–1)

Sol. Al2O3 (2Al3+ + 3O2–) ⎯⎯→ 2Al + 32 O2 (n = 6e

–)

23 Al2O3 ⎯⎯→

43 Al + O2, (n =

23 × 6e

– = 4e–)

ΔrG = 960 × 1000 = 960000 JNow, ΔrG = – n FEcell

Ecell = – ΔrGnF =

– 9600004 × 96500

Ecell = – 2·487 V∴ Minimum potential difference needed to reduce Al2O3 is – 2·487 V


Q. 53. How many moles of mercury will be produced by electrolysing 1·0 M Hg (NO3)2 solution with acurrents of 2·00 A for 3 hours ?

Sol. Mass of mercury produced at the cathode,

m = 2 × I × t =M × I × t

n × F

m =M × g mol–1 × 2A × 3 × 60 × 60 s

2 × 96500 C mol–1 = 0·112 × M × g

= 0·112 × M × g

No. of moles of mercury produced =0·112 × M × g

M g mol–1

= 0·112 mol.

Q. 54. Consider the reaction 2Ag+ + Cd → 2Ag + Cd2+. The standard reduction potentials of Ag+/Agand Cd2+/Cd are + 0·80 volt and – 0·40 volt, respectively.

(i) Give the cell representation.

(ii) What is the standard cell e.m.f. E° ?

(iii) What will be the e.m.f. of the cell if concentration of Cd2+ is 0·1 M and Ag+ is 0·2 M ?

(iv) Will the cell work spontaneously for the condition given in (iii) above ?

Sol. (i) Cd/Cd2+(aq.)||Ag+

(aq.)/Ag

(ii) Given : E°Ag+/Ag = + 0·80 V, E°Cd2+/Cd = – 0·40 V.

Now, E°Cell = E°

cathode – E°anode = 0·80 – (– 0·40) = 1·20 V.

(iii) E°Cell = E°

Cell – 0·0591

n log

[Product][Reactant]

= 1·20 – 0·0591

2 log [0·1][0·2]2

= 1·20 – 0·0591

2 × ·3979

= 1·20 – 0·01175 = 1·188 V

(iv) Cell will work spontaneously because Ecell has a positive value.

Q. 55. When a current of 11 ampere is passed through an aqueous solution of CrCl3, it is observed that6·2 gm of Cr is deposited. For how long was the current passed ?

Sol. CrCl3 solution consists of Cr3+ ions

Cr3+ + 3e– ⎯→ Cr

3 mol 1 mol

or 3F or 52 gm

or 3 × 96500 C

·.· 52 gm Cr is deposited by 3 × 96500 C

∴ 6·2 gm Cr is deposited by 3 × 96500 × 6·2

52 C = 34517·3 C

We know that,

Quantity of electricity passed = Current (amp) × Time (sec)

or 34517·3 = 11 × Time (sec)

or t = 3137·94 sec.

t = 52 min. 18 sec.


Q. 56. The same amount of electricity is passed between electrodes dipped in the following solutionkept in separate beakers. (A) 0·03 molar aluminium chloride, (B) 0·01 molar aluminium chloride(C) 0·1 molar silver nitrate and (D) 0·1 molar sodium hydroxide. If 0·36 gm of silver is depositedin beaker C, what is the mass of aluminium deposited in beakers A and B and volume of oxygengas evolved in cell D at STP ?

Sol. According to Faraday law of electrolysisWt. of AgWt. of Al =

Eq. wt. of AgEq. wt. of Al

Given wt. of silver deposited = 0·36 gm,Eq. wt. of Ag = 108 gm and Eq. wt. of aluminium = 27/3 = 9

Hence 0·36

Wt. of Al =1089

Wt. of Al = 0·03 gm.Therefore, the wt. of Al deposited in beaker A and B is 0·03 gm.

Further, Wt. of Ag depositedWt. of O2 liberated =

Eq. wt. of AgEq. wt. of O2

or0·36

Wt. of O2=

10816

or Wt. of O2 = 0·053 gmWe know that 32 gm of O2 at S.T.P. occupies = 22400 ml

Hence, 0·053 gm of O2 at S.T.P., occupies = 22400

32 × 0·053 ml.

= 37·1 ml.Q. 57. Chromium metal can be plated out from an acidic solution containing CrO3, according to the

following equation.CrO3(aq.) + 6H+(aq.) + 6e– ⎯→ Cr(s) + 3H2O

(i) Calculate, how many gram of Cr (At. wt. 52) will be plated out by passing 24000 C ofelectricity ?

(ii) If a current of 24 amp is being used, how long will it take to plate out the amount of Crcalculated inpart (i)

Sol. (i) According to the equation one mol of Cr is plated out by 6 mol of electrons.

Cr6+(aq.) + 6e– ⎯→ Cr(s)

6 mol 1 mol

or 6 F 52 g

or 6 × 96500 C

·.· 6 × 96500 C are used to plate out = 52 g Cr

∴ 24000 C is used to plate out =52 × 240006 × 96 500 g Cr

= 2·155 g Cr.

(ii) Where, Q = Quantity of electricity = 24000 C (given)

C = Current 24 A (given)

t = Time in sec = ?

We know that Q = C × t

Hence, t = 24000

24 = 1000 sec.

Hence to plate out 2·155 g of Cr, 24 A current should be passed for 1000 sec. through CrO3solution.


Q. 58. A direct current deposit 19·5 g of K (At. wt. 39) in one minute. How many g, of Al (At. wt. 27)will be deposited by the same current in the same time ?

Sol. For deposition of K, reaction is

K+ + e– ⎯→ K

Thus one mole = 39 gm K is deposited by 1 F = 96500 C.

∴ 19·5 gm K will be deposited by = 96500

39 × 19·5 = 48250 C.

For deposition of Al, the reaction is

Al3+ + 3e– ⎯→ Al

Thus, 3F = 3 × 96500 C deposit 1 mol = 27 g Al.

Hence, 48250 C will deposit = 27 × 482503 × 96500 = 4·5 g of Al .

Q. 59. A conductivity cell constant is 0·5 cm–1. This cell when filled with 0·01 M sodium chloride has aresistance of 384 Ω at 25°C. Calculate the equivalent conductivity of 0·01 M NaCl at 25°C.

Sol. Given cell constantla

= 0·5 cm–1

R = 384 ohm

Thus, the conductance C =1R

=1

384 ohm–1

∴ Specific conductivity, K = Observed Conductance × Cell constant

=1

384 × 0·5

= 1·3 × 10–3 ohm–1 cm–1

Further, equivalent conductivity, Λeq. =K × 1000

Ceq.

=1·3 × 10–3 × 1000

0·01 (For NaCl 0·01 M = 0·01 N)

= 130 ohm–1 cm2 eq.–1

Q. 60. For the following cell, calculate the emf :

Al/Al3+ (0·01 M) // Fe2+ (0·02M) /Fe

Given : EoAl

3+ / Al = – 1·66 V

EoFe

2+ / Fe = – 0·44 V

Sol. For the cell

Al | Al3+ (0·01 M) || Fe2+ (0·02M) | FeAnode Cathode

EoAl3+ / Al = – 1·66 V

EoFe2+ / Fe = – 0·44 V

E°Cell = E

oFe – E

oAl = – 0·44 – (– 1·66) = + 1·22 V

Nernst equation for the cell reaction

2Al + 3Fe2+ ⎯→ 2Al3+ + 3Fe is

Ecell = E°Cell +

0·05916 log

[Fe2+]3

[Al3+]2


= 1·22 + 0·0591

6 log (0·02)3

(0·01)2

= 1·22 + 0·0591

6 log 8 × 10– 2

= 1·22 + 0·00985 [log 8 – 2 log 10]= 1·22 + 0·00985 [0·8451 – 2 × 1]= 1·22 + 0·00985 [– 1·1549]= 1·220 – 0·011= 1·209 V

Q. 61. How many hours does it take to reduce 3 mole of Fe3+ to Fe2+ with 2·0 A current.Sol. Given, Current = 2·0 ampere

3Fe3+ + 3e– ⎯⎯→ 3Fe2+

Charge required = 3 faraday = 3 × 96500 coulomb= 289500 coulomb

We know, Charge = Current × Time

or Time =289500

2= 144750 sec.= 40·20 hrs.

Q. 62. Calculate the number of coulombs required to deposit 5·4 g of Al when the electrode reaction isAl3+ + 3e

– → Al [Atomic weight of Al = 27 g/mol.]

Sol. Al3+ + 3e– → Al

For 1 mol (27 g) of Al requires = 3 × 96500 C

1 g of Al requires =3 × 96500

27

5·4 g of Al requires =3 × 96500

27 × 5·4 = 57900 C

Q. 63. If the Faraday were to be 60230 coulomb instead of 96500 coulomb what will be the charge of anelectron ?

Sol. The charge on the electron is given by :

e =FN =

602306·023 × 1023

where F is value of 1 faraday in coulomb and N is Avogadro’s number

e =60230

60230 × 1019 = 1 × 10–19 coulomb

∴ Charge on an electron = 1 × 10–19 coulomb.


Q. 1. In a first order reaction, 10% of the reactant is consumed in 25 minute. Calculate :(1) The half-life period of the reaction.(2) The time required for completing 87·5% of the reaction.

Sol. t = 10% reactant → 25 minute

(1) t1/2 =0·693

k

k =2·303

t log

10090

k = 0·004215

t1/2 =0·693

k


t1/2 = 164·412 minute.

(2) t =2·303

k log

10012·5

t = 493·43 minute.

Q. 2. A substance decomposes by following first order kinetics. If 50% of the compound isdecomposed in 120 minute, how long will it take for 90% of the compound to decompose ?

Sol. Given : Half life (T1/2) = 120 minute

For 1st order reaction, half life (T1/2) =0·693

Rate constant (k)

Here, k =0·693120 = 0·00578 min– 1

Let initial concentration ‘a’ mol and time t in minutes required for 90% decomposition of thecompound.

Now, t =2·303

k log

aa – 0·9a

=2·303

0·00578 log 10 = 398 minute

Thus, 398 minute are required for 90% decomposition of the compound.

Q. 3. Show that the time required for the completion of 75% of a reaction of first order is twice the timerequired for the completion of 50% of the reaction.

Sol. For a first order reaction, rate constant

k =2·303

t log

a(a – x)

For 75% completion, k =2·303t(75%)

log a

(a – x) = 1·403t75%

For 50% completion, k =0·693t(50%)

Because k is constant for a given reaction, so

1·403t(75%)

=0·693t(50%)

⇒ 2·02 t50% = t(75%)

or 2t(50%) = t(75%)

This shows that the time required for 75% completion of a reaction is nearly twice the timerequired for 50% completion.

Q. 4. The rate constant of a first order reaction is 4·5 × 10 – 2 sec – 1. What will be the time required forthe initial concentration of 0.4 M of the reactant to be reduced to 0·2 M ?

Sol. t1/2 = 0·693

k =

0·6934·5 × 10 – 2 = 15·4 sec.

Q. 5. If the half life period for a first order reaction is 69·3 seconds, what is the value of its rateconstant ?

Sol. If half life of first order reaction is t1/2 then its rate constant,

k =0·693t1/2

=0·69369·3

= 0·01 sec– 1


Q. 6. The rate constant of a first order reaction is 2·31 × 10–2 sec–1. What will be the timerequired for the initial concentration 0·1 M, of the reactant to be reduced to 0·05 M ?

Sol. : We know that, k =2·303

t log

a(a – e)

Given : k = 2·31 × 10–2 sec– 1

Hence, 2·31 × 10–2 =2·303

t log

0·10·05

t =2·303

2·31 × 10–2 log 2

t =2·303

2·31 × 10–2 × 0·3010

t = 30·0 sec.Q. 7. The rate of first order reaction is 3 × 10 –4 mol –1 sec –1 when the concentration of reactant is 10 –1

mol –1 . What will be the rate of the reaction when the concentration of reactant is 10–2 mol L–1 ?Sol. We know that, Rate = k [A]' for 1st order reaction

Hence, k =Rate[A]

Given : Rate = 3 × 10–4 mol L–1 sec–1 and [A] = 10 –1 mol L–1

Hence , k =3 × 10 –4

10 –1 = 3 × 10–3 sec –1

Now, the concentration of the reactant is 10–2 mol L–1

Hence, Rate = k [A]= 3 × 10–3 × 10–2

= 3 × 10–5 mol L–1 sec–1.

Q. 8. A first order reaction is completed 50% in 30 minute at 27°C. Calculate the rate constant of thereaction at 27° C.

Sol. Given : t1/2 = 30 minute

Hence, Rate constant, k =0·693t1/2

or k =0·693

30= 0·023 min–1.

Q. 9. A first order reaction is 20% complete in 10 minute. Calculate :(i) The rate constant of the reaction(ii) The time required for the completion of 75% of reaction

Sol. (i) For the first order reaction, we have

K = 2·303

t log

[A]0

[A] =2·303

10 min log 10080 = 0·0223 min– 1

(ii) We have, t =2·303

k log

[A]0

[A]

=2·3030·0223 log

10025 = 62·18 min.

Q. 10. The rate constant for an isomerisation reaction, A → B is 4·5 × 10–3 min–1. If the initialconcentration of A is 1 M, calculate the rate of reaction after 1 hour.

Sol. We know that, K =2·303

t log

[A]0

[A]

4·5 × 10–3 =2·303

60 log 1

[A]


log [A] = – 0·1172[A] = 0·7635 M

Rate after 1 hour or 60 minute = K [A]Rate = 4·5 × 10 –3 × 0·7635

= 3·45 × 10 –3 mol L–1 min–1.Q. 11. At 300 K, a first order reaction is 50% complete in 20 minute. At 350 K, the same reaction is 50%

complete in 5 minute. Calculate the activation energy of the reaction.

Sol. Rate constant at 300 K (k1) =0·693

20 min–1

Rate constant at 350 K (k2) =0·693

5 min–1

According to Arrhenius equation,

log k1

k2= –

Ea

2·303 R ⎣⎢⎢⎡

⎦⎥⎥⎤1

T1 –

1T2

…(1)

Given : T1 = 300 K, T2 = 350 KPutting the values in equation (1),

log

0·69320

0·6935

=– Ea

2·303 × 8·314 ⎣⎢⎢⎡

⎦⎥⎥⎤1

300 – 1

350

– 0·6020 = – Ea

19.147 × 4·76 × 10–4

∴ Ea =0·6020 × 19·147

4.76 × 10–4 = 2·42 × 104 J

Ea = 24·3 kJQ. 12. For the thermal decomposition of acetaldehyde

CH3CHO(g) ⎯→ CH4(g) + CO(g)The following rate data are obtained :

Experiment Initial pressure Initial rate of increase in total pressure(torr) (torr)

1 300 0·61 (r1)2 200 0·27 (r2)

Predict the order of reaction.

Sol. For the given reaction,

Rate = k [CH3CHO] n-order of reaction

log rate = log k + n log [CH3CHO]

For the two rates, given log r1 = log k + nlog [CH3CHO]1

log r2 = log k + n log [CH3CHO]2

∴ log r1

r2= nlog

[CH3CHO]1

[CH3CHO]2

Hence, the order of reaction n =log r1/r2

log [CH3CHO]1/log [CH3CHO]2

=log (0·61/0·27)log (300/200)

=log 2·26log 1·50 =

0·3540·176 = 2·0

Q. 13. In a first order reaction, 10% of the reactant is consumed in 25 minute. Calculate :(i) The half life of the reaction.(ii) The time required for completing 17% of the reaction.

Sol. (i) For 1st order reaction,


k =2·303

t log

aa – x =

2·30325 log

100100 – 10 = 0·0042145

Half life, t1/2 =0·693

k =

0·6930·0042145

t1/2 = 164·43 minutes.(ii) Again, for the given reaction

k =2·303

t log

aa – x

t =2·303

0·0042145 log 100

100 – 17∴ t = 44·21 minute.∴ Time required for completing 17% of the reaction is 44·21 minutes.

Q. 14. The rate of reaction triples when the temperature changes from 20°C to 50°C. Calculate theenergy of activation. (R = 8·314 JK –1 mol –1)

Hint : We know that log k2

k1=

Ea

2·303 R ⎣⎢⎢⎡

⎦⎥⎥⎤T2 – T1

T1T2

Sol. Here k2

k1= 3, T1 = 20 + 273 = 293 K

T2 = 50 + 273 = 323 KR = 8·314 JK–1 mol–1

On substituting the values in above expression.

log 3 =Ea

2·303 × 8·314 ⎣⎢⎢⎡

⎦⎥⎥⎤323 – 293

323 × 293

∴ Ea = 28·8 kJ mol–1.

Q. 15. The half-life period for the decomposition of a substance is 2·5 hours. If the initial weight of thesubstance is 160 g, how much of the substances will be left after 10 hours ?

Sol. We know that , t = 2.303

k log

C°ACA

t = Time = 10 hrs.

k = Rate constant = 0·693t1/2

= 0·6932.5 = 0·2772 hr–1

C°A = Initial amount = 160 g

CA = Amount left after time t (10 hrs.)

Putting the values in equation (i)

10 =2·3030·2772 log

160CA

160CA

= 15·98

∴ CA =160

15·98 = 10·01 g

∴ Amount of substance left = 10.01 g.

Q. 16. Find the two-third life t2/3 of a first order reaction in which k = 5·4 × 10–14 s–1.Sol. The two third life of a first order reaction is the time in which the concentration of the reactant is

reduced to one-third (1/3) of its initial [A]° concentration.Let initial concentration, [A]° = 1 mol L–1

Concentration at t2/3 , [A]t = [A]0 – 23 [A]° = 1/3 [A]° mol L–1


t2/3 =2·303

k log

[A]°[A]t

=2·303

5·4 × 10–14 s–1 log 1

1/3 = 2·303

5·4 × 10–14 s–1 log 3

=2·303

5·4 × 10–14 s–1 × 0·4771 = 2·03 × 1013 s.

Q. 17. The rate constant of a reaction is 1·5 × 107 s–1 at 50°C and 4·5 × 107 s– 1 at 100°C. EvaluateArrhenius Parameters A and Ea

Sol. log ⎝⎜⎛

⎠⎟⎞4·5 × 107

1·5 × 107 = Ea/(2·303 × 1·987 cal k–1 mol– 1)

log 4·5 × 107

1·5 × 107 =Ea

2·303 × 1·987 ⎣⎢⎢⎡

⎦⎥⎥⎤1

323 – 1

373

Ea = 5·26 k cal1·5 × 107 = Aexp (526 × 103/1·987 × 323)

A = 5·46 × 1010 s– 1.Q. 18. A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the

presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activationenergy of the reaction if the catalyst lowers the activation barrier by 20 kJ mol– 1.

Sol. Arrhenius equation for the reaction under two different conditions can be written as :k = Ae– E1/RT1 …(1)k = Ae– E2/RT2 …(2)

From (1) and (2) Ae– E1/RT1 = Ae– E2/RT2

E1

RT1=

E2

RT2

orE1

T1=

E2

T2…(3)

As E2 = E1 – 20, we can write equation (3) as

E1

T1=

E1 – 20T2

or E1 ⎝⎜⎛

⎠⎟⎞1

T2 –

1T1

=

20T2

or E1 =20 × T1

T1 – T2

Substituting the values of T1 and T2, we have

E1 =20 × 500500 – 400

= 100 kJ mol– 1.Q. 19. Prove that :

AA0

= ⎝⎜⎛

⎠⎟⎞1

2 n

Sol. For first order reaction, t =2·303

k log

[A]°

[A]

and k =0.693t1/2

= 2·303 log 2

t1/2

t =2·303 × t1/2

2·303 × log 2 · log [A]°

[A]

ort

t1/2= log

[A]°

[A] · 1

log 2

or log [A]°

[A] =t

t1/2· log 2


Q. 20. The slope of the line in the graph of log k (k = rate constant) versus 1T is – 5841. Calculate the

activation energy of the reaction.

Sol. Slope = – Ea

2·303 Ror – Ea = – 5841 × 2·303 × 8·314

Ea = 111838·45 J mol–1

= 111·838 kJ mol–1

Q. 21. If the half life period for a first order reaction is 69·3 second, what is the value of its rateconstant ?

Ans. If half life of first order reaction is T then its rate constant,

k =0·693t1/2

= 0·69369·3 = 0·01 sec– 1

Chapter 5.Surface Chemistry

Q. 1. 100 ml of a colloidal solution is completely precipitated by addition of 5 ml of 1 M NaCl solution.Calculate the coagulation value of NaCl.

Ans. Coagulation value is the milli moles of an electrolytes that must be added to 1 L of a colloidalsolution for complete coagulation.

∴ 5 ml of 1 M NaCl = 1

1000 × 5 = 0·005 or 5 milli moles

100 mL of a colloidal solution requires. NaCl for complete coagulation = 5 m moles.∴ 1 L of colloidal solution requires NaCl for complete coagulation = 5 m moles.∴ Coagulation value of NaCl = 50

Q. 2. 1 g of charcoal adsorbs 100 ml of 0·5 M CH3COOH to form a monolayer and thereby molarity ofacetic acid is reduced to 0·49 M. Calculate the surface area of the charcoal adsorbed by eachmolecule of acetic acid, surface area of charcoal = 3·01 × 102 m2/g.

Sol. No. of moles of acetic acid initially present

=0·5

1000 × 100 = 0·05 mol.

No. of moles of acetic acid left after adsoption

=0·491000 × 100 = 0·049

Moles of acetic acid adsorbed = 0·05 – 0·049 = 0·001 mol.= 1 × 10–3 mol.

No. of molecules of acetic acid adsorbed= 1 × 10–3 × 6·022 × 1023

= 6·022 × 1020 molecule.Now, 1 g of charcoal has area = 3·01 × 102 m2

∴ 6·022 × 1020 molecule of acetic acid gets adsorbed on surface area= 3·01 × 102 m2

∴ 1 molecule of acetic acid gets adsorbed on surface area

=3·01 × 102

6·02 × 1020 = 5·0 × 10–19 m2

Q. 3. 20% surface sites have adsorbed N2. On heating N2 gas is evolved from sites and were collectedat 0·001 atm and 298 K in a container of volume 2·46 cm3. Density of surface sites is 6·023 × 1014

cm–2 and surface area is 1000 cm2. Calculate the number of surface sites occupied per molecule ofN2.

Sol. For adsorbed N2 on surface sites p(N2) = 0·001 atmV = 2·46 cm3, T = 298 K, R = 82·0 atm cm3 K–1 mol–1

n (N2) =PVRT =

0·001 × 2·4682·0 × 298 = 1·006 × 10–7

Molecules of N2 adsorbed = 1·006 × 10–7 × 6·022 × 1023


= 6·022 × 1016

Total surface sites available = No. of sites per cm2 × Area= 6·023 × 1014 × 1000= 6·023 × 1017

Surface sites on which N2 is adsorbed =6·023 × 1917 × 20

100= 12·046 × 1016

No. of sites adsorbed per molecule of N2 =12·046 × 1016

6·023 × 1016 = 2

Q. 4. The following data were obtained for the adsorption of carbon monoxide gas on 3·0g of charcoalat 0° and 1 atm pressure.Pressure (mm Hg) 200 400Volume of gas adsorbed, x 18·6 31·4(reduced to STP)Calculate the values of the constants k and n using Freundlich adsorption equation.

Sol. According to Freundlich adsorption isotherms.xm

= kP1/3

log xm

= log k + 1n

log P

Substituting the values of the two given sets, we get

log 18·6

3 = log k + 1n

log 200 …(i)

log 31·4

3 = log k + 1n

log 400 …(ii)

Subtracting equation (i) from equation (ii), we get

log 31·418·6 =

1n

log 400200

log 1·688 =1n

log 2

0·2274 =1n

× 0·3010

n =0·30100·2274 = 1·32

Substituting the value of n in equation (i)

log 6·2 = log k + 1

1·32 log 200

0·7924 = log k + 1

1·32 × 2·301

0·7924 = log k + 1·7432

log k = – 0·9508 = –2·8212

k = antilog ( –2·8212) = 0·0662 cm3 g–1

Q. 5. In a coagulation experiment, 5 ml of As2S3 is mixed with distilled water and 0·1 M solution of anelectrolyte AB so that total volume is 10 ml. It was found that all solutions containing more than4·6 ml of AB coagulate with in 5 minutes. What is the flocculation value of AB for As2S3 Sol. ?

Ans. A minimum of 4·6 mL of AB is required to coagulate the sol. The moles of AB in the sol is

=4·6 × 0·1

10 = 0·046 mole

This means that a minimum 0·046 mole of 0·046 × 1000 = 46 milli moles required for coagulating 1litre of sol.∴ Flocculation value of AB for As2S3 sol = 46.


Q. 6. The coagulation of 100 ml of a colloidal solution of gold is completely prevented by the additionof 0·25g of starch to it before adding 1 ml of 10% NaCl solution. Calculate the gold number ofstarch.

Sol. Amount of starch added to 100 ml of gold sol required to prevent coagulation of 1 ml of 10%NaCl solutioon

= 0·25 gor = 250 mgStarch required to be added to 10 ml of gold sol to completely prevent coagulation by 1 ml of 10%

NaCl solution =250100 × 10 = 25 mg

Gold number of starch = 25

Chapter 9.Coordination Compounds

Q. 1. Calculate the oxidation state of the central metal atom in the following coordination compounds :

(a) Na[AlH4] (b) [Ni (CO)4]

(c) [Pt Cl2 (NH3)2]2+ (d) [Co(en)2 (ONO) Cl] Cl

(e) K3 [Co(OX)3] (f) [Co(en)2 Cl(ONO)]

(g) [Co(NH3)2 Cl4]2+ (h) [Fe(CN)6]4–

Ans. (a) Na[AlH4] → Na+ + [AlH4]–

x + 4 (1) = – 1

x + 4 = – 1∴ Oxidation state of Al, x = + 3(b) [Ni (CO)4] → as CO is a mutual ligand

x + 4 (0) = 0∴ Oxidation state of Ni, x = 0

(c) [Pt Cl2 (NH3)2]2+→ x + 2 (–1) + 2(0) = + 2 (as Cl– = – 1 and NH3 = 0)

(d) [Co(en)2 (ONO) Cl] Cl → [Co(en)2 (ONO) Cl]+ + Cl–

→ x + 2(0) + (–1) + (–1) = + 1 ⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤(en) = 0

ONO = –1Cl = –1

x – 2 = + 1∴ Oxidation state of Co, x = + 3

(e) K3 [Co(OX)3] → 3K+ + [Co(OX)3]3

x + 3 (–2) = – 3∴ Oxidation state of Co, x = + 3

(f) [Co(en)2 Cl(ONO)]x + 2(0) + (–1) + (–1) = 0

∴ Oxidation state of Co, x = + 2(g) [Co(NH3)2 Cl4]2+ [Co(NH3)2Cl4]2–

x + 2(0) + 4 (–1) = – 2∴ oxidation state of Co, x = + 6

(h) [Fe(CN)6]4–

x + 6 (–1) = – 4x = – 4 + 6

∴ Oxidation state of Fe, x = + 2❐

Question

11Set Differentiate Between

Chapter 1. Solid StateQ. 1. Differentiate between crystalline and amorphous solids.

Ans. Crystalline Solid Amorphous Solid

1. They have a regular three-dimensionalarrangement of ions, atoms or moleculesdue to which they have well definedgeometrical shape.

1. They do not have a regular arrangementof particles, therefore, do, not have welldefined particles.

2. They have sharp melting point. 2. They do not have sharp melting point.

3. They are true solids i.e., they show all thecharacteristic properties of solids.

3. They are pseudo-solids they do not showall the characteristic properties of solids.

4. They are anisotropic in nature 4. They are isotropic in nature.

Q. 2. Differentiate between Isotropic and Anisotropic substance.

Ans. Isotropic Substance : A substance, which possesses the same magnitude of properties such aselectrical conductivity, refractive index, electrical resistance etc. along all directions, is calledisotropic in nature and this property is referred as isotropy. Amorphous substance are isotropicin nature.

Anisotropic Substance : Substance which have different magnitudes of physical propertiesalong different directions is called anisotropic substance. The crystalline substances areanisotropic in nature.

Q. 3. How will you distinguish between the following pair of terms :

(i) Hexagonal close packing and cubic close packing.

(ii) Crystal lattice and unit cell.

(iii) Tetrahedral void and octahedral void.Ans. (i)

Hexagonal close packing Cubic close packing

1. In hcp, the spheres of the third layer areexactly aligned which those of the firstlayer. This arrangement is represented asAB AB …… type.

1. In ccp, the sphere of the third layer are notaligned with those of the first layer orsecond layer. The layers of fourth layer arealigned with those of the first layer. Thispattern is represented as ABC ABC ……type.

2. In hcp, the tetrahedral void of the secondlayer may be covered by the sphere of thethird layer.

2. In ccp, the third layer may be placed abovethe second layer in a manner such that itssphere covers the octahedral void.

(ii) The three-dimensional arrangement of constituent particles of a substance (atom, ion ormolecule) is called crystal lattice. The smallest repeating pattern in a crystal lattice whichwhen repeated in three-dimensions gives the crystal is called unit cell.

(iii) A void surrounded by four spheres is called a tetrahedral void while a void surrounded bysix spheres is called an octahedral void.

Q. 4. Differentiate between :(i) Hexagonal and monoclinic unit cell.

(ii) Face centred and end centred unit cell.


Ans. (i) Hexagonal unit cell has two edges of equal length (a = b ≠ c) while monoclinic unit cell hasall the three edges unequal.

In hexagonal two angles are of 90° and one angle of 120° (α = β = 90°, γ = 120°). Inmonoclinic two angles are of 90° but one is not of 90° (α = γ = 90°, β ≠ 90°).

(ii) The face centred unit cell has points at all the corner as well as on all the faces of cube. It has4 atoms per unit cell.

End centred unit cell has points at all the corners and at the centre of two opposite faces. Ithas 2 atoms per unit cell.

Q. 5. What is the difference between 13-15 and 12-16 compounds ?

Ans. Group 13-15 compounds are formed by combination of elements of group 13 and group 15 likeAlP, GaAs etc. They have large covalent character whereas group 12-16 compounds are thosewhich are formed by the combination of elements of group 12 and group 16 like ZnS, CS, HgSeetc. They do not have covalent character, but they have ionic character.

Q. 6. In terms of band theory what is the difference :(i) between a conductor and an insulator.(ii) between a conductor and a semiconductor.

Ans. (i) The energy gap (forbidden energy gap) between the valence band and conduction band inan insulator is very large so that electron cannot jump from valence band to conductionband hence cannot conduct electricity.While in case of conductor the energy gap is very small or there is a overlapping betweenvalence band and conduction band so that electrons can move through it and conductelectricity.

(ii) The energy gap is small between valence band and conduction band in case ofsemiconductor whereas in case of conductors energy gap between valence band andconduction band is very-very small they can overlap with each other.

Q. 7. What is the difference between metallic and ionic crystals ?

Ans. Ionic crystals are made up of cations and anions. These ions are arranged in a three-dimensionalarray to form an aggregate type. In this crystal, each ions are surrounded by definite number ofions of opposite charge. Metallic crystals are those in which the particles forming the crystals aremetallic, positive ions called Kernels, which are surrounded by a sea of electrons and are heldtogether by metallic bond, the positive metal ions are formed when the metal atoms lose theirvalence shell electrons.

Q. 8. What are the differences between Schottky and Frenkel Defects ?

Ans. Schottky Defect Frenkel Defect

1. This defect arises due to missing of equalno. of cations and anions from their normalcrystal sites.

1. This defect arises when some atoms or ionsare displaced from their normal site andoccupy interstitial sites.

2. Density of crystal decreases. 2. Density of crystal remains same.3. It is shown by ionic solids which have

high-coordination number.3. It is shown by ionic solids which have low-

coordination number.4. The size of cations and anions are equal. 4. The size of anions are larger than the

cations.5. e.g. : NaCl, CsCl etc. 5. e.g. : AgCl, Zns etc.

Q. 9. Differentiate between n-type and p-type semiconductors.

Ans. (i) p-type semiconductor : It is formed when doping is done in semiconductor from theelement of previous group e.g. Si is doped by Al or As is doped with Ge.

(ii) n-type semiconductor : It is formed when doping is done in a semiconductor from theelement of next group e.g. when P is doped with Si or when Ge is doped with As.

Differentiate Between ■ 281

Q. 10. Differentiate between paramagnetic substance and diamagnetic substance.

Ans. Diamagnetic Substance : The substances which are weakly repelled by magnetic field are knownas diamagnetic substances for e.g. NaCl, H2O etc. In diamagnetic substance all electrons arepaired and hence does not conducts electricity.

Paramagnetic Substance : The substances which have permanent magnetic dipoles and areattracted by the magnetic field are known as paramagnetic substances e.g.. Cu2+, Fe3+, TiO, Ti2O3

etc. Paramagnetic electrons have one or more unpaired electrons and hence conduct electricity.Q. 11. Differentiate among ferromagnetism, anti-ferromagnetism and ferrimagnetism.

Ans. Ferromagnetism : When all the domains are aligned in the same direction then this condition iscalled ferromagnetism.

↑ ↑ ↑ ↑ ↑

Ferromagnetism

Anti-ferromagnetism : When equal number of domains are arranged in opposite direction then itwill result in net zero value of magnetic moments then this situation is calledantiferromagnetism.

↑ ↓ ↑ ↓ ↑ ↓

Anti-ferromagnetism

Ferrimagnetism : When domains are arranged in opposite direction but not in equal numberwhich would result in net value of magnetic movement this situation is called ferrimagnetism.

↑ ↑ ↑ ↓ ↓ ↑ ↑

Ferrimagnetism

Q. 12. Differentiate between piezo-electricity and pyro-electricity.

Ans. Pyro-electricity : The electricity produced on heating a polar crystal is called pyro electricity.

Piezo-electricity : The electricity produced on applying mechanical stress on a polar crystal iscalled piezo-electricity.

Q. 13. Give basic differences between diamond and graphite.

Ans. (1) In diamond the carbon atom is sp3 hybridized while in graphite the carbon atom is sp2-hybridized.

(2) Diamond does not conducts electricity while graphite does.


Q. 1. Differentiate between the term Molarity and Molality.

Ans. Molarity (M) : It is defined as the number of moles of solute present in one litre of solution. It’sunit is mol/litre. It depends on temperature.

Molality (m) : It is defined as the number of moles of solute present in one kilogram of solvent.It’s unit is mol/kg. It does not depends on temperature.

Q. 2. Give difference between Osmosis and Reverse osmosis.

Ans. Osmosis : The movement of solvent particles from dilute solution to viscous solution sidethrough semi-permeable membrane is called osmosis.

Reverse Osmosis : If a pressure more than osmotic pressure is applied on the solution side, thesolvent will flow from the solution side to solvent side through semi-permeable membrane calledreverse osmosis.

Q. 3. Differentiate between Diffusion and Osmosis.

Ans. Diffusion : Inter mixing of particles from higher concentration to lower concentration is calleddiffusion. In diffusion there is no need of semi-permeable membrane (SPM).

Osmosis : Movement of solvent particles from lower concentration to higher concentrationthrough SPM is called osmosis. In this process SPM is needed.


Q. 4. Differentiate between ideal and non-ideal solutions.

Ans. Ideal solution : A solution is called an ideal solution if it obey’s Raoult’s law over a wide range ofconcentration at a specified temperature.

For an ideal solution :

(i) Raoult’s law is obeyed i.e. PA = P°A xA and PB = P°

B xB

(ii) ΔH mix = 0 and

(iii) ΔV mix = 0

Non-ideal solution : A solution which does not obey’s Raoult’s law for all concentration is calleda non-ideal solution.

For a non-ideal solution :

(i) Raoult’s law is not obeyed i.e. PA ≠ P°A xA and PB ≠ P°

B xB

(ii) ΔH mix ≠ 0

(iii) ΔV mix ≠ 0Q. 5. Differentiate between positive or negative deviation from Raoult’s law.Ans. Positive deviation : When A–B molecular interactions are weaker than A–A and B–B molecular

interactions. For example, a mixture of ethanol and acetone.Negative deviation : When A–B molecular interactions are stronger than A–A and B–B molecularinteractions. For example, mixture of chloroform and acetone.

Q. 6. Differentiate between hypotonic and hypertonic solution.Ans. Hypotonic solution Hypertonic solution

A solution having lower osmotic pressure thanthe other solution is said to be hypotonic withrespect to the other solution.

A solution having higher osmotic pressurethan the other solution is said to be hypertonicsolution with respect to the other solution.

Chapter 3. ElectrochemistryQ. 1. What are the main differences between galvanic cell and electrolytic cell ?Ans. Galvanic Cell Electrolytic Cell

1. Converts chemical energy to electricalenergy.

1. Converts electrical energy to chemicalenergy.

2. Reaction takes place spontaneously. 2. Reactions are not spontaneous.3. Salt-bridge is essential. 3. No such arrangement is required.4. Electrons flow from anode to cathode in

the external circuit.4. Electrons are supplied by the external

source. They enter the cell at cathode andleave the cell at anode.

Q. 2. What are the differences between the emf of the cell and cell potential ?Ans. EMF Cell Potential

1. It is the maximum voltage obtained fromthe cell.

1. It is always less than the maximum voltageobtained.

2. It is the potential difference between twoelectrodes when no current is flowing inthe circuit.

2. It is the difference of the electrodepotentials between the two electrodes whenin under operation.

3. It may be +ve or –ve. 3. It is always +ve.Q. 3. Differentiate between metallic conductance and electrolytic conductance. Ans. Metallic Conductance Electrolytic Conductance

1. It is caused due to mobility of freeelectrons.

1. It is due to mobility of ions.

2. There is no change in physical state ofsubstance.

2. Physical state gets changed.

3. It decreases with rise in temperature. e.g.Different metals.

3. It increases with temperature. e.g.,Different ionic substances.


Q. 4. What is the difference between equivalent conductivity and molar conductivity.Ans. Equivalent conductivity (ΛΛΛΛeq) : It is defined as the conductance offered by all the ions formed by

dissolving one gram equivalent of electrolyte in solution.

Λeq =K × 1000

Normality

It’s unit are ohm– 1 cm2 (g equiv)– 1 or Sm2 eq– 1

Molar conductivity (ΛΛΛΛm) : It is defined as the conductance offered by all the ions formed bydissolving one mole of an electrolyte in solution.

Λm =K × 1000Molarity

It’s unit is S m2 mol– 1 or ohm– 1 cm2 mol– 1.Q. 5. Differentiate between primary cell and secondary cell.Ans. Primary cell Secondary cell

1. In these cells, electrode reactions cannotbe reversed by external electric source.

1. In these cells, the electrode reaction can bereversed by external source.

2. Ex. : Dry cell, mercury cell. 2. Ex. : Lead storage battery, nickel cadmiumstorage cell.

Q. 6. Explain the term weak and strong electrolytes with examples. How can these be distinguished ?Ans. Weak electrolyte :

(i) These are those substance which are partially dissociated in their aqueous solution e.g.CH3COOH, NH4OH.

(ii) In case of weak electrolytes, molar conductivity increases with dilution and we get a curve ifwe plot graph between (Λm) and √⎯ c.

Strong electrolytes : (a) These are those substances which are almost completely dissociatedeither in molten state or in aqueous solution e.g. HNO3, HCl etc.(b) In case of strong electrolytes, Λm increases with dilution and we get straight line on plottinggraph between (Λeq) and √⎯ c.


Q. 1. Differentiate between the rate of a chemical reaction and the rate constant for the reaction.Ans. The rate is variable and is proportional to the concentrations of reactants for the reaction.

The rate constant is a constant which depends only on temperature and not on the concentration.Q. 2. In what respects specific reaction rate differs from rate of the reaction ?Ans. Specific reaction rate is the rate of reaction at unit concentration of each of the reactant. It is

always constant for a particular reaction at a given temperature.Q. 3. What is the difference between average and instantaneous rate of a chemical reaction ?Ans. Change in the concentration of a reactant in a given interval of time is average rate while

concentration change in a very small interval of time so that the rate is almost constant is calledinstantaneous rate.

Q. 4. List three differences between reaction rate and specific reaction rate. Ans. Reaction rate (rate of reaction) Specific reaction rate (rate constant)

1. It is the rate of change of concentration of areactant or a product with respect to time.

1. It is a constant of proportionality and isequal to the rate of reaction when the molarconcentration of each of the reactants isunity.

2. The reaction rate at a particular instantdepends upon the concentration ofreactants at that instant.

2. The specific reaction rate remains constantfor a particular reaction at particulartemperature and does not depends uponthe concentration of reactants.

3. Its unit is mol L– 1 s– 1. 3. Its unit depends upon the order of reaction.For a reaction of order n , the units ofspecific reaction rate are mol1 – n Ln – 1 s– 1.


Q. 5. Differentiate between Molecularity and Order of a reaction. Ans. Molecularity Order

1. It is the number of reacting speciesundergoing simultaneous collision in anelementary reaction.

1. It is the sum of power of concentrationterms in rate law expression.

2. It is a theoretical concept. 2. It is determined experimentally.3. It is a whole number (cannot be fractional). 3. It can be whole number or fraction.4. It cannot be zero. 4. It can be zero.

Q. 6. Differentiate between threshold energy and activation energy.

Ans. Threshold energy : The minimum amount of energy which the colliding molecules must possessto give the product.Activation energy : The excess amount of energy that must be supplied to the reactant inaddition to energy already possessed by them to undergo chemical reactions.

Q. 7. Differentiate between homogeneous and heterogeneous catalysis.Ans. In homogeneous catalysis the catalyst will be in the same phase as that of the reactants while in

heterogeneous catalysis the catalyst are used in different phases.Heterogeneous catalysis : Contact process

2SO2(g) + O2(g)NO

⎯→ 2SO3(g)

Homogeneous catalysis : Hydrolysis of cane sugar.

C12H22O11(l) + H2OH2SO4(l)

⎯⎯⎯⎯→ C6H12O6(l) + C6H12O6(l)


Q. 1. What are the major differences between adsorption and absorption.

Ans. Adsorption Absorption1. It is a surface phenomenon. 1. It occurs throughout the surface of the

material.2. It is rapid in the beginning and slows down

later.2. It occurs at a uniform rate.

Q. 2. Differentiate between positively charged sols and negatively charged sols.

Ans. Positively charged sols Negatively charged sols1. Hydrated metallic oxides e.g.

Al2O3.xH2O, CrO3·xH2O and Fe2O3. x H2Oetc.

1. Metal particle e.g. gold, silver, copper sols.

2. Basic dye like methylene blue sol. 2. Acidic dyes like eosin, congo red sols.3. Haemoglobin. 3. Sol of starch, gum, gelatin, clay, charcoal,

etc.4. Oxides, e.g. TiO2 sol. 4. Metallic sulphide e.g. As2S3, Sb2S3, CdS sols.

Q. 3. Differentiate between physisorption and chemisorption.Ans. Physisorption Chemisorption

1. It arises because of van der Waals forces. 1. It is caused by chemical bond formation.2. It is not specific in nature. 2. It is highly specific in nature.3. It is reversible in nature. 3. It is irreversible.4. It depends on the nature of gas. More easily

liquificable gases are adsorbed readily.4. It also depends upon the nature of gas.

Gases which can react with the adsorbentshows chemisorption.

5. Enthalpy of adsorption is low (20–40 kJmol–1) in this case.

5. Enthalpy of adsorption is high (80–240 kJmol–1) in this case.

6. Low temperature is favourable foradsorption. It decreases with increases oftemperature.

6. High temperature is favourable foradsorption. It increases with the increase oftemperature.


Q. 4. What is the difference between multimolecular and macromolecular colloids ? Give one exampleof each. How are associated colloids different from these two types of colloids ?

Ans. Multimolecular colloids : In this type of colloid, particles are aggregates of atoms or moleculeseach having size less than 1 nm e.g. sulphur sol, gol sol. Multimolecular colloids are generallylyophobic in nature.

Macromolecular colloids : In this type of colloids, colloidal particles are themselves largemolecules of colloidal dimensions e.g. Starch, Proteins, Polyethene etc. Macro-molecular colloidsare generally lyophilic in nature.

Associated colloids : There are certain substances which at low concentration behaves as normalelectrolyte but at higher concentrations exhibits colloidal behaviour due to the formation ofaggregates. Such colloids are known as associated colloids e.g. soaps and detergents.

Q. 5. Differentiate among a homogeneous solution, a suspension and colloidal solution, givingsuitable example of each.

Ans.

S. No. Property Homogeneous solution Colloidal solution Suspension

(i) Particle size Less than 1 nm between 1 nm to1000 nm

more than 1000 nm

(ii) Separation by* Ordinary filteration* Ultra filtration

not possiblenot possible

not possiblepossible

possiblepossible

(iii) Settling of particles Do not settle Settle only oncoagulation

Settle under gravity

(iv) Appearance Transparent Translucent Opaque(v) Example Glucose dissolved in

water.Smoke, milk, gold,gold sol.

Solid in water.

Q. 6. Differentiate between lyophilic and lyophobic sols.

Ans. Lyophilic sols Lyophobic sols1. These are those sols in which the particles

of dispersed phase have greater affinityfor the dispersion medium e.g. sols of gum,gelatins, starch etc.

1. In this type of sols, the particles ofdispersed phase have little or no afffinityfor the dispersion medium e.g. gold sols,Fe(OH)3 sols, As2O3 sol etc.


Q. 1. Differentiate between ‘mineral’ and ‘ores’.

Ans. The naturally occurring chemical substance in the form of which metals occurs in the earth’s crustalong with impurities are called minerals.

The minerals from which the metals can be extracted conveniently and profitably is called an ore.

Q. 2. Giving examples, Differentiate between ‘roasting’ and ‘calcination’.

Ans. Roasting Calcination

1. It is the process of heating the ore below itsmelting point with excess of air.

1. It is the process of heating the ore below itsmelting point in the absence or limitedsupply of air.

2. Sulphide ores are converted into oxide form.

2ZnS + 3O2 Heat

⎯⎯⎯⎯→ 2ZnO + 2SO2 ↑

2. Carbonates ores are converted into oxide.

ZnCO3 Heat

⎯⎯⎯⎯→ ZnO + CO2 ↑

3. Volatile impurities are removed as oxidesSO2, As2O3 etc.

3. Water and organic impurities are removed.


Chapter 7. p-Block ElementsQ. 1. Differentiate between white and red phosphorus.Ans.

Property White Phosphorus Red Phosphorus1. State Translucent white waxy solid. Iron grey lustrous powder.2. Odour Garlic smell Odourless3. Physiological

actionPoisonous Non-poisonous

4. Solubility Insoluble in water but soluble in CS2 Insoluble in water as well as in CS2

5. Stability Less stable More stable6. Action of air Readily catches fire in air with greenish

glow which is visible in dark.Does not glow

7. Effect of heat Changes to β-Black phosphorus whenheated at 473 K under high pressureand changes to red phosphorus whenheated at 573 K.

Changes to α-Black phosphorus whenheated at 803 K in solid tube.

8. Structure Consists of discrete tetrahedral P4molecules.

P

P P

P

60

Tetrahedral units of P4 joined togetherthrough covalent bond to givepolymeric structure.

P

P P

P

P

P P

P

P

P P

P

Chapter 8. d-and f-Block ElementsQ. 1. What is the basic difference between transition and inner transition elements.Ans. The elements of d-block are known as transition elements as they possess properties that are

transitional between the s-and p-block elements.The elements of f-block are known as inner transition elements as they are characterised by fillingof the ‘f ’ orbitals. The inner transition elements consist of lanthanoids and actinoids.

Q. 2. Differentiate between lanthanoids and actinoids.Ans. Lanthanoids Actinoids

1. + 3 oxidation state is most common alongwith + 2 and + 4.

1. They show + 2, + 3, + 4, + 5, + 6, + 7oxidation states.

2. Among lanthanoids only promethium (Pm)is radioactive element.

2. All are radioactive elements.

3. They are less reactive than actinoids 3. They are highly reactive.4. Their magnetic properties are less complex. 4. They have more complex magnetic

properties.

Chapter 9. Coordination CompoundsQ. 1. What is the difference between Homoleptic and Heteroleptic complexes ?Ans. Homoleptic complexes : The complexes in which the metal atom or ion is linked to only one type

of ligands are called homoleptic complexes e.g. [Co(NH3)6]3+, [Fe(CN)6]4– etc.Heteroleptic complexes : The complexes in which the metal atom or ion is linked to more thanone kind of ligand are called heteroleptic complexes e.g. [Co(NH4)Cl2]+.

Q. 2. Differentiate between homonuclear and polynuclear complexes.Ans. The complexes containing only one metal atom in the coordination sphere are called homo

nuclear complexes e.g., [Co(NH3)6] Cl3. On the other hand the complexes which contains morethan one metal atom in the coordination sphere are called poly nuclear complexes,

e.g. : ⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

(NH3)4 CoNH

NH

Co (NH3)4

3–


Q. 3. Differentiate between mono-dentate, bidentate and poly-dentate ligands.Ans. Ligands : An atom or molecule which can donate a pair of electron to the central metal atom or

ion and forms a coordinate bond with it is called ligand.

Ligand may be classified as :

(1) Unidentate Ligand : Having one donor atom e.g., NH3 — amine, CO-carbonyl

(2) Bidentate Ligand : Having two donor atoms e.g.,

CH2 — NH2|CH2 — NH2Ethylendiamine(en)

(3) Polydentate Ligand : Having several donor atoms e.g., CH2 — N|CH2 — N

CH2COO–

CH2COO–

CH2COO–

CH2COO–

Ethylene amine tetra acetate ion (EDTA4–) Hexa dentate.

Q. 4. Differentiate between weak field and strong field coordination entity.

Ans. Weak field coordination entity Strong field coordination entity

1. They are formed when the crystal fieldstabilisation energy (Δo) in octahedralcomplexes is less than the energy requiredfor an electron pairing in a single orbital(P).

1. They are formed when crystal fieldstabilisation energy (Δo) is greater than P.

2. They are also called high spin complexes. 2. They are also called low spin complexes.

3. They are paramagnetic in nature. 3. They are mostly diamagnetic in nature.

4. Never formed by CN– ligands. 4. Formed by CN– ligands.

Q. 5. Differentiate between (a) Ionisation isomerism and Hydrate or solvate isomerism (b) Coordinateisomerism and linkage isomerism.

Ans. (a)

Ionisation Isomerism Hydrate or Solvate Isomerism

1. The compounds which have same molecularformula but give different ions in solutionare called ionisation isomerism andphenomenon is called ionisation isomerism.

1. The compounds which have samemolecular formula but differ in thenumber of water molecules present asligands or as molecules of hydration arecalled hydrate isomers and thisphenomenon is called hydrate isomerism.

2. e.g. [CoBr (NH3)5]SO4 and [CoSO4 (NH3)5]Br. 2. e.g. [Cr(H2O)6] Cl3

and [CrCl(H2O)5] Cl2.H2O.

(b)

Coordinate Isomerism Linkage Isomerism

1. This type of isomerism arises due to interchange of ligands between cationic andanionic entities of different metal ionpresent in complex.

1. This type of isomerism arises due to thepresence of an ambidentate ligand in a co-ordination compound.

2. e. g. [Co(NH3)6] [Cr(CN)6]

and [Cr(NH3)6] [Co(CN)6]

2. e.g. [Cr(NH3)5] NO2Cl2

and [Co(NH3)5 ONO]Cl2


Q. 6. Write the differences between double salt and a complex or coordination compound. Ans. Double Salt Coordination Compound

1. When two salts in stoichiometric ratio arecrystallised together from their saturatedsolution. They are called double salts.

1. A coordination compound contains acentral metal atom or ion surrounded bynumber of oppositely charged ions orneutral molecules.

2. They dissociate into simple ions whendissolved in water.

2. They do not dissociate into simple ionswhen dissolved in water.

3. They exists in solid state only. 3. They exists in solid state as well as inaqueous solution.

4. Example : FeSO4.(NH4)2SO4·6H2O(Mohr’s salt)

4. Example : K4[Fe (CN)6]

Chapter 10. Haloalkanes and HaloarenesQ. 1. Differentiate between compounds and configuration in open chain molecule by giving one

example each.Ans. Conformation isomers : These are compounds having different spatial arrangement of atoms or

groups attached to carbon atom bonded by single bond and are obtained by the rotation of singlebond. These isomers are called conformers or rotational isomers and have different energy. Theconformation isomerism is exhibited by alkanes and cyclo alkanes.Configurational isomers : It is due to certain type of rigidity within the molecule.Configurational isomers can be interconverted only by breaking and remaking of covalent bonds.These are of two types :(1) Geometrical isomerism (2) Optial isomerism.

Q. 2. Differentiate between retention and inversion.Ans. If the relative configuration of the atom or groups around a chiral centre in an optically active

molecule remains the same before and after the reaction, the reaction is said to proceed withretention of configuration.On the other hand, if the relative configuration of the atoms or groups around a stereocentre inthe product is opposite to that in the reactant, the reaction is said to proceed with inversion ofconfiguration, for example

Chapter 11. Alcohols, Phenols and EthersQ. 1. Distinguish between 1-propanol and 2-propanol.Ans. To distinguish between 1-propanol and 2-propanol Lucas test can be used.

When Lucas reagent (solution of HCl and ZnCl2) is added to 2-propanol, cloudiness appearswithin 5 minutes. When Lucas reagent is added to 1-propanol, solution remains clear.

Q. 2. Distinguish between Ethanol and Phenol.Ans. Ethanol gives red colour with ferric ammonium nitrate while phenol gives violet colour.

Q. 3. How will you distinguish alcohols.Ans. We can distinguish between primary (1°), secondary (2°) and Tertiary (3°) alcohols with the help

of following Test :(a) Lucas test : In this test, the alcohol is treated with Lucas reagent which is an equimolarmixture of hydrochloric acid (HCl) and zinc chloride (ZnCl2). Alcohols are soluble in Lucasreagent and forms a clear solution. On reaction alkyl chlorides are formed which being insolubleresults in turbidity in the solution.


AlcoholHCl

ZnCl2

⎯⎯⎯→ Alkyl chloride + H2O

(Anhyd.)

(i) If turbidity appears immediately. It is tertiary alcohol.

(ii) If turbidity appears within five minutes it is secondary alcohol.

(iii) If turbidity appears upon heating, it is primary alcohol.

(b) Victor Meyer’s test : It involves treatment of alcohol with (i) red P + I2 (ii) AgNO2 (iii) HNO3

and (iv) aqueous KOH and NaOH.

(i) Primary alcohol (1°) gives red blood colour.

(ii) Secondary alcohol (2°) gives blue colour.

(iii) Tertiary alcohol (3°) do not give any colour.

3. By Oxidation :

(i) Primary (1°) alcohol : A primary alcohol is easily oxidised to form first an aldehyde andthen a carboxylic acid.

CH2CH2OH

(2C)

Ethyl Alcohol K2Cr2O7 + dil. H2SO4

⎯⎯⎯⎯⎯⎯⎯⎯⎯→ CH3 —

H|C = O

(2C)

Acetaldehyde

[O]⎯→ CH3 —

O||C —OH

(2C)

Acetic acid

(ii) Secondary (2°) alcohol : A secondary alcohol is oxidised to ketone which further oxidises toform a mixture of acid

CH3 —�C —�CH2CH2CH3

Ethyl alcohol(5C)

Pentan-2-one(5C)

Acetic acid(2C)

Propanoic acid(2C)

O

CH3COOH + CH2CH2COOH

OH

CH3 — CH —�CH2CH2CH3[O]⎯→

[O]⎯→

(iii) Tertiary (3°) alcohol : It is very difficult to oxidise because they do not have hydrogen oncarbon bearing —OH group. However on treatment with K2Cr2O7/H2SO4 it oxidises into amixture of ketone and carboxylic acid.

CH3—

CH3|C—|CH3

OH

tert. butyl alcohol

(4C)

(Tertiary)

4 [O]

⎯⎯→ CH3 —

CH3|C = O

(3C)

4 [O]

⎯⎯→ CH3COOH

(2C)

Acetic acid

CH3—

CH3|C—|OH

CH2CH3

2-methyl but-2-ol

[O]⎯→

CH3 CH3

C = O

(3C)

Acetone

+ CH3COOH

(2C)

Acetic acid

4. By Dehydration : When alcohols are treated with protonic acid such as conc. H2SO4 or H3PO4

at 443 K they get dehydrated to form alkene.

The relative ease of dehydration of alcohol follows the following order :

Tertiary Alcohol > Secondary Alcohol > Primary Alcohol.


Q. 4. Give distinction between alcohols and phenols.Ans. Alcohols Phenols

1. Alcohols do not have any effect on litmussolution.

1. Phenols turn blue litmus to red.

2. Alcohols do not give any characteristiccolour with FeCl3 solution.

2. Phenols give characteristic colours such asviolet, red, etc., with FeCl3 solution.

3. Alcohols do not react with NaOH solution. 3. Phenols react with NaOH solution andforms sodium salts.


Q. 1. Give a chemical test to distinguish :

(i) Formaldehyde and Acetaldehyde

(ii) Dimethyl ether and Ethyl alcohol

Ans. (i) Formaldehyde and Acetaldehyde

(a) With freshly prepared sodium nitroprusside solution and excess of dil. NaOHacetaldehyde produces a deep blue colour.

(b) Formaldehyde does not give the above test.

(ii) Dimethyl ether and Ethyl alcohol

(a) Ethyl alcohol gives iodoform test a pale yellow precipitate of iodoform is formed.

CH3CH2OH + 4I2 + 6NaOH → CHI3 + HCOONa + 5NaI + 5H2OEthyl alcohol Iodoform

(b) Dimethyl ether does not gives iodoform test.

Q. 2. Distinguish between Acetaldehyde and Benzaldehyde

Ans. When acetaldehyde is heated with Fehling's solution, red precipitate of cuprous oxide is formed.

Acetaldehyde

CH3CHO + 2CuO

Fehling’ssolution

↓ ⎯→

⎝⎜⎛

⎠⎟⎞

Redppt

Cu2O(s) ↓ + CH3COOH

On the other hand, benzaldehyde does not give red precipitate with Fehling’s solution.

Q. 3. Distinguish between acetone and phenol.

Ans. When acetone is heated with iodine solution and aqueous NaOH in a water bath, yellow crystalsof iodoform are obtained having a characteristic odour.

CH3COCH3 + 3I2 + 4NaOH ⎯→ CHI3 ↓ + CH3COONa + 3NaI + 3H2O

Acetone Iodoform

Phenol does not give this test.

Q. 4. Distinguish between Methanal and Ethanal.

Ans. Ethanal, on warming with iodine and NaOH solution gives yellow crystalline precipitate ofiodoform while, methanal does not give this test.

Q. 5. Distinguish between Benzaldehyde and Acetone.

Ans. Benzaldehyde gives silver mirror when heated with Tollen’s reagent. Acetone does not give thistest.

Q. 6. Distinguish between Acetaldehyde and Acetone.

Ans. Acetaldehyde gives a silver mirror with Tollen’s reagent but acetone gives no such test withTollen’s reagent.


Q. 7. Distinguish between Aldehydes and Ketones.

Ans. Test Aldehydes Ketones

1. Tollen's reagent test 1. Gives silver mirror. 1. No Silver Mirror.

2. Fehling’s solution test 2. Gives red precipitate. 2. No precipitate.

3. Schiff’s reagent test 3. Gives pink colour. 3. No colour.

4. With NaOH. 4. Gives brown resinous

mass (HCHO exception).

4. No change.

5. With alcohols in presenceof dry HCl.

5. Forms acetals easily. 5. Do not form acetals easily.

Q. 8. Distinguish between Aliphatic aldehydes and Aromatic aldehydes.

Ans. Test Aliphatic aldehydes Aromatic Aldehydes

1. Iodoform 1. Positive. 1. Negative.

2. Aldol condensation 2. Given by aldehydes. 2. No.

3. Cannizzaro’s reaction 3. No reaction. 3. Positive.

4. Reaction with NH3 4. They give addition reaction. 4. Condenses with NH3.

5. Fehling’s solution 5. Give red precipitate. 5. No reaction.Q. 9. Distinguish between Formic acid and Acetic acid.Ans. Formic acid gives a silver mirror with Tollen’s reagent.

HCOOH + Ag2O ⎯→ H2O + CO2 + 2Ag ↓Formic acid Silver mirror

Acetic acid does not give this test.

Q. 10. Distinguish between Oxalic acid and Benzoic acid.

Ans. Oxalic acid decolourises hot solution of acidified KMnO4. While benzoic acid does not gives thistest.

Q. 11. Distinguish between Oxalic acid and Acetic acid.

Ans. Oxalic acid when treated with acidified KMnO4 solution, gets decolourised. While acetic aciddoes not gives this test.

Chapter 13. Organic Compounds Containing NitrogenQ. 1. Distinguish between Ethylamine and Diethylamine.Ans. These can be distinguished by the carbylamine test. Ethylamine is primary (1°) amine therefore, it

gives carbylamine test i.e., when heated with an alcoholic solution of KOH and CHCl3, it gives anunpleasant foul smell of isocyanide which is easily detectable.

C2H5NH2 + CHCl3 + alc. KOH ⎯→ C2H5NC + 3KCl + 3H2O

Ethylisocyanide

On the other hand, diethylamine is a secondary (2°) amine and hence does not give this test.Q. 2. Distinguish between Aniline and Ethylamine.Ans. Aniline and Ethylamine can be distinguished using azo-dye test.

Aniline Ethylamine

Aniline reacts with HNO2 (NaNO2 + dil. HCl)at 0– 5°C to give benzene diazonium chloride,which on further heating with β-naphtholgives red dye.

Ethylamine gives a brisk effervescence due toevolution of N2 gas under similar conditions.

Q. 3. Distinguish between Methylamine and dimethylamine.Ans. Methylamine gives offensive smell of methyl isocyanide when heated with chloroform and

alcoholic KOH. Dimethylamine does not gives this test.


Q. 4. How will you distinguish between primary (1°), secondary (2°) and tertiary (3°) amines.Ans. Hinsberg’s test for amines : This test is used to distinguish primary, secondary and tertiary

amines. In this test, different amines are heated with benzene sulphonyl chloride C6H5SO2Cl(Known as Hinsberg’s reagent) in presence of excess of alkali. The following observations arefound.(i) Primary amine gives clear solution which on acidification yields an insoluble material.

RNH2 + C6H5 SO2Cl ⎯→ RNH SO3C6H5 KOH

HCl RNKSO2C6H5

Isoluble(ii) Secondary amine gives an insoluble substance which remains unaffected on addition of

acid.R2NH + C6H5SO2Cl ⎯→ R2NSO2C6H5

Insoluble(iii) Tertiary amines do not react and remains insoluble in alkali and can be dissolved in acid.

Q. 5. What are the main differences between ethyl cyanide and ethyl isocyanides ?Ans. Characteristics Ethyl cyanide Ethyl isocyanide

1. HydrolysisC2H5CN

H2O/H+⎯⎯→ C2H5COOH

Propanoic acid + NH4

+ C2H5NC H2O/H+⎯⎯→ C2H5NH2

Ethylamine

+ HCOOH2. Reduction

C2H5CN Zn–Hg⎯⎯→HCl C2H5CH2NH2

n-propyl amineC2H5NC

Zn–Hg⎯⎯→HCl C2H5NCH3

Ethylmethylamine

3. Boiling point B. P. = 344 K B. P. = 352 K4. Solubility in water. Appreciably soluble. Sparingly soluble .


Q. 1. Distinguish between Glucose and Sucrose.Ans. To 2 ml of aqueous solution of Glucose add 2 ml of Tollen’s reagent – a silver mirror is obtained

while sucrose does not gives this test. To 2 ml of aqueous solution of the sucrose add a trace ofresorcinol, a deep wine red colour precipitate is obtained, sucrose is confirmed.

Q. 2. Mention two structural differences between Amylopectin and Cellulose.Ans. Amylopectin Cellulose

1. It is a linear polymer of α-glucose. 1. It is a linear polymer of β-glucose.2. It consists of branched chain of α-glucose. 2. In cellulose, the chains are arranged to

form bundles and held together byhydrogen bonds between glucose andadjacent strands.

Q. 3. Explain the structural differences between α-glucose and β-glucose. What effect does thesedifferences makes on utilization of glucose molecule by the cell.

Ans. The α-glucose and β-glucose differ in the orientation of hydroxyl group at C2 carbon atom. Thesesmall differences determine whether or not a cell will be able to utilize a molecule. For example :glycogen is a polymer of α-glucose which is easily digested by human being; on the other handcellulose a polymer of β-glucose is not digested by human beings.

Q. 4. What is the difference between reducing and non-reducing sugars ?

Ans. All those carbohydrates which contains free aldehyde or ketonic group and reduces Fehling’ssolution and Tollen’s reagent are called reducing sugars. e.g.: D-glucose or D-fructose.

On the other hand those carbohydrates (monosaccharide) which do not reduce Fehling’s solutionand Tollen’s reagent are called non-reducing sugars. e.g. : Starch, Cellulose etc.

Q. 5. Differentiate between Non-essential and Essential amino acids.

Ans. Those amino acids which can be synthesise by our body and therefore, we do not require them inour diet are called non-essential amino acids. e.g. : Glycine, alanine etc. They are ten in number.


On the other hand those amino acids which cannot be synthesize by our body and therefore, werequire them in our diet are called essential amino acids.

Ex. : Tyrosine, Valine etc.Q. 6. Differentiate between Fibrous proteins and Globular proteins.

Ans. Fibrous Protein Globular Protein

1. These types of proteins consist of linearthread like molecules which tend to lieside by side to form fibre.

1. In this type of proteins the molecules arefolded together into compact form.

2. They are insoluble in water. 2. They are soluble in water.

3. e.g. Keratin, Collagen. 3. e.g. Enzyme, Hormones.

Q. 7. What is difference between amylose and amylopectin ?

Ans. Amylose is water soluble linear polymer of α-glucose. Amylopectin is water insoluble branchedchain polymer of α-glucose.

Q. 8. What is the structural difference between a nucleoside and nucleotide ?

Ans. A nucleoside is formed by pyrimidine or purine base connected to C-1 of sugar (ribose or deoxyribose) by a β-linkage.

A nucleotide contains all the three basic components of nucleic acid i.e. , a phosphoric acid group,a pentose sugar and nitrogenous base.

Q. 9. Differentiate between Peptide linkage and Glycosidic linkage.

Ans. A peptide linkage is an amide ⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

—

O||

C—NH— . Linkage formed between —COOH group of one

α-amino acid and —NH2 group of other α-amino acid by loss of a water molecule. Whereas thelinkage between two monosaccharide units through oxygen atom in an oligosaccharide or apolysaccharide is known as glycosidic linkage.

Q. 10. Differentiate between monosaccharides and polysaccharides.

Ans. The simple carbohydrates that cannot be broken further into smaller units on hydrolysis arecalled monosaccharides. e.g. glucose and fructose.

On the other hand the carbohydrates which produces a large number of monosaccharide units onhydrolysis e.g. starch, cellulose etc. are called polysaccharides.


Q. 11. What are the differences between Vitamins and Hormones.

Ans. Hormones Vitamins

1. These are chemical substances which areproduced in the ductless glands in thebody.

1. These are not produced in the body(except vit. D) but have to be supplied indiet and are essential for properfunctioning of the different organs.

2. These are not stored in the body but arecontinuously produced.

2. These may be stored in the body to fightout disease.

Q. 12. Differentiate between the nucleotide of DNA and RNA.Ans. Nucleotide of DNA contains deoxyribose sugar, RNA contains ribose sugar. They differ in one

base, DNA contains thymine and RNA contains uracil.Q. 13. What is the basic structural difference between starch and cellulose ?

Ans. Both starch and cellulose contains a large number of α-D(+) glucose units. Starch consists of twocomponents (a) amylose which is a linear polymer and (b) amylopectin which is a branchedpolymer but in both the D-glucose units are linked through α-glycosidic linkage between C1 ofone glucose with C4 of next glucose unit.

Cellulose is only a linear polymer of D-glucose units joined through β-glycosidic linkage betweenC1 of one glucose with C1 of next glucose unit.

Q. 14. Give tests to distinguish between glucose and fructose.Ans.

S. No. Test Glucose Fructose1. Bromine water test Bromine water oxidizes glucose

to gluconic acid and itself getsdecolourized to HBr.

No characteristic observation.

Br2 + H 2O

Bromine water↓

CHO |(CHOH)4 | CH2OH

+

Reddish Brown

Br2 + H2O

↓

⎯⎯⎯→

COOH |(CHOH)4 | CH2OHgluconicacid

+ 2HBrColourlesssolution

2. Reduction : Reducingagent i s sodiumamalgam and water.

On reduction sorbitol isformed. CHO |(CHOH)4 | CH2OH

+ 2[H] →

CH2OH |(CHOH)4 | CH2OHD-sorbitaol

On reduction sorbitol and mannitolare formed. CH2OH | CO |(CHOH)4 | CH2OH

+ 2[H] → H—

CH2OH |C—OH

|(CHOH)3 | CH

2OH

D-sorbitol

+ HO—

CH2OH | C—H |(CHOH)3 | CH2OH

L-mannitol3. Warm with NaOH

solution (concentrated)Brown resinous mass isformed.

Do not form a resin.


Q. 15. Differentiate between enzymes and catalysts.

Ans. Enzymes Catalysts

1. Enzymes generally have complexstructure.

1. Catalysts are very simple, such as metalions, H+ ions.

2. Enzymes are produced by living beings. 2. Catalysts are not produced by living beings.

3. Enzymes are highly specific in action. 3. Catalysts are less specific.

4. Enzymes are active at lower temperatures,close to body temperature.

4. Catalysts are generally active at highertemperature.

Q. 16. Compare the components that constitute DNA and RNA.

Ans. DNA RNA

1. It contains deoxyribose sugar. 1. It contains ribose sugar.

2. It has thymine. 2. It lacks thymine.

3. It lacks uracil. 3. It has uracil.

4. It is a double chain of polynucleotides. 4. It is a single chain of polynucleotides.

5. It is the main constitute of chromosomes. 5. It is mainly a component of ribosome's.

6. It can replicate. 6. It is formed by DNA and cannot replicateitself.

Q. 17. Give differences between Globular and fibrous proteins.Ans. Difference between Globular and Fibrous Proteins

Globular Proteins Fibrous Proteins

1. They are cross-linked condensation polymersof acidic and basic amino acids.

1. They are linear condensation products.

2. They are soluble in water or aqueous solutionsof acids, bases or salts.

2. They are insoluble in common solvents butsoluble in strong acidic or basic solutions.

3. T h e y are f o l d e d t o give r i s e to t h r e e-

dimensional spheroidal shape.3. The long linear protein chains are held together

by intermolecular H-bonds.

4. Examples : Albumins in egg, enzymes andsome hormones, etc.

4. Examples : fibrous in silk ; collagen in tendons;myosin in muscles, etc.

Chapter 15. PolymersQ. 1. Differentiate between Homopolymers and Co-polymers.

Ans. Homopolymers Co-polymers

They are formed by the polymerization ofonly one type of monomers.

Ex. : Polythene, PVC

They are formed by the polymerization of twoor more type of monomers.

Ex. : Nylon-6, 6, Buna-S

Q. 2. What is the difference between Branched chain polymers and Cross-linked polymers.

Ans. Branched chain polymers Cross linked polymers

In these polymers monomer units are link togive long chain with side chain of differentlengths.

Ex. : Amylopectin, Glycogen.

In these polymers monomer units are linkedto give a three-dimensional solid networkwith cross linking.

Ex. : Bakelite, Melamine.


Q. 3. Differentiate between Addition polymers and Condensation polymers.

Ans. Addition polymers : In these polymers, monomer units are added to form long chains withoutthe elimination of any by product.

e.g. : Polythene, Teflon etc.

Condensation polymers : In these type of polymers, monomer units are added to form longchain, with the elimination of some by products like water, ammonia etc.

e.g. : Nylon-6, 6, and Nylon-6,10 etc.

Q. 4. Explain the difference between Chain growth polymers and Step growth polymers.

Ans. Chain growth polymers : In these polymers monomer units gets converted into some activeintermediate by some initiator. So monomer units are added one by one to this activeintermediate by a chain process. e.g. : PTFE, Polythene.

Step growth polymers : In these polymers, monomer units are condensed stepwise throughindependent reaction. Each independent reaction involves a condensation process by the loss ofsome simple molecules e.g. : Nylon-6, 6, Nylon-6 etc.

Q. 5. Distinguish between Chain growth and Step growth polymerization.

Ans. Chain Growth Polymerization Step Growth Polymerization

1. In this type of polymerization, themonomers having one or more doublebonds undergo repeated addition inpresence of an initiator to form a chain ofpolymer.

1. In this type of polymerization twodifferent types of monomer units or twodifferent monomer molecules undergo aseries of condensation reaction with theloss of simple molecules like H2O, NH3,HCl or alcohol.

2. The rate of reaction is fast and polymer israpidly formed.

2. Rate of reaction is slow. Comparativelymuch more time is required for thisprocess.

3. Example : Formation of polyethylenefrom ethylene molecules.

3. Example : Formation of nylon-6, 6 fromthe molecules of hexamethylenediamineand adipic acid.

Q. 6. Write the differences between Thermosetting and Thermoplastic polymers.

Ans. Thermosetting Polymers Thermoplastic Polymers

1. These are formed by condensation poly-merization.

1. These are formed by addition poly-merization.

2. These do not soften on heating but ratherbecome hard.

2. These soften and melt on heating.

3. These are more brittle and are insoluble inorganic solvents.

3. These are less brittle and soluble in someorganic solvents.

4. These cannot be remoulded or reshaped. 4. These can be remoulded, recast and res-haped.

5. These have three-dimensional cross-linkedstructures.

5. These usually have linear structures.

Examples : Bakelite, Terylene, Ureaform-aldehyde resin etc.

Examples : Polyethylene, PVC, Teflon, Nylonetc.


Chapter 16. Chemistry in Everyday LifeQ. 1. What are Biodegradable and Non-biodegradable detergents ? Give one example of each.

Ans. Biodegradable Detergents Non-Biodegradable Detergents

1. Detergents having straight hydrocarbonchains are easily degraded by microorganisms. are called biodegradabledetergents.

1. Detergents containing branchedhydrocarbon chains which are not easilydegraded by microorganisms are calledNon-biodegradable detergents.

2. They do not cause pollution. 2. They cause severe pollution.

Example : Sodium lauryl sulphate. Example : Sodium 4-(1,3,5,7–tetramethyl-octyl) benzene sulphonate)

Q. 2. How do Antiseptics differ from Disinfectants ? Give one example of each.

Ans. Antiseptic Disinfectant

1. Antiseptics are chemical substances whichprevents the growth of microorganismand may even kill them but are notharmful to living tissue.

1. Disinfectants are chemical substanceswhich kill microorganism or stop theirgrowth but are harmful to human tissues.

2. Antiseptic are generally applied to livingtissues such as wounds, cuts, ulcers anddisease skin surfaces.

2. Disinfectants are applied to inanimateobjects such as floor, drainage system,instruments etc.

3. Examples : Dettol, Fenazine. 3. Examples : 1% solution of phenol.

Q. 3. Differentiate between Anionic, Cationic and Non-ionic detergents.

Ans. Anionic Detergents Cationic Detergents Non-Ionic detergent

These are so named becauselarge part of their moleculesare anions and it is the anionicpart of the molecule which isinvolve in the cleaning action.For example sodium lauryls u l p h a t e , s o d i u mdodecylbenzene sulphonate,etc.

These are so called becauselarge part of their moleculesare cations and it is thecationic part of the moleculewhich is involved in thecleaning action. For examples :Cetyltrimethyl-ammoniumbromide.

Non-ionic detergent donotcontain any ion. These areesters of high molecular massalcohol.

Example : Liquid dishwashingdetergent are non-ionic type.

Q. 4. Differentiate between Broad spectrum and Narrow spectrum antibiotic.

Ans. Broad spectrum antibiotics Narrow spectrum antibiotics

Antibiotics which kills or inhibits a widerange of Gram positive and Gram-negativebacteria are called Broad specrtrumantibiotics.

Those antibiotics which inhibits only certainclass of Gram-positive or Gram-negativebacteria are called narrow spectrumantibiotics.

Example : Ampicillin, Tetracycline. Examples : Penicillin G❐

Question

12Set Laws and Naming Reactions

Chapter 2. SolutionsQ. 1. State Henry’s law. What is the significance of KH ?Ans. Henry’s law : It states that the partial pressure of the gas in vapour phase (p) is proportional to

the mole fraction of the gas (x) in the solution.p = KHx

where, KH = Henry’s constant.Significance of KH : Solubility is inversely proportional to Henry’s constant. Higher the value ofKH lower is the solubility of gas.

Q. 2. State and explain Raoult’s second law.Ans. Raoults second law/Raoult’s law in modified form : It states that the relative lowering of vapour

pressure of an ideal solution containing the non-volatile solute is equal to the mole fraction of thesolute at a given temperature

p°A – p

P°A= xB

where P°A → Vapour pressure of pure componentsPA → Vapour pressure of solutionxB → Mole fraction of solute

Q. 3. State Raoults first law ?Ans. French chemist Franco’s Mante Raoult (1886) gave a relationship between the partial pressure

and mole fraction of two components. It states that :“At a given temperature for a solution of volatile liquids the partial vapour pressure of eachcomponent in a solution is equal to the product of the vapour pressure of the pure componentsand its mole fraction.”

pA = p°A.xA or pB = p°B.xB

where pA and pB are the vapour pressure of the component A and B in the solution, p°A and p°Bare vapour pressure of pure component A and B, xA and xB are mole fraction.

Chapter 3. ElectrochemistryQ. 1. State and explain Ostwald’s dilution law.Ans. Ostwald’s dilution law (for weak electrolytes)

Let us consider a weak electrolyte acetic acid with concentration ‘c’ and degree of dissociation ‘α’CH3COOH (aq)

cc – cα

CH3COO–(aq)0cα

+ H+(aq)

0cα

(Initial)(At equilibrium)

Applying law of mass action

K =[CH3COO–] [H+]

[CH3COOH]

where, K = Dissociation constant or ionisation constant

K =cα·cαc – cα

K =cα·αc

c(1 – α)

K =cα2

(1 – α)for weak electrolyte α <<< 1, 1 – α ≈ 1

Laws and Naming Reactions ■ 299

K = cα2

α2 =Kc

α =

Kc

where α = degree of dissociation, c = concentration/molarity and K = dissociation constant.Hence, according to Ostwald dilution law, “At a particular temperature the degree ofdissociation of weak electrolytes is inversely proportional to the under root of concentration”.

Q. 2. State Kohlrausch’s law and give its mathematical expression mentioning the terms involved in it.Ans. Kohlrausch’s law : Molar conductance of an electrolyte, at infinite dilution is equal to the sum of

molar conductance of its cations and anions, with each conductance term multiplied by thenumber of respective ions present in the formula unit of the electrolyte. If m and n are thenumber of cations and anions respectively, as furnished by one formula unit and λ∞

c and λ∞a are

molar conductance of cations and anions at infinite dilution then according to Kohlrausch’s law :

Λ∞molar = mΛ

∞(c) + nΛ

∞(a)

Q. 3. Write down Faraday’s laws of electrolysis.Ans. Faradays’s first law of electrolysis : It states that, “The amount of any substance deposited or

liberated at the electrode is directly proportional to the quantity of electricity passed through it.”i.e. let W gram of the substance deposited on passing Q coulomb of electricity then

W ∝ Q or W ∝ I × t

W = Z × I × t

Where Z is a constant of proportionality known as electrochemical equivalent of the substancedeposited.Faraday’s second law of Electrolysis : It states that in an electrolyte the amount of differentsubstance deposited at the electrode is directly proportional to their equivalent masses.

Mathematically,W1

W2=

E1

E2


Q. 1. What is Rate law ?

Ans. Rate law states that the rate of chemical reaction is directly proportional to the molarconcentration of reactant.

Q. 2. State law of mass action.

Ans. Law of mass action : “The rate of a chemical reaction at any particular temperature isproportional to the product of the molar concentrations of reactants with each concentration therespective reactant taking part in the reaction”.

aA + bB → Product

Thus law of mass action may be written as :

Rate of reaction, r = k [A]a [B]b

Q. 3. What is law of chemical equilibrium ?

Ans. Law of Chemical equilibrium states that the ratio of product of concentration to that of thereactant has a constant value for a reaction at a given temperature.

Q. 4. State Le Chatelier’s principle.

Ans. Le-chatelier’s principle states that, “Change in any of the factor that determines the equilibriumconditions of a system will shift the equilibrium in such a manner so as to reduce or to counteractthe effect of the change.



Q. 1. Explain haloform reaction.

Ans. Aldehydes and ketones containing CH3—

O||C— group are oxidized with sodium hypohalite to give

haloform and salt of carboxylic acid. Sodium hypohalite is obtained by the reaction of halogenswith an alkali.

2NaOH + X2 ⎯⎯→ NaX + NaOX + H2O Sod. hypohalite

R—

O||C—CH3 + 3NaOX ⎯⎯→ CHX3 + R–COONa + 2NaOH

Q. 2. Explain Finkelstein reaction.

Ans. Finkelstein rReaction : When chloro alkanes or bromo alkanes are converted into correspondingiodoalkanes by treating with sodium iodide in acetone.

CH3CH2Cl + NaIEthyl chloride

Dry acetone⎯⎯⎯⎯⎯→ CH3CH2l

Ethyl iodide + NaCl

CH3CH2Br + NaIDry acetone

⎯⎯⎯⎯⎯→ CH3CH2I + NaBr Ethyl bromide Ethyl iodide

Q. 3. Explain the following : (a) Wurtz reaction, (b) Wurtz-Fittig reaction and (c) Fittig reaction.

Ans. (a) Wurtz reaction : When alkyl halide reacts with metallic sodium in the presence of dry oranhydrous ether to form alkanes the reaction is called as Wurtz reaction.

CH3Br + 2Na + Br–CH3Dry ether

⎯⎯⎯⎯→ CH3—CH3 + 2NaBr

Methylbromide Sodium Methyl

bromide Alkane

By this reaction only higher alkanes are formed. This reaction is also useful for the preparation ofalkane containing even number of carbon atoms, and not for alkane having odd number ofcarbon atoms.

(b) Wurtz-fittig reaction : When aryl halides are treated with alkyl halide and sodium inpresence of dry ether to form alkyl benzene the reaction is called as Wurtz-Fittig reaction.

Chlorobenzene

— Cl + 2Na + Cl — CH3 ⎯⎯⎯→ — CH3 + 2NaCl

Sodium Methylchloride

Methylbenzene

(c) Fittig’s reaction : The reaction in which two molecules of haloarenes combines with metallicsodium in the presence of dry ether to give diphenyl or biphenyls the reaction is called as Fittig’sreaction.

Chlorobenzene

Chloro benzene

— Cl + 2Na + Cl — ⎯⎯⎯⎯⎯→ — + 2NaCl Sodium

Diphenylor biphenyl

Dry ether

Q. 4. Explain the Friedel-Craft alkylation reaction with an example.

Ans. Friedel-Craft alkylation reaction : In this reaction benzene and other aromatic compounds reactswith alkyl halides in presence of anhydrous AlCl3 to form alkyl benzene.

In this reaction alkyl group (from alkyl halide) is added to benzene ring.


For example :

—

——

+ CH3Cl ⎯⎯⎯⎯⎯→ + HCl

⎯⎯⎯⎯⎯→ +

+ CH3CH2Br ⎯⎯⎯⎯→ + HBr

CH3

—

CH2 CH3

Benzene

Benzene Ethylbromide

Methylchloride Toluene

—

CH3

Toluene

—

CH3

o-Xylene

—

CH3

CH3

CH3

p-Xylene

Anhyd. AlCl3

Anhyd. AlCl3

Anhyd.

AlCl3

CH3Cl

Ethyl benzene

Q. 5. Explain the following : (a) Dow’s process (b) Hunsdieker’s Reaction.Ans. (a) Dow’s process : When chloro benzene is treated with an aqueous solution of NaOH at 623K,

300 atm pressure then sodium phenoxide is formed which on acidification gives phenol.

+ NaOH ⎯⎯⎯⎯⎯⎯⎯→ ⎯⎯⎯→

—

Cl

—

ONa

—

OH

Chlorobenzene

623K, 300 atm H+

(– NaCl, – H2O)Sodium

phenoxidePhenol

(b) Hunsdiecker’s reaction : In this reaction bromo alkanes are obtained by refluxing silver saltof fatty acids with Br2 in CCl4.

R—COOAg + Br2CCl4

Reflux⎯⎯⎯→ R—Br + AgBr + CO2

CH3CH2COOAg + Br2CCl4

Reflux⎯⎯⎯→ CH2CH2Br + AgBr + CO2

This method is used to decrease the number of carbon atoms.Q. 6. What do you mean by Etard reaction ?Ans. The oxidation of toluene to benzaldehyde with chromyl chloride (CrO2Cl2) dissolved in CCl4 or

CS2.

——

⎯⎯⎯⎯⎯→ ⎯⎯→

OCr(OH)Cl2

OCr(OH)Cl—

CH3

Toluene

—

CH

Brown complex

—

CHO

Benzaldehyde

2 CrO2Cl2

CCl4

H2O

Q. 7. Write Ulmann reaction.Ans. Iodo benzene is heated with copper powder in a sealed tube forming biphenyl the recation is

called Ulmann reaction.

— ——I + 2 Cu + I ⎯⎯→ + 2 CuI

Iodo benzene Iodo benzene Diphenyl



Q. 1. Explain the following name reactions :

(a) Kolbe’s reaction (b) Reimer-Tiemann reaction.

Ans. (a) Kolbe’s reaction : When sodium phenoxide is heated with CO2 at 400K under a pressure of6–7 atmosphere which on further acidification gives salicylic acid is called Kolbe’s reaction orKolbe’s Schmidt reaction.

— — ⎯⎯⎯⎯⎯→ ⎯⎯⎯→

—

OH

—

ONa

—

OH

—

OH

Phenol

NaOHCOOHCOONa

400K, 6-7 atm

+ CO2Sodium

phenoxideSodium

salicylateSalicylic acid

(2-hydroxy benzoic acid)

⎯⎯→H2SO4

(b) Reimer-Tiemann Reaction : In this reaction phenol is treated with chloroform in thepresence of aqueous alkali [NaOH(aq)] at 340 K temperature followed by hydrolysis to givesalicyaldehyde.

⎯⎯⎯⎯⎯⎯⎯⎯→

Phenol

—

OH

CHCl3 + NaOH (aq) ⎯⎯⎯→

NaOH—

—

ONaCHCl2

——

ONa

—

—

OHCHO CHO

⎯⎯→

Salicyldehyde(2-Hydroxy benzaldehyde)

H3O+

340 K

Q. 2. Explain briefly oxymercuration and demercuration of alkenes.Ans. Oxymercuration and demercuration of alkenes : When alkenes reacts with mercuric acetate in a

mixture of tetrahydrofuran and water to give hydroxy mercurial compound, involving additionof – OH and –HgOAc to the double bond. This is called oxymercuration. Then NaBH4 reduces– HgOAc and replaces hydrogen. This process is known as demercuration.

CH3—CH2 — CH = CH2Butene

Hg(OAc)2

THF‚ H2O⎯⎯⎯⎯⎯→ CH3—CH2—CH

|OH

—CH|HgOAc

NaBH4⎯⎯⎯→ CH3—CH2—CH|OH

—CH3

Sec-butyl alcoholQ. 3. Explain the following Fries rearrangement and Williamson synthesis.Ans. Fries rearrangement : The conversion of an aryl ester into o- and p- hydroxy ketone or a mixture

of both by treatment with anhydrous AlCl3.

—

— ⎯⎯⎯⎯→ +

—

OCOCH3

Phenyl acetate

—

OH

o-hydroxyacetophenone

—

OH

COCH3

COCH3

p-hydroxyacetophenone

AlCl3

Heat

The reaction involves migration of acyl group from phenolic oxygen to ortho or para position ofaromatic ring.Williamson synthesis : It consists of reaction of any alkyl halide with sodium alkoxide onsodium phenoxide to form ether.

R—X + Na—O—R ⎯⎯→ ROR + NaXAlkyl Sodium Etherhalide alkoxide

This method is used in the preparation of both symmetrical as well as unsymmetrical ethers.


Q. 4. Write Elb’s reaction.

Ans. The oxidation of phenol by potassium persulphate (K2S2O8) in alkaline medium to form amixture of catechol and p-quinol is known as Elb’s reaction.

—

— ⎯⎯⎯⎯→ +

—

OH

Phenol

—

OH

Catechol

—

OH

OH

OH

p-quinol

K2S2O6

Chapter 12. Aldehyde, Ketones and Carboxylic Acids

Q. 1. Explain the following named reactions (a) Rosenmund reduction and (b) Stephen reaction.

Ans. (a) Rosenmund reduction : Acyl chlorides are converted to corresponding aldehydes by catalyticreduction. The reaction of acyl chloride in the presence of palladium deposits over BaSO4(partially poisoned with sulphur or quinoline).

R—C||O

—ClAcyl chloride

+ H2 Pd/BaSO4

S⎯⎯⎯⎯⎯→ R—C

||O

—HAldehyde

+ HCl

CH3—C||O

—ClAcetyl chloride

+ H2 Pd/BaSO4

S⎯⎯⎯⎯⎯→ CH3—C

||O

—HAcetaldehyde

+ HCl

— — — —

O

Benzaldehyde

O

C—Cl

⎯⎯⎯⎯⎯→+ H2 Pd/BaSO4

+ HClS

C—H

(b) Stephen reaction : The reaction in which Nitriles are reduced to corresponding imines withSnCl2 in the presence of hydrochloric acid, which on hydrolysis gives corresponding aldehydes iscalled Stephen reaction.

SnCl2 + 2HCl ⎯→ SnCl4 + 2[H]

CH3—C ≡ N + 2[H] + HCl ⎯→ CH3—CH = NH·HClAcetaldoxime hydrochloride

↓ H2O (boil)

CH3CHO + NH4Cl

Acetaldehyde

C ≡ N

⎯⎯⎯⎯→

Phenylcyanide

+ HCl+ 2 [H] + HCl ⎯→H2O (boil)

CH = NH.HCl CHO

Benzaldoximehydrochloride

Benzaldehyde

Q. 2. Explain Etard’s reaction and Gattermann-Koch reaction.

Ans. Etard reaction : When chromyl chloride oxidises toluene to chrominium complex which onhydrolysis gives benzaldehyde is called as Etard’s reaction.


For example :

——CH3 HC

⎯⎯→

Benzaldehyde

+ CrO2Cl2 ⎯⎯→CS2

CHO

Chromylchloride

Toluene

OCr (OH) Cl2

OCr (OH) Cl2H3O+

Gattermann-Koch reaction : In Gattermann-Koch reaction hydrogen chloride in the presence ofanhydrous AlCl3 and CuCl with benzene, gives benzaldehyde or a substituted benzaldehyde.

⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Benzene

CO, HCl

CHO

Anhydrous AlCl3 /CuCl

BenzaldehydeQ. 3. Write a short note on :

(a) Clemmensen’s reduction, (b) Wolff-Kishner reduction, (c) Cannizzaro’s reaction.Ans. (a) Clemmensen’s reduction : When the carbonyl group of aldehyde and ketone is reduced to

CH2 group on treatment with zinc amalgam and concentrated hydrochloric acid then reaction iscalled Clemmensen’s reduction.

——C = O

Carbonylgroup

+ 4[H]

Zn–Hg.HCl (conc.)⎯⎯⎯⎯→

——CH2 + H2O

ButanalCH3—CH2—CH2—CHO + [H]

Zn–Hg.HCl (conc.)⎯⎯⎯⎯→

n-butaneCH3—CH2—CH2—CH3

⎯⎯⎯⎯→

CH2CH3

+ H2O+ 4 [H]

Acetophenone Ethyl benzene

COCH3

Zn—Hg

HCl (conc.)

(b) Wolff-Kishner reduction : When the carbonyl group of aldehyde and ketone is reduced to –CH2 group on treatment with hydrazine followed by heating with potassium or sodiumhydrazine in a high boiling solvent such as ethylene glycol is called Wolff-Kishner reduction.

——C = O +

group Carbonyl

Hydrazine

NH2—NH2 –H2O⎯⎯→

——Hydrazone

C = NNH2 KOH

Ethyleneglycol

⎯⎯⎯⎯→ > CH2 + N2

CH3 CH3

——Acetone

C = O + Hydrazine

NH2—NH2 –H2O⎯⎯→

CH3 CH3

—— C = NNH2

KOH

Ethyleneglycol

⎯⎯⎯⎯→ Propane

CH3—CH2—CH3

(c) Cannizzaro’s reaction : In this reaction those aldehydes which do not have an α-hydrogen,undergoes self oxidation and reduction (disproportionation) reaction on treatement withconcentrated alkali. In this reaction one molecule of aldehyde undergoes reduction to alcoholwhile another molecule of aldehyde undergoes oxidation to carboxylic acid salt.

2HCHO Conc. KOH⎯⎯⎯⎯→ CH3—OH + HCOO– K+

FormaldehydealcoholMethyl

formatePotassium

2C6H5CHO Conc. KOH⎯⎯⎯⎯→ C6H5CH2OH + C6H5COO+ Na+

BenzaldehydealcoholBenzyl

benzoateSodium


Q. 4. Explain Aldol and Cross Aldol condensation.

Ans. Aldol condensation : When two molecules of aldehydes or ketones containing atleast oneα-hydrogen atom on treatment with dilute alkali undergoes condensation to form β-hydroxyaldehyde (aldol) or β-hydroxy ketone (ketol) is known as Aldol condensation.

⎯→

— — —

O OH

– H2O

CH3—C—H + H—CH2—CHO CH3—CH—CH2—CHO

CH3—CH = CH—CHO

Acetaldehyde(2 molecules)

But - 2 - enal

dil. NaOH

3-hydroxy butanal(Aldol)

— —— — — —

——

——

CH3—C + H—CH2—C—CH3

CH3

⎯→ – H2O

2 - molecules of acetone

Ba(OH)2CH3—C—CH2—C—CH3

CH3—C = CH—C—CH3

OH OO O

O

CH3

CH3

4-hydroxy-4-methyl- pentan - 2 - one (Ketol)

4-methyl pent-3-ene-2-one— —

Q. 5. Write down Diel’s Alder reaction.Ans. The reaction between α , β-unsaturated carbonyl compound and a conjugate diene to form an

addition product is known as Diel’s Alder reaction.

+

CH2

CH2

CH2

CH2

CH2

CH2

—

———

— — — ——

—

———

373 KCH

CH

—

CH

CH

1, 3-butadiene(Conjugate diene)

1, 2, 3, 6-tetrahydrobenzaldehyde

⎯⎯→

CH—CHO CH—CHO

Aeroteinα, β-unsaturated

carbonyl compound

It is also called cyclo addition reaction.

Q. 6. Explain Perkin’s reaction.

Ans. Perkin’s reaction : Benzaldehyde on heating with acetic anhydride in the presence of sodiumacetate and hydrolysis of the product obtained gives α, β-unsaturated acid, cinnamic acid.

Benzaldehyde

C6H5CH = O +

Aceticanhydride

H2CHCO

CH3COO

CH3COONa

453K‚ –H2O⎯⎯⎯⎯⎯→

Cinnamic acidC6H5CH = CH COOH +

Acetic acid

CH3COOH

Q. 7. Give iodoform or haloform reaction.

Ans. Haloform reaction : Methyl ketones (CH3—C||O

—) are oxidised to sodium salts of the acids andgives precipitate of iodoform (when oxidation is carried out with I2 and NaOH). The reaction iscalled Haloform reaction and is used for the identification of methyl ketone.


CH3

CH3

C = O + 3I2 + 3NaOH ⎯⎯→

I3C

CH3

C = O + 3NaI + 3H2O

I3C

CH3

C = O + NaOH ⎯⎯→

NaO

CH3

C = O

Sodium acetate

+ CHI3Iodoform

Q. 8. Complete and write the name of the following equations :(i) CH3COOC2H5 + NaOH ⎯⎯→

(ii) CH3COOC2H52 molecules

C2H5ONa⎯⎯⎯→

(iii) CH3COCl + H2Pd/BaSO4⎯⎯⎯→

Ans. (i) CH3COOC2H5 + NaOH ⎯⎯→ CH3COONa + C2H5OHEthyl acetate Sodium acetate Ethanol

This is alkaline hydrolysis of esters, known as saponification

(ii) CH3COOC2H5 + CH3COOC2H5C2H5ONa

⎯⎯⎯⎯→– C2H5OH

CH3COCH2COOC2H5Acetoacetic ester Ethylacetoacetate

This reaction is known as Claisen condensation.

(iii) CH3COCl + H2 Pd

⎯⎯⎯⎯→BaSO4

CH3CHOAcetaldehyde

+ HCl

This reaction is called Rosenmund reduction.

Q. 9. Explain Benzoin condensation.

Ans. Benzoin condensation : When two molecules of aromatic aldehyde (such as benzaldehyde) onheating in the presence of ethanolic KOH gets converted to form benzoin and this reaction iscalled Benzoin condensation.

2 —CHO ⎯⎯⎯⎯⎯→ —CH — C —

—

OHAlc. KCN — —

O

Benzaldehyde(2-molecule)

Heat Benzoin

Q. 10. Explain Hell Volhard Zelinsky (HVZ) reaction.

Ans. Hell Volhard Zelinsky (HVZ) reaction : The HVZ reaction is an organic reaction used to converta carboxylic acid with an α-hydrogen and a halogen, to an α-halo carboxylic acid, usingphosphorus catalyst and water.

Example : When chlorine reacts with acetic acid in presence of traces of phosphorus, α-H atomsof the acid gets substituted by chlorine.

CH3COOHAcetic acid

+ Cl2 Red P

– HCl⎯⎯⎯⎯→ CH2Cl·COOH

Monochloro acetic acid + Cl2 Red P

– HCl⎯⎯⎯⎯→ CHCl2·COOH

Dichloro acetic acid

+ Cl2 Red P– HCl

⎯⎯⎯⎯→ CCl3·COOHTrichloro acetic acid


Q. 1. What is Gabriel phthalimide synthesis ?Ans. In this reaction, phthalimide is converted into potassium salt when treated with alc. solution of

KOH. The salt is then reacted with alkyl halide to give N-alkyl phthalimide, which on hydrolysiswith dilute hydrochloric acid gives a primary amine as the product. This reaction is used toprepare primary amines.


⎯→

—

—

— —— —

—

—

C

C

——

COOH

COOH

NHKOH (alc.)

Phthalic acid

+ H2O/H+

⎯⎯⎯⎯→

O

O

—

—

— —— —

—

—

C

CNK

+ R—X⎯⎯→

O

O

—

—

— —— —

—

—

C

CNR

O

O

+ R — NH21° Amine

Q. 2. Write a short note on (a) Hofmann bromamide reaction (b) Carbylamine reaction.Ans. (a) Hofmann bromamide reaction : When a primary acid amide is heated with an aqueous or

ethanolic solution of NaOH or KOH and bromine; it gives primary amines which has one carbonatom less than the number of carbon atoms in that amide.

R—CONH2 + Br2 + 4NaOH ⎯→ R—NH2 + Na2CO3 + 2NaBr + 2H2OAcid amide 1° Amine

C6H5—CONH2 + Br2 + 4KOH ⎯→ C6H5NH2 + K2CO3 + 2KBr + 2H2OBenzamide Aniline

(b) Carbylamine reaction : When aliphatic and aromatic primary amines are heated withchloroform and alcoholic solution of KOH it gives isocyanides (carbylamines) which haveextremely unpleasant smell.

CH3—NH2 + CHCl3 + 3KOH (alc.)Δ

⎯→ C6H5—N ≡ C + 3KCl + 3H2O

(1° amine)Methyl amine

isocyanidephenyl

Q. 3. Explain briefly Sandmeyer’s reaction and Gattermann’s reaction.

Ans. Sandmeyer’s reaction : Nucleophiles like Cl–, Br– and CN– can be easily introduced in benzenering in the presence of Cu(I) ion. This reaction is called Sandmeyer’s reaction.

For example :

Chloro benzene

Bromo benzene

CuCl/HCl

Benzonitrile

Cl

N+ ≡ NCl– Br

CN

+ N2

+ N2

+ N2

CuBr/HBr

CuCN/KCN

Benzene diazoniumchloride

Gattermann’s reaction : Introduction of chlorine or bromine in benzene ring by treating thediazonium salt solution with corresponding halogen acid in presence of copper powder is knownas Gattermann’s reaction.


Chloro benzene

Bromo benzene

Cu/HCl

Cl

N2Cl

Br

+ N2 + CuCl

+ N2 + CuClCu/HBrBenzene diazonium

chloride

Cross Aldol condensation : When aldol condensation is carried out between two differentaldehydes and/or ketones, it is known as cross aldol condensation.

H—C||O

—H + H—CH2—C||O

—H dil. NaOH⎯⎯⎯⎯→ HO—CH2—CH2—C

||O

—H + CH3—

OH|CH—CH2—C

||O

—H Formaldehyde Acetaldehyde Cross aldol product Simple aldol product

—CHO + CH3 — C — ⎯⎯⎯⎯→ —CH = CH — C —

— — — —

OO

Benzaldehyde

alc. NaOH

ΔAcetophenone Benzalacetophenone

(1, 3-diphenyl prop-2-ene-1-one)

Q. 4. Write down Balz-Schiemann reaction.

Ans. Balz-Schiemann reaction : This reaction involves the decomposition of diazonium fluorborate toaryl fluoride

+ HBF4

N2Cl

Benzenediazonium chloride

273-283 K⎯⎯⎯⎯⎯→ ⎯⎯⎯→

N2BF4 F

Fluroboricacid

Benzene diazoniumfluoroborate

Fluoro benzene

Q. 5. Explain Diazotisation reaction.

Ans. The formation of diazonium salt from primary amines in a dilute mineral acid (HCl or H2SO4)and treatment with cold solution of nitrous acid (NaNO2 + dil. HCl) at 273–278K is known asdiazotisation reaction

NH2

NaNO2, HCl⎯⎯⎯⎯⎯→

Aniline

+ HONO

N ≡ NCl

+ 2 H2O273-278 K

Benzenediazonium chloride

Q. 6. Explain Exhaustive alkylation reaction.

Ans. The process of converting an amine (1°, 2° or 3°) into its quaternary ammonium salt on treatmentwith excess of alkyl halide is called exhaustive alkylation. If alkyl halide is methyl halide then itis called exhaustive methylation.

CH3NH2

amineMethyl

CH3Cl

–HCl⎯⎯⎯→ (CH3)2NH

amineDimethyl

CH3Cl

–HCl⎯⎯⎯→ (CH3)3N

amineTrimethyl

CH3Cl

⎯⎯⎯→ (CH3)4N+Cl–

chloride

Tetramethylammonium

Q. 7. Write down the Hofmann mustard oil reaction.Ans. When a mixture of primary amine and carbon disulphide are treated with HgCl2 characteristic

smell of mustard oil is formed. Hence it is called Hofmann mustard oil reaction.


Ethyl amineCH3CH2NH2 +

Carbondisulphide

S = C = S HgCl2⎯⎯→ CH3CH2—NH—C

||S

—SH Heat

HgCl2⎯⎯→ CH3CH2N = C = S

Ethyl isothiocyanate(Mustard oil)

+ HgS + 2HCl

Q. 8. What is Hoffmann’s reaction ?Ans. When acetamide is treated with liquid bromine and aqueous caustic soda or caustic potash, it is

degradated to methyl amine. This reaction is known as Hofmann’s reaction.

CH3CONH2Acetamide

Br2/NaOH⎯⎯⎯→

Δ CH3NH2

MethylamineQ. 9. What is acylation ? Give one example.Ans. Acylation is the addition of acyl group to an organic substance. In this the hydrogen atom of the

amine is replaced by the acyl group.R—NH2 + CH3COCl ⎯→ R—NH.CO.CH3 + HCl1°-amine Acylchloride N-alkylacetamide

Q. 10. What is Mendius reaction ? Give equation.Ans. Chemical reduction of alkyl or aryl cyanides using LiAlH4 or Na or ethanol is referred to as

Mendius reaction.

RC ≡ N + 4 [H] LiAlH4 or

⎯⎯⎯⎯→Na/C2H5OH

Amine

RCH2NH2

CH2NH2 LiAlH4 ⎯⎯⎯→ + 4 [H]

C ≡ N

Benzenenitrile Benzylamine

❐

Question

13Set Conversions

Chapter 10. Haloalkanes and HaloarenesQ. 1. How the following conversions can be brought about :

(i) Methyl chloride to acetone.(ii) Benzene to benzene diazonium chloride.

Ans. (i) Methyl chloride to Acetone

CH3Cl(methylchloride)

+ HOH –HCl

⎯⎯→ CH3OH I2‚P

⎯⎯→ CH3I KCN

⎯⎯→ CH3CN H2O

⎯⎯→ CH3CONH2(Acetamide)

H+

⎯→

CH3COCH3(acetone)

– CaCO3

←⎯⎯⎯ (CH3COO)2Ca Ca(OH2)

←⎯⎯⎯ CH3COOH(acetic acid)

(ii) Benzene to benzene diazonium chloride

N+ ≡ NCl–

Q. 2. How will you convert the following :

(a) n-Propyl bromide to ethyl amine

(b) Methyl chloride to chloro acetic acid.

Ans. (a)n-Propyl bromide

CH3CH2CH2Br aq.KOH

Heat⎯⎯⎯→

Propanol

CH3CH2CH2OH K2Cr2O7

H+⎯⎯⎯→

||↓

SOCl2

CH3—CH2COOH

Ethyl amine

CH3CH2NH2 Br2

NaOH⎯⎯→ CH3CH2CONH2

NH3←⎯⎯ CH3CH2COCl

(b)Methylchloride

CH2Cl2 Mg

⎯→ Methyl

magnesiumchloride

CH3MgCl CO2⎯⎯→ C—

||O

|CH3

OMgCl H+‚ H2O

⎯⎯⎯→ CH3COOH P‚ Cl2⎯⎯→ CH2COOH

|ClChloroacetic acid

Q. 3. How will you make the following conversions :

(i) Ethyl benzene to 2-phenyl propionic acid

(ii) Bromo benzene to m-nitro benzoic acid

(iii) Toluene to o-cresol.

Ans. (i) ⎯⎯⎯⎯→⎯⎯⎯→⎯⎯⎯→

CH2CH3

Br

Br2/hv

CHCH3

KCN

CN

CH—CH3 HOOC—CH—CH3

Ethyl

benzene

2-phenyl

propionic acid

| |

Hydrolysis

Conversions ■ 311

(ii) ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯→⎯⎯⎯→

COOH COOHBr

CuCN

CN

NO2

+

Pyridine, 457KBromo benzene

m–nitro benzoic acid

H, H2O HNO3 , H2SO4

(iii) ⎯⎯⎯⎯→ ⎯⎯⎯⎯→ ⎯⎯⎯⎯→⎯⎯⎯→

CH3 CH3 CH3 CH3 CH3

OHHNO3

H2SO4

Sn–HCl NaNO2

HCl

boil

H2O

NO2 NH2 N2Cl

Toluene o-cresol

Q. 4. How will you convert Benzene into the following :

(a) m-dichlorobenzene (b) 4-nitro benzaldehyde

(c) Cinnamic acid (d) α-(p-nitrophenyl) ethylamine (e) m-chlorotoluene.

Ans. (a) ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→ ⎯⎯⎯⎯→

⎯⎯⎯→

⎯⎯→

Cl

Cl

HNO3Conc. HNO3

H2SO4 H2SO4

Sn–HCl

CuCl

NaNO2HCl

HCl

NO2 NO2

NO2

NH2

NH2

N2Cl

N2Cl

Benzene

m-Dichlorobenzene

(b)

||

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→CH3Cl

AlCl3

, H2SO4HNO3

CH3

⎯⎯⎯⎯→CrO3

CH3

NO2NO2

NO2

OCOCH3

OCOCH3CH

⎯⎯→H2O

CHO

Benzene

4-Nitro benzaldehyde

OH–

(CH3CO)2 O

(c)

Sunlight⎯⎯⎯⎯⎯→

2Cl2⎯⎯⎯⎯→

CH3Cl

Anhy. AlCl3⎯⎯⎯⎯→

NaOH

H2O

CH3 CHCl2 CHO CH = CHCOOH

⎯⎯⎯⎯⎯⎯→CH3COONa, Heat

(CH3CO)4 O

Benzene Cinnamic Acid

(d)AlCl3

⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→C2H5Cl

NO2NO2 NO2

CH2CH3 CH2CH3

Cl2

H2SO4

CH—CH3 NH2 — CH2 — CH3

NH3HNO3

U.V.Benzene

α–(p-nitrophenyl)ethylamine

Cl


(e) ⎯⎯⎯⎯⎯→

CH3CH3CH3

Cl

HNO3, H2SO4

Benzene

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→

NO2NO2 N2Cl

CH3Cl, AlCl3 H2/Ni

NaNO2, HClHCl, CuCl

Friedal-Craft’sReaction

Nitration

m-chlorotoluene

Q. 5. How will you convert Toluene into the following :

(a) 1, 3, 5-trinitrobenzene

(b) o-cresol

(c) m-cresol

(d) 3-nitro-4-bromobenzoic acid

(e) o-bromo-p-hydroxy toluene.

Ans. (a) ⎯⎯⎯→ ⎯⎯⎯→

CH3 CH3

+ 2HNO3H2SO4 H2SO4

K2Cr2O7

⎯→

Toluene

Fusing NO2

NO2

NO2

NO2

NO2

NO2

NO2COOH

NO2NO2

Sodalime

1, 3, 5-Trinitrobenzene

(b) ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯→

CH3 CH3 CH3 CH3 CH3

Toluene

Sn–HCl BoilHNO3 NaNO3

H2SO4 H2O

NO2 NH2 N2Cl OH

o-cresol

(c)

⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯→ ⎯⎯→

⎯⎯⎯⎯→ ⎯⎯⎯⎯→

(CH3CO)2O

CH3 CH3

Toluene

Sn–HCl

⎯⎯⎯→Sn–HCl

Heat

⎯⎯⎯⎯⎯→HNO3 , H2SO4

H2O (Heat)

NO2

NO2 NO2 NO2

CH3

NH2

NH2

⎯⎯⎯→

CH3

CH3CH3

NH3 NHCOCH3

CH3CH3CH3

HNO3

NHCOCH3

OH NaOHm-Cresol

(d) ⎯⎯⎯→H2SO4

K2Cr2O7⎯⎯⎯→

HNO3

H2SO4

⎯⎯→Br2

FeNO2

CH3 CH3

3-nitro-4-bromobenzoic acid

Toluene Br

COOH

Br

COOH

Br

Conversions ■ 313

(e)Toluene

o-bromo-p-hydroxy toluene

⎯⎯⎯→HNO3

H2SO4⎯⎯⎯→HCl, 0°C

NaNO2⎯⎯→

Br2

Fe⎯⎯→

Zn

HCl

NO2NO2

CH3 CH3 CH3

Br

NH2

CH3

Br

OH

CH3

Br

N2Cl

CH3

Br

⎯→ WaterBoil

Q. 6. How will you make the following conversions :

(i) Bromobenzene into Grignard reagent.

(ii) Chlorobenzene into phenol.

Ans. (i)

Br

Bromobenzene

+ Mg Dry

ether⎯⎯⎯→

MgBr

Aryl magnesium bromideor

(Grignard Reagent)

(ii)

Cl

aq. NaOH

623 K‚ 300 atm⎯⎯⎯⎯⎯⎯⎯⎯→

O– Na+

dil. HCl⎯⎯⎯→

OH

Chlorobenzene Phenol


(i) Chloroform to phosgene (ii) Chloroform to chloropicrin

(iii) Chloroform to chloretone (iv) Dimethyl ketone to tri-iodomethane

Ans. (i)ChloroformCHCl3 + 1

2 O2 ⎯→phosgene

Cl—C||O

—Cl + HCl

(ii)Chloroform

CHCl3O + Nitricacid

HO—NO2 ⎯→ Chloropicrin

Cl3C—NO2 + H2O

(iii) chloroformCHCl3 +

AcetoneCH3—C

||O

—CH3 ⎯→ CH3 —

OH|C—|CCl3

CH3

Chloretone

(iv) Acetone

CH3COCH3 + 3I2 NaOH

Δ⎯⎯⎯→

Iodoform

CHI3 + CH3COONa + 3 NaI + 3 H2O

Q. 8. How will you convert benzene into :

(i) DDT

(ii) BHC


Ans. (i) DDT

Benzene

Cl2

FeCl3⎯⎯⎯→

Cl

CCl3CHO

H2SO4⎯⎯⎯⎯⎯→ CH—CCl3

ClDDT

Cl

(ii) BHC

Benzene

+ 3Cl2 UV light

⎯⎯⎯⎯→ Cl

ClCl

Cl

Cl

Cl

BHC

Q. 9. How will you convert chlorobenzene to cyanobenzene :

Ans. + C–

N

Cl

Chlorobenzene

NaCN + CuCN

473 K‚ pressure⎯⎯⎯⎯⎯→ + Cl

CN

Cyanobenzene

–

Chapter 11. Alcohols, Phenols and EthersQ. 1. How will you convert chlorobenzene to phenol :Ans.

Cl

+ NaOH

Cu-catalyst

Chlorobenzene

623K

320 atm_ HCl

_

_ NaCl

O

HCl

Na+ OH

Phenol

Q. 2. How will you convert diethyl ether to ethanol :Ans. C2H5OC2H5 + H2O ⎯⎯⎯⎯⎯→ 2C2H5OH

Diethyl Superheated Ethanolether steam

Q. 3. How will you convert Benzene to phenol :Ans. Benzene to phenol.

+ conc.

H2SO4 – H2O⎯→

SO3H

NaOH

– H2O⎯⎯→

SO3Na

heated with

fused NaOH⎯⎯⎯⎯→

ONa

+ HCl

– NaCl⎯⎯→

OH

Benzene Benzene sulphonic Sod. salt of benzene Sod. phenoxide Phenolacid sulphonic acid

Q. 4. How will you convert 2 propanol to acetoxime :Ans. 2-Propanol to acetoxime :

CH3CHOHCH32-propanol

+ (O)

– H2O⎯⎯⎯→ CH3COCH3

acetone

+ NH2OHpH 3·5 – H2O⎯⎯⎯⎯⎯→ (CH3)2C = NOH

acetoxime

Q. 5. How will you convert the following :(i) Ethanol to ethane diol (ethylene glycol) (ii) Ethyl alcohol to vinyl acetate(iii) n-propyl alcohol to isopropyl alcohol (iv) Methyl alcohol to ethylene glycol(v) 1-propanol to 1-chloro-2-propanol

Conversions ■ 315

Ans. (i)Ethanol

CH3CH2OH conc.H2SO4

443K⎯⎯⎯⎯⎯→ CH2 = CH2

alk.KMnO4

H2O; O⎯⎯⎯⎯→ CH2

|OH

—CH2|OH

Ethylene glycol

(ii) CH3CH2OH

alcohol

Ethyl

conc.H2SO4

443K⎯⎯⎯⎯⎯→

Ethylene

CH2 = CH2 Cl2⎯→ ClCH2CH2Cl

dichloride

Ethylene

alc.KOH

⎯⎯⎯→ Ethyne

CH ≡ CH

||↓

CH3COOH

Vinyl acetate

CH2 = CHCOOCH3

(iii)n–PropylAlcohol

CH3CH2CH2OH conc.H2SO4

– H2O‚ heat⎯⎯⎯⎯⎯→

Propene

CH2CH = CH2 HBr

⎯⎯→ CH3—CH|Br

—CH3

bromide

Isopropyl

||↓

AgOH

CH3—CH|OH

—CH3Isopropyl Alcohol

(iv) CH3OH

alcohol

Methyl

PCl5⎯⎯→ CH3Cl

Na‚ ether⎯⎯⎯⎯→ CH3CH3

Cl2‚ hν⎯⎯→ C2H5Cl

||↓

Alc.KOH

CH2|OH

—CH2|OH

2AgOH

⎯⎯⎯→ CH2|Br

—CH2|Br

Br2

CCl4←⎯⎯ CH2 = CH2

Ethylene glycol

(v) 1-propanol to 1-chloro-2-propanol

1-propanol

CH3CH2CH2OH H+

443 K⎯⎯→ CH3CH = CH2

Cl2

H2O⎯⎯→ CH3—CH

|OH

CH2Cl

1-chloro-2-propanol

Q. 6. How will you convert ethylene glycol to dioxane :

Ans. CH2—OH|CH2—OH

conc.H2SO4

Heat⎯⎯⎯⎯⎯→ O

CH2—CH2 CH2—CH2

O

Ethylene glycol Dioxane

Q. 7. Convert the following :

(a) Primary alcohol into secondary alcohol (b) Secondary alcohol into tertiary alcohol

(c) Primary into tertiary alcohol (d) Higher into lower alcohol

(e) Lower into higher alcohol

Ans. (a) CH3CH2CH2OHn-Propyl alcohol

(1° alcohol)

SOCl2⎯⎯→ CH3CH2CH2Cln-Propyl chloride

alc KOH⎯⎯⎯→ CH3CH = CH2Propylene

HBr⎯⎯⎯⎯⎯⎯⎯→(Markovnikof’s rule)

CH3—CH|Br

—CH3 Isopropyl bromide

aq. KOH⎯⎯⎯⎯→ CH3—CH|OH

—CH3 Isopropyl alcohol

(2° alcohol)

(b) CH3—CH|OH

—CH3 Isopropyl alcohol

K2Cr2O7/H+⎯⎯⎯⎯⎯→

oxidationCH3—C

||O

—CH2acetone

(2° alcohol)


CH3MgBr⎯⎯⎯⎯→ CH3—

OMgBr|C—CH3|CH3

H2O/H+⎯⎯⎯⎯⎯→

(Hydrolysis)CH3—

OH|C—CH3|CH3

tert-Butyl alcohol(3° alcohol)

(c) CH3—CH|CH3

—CH2 Isobutyl alcohol

(1° Alcohol)

OH H2SO4/Δ⎯⎯⎯⎯⎯→(Dehydration)

CH3—CH|CH3

= CH2

Isobutylene HBr⎯⎯⎯⎯⎯⎯⎯⎯→

Markovnikof’ s rule

CH3—

CH3|C—CH3|Br

tert-Butyl bromide

aq. KOH⎯⎯⎯⎯→ CH3—

CH3|C—CH3|OH

tert-Butyl alcohol(3° Alcohol)

(d) CH3CH2OHEthyl alcohol(two carbons)

K2Cr2O7/H+⎯⎯⎯⎯⎯→

[O]CH3—C

||O

—HAcetaldehyde

K2Cr2O7/H–⎯⎯⎯⎯⎯→

[O]

CH3—C||O

—OHAcetic acid

NaOH⎯⎯⎯⎯⎯→ CH3—C||O

—ONa NaOH + CaO⎯⎯⎯⎯⎯⎯→Δ

CH4Methane

Cl2⎯⎯⎯⎯⎯→UV

CH3ClMethyl chloride

aq. NaOH⎯⎯⎯⎯→ CH3OHMethyl alcohol

(1 carbon)

(e) CH3CH2OHEthyl alcohol(two carbons)

SOCl2⎯⎯⎯→ CH3CH2ClEthyl chloride

Mg

⎯⎯→ CH3CH2MgClEthylmagnesium

chloride

(1) H—C

||O

—H⎯⎯⎯→

(2) H2O/H+ CH3CH2CH2OHn-Propyl alcohol

(three carbons)

Chapter 12. Aldehydes, Ketones and Carboxylic acids

Q. 1. How will you convert acetaldehyde to acetaldehyde phenyl hydrazone ?

Ans. CH3CHO CH3COOH

←⎯⎯⎯⎯⎯⎯–H2O

C6H5HNNH2

Acetaldehyde Phenylhydrazone

←⎯⎯⎯⎯Dry distill

Acetone

Acetaldehyde Acetic (acid)

[O]⎯⎯⎯⎯⎯⎯⎯→KrCr2O7/H2SO4

Ca (OH)2⎯⎯⎯⎯⎯⎯→(CH3COO)2Ca

O

CH3—C—CH3NNHC6H5

CH3—C—CH3

Q. 2. How will you convert acetaldehyde to formaldehyde ?

Ans. CH3 CHO (O) CH3 COOH

+ NH3

Δ CH3 CONH2

Acetaldehyde Acetamide (– H2O)

Acetic acid⎯→ ⎯⎯→

↓

HCHO (O)

←⎯⎯⎯ CH3OH NaNO2 + HCl←⎯⎯⎯⎯⎯ CH3NH2

Formaldehyde Methyl alcohol Methylamine

Conversions ■ 317

Q. 3. How will you convert benzoic acid to benzaldehyde ?

Ans.

COOH COCl CHO

Benzoic acid Benzaldehyde

– SO2

– HCl+ SOCl2

+ H2, – HCl Pd/BaSO4

Q. 4. How will you convert ethanol to acetone ?

Ans.Ethanol

2C2H5OH + 2O

– 2H2O⎯⎯→ 2CH3COOH

+ Ca(OH)2

– 2H2O⎯⎯⎯→ (CH3COO)2Ca

Δ⎯→

Acetone

CH3COCH3 + CaCO3

Q. 5. How will you convert diethyl ether to ethanol ?Ans. Ethers boiled with dil. H2SO4

C2H5OC2H5 + H2O⎯→

Δ 2C2H5OH

Q. 6. How will you bring the following conversions :

(i) Propanal to 2, 3-dihydroxy propanal

(ii) Ethanal to 2-hydroxy-3-butanoic acid

(iii) Acetaldehyde to malonic acid

Ans. (i)2-propanal

CH2 = CHCHO C2H5OH

⎯⎯⎯→ CH2 = CHCH (OC2H5)2 alk.KMnO4⎯⎯⎯⎯→ CH2

|OH

—CH|OH

—CH (OC2H5)2

|↓H3O+

CH2|OH

—CH|OH

CHO

2, 3-Dihydroxy propanal

(ii)Ethanal

CH3CHO HCHO‚ OH–

⎯⎯⎯⎯⎯→ HO CH2CH2CHO Heat

⎯⎯→ |↓HCN

CH2 = CHCHO

CH2 = CH—CH|OH

—COOH H+/H2O

←⎯⎯⎯ CH2 = CH—CH|OH

—CN

2-Hydroxy-3-butanoic acid

(iii)Acetaldehyde

CH3CHO [O]

⎯→ Acetic acid

CH3COOH Cl2‚ P

⎯⎯→ CH|Cl

2COOH KCN

⎯⎯→ CH|CN

2—COOH H+/H2O⎯⎯→ CH2—COOH

|COOH

2-chloro aceticacid

Malonic acid

Q. 7. How will you convert :

(a) Acetaldehyde to acetamide

(b) Acetaldehyde to acetoacetic ester

Ans. (a)Acetaldehyde

CH3CHO K2Cr2O7

H+⎯⎯⎯→ CH3COOH

SOCl2⎯⎯→ CH3COCl NH3⎯⎯→

Acetamide

CH3CONH2

(b)Acetaldehyde

CH2CHO [O]

⎯⎯→ CH3COOH C2H5OH

⎯⎯⎯→ CH3COOC2H5 C2H5OH

Claisencondensation

⎯⎯⎯⎯→ CH3COCH2COOC2H5

esterAcetoacetic

Q. 8. How will you convert Benzaldehyde into :

(a) Cyanobenzene (b) Aniline

(c) Benzoic acid (d) Phenyl acetic acid


Ans. (a)

Benzaldehyde Cyanobenzene

⎯⎯⎯⎯→Oxidation

⎯⎯⎯→Br2 , KOH

HONO

0°C⎯⎯→

NH3

Heat⎯⎯→HONO

⎯⎯→

CONH2CHO NH2 N2Cl

CuCN

COOH CN

(b)

Benzaldehyde Aniline


⎯⎯⎯→Br, KOH

⎯⎯→NH3

Heat

CONH2CHO NH2COOH

(c)

Benzaldehyde Benzoic acid


⎯⎯⎯→H+, H2O

Heat

COOHCHO CN

(d)

Benzaldehyde Phenyl acetic acid

⎯⎯⎯→LiAlH4

CHO

⎯⎯⎯→SOCl2

CH2OH

⎯⎯⎯→KCN

CH2Cl

⎯⎯⎯→H+, H2O

CH2CN CH2COOH

Q. 9. Write a chemical reaction for the conversion of acetaldehyde to dimethyl phenyl carbinol.

Ans.Acetaldehyde

CH3CHO [O]

⎯⎯→ CH3COOH Ca(OH)2⎯⎯⎯→ (CH3COO)2Ca

Distillation⎯⎯⎯⎯→ CH3—C

||O

—CH3 H+‚H2O

C6H5MgBr⎯⎯⎯⎯→

CH3—C—C6

|OH

|CH3

H5

Dimethyl phenylcarbinol

Q. 10. How will you made the following conversions :(a) Butanone to 3, 4-dimethyl hexane(b) Propanone to mesityl oxide.

Ans. (a)Butanone

CH3—C||O

—CH2—CH3 2[H]

Na/Hc⎯⎯→ CH3CH

|OH

—CH2—CH3 PCl5⎯⎯→ CH3—CH

|Cl

—CH CH2CH3

Na

Dry ether⎯⎯⎯→

CH3—

CH3—

CH—CH2—CH3|CH—CH2—CH3

3, 4-Dimethyl hexane

(b)Propanone

CH3COCH3 Ba(OH)2⎯⎯⎯→ (CH3)2 C—CH2COCH3

|OH

Heat⎯⎯→

Mesityl chloride

(CH3)2 C = CHCOCH3

Q. 11. Give balanced equation for the reaction of Aldehydes and Ketones with (give reactions) :(a) HCN (b) NH3(c) NH2OH (d) PCl5

Ans. (a) Reaction with hydrogen cyanide (HCN) :

CH3

H

C = O + HCN ⎯⎯⎯→

CH3

H

C

OH CN

Acetaldehyde Acetaldehyde cyanohydrin

Conversions ■ 319

CH3

CH3

C = O + HCN ⎯⎯⎯→

CH3

CH3

C

OH CN

Acetone Acetone cyanohydrin

(b) Reaction with NH3 :

CH3

H

C = O + NH3 ⎯⎯⎯→

CH3

H

C

OH NH2

– H2O

⎯⎯→

CH3

H

C = NH

Acetaldehyde Acetaldehyde ammonia Acetaldemine

Formaldehyde

CH2O + 4NH3– 6H2⎯⎯⎯→

N

N

N

N

CH2H2C CH2

CH2 CH2

CH2

Hexamethylenetetramine

or Urotropine

(c) Reaction with NH2OH :

CH3

H

C = O + NH2OHH+

⎯⎯⎯→

CH3

H

C = NOH + H2O

Acetaldehyde Acetaldoxime

CH3

CH3

C = O + NH2OHH+

⎯⎯⎯→

CH3

CH3

C= NOH + H2O

Acetone Acetoxime(d) Reaction with PCl5 :

CH3

H

C = O + PCl5 ⎯⎯⎯→

CH3

H

C

Cl

Cl

+ POCl3

Acetaldehyde Ethylidine chloride

R

R′

C = O + PCl5 ⎯⎯⎯→

R

R′

C

Cl

Cl

+ POCl3

Ketone Geminal dihalide

Q. 12. How will you bring about the following conversions ( not more than two steps) :

(a) Propanone to Propene

(b) Propanal to Butanone

(c) Benzaldehyde to Benzophenone

(d) Benzaldehyde to α-Hydroxyphenylacetic acid

Ans. (a) CH3COCH3Propanone

Reduction⎯⎯⎯→

[2H] CH3—CH

|OH

—CH3

(Sec. alcohol)

Dehydration

⎯⎯⎯⎯⎯→Al2O3/H3PO4

CH3—CH = CH2Propene


(b) CH3CH2CHOPropanal

(i) CH3MgBr

⎯⎯⎯⎯⎯→(ii) hydrolysis

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

CH3CH2CHCH3

|OMgBr

⎯⎯→ CH3CH2CH—CH3

|OH

sec. alcohol

[O] ↓ KMnO4/H2SO4CH3CH2COCH3

Butanone

(c) CHO

Benzaldehyde

oxidation

⎯⎯⎯⎯→KMnO4/H+

COOH

benzoic acid

decarboxylation

Soda limeand heat

⎯⎯⎯⎯⎯⎯⎯⎯→

C||O

—

Benzophenone

benzoylation←⎯⎯⎯⎯⎯

C6H5COCl

benzene

(d)

CHO

HCN

⎯⎯⎯→addition

CH—CN|OH

hydrolysis

⎯⎯⎯⎯→ CH—COOH|OH

Benzaldehyde cyanohydrin α-hydroxyphenylacetic acidQ. 13. How will you convert acetaldehyde into the following compounds ?

(a) Butane-2-one (b) Butane-1, 3-diol(c) But-2-enal

Ans. (a) CH3CHO (i) C2H5MgBr⎯⎯⎯⎯⎯→

(ii) H+

/H2O CH3—CH

|OH

—C2H5 oxidation ⎯⎯⎯→ CH3—C

||O

—CH2CH3

(b) 2CH3CHO Aldol condensation⎯⎯⎯⎯⎯⎯⎯→

aq. NaOH CH3—CH

|OH

—CH2—CHO reduction

⎯⎯⎯⎯→LiAlH4

CH3—CH—|OH

CH2—OH

(c) 2CH3CHO Aldol condensation followed⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

by reduction CH3—CH

|OH

—CH2—CH2OH

oxidation by PCC⎯⎯⎯⎯⎯⎯⎯⎯→followed by dehydration

CH3 CH = CH—CHOBut-2-enal

Q. 14. How will you convert(i) Benzoyl chloride to benzaldehyde(ii) Propanone to 2-propanol

Ans. (i) Benzoyl chloride is subjected to Rosenmund’s reduction.

Pd/BaSO4 quinoline ⎯⎯⎯⎯→

COCl

+ H2 Benzoyl chloride

CHO

+ HCl Benzal dehyde

(ii) Propanone is subjected to reduction in presence of LiAlH4.

CH3COCH3Propanone

+ H2LiAlH4⎯⎯⎯⎯→

or Ni/H2 CH3—CH

|OH

—CH3

2-propanol

Q. 15. How Benzoic acid is converted to aniline ?

Ans.Benzoic acid

C6H5COOH +NH3⎯⎯→ C6H5COONH4

– H2O⎯⎯→ C6H5CONH2

Br2 /KOH⎯⎯⎯⎯⎯⎯→

Hoffman Bromamide Reaction C6H5NH2Aniline

Q. 16. How will you convert glycerol to formic acid ?

Ans.

CH2OH|CHOH|CH2OHGlycerol

+ COOH|

COOHOxalic acid

110°C

Δ⎯⎯⎯→

CH2OOC

CHOH |

CH2OHGlycerol

+

| |COOH

mono-oxalate

H2 O

Conversions ■ 321

CH2O·OC·COOH|CHOH|CH2OH

⎯⎯→–CO HCOOH

Formic acid +

Glycerolmono-oxalate

2

CH2O·OCH|CHOH|CH2OH

⎯⎯→+ H O2

|CHOH|CH2OH

CH 2OH

Q. 17. How will you convert methyl chloride to acetic acid ?

Ans.Methyl chloride

CH3Cl + KCN– KCl

⎯⎯→ CH3CN + 2H2O⎯⎯→ CH3COONH4

+ HCl– NH4Cl⎯⎯⎯→

Acetic acid

CH3COOH

Q. 18. How will you convert benzene to benzoic acid ?

Ans.Benzene

C6H6 + CH3Cl anhydrous AlCl3

– HCl⎯⎯⎯⎯⎯⎯→ C6H5CH3

2C6H5CH3 + 3O2 443 – 573 K

(CH3COO)2Mn⎯⎯⎯⎯⎯→

Benzoic acid

2C6H5COOH + 2H2O

Q. 19. How will you convert phenol to benzoic acid ?

Ans. Zn dust

Δ⎯⎯⎯→

CH3Cl

Anhy. AlCl3⎯⎯⎯⎯⎯→

KMnO4

Δ⎯⎯⎯→

Benzoic acid

OH

Phenol Benzene

CH3 COOH

Toluene

Q. 20. How will you bring about the following conversion ?

Ans.NaOOC –�COONa + H22HCOONa⎯⎯⎯→

– 2H2O2HCOOH + 2NaOH

+ CaSO4

COOH

COOH

⎯→ Ca(OH)2

⎯→ H2SO4

⎯⎯→400°C

Oxalic Acid

– – –+ + +

Sodiumformate

Formic Acid

Ca2+ + 2NaOHCOO–

COO–

Calcium oxalate

Q. 21. How would you convert Benzoic acid to benzene ?

Ans.Benzoic acid

C6H5COOH + NaOH ⎯→ C6H5COONa + H2O

C6H5COONa + NaOH CaOHeat

⎯⎯→ Benzene

C6H6 + Na2CO3


(a) Ethanoic acid to a mixture of methanoic acid and diphenyl ketone

(b) Propanoic acid to 2-propanol

(c) 2-chloro butanoic acid to 3-chloro butanoic acid


Ans. (a)Ethanoic

acid

CH3COOH SOCl2⎯⎯→ CH3COCl

C6H5MgBr3⎯⎯⎯⎯⎯→ C6H5COCH3 C2O/H+

C6H5MgBr⎯⎯⎯⎯→ C6H5—C—OH

|C6H5

|CH3

|↓Heat

HCOOH

acidMethanoic

+

C6H5

C6H5

Diphenyl ketone

C = O CrO3←⎯⎯ C = CH2

|C6H5

|C6H5

(b)Propanoic

acid

CH3CH2COOH Reduction

LiAlH4⎯⎯⎯⎯→ CH3CH2CH2OH

H+

443 K⎯⎯→ CH3CH = CH2

H2O/H+⎯⎯⎯→ CH3CH—CH3

|OH

2-propanol

(c)2-chlorobutanoic

acid

CH3CH2—CH|Cl

—COOH alk.KOH

⎯⎯⎯→ 2-Butanoic acid

CH3CH = CHCOOH HCl

⎯→ CH3CH—CH2|Cl

—COOH

3-chloro butanoic acid

Q. 23. How will you convert acetic acid into the following :(a) Succinic acid(b) Glycine(c) Acetaldehyde

Ans. (a)Acetic Acid

CH3COOH PCl5⎯⎯→ CH2—COOH

|Cl

Na/Dry ether

Wurtz reaction⎯⎯⎯⎯⎯⎯→ CH2—COOH

|CH2—COOH

Succinic acid

(b)Acetic acid

CH3COOH PCl5⎯⎯→ CH2—COOH

|Cl

NH3⎯⎯→ CH2—COOH

|NH2

2-chloro ethanoicacid Glycine

(c)Acetic acid

CH3COOH SOCl2⎯⎯→ CH3COCl

H2

Pd‚ BaSO⎯⎯→

Acetaldehyde

CH3CHO

Q. 24. How will you convert propanoic acid into Lactic acid ?

Ans.Propanoic acid

CH3CH2COOH Br2‚ P

⎯⎯→ CH3CHClCOOH + HOH(α-Chloropropionic acid)

⎯⎯→ CH3CHOHCOOH + HCl

(Lactic acid)

CH3CHOHCOOH + HCl

Q. 25. How will you convert Acetic acid into :(a) Acetonitrile(b) Propanoic acid.

Ans. (a)Acetic Acid

CH3COOH + NH3 ⎯→ CH3COONH4 Heat

P2O5⎯⎯→

Acetonitrile

CH3CN + 2H2O

(b)Acetic acid

CH3COOH LiAlH

⎯⎯→ CH3CH2OH SOCl2⎯⎯→ CH3CH2Cl

Mg⎯→ CH3CH2MgCl

(i) CO2

(ii) H+/H2O⎯⎯⎯⎯→

CH3CH2COOHPropanoic acid

Q. 26. How will you convert formic acid into glycolic acid ?

Ans.Formic

acid

HCOOH Ca(OH)2⎯⎯⎯→ (HCOO)2Ca

Distil⎯⎯→ HCHO

HCN⎯⎯→ H—C—CN

|OH

|H

Hydrolysis⎯⎯⎯⎯→

Glycolic acid

H—C—COOH|OH

|H

Conversions ■ 323

Q. 27. How will you convert phthalic acid to anthranilic acid ?

Ans.COOH

COOH

CO

COO

CO

CONH

COONa

CONH2

COONa

NH2

COOH

NH2

⎯⎯→

Anthranilic acid

Phthalic acid

⎯⎯⎯→Br2 , KOH⎯→H+

⎯⎯→ ⎯⎯→NH3

– H2O

H2O NaOH

Heat

Q. 28. How will you convert benzoic acid to aniline ?

Ans. C6H5COOHBenzoic acid

+ NH3⎯⎯→ C6H5COONH4 – H2O

⎯⎯→ C6H5CONH2 Hoffman Bromide reaction⎯⎯⎯⎯⎯⎯⎯→

Br2/KOH

Aniline

C6H5NH2


Q. 1. How will you convert phenol to aniline ?

Ans.

Phenol Aniline

ZnCl

30°C

OH NH2

NH3+ + H2O

Q. 2. How will you convert acetic acid to methyl cyanide ?Ans. CH3COOH + NH3 ⎯→ CH3COONH4

Δ– H2O⎯→ CH3CONH2

Acetic acid Ammonium Acetamideacetate

H2O + CH3CN Δ

P2O5 Methylcyanide

⎯⎯⎯→

Q. 3. How will you convert nitrobenzene to 2, 4, 6-tribromoaniline ?

Ans.

NO2

+ 6H Sn/HCl– 2H2O⎯⎯→

NH2

3Br2

– 3HBr⎯⎯→

Br Br

Br

NH2

Nitrobenzene Aniline

2, 4, 6-tribromoaniline

Q. 4. How will you convert propanoic acid to ethylamine ?

Ans. CH3CH2.COOHPropanoic acid

+ NH3 ⎯⎯→ CH3CH2COONH4 Δ

⎯⎯⎯→– H2O

CH2CH2CONH2

Br2 + 4NaOH⎯⎯⎯⎯⎯⎯→

Δ CH3CH2NH2

Ethylamine

+ Na2CO3 + 2NaBr + 2H2O

Q. 5. How will you convert ethanol to methylamine ?

Ans.Ethanol

C2H5OH + (O)– H2O⎯⎯→

Acetic acid

CH3COOH + NH3⎯⎯→

Ammonium acetate

CH3COONH4 Δ

– H2O⎯⎯→

Acetamide

CH3CONH2

Br2‚ NaOH⎯⎯⎯⎯→ CH3NH2

Methylamine

Q. 6. How will you convert ethylamine to methylamine ?

Ans. C2H5NH2Ethylamine

+ HNO2

– N2‚ – H2O⎯⎯⎯⎯→ C2H5OH

ethanol

+ 2(O)– H2O

⎯⎯⎯→ CH3COOHacetic acid


+ NH3

– H2O⎯⎯→ CH3CONH2

acetamide Br2 + 4NaOH

Δ⎯⎯⎯⎯⎯→

Methyl amine CH3NH2

+ 2NaBr + Na2CO3 + 2H2O

Q. 7. How can the following conversion be brought about Benzene to acetanilide ?

Ans. C6H6Benzene

conc. HNO3 + conc. H2SO4

– H2O⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ C6H5NO2

nitrobenzene + 4[H]– 2H2O⎯⎯→ C6H5NH2

aniline

+ CH3COCl⎯⎯⎯⎯⎯→ C6H5NHCOCH3 + HCl

Acetanilide

Q. 8. How would you convert Methyl amine to ethyl amine ?

Ans.Methyl amine

CH3NH2 HNO2⎯⎯→

Methanol

CH3OH PCl5

⎯⎯→ Methylchloride

CH3Cl NaCN⎯⎯→

Methylcyanide

CH3CN LiAlH4

⎯⎯⎯→2H2

Ethyl amine

CH3CH2NH2

Q. 9. How the following conversions can be brought about :

(a) Acetamide to propionamide(b) p-nitro aniline to 1, 2, 3 tribromobenzene

Ans. (a)Actamide

CH3CONH2 H+

⎯→ CH3COOH LiAlH4⎯⎯⎯→ CH3CH2OH

P/I2⎯⎯→ CH3CH2I

||↓HCN

Propionamide

CH3CH2CONH2 Partial

hydrolysis←⎯⎯ CH3CH2CN

(b) ⎯⎯⎯⎯⎯→BrBr

HCl

⎯⎯→

1, 2, 3 -Tribromobenzene

p-nitro aniline

⎯⎯⎯→H3PO2

Br2 NaNO2/HCl⎯⎯→

CuBr⎯⎯→

NaNO2 Sn/HCl

NH2 NH2

NO2 NO2

BrBrN2Cl

N2Cl

NO2

BrBr

Br

NO2

BrBr

BrBrBr

Br

Q. 10. Convert aniline into the following compounds :(a) Benzaldehyde (b) Phenyl hydrazine(c) Fluorobenzene (d) Benzylamine (e) Benzoic acid

Ans. (a) ⎯⎯⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→

Aniline

CuCNNaNO2 , HCl

NH2 N2Cl CN CHO

BenzaldehydeKCN

Sn/HCl

(b) ⎯⎯⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯⎯→

Aniline

SnCl2NaNO2 , HCl

NH2 N2Cl NHNH2HCl NHNH2

HCl

KOH, Heat

Phenyl hydrazine

(c)0°C

⎯⎯⎯⎯⎯→ ⎯⎯⎯→

Aniline Fluorobenzene

HBF4NaNO2

NH2 N2Cl F

HCl

Conversions ■ 325

(d) CuCN⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯⎯→

Benzylamine

NH2 N2Cl CN

LiAlH4

Aniline

NaNO2

CH2NH2

(e) ⎯⎯⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→CuCl2NaNO2 , HCl

NH2 N2Cl CN COOH

Benzoic acidAnilineKCN

H+, H2O

Q. 11. How will you convert nitrobenzene into the following :(a) Anisole (b) Benzoyl alcohol

Ans. (a)(NaNO2 + HCl)⎯⎯⎯⎯⎯→HNO2 , Warm

NO2 NH2 OH ONa OCH3

NaOH⎯⎯⎯→

CH3⎯⎯→

Sn/HCl⎯⎯⎯→

Nitrobenzene Aniline Anisole

(b) NaNO2 + HCl

CNN2Cl

CH2NH2 CH2OH

⎯⎯⎯⎯⎯→

NO2 NH2

⎯⎯⎯⎯→Na/Alcohol

HNO2

CuCN⎯⎯→

⎯⎯→

Sn/HCl⎯⎯⎯→

Nitrobenzene Aniline

Benzoyl AlcoholQ. 12. Bring out the following conversions :

(i) Benzene to m-nitroaniline (ii) Propanenitrile to ethanamine(iii) Benzene to 3-bromophenol (iv) Ethylamine to methyl amine(v) Acetamide to methyl amine

Ans. (i)HNO3/H2SO4

413°K ⎯⎯⎯⎯⎯⎯→

Partial Reduction

Na2S2/H2O ⎯⎯⎯⎯⎯⎯⎯→

NO2

NO2 m-dinitro benzene

NO2

NO2 m-nitroaniline

(ii) Partial hydrolysis

H2O/OH–⎯⎯⎯⎯⎯⎯⎯⎯→

O Br2/NaOH ⎯⎯⎯⎯⎯→

OH3CH2CN

Propanenitrile

CH3CH2NH2

Ethanamine

CH3CH2C—NH2

Propanamide

(iii)FeBr3

Br2⎯⎯⎯→

HNO3/H2SO4

335°K ⎯⎯⎯⎯⎯⎯→

NaNO2/HCl

273°K ⎯⎯⎯⎯⎯⎯→

Sn/HCl

Reduction ⎯⎯⎯⎯⎯→

NO2

Br m-bromonitrobenzene

N2+Cl–

Br H2O/H +

Warm ⎯⎯⎯⎯→

OH

Br

NH2

Br

NO2

3-bromophenol

(iv)

CH3CH2CH2

Ethylamine

CH3COONH4

Ammonium acetate

CH3CH2OHEthyl alcohol

CH3COOHAcetic acid

–H2O

NH4OH⎯⎯⎯→

Δ

–H2O←⎯⎯⎯

HNO2⎯⎯⎯→

K2Cr2O7, H+

⎯⎯⎯⎯⎯→

CH3NH2

Methyl amine

CH3CONH2

Acetamide

Br2/KOH←⎯⎯⎯⎯

(v) + Br2 + 4KOH ⎯⎯→ CH3CONH2

Acetamide

CH3NH2

Methyl amine

+ 2 KBr + K2CO3 + 2H2O

❐

Question

14Set Mechanism of Reactions

Chapter 10. Haloalkanes and HaloarenesQ. 1. Write the mechanism of the following reaction :

CH3CH2OH HBr

⎯⎯→ CH3CH2Br + H2OAns. This reaction occurs by SN2 mechanism.

(i) H —Br H+ + : Br–

CH3CH2OH + H2 CH3CH2OH2 + H2

Ethyl oxonium ion

(Protonated 1° Alcohol)

(ii)δ

⎯⎯→ ⎯⎯→

++

Br– + CH2 — OH2 Br– …… CH2 …… OH2 Br —CH2—CH3 + H2OTransition state

CH3

—

CH3—

Ethylbromide

Q. 2. Explain the mechanism of the following reaction :

n-BuBr + KCN EtOH–H2O

⎯⎯⎯⎯⎯→ n - BuCNAns. This reaction undergoes bimolecular nucleophilic substitution reaction (SN2)

K+ [: C ≡ N : ←→ : C = ˙N:

–]

CN– ion is an ambident nucleophile so it can attack the carbon atom of C—Br bond in n-BuBrthrough C or N as C—C bond is stronger than C—N bond, therefore, attack occurs through C to Nfor n-BuCN.

N– CK+ + CH3CH2CH2CH2Br ⎯⎯→ CH3CH2CH2CH2CN + KBrδ

+δ

–

Q. 3. An optically active compound having molecular formula C7H15Br reacts with aq. KOH to give aracemic mixture of products. Write the mechanism involve in this reaction :

Ans. This reaction proceeds by unimolecular substitution reaction.

CH3CH2 — C — Br C+ — CH2CH2CH3 + Br–Slow

CH3 CH3—

——

—

CH2CH3CH2CH2CH3

H3C

OH– + C+ — CH2 — CH2 — CH3 ⎯⎯⎯→ CH3CH2 — C — OH + HO — C — CH2CH3

CH3—

—

CH2CH3

——

CH2CH2CH3

CH3—

—

CH2CH2CH3

Racemic mixture

Fast

Step

50% 50%

Chapter 11. Alcohols, Phenols and EthersQ. 1. Alcohol reacts both as nucleophiles as well as electrophiles. Write one reaction of each type and

describe its mechanism :Ans. When alcohol acts as Nucleophiles : When alcohols react as nucleophiles the bond between

O—H of alcohol is broken.

Mechanism of reactions ■ 327

——R — O — H + C+ — ⎯⎯→ R —+ O — C — ⎯⎯→ R — O — C — + H+

— — ———

H

When alcohol acts as electrophiles : When alcohols react as electrophiles the bond betweenC—O is broken. Protonated alcohols react in this manner.

R — CH2 — O + H+ ⎯⎯→ R — CH

R R

2 — OH2+

Br– + CH2 — OH2 ⎯⎯→ Br — CH2 + H2O+

— —

ElectrophileQ. 2. Explain the mechanism of the following reactions :

(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adductfollowed by hydrolysis.

(ii) Acid catalysed dehydration of an alcohol forming an alkene.(iii) Acid catalysed hydration of an alkene forming an alcohol.

Ans. (i) Step I : Nucleophilic addition of Grignard reagent to carbonyl group

Carbonyl group

δ+

δ+

δ–

δ–—

——

—C = O + R — Mg — X ⎯⎯→ C — O– Mg+ — X—

RGrignard reagent

Adduct

Step II : Hydrolysis of Adduct

H2O ——

——C — O–Mg+ — X ⎯⎯→ C — OH + Mg (OH) — X

—

R

—

RAdduct

(ii) CH3 — CH2 — OH H

+

⎯→Δ

CH2 = CH2 + H2O

Mechanism :Step I : Formation of protonated alcohol.

Ethyl alcohol

Fast

—

H

—

H

—

H

—

H

—

H

—

H

—

H

—

H

—

H

H—C—C—O—H ⎯⎯→ H—C—C—OH

Ethyl oxonium ion

+

Step II : Formation of carbocation. It is due to the slowest step and hence, the rate determing step.

Slow

—

H

—

H

—

H

—

H

—

H

—

H

—

H

—

H

—

H

H—C—C—O—H H—C—C + H2O

Ethyl oxonium ion

+ +

Step III : Formation of ethylene by elimination of a proton.

———

—H

H

H

H

+

Ethyl oxonium ion

—

H

—

H

—

H

—

H

H—C—C C = C + H

Ethylene

+

To derive the equilibrium to the right ethylene is removed as it is formed.


(iii)

———

— ———

—H+

—

H

—

OH

C = C + H2O

Alkene

C — C

Mechanism of the reaction

Step I : Protonation of alkene to form carbocation by electrophilic attack of H3O+

H2O + H+ ⎯⎯→ H3O+

+ +

——

———

—

—

——

H H

C = C + H—O—H —C—C + H2O

Step II : Nucleophilic attack of water on carbocation

+ +

——

—

——

—C—C + H2O —C—C—O—H

—

H

—

H

—

H

Step III : Deprotonation to form an alcohol

+

— — — —

—

—C—C—O—H + H2O ⎯⎯→ —C—C— + H3O+

—

H

—

H

—

H—

O—H

Q. 3. Give the mechanism for the reaction, when 3-methyl butan-2-ol, is treated with HBr.

CH3—CH—CH—CH3

CH3 OH

HBrCH3—C—CH2—CH3

CH3

Br

|⎯⎯⎯→

| |

|

Ans. Step I : Formation of protonated alcohol

FastCH3—CH—CH—CH3 + H+ CH3—CH—CH—CH3

CH3—O—H+O—HCH3

— —

——

H

Step II : Formation of carbocation

2° Carbocation (less stable)

CH3—CH—CH—CH3 CH3—CH—CH—CH3 + H2O+

O—HCH3

— —

CH3

—

—

H

Slow

Secondary (2°) carbocation being less stable undergoes hydride shift to form more stable3° carbocation.

3° carbocation

CH3—CH—CH—CH3 + H2O CH3—C—CH2—CH3 +

+

CH3

—

CH3

—

Hydride shift


Step III : Attack of Nucleophile

CH3—C—CH2—CH3 + Br ⎯⎯⎯→ CH3—C—CH2—CH3

+ –

CH3

—

CH3

——

Br

Q. 4. What is the mechanism of acid catalysed dehydration of alcohols ? What is the order of reactivityof different types of alcohols towards this reaction ?

Ans. The mechanism of this reaction involves protonation of the alcohol followed by loss of a watermolecule and a proton.

1. H—

H|C|H

—

H|C|H

—OH + H+ Protonation

H—

H|C|H

—

H|C|H

—+OH2

Oxonium ion

2. H—

H|C|H

—

H|C|H

—+OH2 + H+

Loss of H2O

H—

H|C|H

—

H|C|H

+

Carbocation

3. H—

H|C|H

—

H|C|H

+ Loss of proton

H—C|H

= C|H

—H

The ease of dehydration of alcohols follows the order :

tertiary alcohol > secondary alcohol > primary alcohol

This order of reactivity can be explained in terms of stability of the intermediate carbocation.Tertiary alcohols reacts via 3° carbocation as intermediate, which being most stable is formed atfastest rate. Hence, 3° alcohols are most reactive, followed by 2° and 1° alcohols.


Q. 1. Give mechanism of preparation of ethoxy ethene from ethanol.

Ans. The formation of ether involving the attack of an alcohol molecule on a protonated alcoholproceeds by SN

2 reaction :

(i)++

Ethyl oxonium ion

CH3—CH2—OH + H ⎯⎯⎯→ CH3—CH2—O—H

—

H

(ii)⊕ +

——

CH3—CH2—O + CH3—CH2—O ⎯⎯→ CH3—CH2—O—CH2—CH3 + H2O

H

H

—

H

—

H

(iii)⊕ +

CH3—CH2—O —CH2—CH3 ⎯⎯→ CH3—CH2—O—CH2—CH3 + H

—

H

Q. 2. Write the mechanism of reaction of HI with methoxy methane.

Ans. The cleavage of methoxymethane with HI follows SN2 mechanism.


Step I :+ –CH3—O—CH3 + H — I CH3—O—CH3 + I

Protonation —

H

FastDiethyl

oxonium ion

Step II :

⎯⎯

→

++

–

–

CH3—I + CH3—OH

I + CH3—O—CH3 ⎯⎯⎯→ I ……… CH3 ……… O ……… CH3

—

H

—

H

Methyliodide

Slow

Methylalcohol

If excess of HI is used methyl alcohol formed in step II further reacts with another molecule of HIto form CH3I.Step III :

+ –CH3—O—H + H—I CH3—O —H + I

—

H

Protonatedmethyl alcohol

Protonation

Fast

Step IV :+–I + CH3—O—H ⎯⎯⎯→ CH3 —I + H2O

—

H

Slow Methyliodide

SN2

Water

Q. 3. Write down the mechanism of Cannizzaro’s reaction.

Ans. Cannizzaro’s reaction is an example of hydride (H:–) transfer as shown below :

Step I : The OH– ion attacks the carbonyl carbon to form hydroxy alkoxide. It is a nucleophilicattack like other nucleophiles.

— —

C6H5 —C—OH

—

(I)

—

H

C6H5—C—HBenzaldehyde

FastO

–OH

O–

+

Step II : The anion acts as a hydride ion donor to the second molecule of aldehyde. The presenceof negative charge on (I) helps in the loss of hydride ion. In the final step of the reaction the acidand the alkoxide ion exchange proton to acquire stability.

⎯⎯

→

C6H5 —C—OH + C—C6H5 ⎯⎯⎯⎯⎯⎯→ C6H5 —C—OH + H—C—C6H5

C6H5 —C—O + C6H5 —CH2OH

— —

O

— —

O

— —

O

–

–

—

O–

—

O

—

H

—

H

—

H

Slow

Benzoate ion

Protonexchange (fast)

Benzyl alcohol

Hydridetransfer


Q. 4. Write the mechanism of Aldol formation reaction.Ans. The α-hydrogen atom of aldehydes and ketones are slightly acidic in nature due to strong electron

withdrawing inductive (– I) effect of the carbonyl group. This acidity of α-hydrogen is due toresonance stabilization of the conjugate base. Due to electron withdrawing inductive effect (– Ieffect) of the carbonyl group, it withdraws electrons from the adjacent carbon-carbon bond. Thismakes α-carbon electron deficient and in turn withdraws electron from Cα–H bond.

Step I : In the first step the base (OH–) removes proton to form an enolate ion from the aldehyde

or ketone. The enolate ion gets stablized by resonance.

––

–

——

- hydrogen

HO + H—CH2—C CH2—C —C + H2O

—

α

———

———

HHH

CH2

OOOSlow ←⎯⎯→

Step II : The enolate ion being a strong nucleophile attacks the carbonyl group of the secondmolecule of acetaldehyde.

— —

–

–CH

O O

3 — C CH2 — C CH3 — C — CH2 — C

—

———

H

O

———

H

O

—

H

—

Hattack

Nucleophilic

Acetaldehyde AnionEnolate ion

+

Step III : The anion formed in step II abstracts a protron from water to form aldol.–

CH3 — C — CH2 — C + H — OH CH3 — C — CH2 — C

—

H

—

H

—

O—

OH

———

H

O

———

H

O

AldolNote : The aldehyde and ketone which do not have α-hydrogen cannot undergo aldolcondensation.

Q. 5. Give the mechanism of Esterification of carboxylic acid.Ans. The esterification of carboxylic acid with alcohol is a kind of nucleophilic acyl substitution

reaction. The mechanism of esterification involves the following steps :Step I : A proton from the protonoic acid attacks the carbonyl oxygen of carboxylic acid.

+

+

+R — C + H R — C R — C—

—

OH

O—H

———

OH———

OH

O O—H

Protonatedcarboxylic acid

Step II : As a result of protonation, the carbonyl carbon gets activated and hence readilyundergoes attack by the lone pair of electron on the oxygen atom of an alcohol to form atetrahedral intermediate.

+R — C + + O — R′ R — C — O — R′

—

H

—

OH

—

OH

——

OH

OH

—

H

Tetrahedral Intermediate

Step III : From the resulting intermediate, a proton shifts to OH group to form anothertetrahedral intermediate. During this proton transfer, the —OH group gets converted into —OH2

+

group.


+

+

R — C— O — R′ R — C — O — R′

—

H

—

OH

—

OH

—

OH2

—

OHProton

Transfer

Step IV : The intermediate obtained in step III loses a water molecule to form a protonated ester.

+

+

R — C— O — R′ R — C

—OH2

—

OH

——

OR′

OH

Protonated Ester

– H2O

Step V : The protonated ester loses a proton to form an ester.

R — C ⎯⎯→ R — C OR′

+

———

O—H

OR′———

O

Protonated ester Ester


Q. 1. Write down the mechanism of diazotization of Amine.

Ans. The diazotization of amines is believed to occur by the following mechanism :

Nitrous acid formed by the reaction of sodium nitrite and mineral acid takes up a proton from theacid and undergoes hydrolysis to form nitrosonium ion.

H—O—N = O + H ⎯⎯→ H2O — N = O ⎯⎯→ N = O + H2O+++

Nitrosonium ion

The electrophilic nitrosonium ion reacts with the nitrogen of the amine and combines with thelone pair of electrons at N to form N-nitroso derivatives, which by protonic shift rearranges todiazohydroxide,

Ar — N + N = O ⎯⎯→ Ar — N — N = O ⎯⎯⎯→ Ar — N — N = O

Ar — N = N — O — H ←⎯⎯ Ar — N — N = OH

+ +

+

H+– H+

– H+

– H+

—

H

—

H

—

H

—

H

—

H

—

H

The diazohydroxide in acidic solution takes up a proton and by the elimination of water moleculeforms diazonium ion, which may take up acid anion X to form diazonium salt.

Ar — N = N — O — H ⎯⎯→ Ar — N = N — OH2 – H2O

+

+

+

+– H+

Diazonium cation

Diazonium salt

Ar — N ≡ N

Ar — N ≡ N + X– ⎯⎯→ Ar — N ≡ N + X–

⎯⎯→



Q. 1. Give the mechanism of Free Radical Addition polymerisation.

Ans. The reaction involves the following steps :

(i) Chain initiation step : Organic peroxides undergoes homolytic fission to form free radicals

which act as initiator (I·n).

Peroxides ⎯→ Rad·

(as an In·)

The initiator adds to the carbon-carbon double bond of an alkene molecule to form a new freeradical.

I·n + CH2 ˙˙⎯⎯ CH2 ⎯→ In—CH2 — ·CH2

Free radical

(ii) Chain propagation step : The new free radicals add to a double bond of monomer to form alarger radical.

In—CH2—·CH2 + CH2

·CH2 ˙˙⎯⎯ CH2 ⎯⎯→ Rad CH2—CH2—CH2—

·CH2

In—CH2—CH2—CH2—·CH2 + CH2 ˙˙⎯⎯CH2 ⎯⎯→ In CH2—CH2)n ( CH2

·CH2

(iii) Chain termination step : The above chain reaction terminates when the two free radicalscombine.

2InCH2CH2CH2CH2 ⎯→ In CH2CH2CH2CH2CH2CH2CH2CH2 In

❐

Question

15Set Description based questions


Q. 1. What type of substance would make better permanent magnets, Ferromagnetic or ferrimagnetic ?Justify your answer.

Ans. Ferromagnetic substance would make better permanent magnets. In solid state, the metal ions offerromagnetic substance are grouped together into small region called domains. Each domainacts as tiny magnet. Since domains are randomly oriented, their magnetic moments get cancelled.When the substance is placed in a magnetic field all the domains gets oriented in the direction ofmagnetic field and a strong magnetic field is produced. This ordering of domain persists evenwhen the magnetic field is removed and the ferromagnetic substance becomes a permanentmagnet.

(a) ↑

↓

↑ ↑ ↑ ↑ ↑

(b) ↓↑ ↑ ↑ ↑

Schematic alignment of magnetic moment in (a) Ferromagnetic substance and (b) Ferrimagneticsubstance.

Q. 2. Discuss the structure of graphite and explain its important properties on the basis of structure.

Or

Why graphite is soft and can be used as a lubricant ?

Ans. In graphite sp3 hybridized carbon atoms are in a series of parallel sheets. In all the sheets carbonatoms are arranged at the corners of the plane of regular hexagon. An infinite number of sheetsare formed. The distance between the adjacent sheets is 3·35Å and these sheets are held togetherby weak van der Waal’s forces. The distance between two carbon atoms in a sheet is 1·42Å. Theenergy required to break the network of graphite into constituent carbon atoms is nearly 836·8kJmol–1, therefore the melting point of graphite is quite high. Out of four valence electrons ofcarbon atom, three are used to form three sigma bonds with other carbon atoms of a hexagonalring and fourth electron on each carbon atom is free to move. Due to this free electron graphite isa good conductor of heat and electricity.

Two successive layers are held together by weak van der Waals forces of attraction, hence onelayer can slip over the other. Due to this reason graphite is soft and a good lubricating agent.

Layer - LayerVan der Waals

forces(Weaker bonds)

Layer No. 1

Carbon-carbon covalent bonds(strong bonds)

Layer No. 2

Layer No. 3

3.35 A

3.35 A

1·42ÅFig. Structure of graphite

Description based questions ■ 335

Q. 3. State the main characteristics of an ionic and a network type of crystal giving a suitable examplefor each.

Fig. Sodium chloride crystal lattices

Ans. Ionic Crystal : Example : Sodium chloride crystal.(i) Lattice sites are occupied by ions. Example : Na+ and

Cl–.

(ii) Ions are generally arranged in an alternate fashion, asin case of NaCl crystal, in three dimensional space.

Na+Cl– Na+Cl–

Cl– Na+ Cl– Na+

(iii) Ions are held together by strong electrostatic forces ofattraction.

(iv) Structures are hard, brittle and have high melting andboiling points.

Network Type Crystal : Example : Diamond

(i) Lattice sites are occupied by atoms.(ii) Atoms are arranged in tetrahedral manner giving rise to intersecting face centered lattices.

Fig. Diamond Crystal

(iii) Atoms are bonded together with strong covalent bonds.

(iv) Structure is hard, brittle and have high melting and boiling points.

Q. 4. Examine the given defected crystal

A+ B– A+ B– A+

B– O B– A+ B–

A+ B– A+ O A+

B– A+ B– A+ B–

Answer the following questions :

(i) What type of stoichiometric defect is shown in the crystal ?

(ii) How is the density of crystal affected by this defect ?

(iii) What type of ionic substance show such defect ?

Ans. (i) Schottky defect

(ii) Decreases

(iii) This type of defect is shown by ionic compounds which have high coordination number andsmall difference in size of cations and anions e.g NaCl, KCl etc.

Q. 5. Non-stoichiometric cuprous oxide, Cr2O can be prepared in the laboratory. In this oxide, copperto oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor.

Ans. The ratio less than 2 : 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced bycupric (Cu2+) ions. For maintaining electrical neutrality every two Cu+ ions will be replaced byone Cu2+ ion there by creating a hole. As conduction will be due to the presence of these positiveholes, hence it is a p-type semiconductor.


Q. 6. If the radius of the octahedral void is ‘r’ and the radius of the atom in the close packing is Rderive relation between ‘r’ and ‘R’.

Ans. A sphere fitting into the octahedral void is shown by shaded circle. The sphere present above andbelow the void are not shown in the figure. As ABC is a right angle triangle, pythagoras theoremis applied.

AC2 = AB2 + BC2

(2R)2 = (R + r)2 + (R + r)2

= 2 (R + r)2

4R2 = 2 (R+ r)2

2R2 = (R + r)2

( )√⎯⎯2R2

= (R + r)2

√⎯ 2R = R + r

r = √⎯ 2 R – R

r = ( )√⎯ 2 – 1 R

= (1·414 –1) Rr = 0·414R

Q. 7. Calculate the efficiency of packing in case of metal crystal for simple cubic (with the assumptionthat atoms are touching each other).

Ans. Packing efficiency =2 × Volume of one atomVolume of cubic unit cell × 100

=2 ×

43 πr3

a3 × 100

for simple cubic lattice a = 2r and Z = 1

∴ Packing efficiency =1 ×

43 πr3

(2r)3 × 100

=π6 × 100 = 52·36%

= 52·4%

Q. 8. (i) What type of stoichiometric defect is shown by KCl and why ?

(ii) What type of semiconductors is formed when silicon is doped with As ?

(iii) Which one of the following is an example of molecular solid CO2 or SiO2 ?

(iv) What type of substance would make better magnet, ferromagnetic or ferrimagnetic ?

Ans. (i) KCl shows Schottky defect as the cation K+ and anion Cl– are of almost similar sizes.

(ii) n-type semiconductor.

(iii) CO2

(iv) Ferromagnetic.

Q. 9. What makes a glass different from solid such as quartz ? Under what conditions it is quartz ?Under what conditions would quartz be converted into glass ?

Ans. Glass is an amorphous solid in which the constituent particles (SiO4 tetrahedra) have only a shortrange order and there is no long range order. In quartz the constituent particles (SiO4 tetrahedra)have both short range as well as long range orders. On melting quartz then cooling rapidly, it isconverted into glass.


Q. 10. Calculate the packing efficiency of bcc structure.Ans. Efficiency of packing in bcc structure : In this case the atom at the centre is in touch with other

two atoms which are diagonally arranged. The sphere along the body diagonal are shown withsolid boundaries.In Δ EFD b2 = a2 + a2 = 2a2

b = √⎯⎯2aIn Δ AFD,

c2 = a2 + b2 = a2 + 2a2

= 3a2

c = √⎯⎯3aThe length of the body diagonal a is equal to 4r. r being the radiusof the sphere (atom). As all the three spheres along the diagonaltouch each other.

a = 4a

Therefore a = 4r = √⎯⎯3a

a =4r

√⎯ 3 or r = √⎯

34 a

As already calculated the total number of atoms associated with a bcc unit cell is 2, the volume(v) is therefore

2 × 43 πr3 =

83 πr3

Volume of the unit cell (v) = a3

= ⎝⎜⎛

⎠⎟⎞4r

√⎯ 3

3 =

64r3

3√⎯ 3

Packing efficiency = vV × 100

= ⎝⎜⎛

⎠⎟⎞8

3 πr3

⎝⎜⎛

⎠⎟⎞64

3 √⎯ 3 × r3

× 100 = √⎯3

8 π × 100 = 68%

Therefore, 68% of unit cell is occupied by atoms and the rest 32% is empty space in bcc structure.Q. 11. Calculate the packing efficiency for ccp arrangements.

Ans. The efficiencies of both types of packing ccp and hcp are equally good since in both, atomsspheres occupy equal fraction (74%) of the available volume. We shall now calculate theefficiency of packing in ccp structure. Let the unit cell length be ‘a’ and face diagonal be ‘b’(represented as AC in figure). In this figure other sides are not shown for the sake of clarity.In triangle ABC, ∠ABC is 90°Therefore

Cubic Close packing

AC2 = b2 = BC2 + AB2

= a2 + a2 = 2a2

b = √⎯⎯2aIf ‘r’ is the radius of sphere, we find

b = 4r = √⎯⎯2a

or a =4r

√⎯ 2 = 2√⎯ 2 or r =

a

2√⎯ 2As ccp structure has 4 atoms per unit cell, therefore the total volume

of 4 spheres (v) are = 4 × 43 πr3

Total volume of the unit cell (v) = a3 = ( )2√⎯ 2 r3


Packing efficiency =vV × 100

=4 ×

43 πr3 × 100

( )2√⎯⎯2r3

=

163 × πr3

16 × √⎯ 2 r3 × 100 =

π

3√⎯ 2 × 100 = 74%

Therefore 74% of unit cell is occupied by atom and the rest 26% is empty in ccp arrangement.Q. 12. What is the coordination number of

(i) Zinc is zinc blende (ZnS)(ii) Sodium in sodium oxide (Na2O)(iii) Oxide ion in sodium oxide (Na2O)(iv) Calcium in calcium fluoride (CaF2)

Ans. (i) 4 (ii) 4 (iii) 8 (iv) 8Q. 13. Write the coordination number of each ion in the following crystals.

(i) NaCl (ii) CsCl(iii) ZnS (iv) CaF2

(v) Na2O

Ans. (i) Na+ = 6, Cl– = 6 (ii) Cs+ = 8, Cl– = 8

(iii) Zn2+ = 4, S2– = 4 (iv) Ca2+ = 8, F– = 4

(v) Na+ = 4, O2– = 8

Chapter 2. SolutionsQ. 1. Given below is the sketch of a plant for carrying out a process.

(i) Name the process occuring in the above plant.(ii) To which container no flow of solvent takes place.(iii) Name one SPM which can be used in this plant.(iv) Give one practical use of the plant.

Ans. (i) Reverse Osmosis.(ii) Fresh water container.(iii) Parchment paper, cellulose acetate.(iv) This plant is used to get pure water from sea water in ships.

Q. 2. Refer to the given diagram and answer the following :

O

(a) What does X represent ?


(b) What do A and B represent ?(c) What is ΔTb ?

Ans. (a) Vapour pressure(b) A = V. P. of pure solvent, B = vapour pressure of solution(c) E.

Q. 3. Derive the formula of (i) degree of association, (ii) degree of dissociation in terms of vant Hoff’sfactor.

Ans. (i) Calculation of degree of association— Fraction of the total molecules which exists asassociated molecules. If the association reaction is :

nA An

Initial mole 1 0No. of mole after association (1 – α) α/n

Total mole after association (1 – α) + (α/n)

∴ vant Hoff’s factor, i =No. of mol. of solute (associated + unassociated) molecules

Total number of molecules taken

=1 – α + (α/n)

1 = 1 – α (n – 1)

n∴ Degree of association

α =(1 – i )1/n

(n – 1)(ii) Calculation of degree of dissociation—The fraction of the total molecule which undergoesdissociation. If the dissociation reaction is,

AxBy xAy+ + yBx–

Initial mole 1 nα–

Let (x + y) = n

No. of mole after dissociation (1 – α)

Total mole after dissociation = (1 – α) + nα = 1 + (n + 1) α

∴ vant Hoff’s factor, i =No. of mol. (dissociated + undissociated) molecules

Total number of mol. taken

=1 + (n – 1) α

1

= 1 + (n – 1) α

∴ Degree of dissociation

α =(i – 1)(n – 1)

Q. 4. (a) Define the term osmotic pressure. Describe how the molecular mass of a substance can bedetermined by method which is based on measurement of osmotic pressure.

(b) How is it that measurement of osmotic pressure is more widely used for determining molarmasses of macromolecules than the elevation in boiling point or depression in freezing point oftheir solution ?

Ans. Osmotic pressure (π) is defined as the extra pressure that must be applied to the solution side inorder to prevent the flow of solvent molecules into it through semi-permeable membrane.According to vant Hoff equation

π = nB

v RT

where π is the osmotic pressure, R is the gas constant and v is the volume of solution in litrescontaining nB moles of the solute.

If WB gram of solute of molar mass MB is present in the solution, then,


nB =WB

MB and we can write

π =WB × R × T

MB × V or MB = WB × R × T

π × V

Thus knowing WB, T, π and V, the molecular mass of solute MB can be calculated.

(b) The osmotic pressure method has the advantage over elevation in boiling point or depressionin freezing point for determining molar mass of macromolecules because :

(i) Osmotic pressure is measured at the room temperature and the molarity of solution is usedinstead of molality.

(ii) Compared to other colligative properties its magnitude is large even for very dilutesolutions.

Q. 5. Suggest the most important type of intermolecular interaction in the following pairs :

(i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone(v) acetonitrile (CH3CN) and acetone (C3H6O).

Ans. (i) Dispersion or London forces as both are non-polar.

(ii) Dispersion or London forces as both are non-polar.

(iii) Ion-dipole interaction.

(iv) Dipole-dipole interaction as both are polar molecules.

(v) Dipole-dipole interaction as both are polar molecules.

Q. 6. Among the following compound identify which are insoluble, partially soluble and highlysoluble in water (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform(vi) pentanol.

Ans. (i) Partially soluble as it has non-polar C6H5 group and polar –OH group which can formhydrogen bond with water.

(ii) Insoluble as toluene is non-polar while water is polar.

(iii) Highly soluble as formic acid can form hydrogen bonds with water.

(iv) Highly soluble as ethylene glycol can form hydrogen bonds with water.

(v) Insoluble as chloroform despite its polarity, cannot form hydrogen bonds with water.

(vi) Partially soluble as –OH group is polar but the bulky C5H11 part is non-polar.

Q. 7. Give three applications of Henry’s law.

Ans. Applications of Henry’s law : (i) To increase the solubility of CO2 in soda water and soft drinks,the bottle is sealed under high pressure.

(ii) To avoid the toxic effect of high concentration of nitrogen in the blood, the tanks used byscuba divers are filled with air diluted with helium (11·7% helium, 56·2% nitrogen and 32·1%oxygen).

(iii) At high altitude, low blood oxygen causes climbers to become weak and make them unableto think clearly, which are symptoms of condition known as anoxia.

Q. 8. Based on solute-solvent interaction, arrange the following in order of increasing solubility in n-octane and explain Cyclo hexane, KCl, CH3OH, CH3CN.

Ans. (i) Cyclohexane and n-octane both are non-polar. Hence they will mix completely in allproportions.

(ii) KCl is an ionic compound whereas n-octane is non-polar. Hence, KCl will not dissolve at all inn-octane.

(iii) CH3OH and CH3CN both are polar but CH3CN is less polar than CH3OH. As n-octane isnon-polar CH3CN will dissolve more than CH3OH in octane.

Therefore, the order of solubility in n-octane will be KCl < CH3OH < CH3CN < Cyclohexane.


Q. 9. The graphical representation of vapour pressure of two component system as a function ofcomposition is given alongside.

By graphical inspection, answer the following question :

(i) Are the A–B interactions weaker, stronger or of the same magnitude as A—A and B—B ?(ii) Name the type of deviation shown by this system from Raoults law.(iii) Predict the sign of Δmix H for this system.(iv) Predict the sign of Δmix V for this system.(v) Give an example of such a system.(vi) What type of azeotrope will this system form, if possible ?

Ans. (i) Stronger (ii) Negative deviation(iii) Negative (iv) Negative(v) A liquid mixture consisting of 20% acetone and 80% chloroform by mass.(vi) Maximum boiling azeotrope.


Q. 1. Discuss reactions taking place in lead storage battery ?Ans. Lead storage battery consists of 6 voltaic cells connected in series. Anode is made up of spongy

lead, cathode is a grid of lead packed with PbO2 and electrolyte is aqueous solution of H2SO4.Anode : Pb + SO4

2 – ⎯→ PbSO4 + 2e–

Cathode : PbO2 + SO42 – + 4H+ + 2e – ⎯→ PbSO4 + 2H2O

Overall reaction : Pb + PbO2 + 4H + + 2SO42 – ⎯→ 2PbSO4 + 2H2O

During recharging operation this cell behaves as electrolytic cell. The recharging reactions are :

Anode : PbSO4 + 2e – ⎯→ Pb + SO42 –

Cathode : PbSO4 + 2H2O ⎯→ PbO2 + SO42 – + 4H + + 2e–

Overall reaction : 2PbSO4 + 2H2O ⎯→ Pb + PbO2 + 4H+ + 2SO4 2 –

Fig. Lead storage battery


Q. 2. Describe the working of Galvanic cell.Ans. The device in which chemical energy is converted into electrical energy is called galvanic cell or

electrochemical cell. In galvanic cell a redox reaction is carried out in indirect manner and thedecrease in free energy during the chemical process appears as electrical energy.e.g., Zn–CuSO4 reaction is the basis of the cell reaction.A Zn strip is dipped in the ZnSO4 solution and a copper strip is dipped in the CuSO4 solutiontaken in separate beakers. The two metallic strips which acts as electrodes are connected by theconducting wires through a voltmeter and the solutions are joined by salt-bridge (made up ofKCl or KNO3 and agar-agar or gelatin).

Fig. An electrochemical or galvanic cell

Working of the cell : Zn undergoes oxidation to form Zn2+ ions.Zn(s) ⎯→ Zn2+(aq.) + 2e– (Oxidation)

The electrons liberated during oxidation are pushed through the connecting wires to copperstrips. Cu2+ ion moves towards copper strip, picks up the electrons and gets reduced to Cuatoms which are deposited at the Cu strip.

Cu2+(aq.) + 2e– ⎯→ Cu(s) (Reduction)Zn + Cu2+ ⎯→ Zn2+ + Cu (Overall reaction)

Q. 3. Consider the figure given below and the answer the question (i) to (vi)

(i) Write the direction of electron flow.(ii) Is silver plate the anode or cathode ?(iii) What will happen if salt bridge is removed ?(iv) When will the cell stop functioning ?(v) How will concentration of Zn2+ ion and Ag+ ions be affected when the cell functions ?(vi) How will the concentration of Zn2+ ion and Ag+ ions be affected after the cell becomesdead ?

Ans. (i) Electrons move from Zn to Ag.(ii) Ag is the cathode.(iii) Cell will stop functioning.(iv) When Ecell = 0.

(v) Concentration of Zn2+ ions will increase and concentration of Ag+ ions will decrease.

(vi) When Ecell = 0, equilibrium is reached and concentration of Zn2+ ions and Ag+ ions will notchange.


Q. 4. Predict the product of electrolysis of the following :

(i) An aqueous solution of AgNO3 with silver electrode.

(ii) An aqueous solution of AgNO3 with platinum electrode.

(iii) A dilute solution of H2SO4 with platinum electrode.

(iv) An aqueous solution of CuCl2 with platinum electrode.

Ans. (i) Electrolysis of aqueous solution of AgNO3 with silver electrode

AgNO3 (s) ⎯⎯⎯⎯→→→→ Ag+ (aq) + NO3– (aq)

H2O H+ + OH

–

At cathode : As Ag+ ions have lower discharge potential than H+ ions. Therefore Ag+ ions willbe deposited as Ag in preference to H+ ions.

At anode : As Ag anode is attacked by NO3– ions Ag of the anode will dissolve to form Ag+ ion in

the solution.

Ag(s) ⎯⎯⎯⎯⎯⎯⎯⎯→→→→ Ag+ (aq) + e–

(ii) Electrolysis of aqueous solution of AgNO3 with platinum electrode.

At cathode : As Ag+ ions have lower discharge potential potential than H+ ions. Therefore Ag+

ions will be deposited as Ag in p[reference to H+ ions.

At anode : As anode is not attackable, out of OH– and NO3

– ion, OH– ion will be discharge inpreference to NO3

– ions, which will then decompose to given out O2.

OH– ⎯⎯⎯⎯→→→→ OH + e

–

4OH ⎯⎯⎯⎯→→→→ 2H2O + O2

(iii) Electrolysis of dilute H2SO4 with platinum electrodes.

H2SO4 ⎯⎯⎯⎯→→→→ 2H+ + SO4

2–

H2O H+ + OH

–

(iv) At cathode :

2H+ + 2e

–⎯⎯⎯⎯→→→→ H2

H + H ⎯⎯⎯⎯→→→→ H2

At anode OH–

⎯⎯⎯⎯→→→→ OH + e–

4OH–

⎯⎯⎯⎯→→→→ 4444OH + 4e–

4OH–

⎯⎯⎯⎯→→→→ 2H2O + O2

Thus H2 is liberated at the cathode and O2 at the anode.

(iv) Electrolysis of aqueous solution of CuCl2 with platinum electrode.

CuCl2 ⎯⎯⎯⎯→→→→ Cu2+ + 2Cl–

H2O H+ + OH

–

At Cathode : Cu2+ ions will be reduced in preference to H+ ions.

Cu2+ + 2e–

⎯⎯⎯⎯→→→→ Cu

At anode : Cl– ions will be oxidised in preference to OH

– ions.

Cl– ⎯⎯⎯⎯→→→→ Cl + e

–

Cl + Cl ⎯⎯⎯⎯→→→→ Cl2

Hence Cu will be deposited on the cathode and Cl2 will be liberated at the anode.


Q. 5. The electrochemical cell given along side converts the chemical energy released during the redoxreaction

Zn(s) + Cu2+ (aq) ⎯→ Zn2+ (aq) + Cu(s)

to electrical energy. It gives an electrical potential of 1·1V when concentration of Zn2+ and Cu2+

ions is unity. State the direction of flow of current and also specify whether zinc and copper aredeposited or dissolved at their respective electrode when :

(i) an external opposite potential of less than 1·1V is applied

(ii) an external potential of 1·1 V is applied.

(iii) an external potential of greater than 1·1 V is applied.

Ans. (i) Reaction continues to take place and electrons flow from Zn electrode to copper electrode,hence current flows from Cu to Zn. Zn dissolves and copper deposits at their respectiveelectrodes.

(ii) The reaction stops and no current flows. A state of equilibrium is achieved and no change isobserved at zinc and copper electrodes.

(iii) Reaction takes place in opposite direction. Electrons flow from copper electrode to zincelectrode and hence current flows from Zn to Cu. Zinc deposits and copper dissolves at theirrespective electrodes. The cell functions as an electrolytic cell.

Chapter 4. Chemical KineticsQ. 1. Draw a graph which is used to calculate the activation energy of a reaction. Give the appropriate

expressions used to calculate the activation energy graphically.

slope =

E

a

2·303 R

Fig. Activation energy of a Reaction

Ans. According to Arrhenius equationK = Ae–Ea/RT

where K is rate constant, A is frequency factor, Ea isenergy of activation, R is gas constant and T istemperature in Kelvin.

Taking log and converting log to base 10,

We get : log10 K = log10 A – Ea

2·303 RT

This is a straight line equation. Thus, if a graph isplotted between log10 K and 1/T, a straight line isobtained with slope which is equal to :

= – Ea

Knowing the value of R we can calculate Ea graphically.

Q. 2. Explain graphically how the rate of a reaction changes with every 10°C rise in temperature.

Ans. The rate of many reactions at room temperature is approximately doubled for every 10°C rise intemperature. When heat is supplied, the energy increases the speed at which reactant moleculestravel. If the molecules travel at a greater speed when the temperature is increased, then thefrequency of collisions would increase and the reaction proceeds at a faster rate.


EEnergy

No

. o

f m

ole

cu

le

s w

ith

re

la

tive

e

ne

rg

y o

f co

llisio

n

t + 10°C

t°C

Fig. Plot of Energy vs. Collision.At t°C the fraction of molecules possesses kinetic energy greater than or equal to thresholdenergy and is capable of undergoing effective collisions. As the temperature is increased from t°Cto (t + 10)°C the whole distribution curve shifts to higher energy.This increases the fraction of molecules having energy greater than or equal to threshold energyE. The curve t°C has been displaced only a little to the right, the number of molecules with kineticenergies exceeding the threshold energy E has approximately doubled as shown by the shadedareas in the figure.

Q. 3. Describe the factors which affects the rate of reaction.Ans. The rate of reaction depends upon the following factors :

(i) Concentration of reactants : An increase in the concentration of reactants increases the rateof reaction.

(ii) Temperature : The reaction rate increases with increase in temperature.(iii) Presence of catalyst : A catalyst increases the rate of reaction without being consumed.(iv) Surface area : If the surface area of the solid reactant or the catalyst is increased, the

reactants reacts at a faster rate.Q. 4. Prove that the time taken to complete the 50% of a first order reaction is independent of the initial

concentration of the reactant.Ans. We know that, the kinetic equation for the first order reaction is :

K =2·303

t log

a(a – x) …(i)

where K = rate constant, a = initial concentration, x = concentration of the reactant which ischanged into product in time t. When 50% of the reaction is completed, then x = a/2 and t = t1/2.The eq. (i) may be written as :

K =2·303t1/2

log a

(a – a/2)

or t1/2 =2·303

K log 2

or t1/2 =2·303

K × 0·301 (·.· log 2 = 0·301)

or t1/2 =0·693

KIt is clear from the above equation that the time taken to complete the half–life of the reaction isindependent of the initial concentration.

Q. 5. Give one example each of homogeneous and heterogeneous catalysis.Ans. Homogeneous Catalysis : Presence of conc. H2SO4 in esterification reaction, is an example of

homogeneous catalysis as reactants, products and the catalyst all are in same phase (liquid).

CH3COOH + C2 H5OHconc. H2SO4

heated⎯⎯⎯⎯⎯→ CH3COOC2H5 + H2O

acetic acid ethyl alcohol ethyl acetate(ester)

Heterogeneous Catalysis : Presence of iron powder in the manufacture of ammonia by Haber’sprocess is an example of heterogeneous catalysis, where catalyst is in solid phase whereasreactants and products are gaseous.

N2 + 3H2Fe Powder / Mo 2NH3


Q. 6. Give three general characteristics of a catalyst. Name an industrial process in which a catalyst isused and name the catalyst.

Ans. There are following three characteristics of a catalyst :(i) Catalyst is unchanged chemically at the end of the reaction.(ii) The catalyst does not changes the position of equilibrium in a reversible reaction.(iii) Catalysts are very specific in respect of reaction.Haber’s process is an industrial process in which a catalyst is used. The catalyst which is used inthis process is Iron (Fe) and Molybdenum (Mo) acts as promoter.

N2 + 3H2

Fe/Mo

Δ2NH3

Ammonia

Q. 7. A graph between ln K and 1T for a reaction is given. Here K is rate constant and T is temperature

in Kelvin. If OA = a and OB = b answer the following :(i) What is the activation energy (Ea) of the reaction ?(ii) What is the frequency factor (A) for the reaction ?

ln K

Ans. (i) Slope = –OBOA =

–ba

= – Ea

K or Ea = ba R.

(ii) Intercept on y-axis = OB = b = In A or A = eb

Q. 8. The following rate data were obtained at 303 K for the following reaction.2A + B ⎯→ C + D

Experiment [A] molL–1 [B] molL–1 Initial rate of formation of D/mol L–1 mol–1

I 0·1 0·1 6·0 × 10–3

II 0·3 0·2 7·2 × 10–2

III 0·3 0·4 2·88 × 10–1

IV 0·4 0·1 2·40 × 10–2

What is the rate law ? What is the order with respect to each reactant and the overall order ? Alsocalculate the rate constant and write its unit.

Ans. Suppose order of reaction w.r.t. reactant A is m and with respect to B is n. Then the rate law willbe

Rate = K[A]m

[B]n

Substituting the values of experiments I to IV, we have

(Rate)expt I = 6·0 × 10–3 = K (0·1)m

(0·1)n

…(1)(Rate)expt II = 7·2 × 10–2 = K (0·3)m (0·2)n …(2)

(Rate)expt III = 2·88 × 10–1 = K (0·3)m

(0·4)n …(3)(Rate)expt IV = 2·4 × 10–2 = K (0·4)m (0·1)m …(4)

∴(Rate)expt I

(Rate)extpIV=

6·0 × 10–3

2·4 × 10–2 = K (0·1)

m (0·1)

n

K (0·4)m

(0·1)n

or14 =

(0·1)m

(0·4)m =

⎝⎜⎛

⎠⎟⎞1

4 m

, m = 1

(Rate)expt-II

(Rate)expt III=

7·2 × 10–2

2·88 × 10–1 = K (0·3)

m (0·2)

n

K(0·3 )m

(0·4)n


or 14 =

(0·2)n

(0·4)n =

⎝⎜⎛

⎠⎟⎞1

2 n

or ⎝⎜⎛

⎠⎟⎞1

2 2

= ⎝⎜⎛

⎠⎟⎞1

2 n

or n = 2

∴ Rate law expression is given byRate = K [A] [B]2

Order of reaction w.r.t. A = 1 Order of reaction w.r.t B = 2Overall order of reaction = 1 + 2 = 3

K =Rate

[A] [B]2

=6·0 × 10–3 mol L

–1 min–1

(0·1 mol L–1

) (0·1 mol L–1

)2

= 6·0 mol–2

L2 min–1

Q. 9. For the reaction :2NO (g) + Cl2 (g) ⎯→ 2NOCl (g)

the following data were collected. All the measurement were taken at 263 K.

Experiment Initial [NO] (M) Initial [Cl2] (M) Initial rate of disappearance of Cl2

(M) min.

1 0·15 0·15 0·60

2 0·15 0·30 1·20

3 0·30 0·15 2·40

4 0·25 0·25 ?

(a) Write the expression for rate law.(b) Calculate the value of rate constant and specify its units.(c) What is the initial rate of disappearance of Cl2 in experiments ?

Ans. Suppose order w.r.t. NO is m and order w.r.t. Cl2 is n.Then the rate will be :

Rate = K [NO]m [Cl2]nk

0·60 = K (0·15)m

(0·15)n

…(1)

1·20 = K (0·15)m

(0·30)n …(2)

2·40 = K (0·30)m

(0·15)n

…(3)Dividing equations (3) by (1), we get

2·400·60 =

K (0·30)m (0·15)n

K(0·15)m (0·15)n

4 = 2m or 22 = 2m or m = 2Dividing eq. (2) by (1), we get

1·200·60 =

K (0·15)m (0·30)n

K (0·15)m (0·15)n

2 = 2n or n = 1

(a) Rate law expression is Rate = K [NO]2 [Cl2]

(b) 0·60 mol L–1 min–1 = K (0·15 mol L–1)2 (0·15 mol L–1)

K = 177·77 mol–2 L2 min–1

(c) Rate = 177·77 mol–1 L2 min–1

× (0·25 mol L–1

)2 × (0·25 mol L–1)

= 2·778 mol L–1 min–1


Q. 10. Consider the reaction R k

⎯⎯→ P. The change in concentration of R with time is shown in thefollowing plot :

(i) Predict the order of the reaction.

(ii) Derive the expression for the time required for the completion ofthe reaction.

(iii) What does the slope of the above line indicates ?Ans. (i) The reaction R ⎯→ P is a zero order reaction.

(ii) For the reaction R k

⎯⎯→ P

rate =– d [R]

dt = K

d [R] = – k dtIntegrating both sides,

[R] = – kt + c where c is constant of integration … (1)At t = 0, [R] = [R]0This in equation (i)

C = [R]0Substituting the value of C in equation (i)

[R] = – kt + [R]0 …(2)

kt = [R]0 – [R] ⇒ t = [R]0 – [R]

k(iii) From equation (2), we have slope of curve

slope =d [R]

dt = – k

Q. 11. For a certain chemical reaction variation in the concentration ln [R] vs time plot is givenalongside(i) What is the order of the reaction ?(ii) What is the unit of rate constant K ?(iii) Give the relationship between K and t1/2 (half-life period)(iv) What is the slope of the curve ?(v) Draw the plot log [R]0 /[R] vs. time t(s)

Ans. (i) First order (ii) time–1 or s–1

(iii) K = 0·693(t1/2)

(iv) slope = – K (rate constant)

(v)


Q. 12. Write applications of Kohlrausch law.Ans. Applications of Kohlrausch’s Law :

(i) Determination of degree of dissociation of weak electrolytes :

Degree of dissociation (α) =λm

C

λ°m(ii) Calculation of molar conductivities of weak electrolyte at infinite dilution :For example : Molar conductivity of acetic acid (weak acid) at infinite dilution can be obtainedfrom the knowledge of molar conductivities of infinite dilution of strong electrolytes like HCl,CH3COONa and NaCl as shown below :

Λ°m(CH3COOH) = λ°CH3COOH– + λ°H+

= [λ°CH3COOH– + λ°Na] + [λ°H+ λ°Cl–] – [λ°Na + λ°Cl–]

Λ°m(CH3COOH) = Λ°m(CH3COOH) + Λ°m(HCl – Λ°m(NaCl)

Q. 13. Write notes on the following :(i) Mechanism of catalysis (ii) Negative catalysis(iii) Induced catalysis (iv) Auto catalysis

Ans. (i) Mechanism of catalysis : Due to the different nature of catalytic reactions, a uniformexplanation of the mechanism is not possible to give. But two main theories have beenput forward to explain the phenomenon of catalysis.(a) Intermediate compound formation theory : This theory postulates the formation

of intermediate compounds often of doubtful composition.This theory finds support from the fact that intermediate compounds in manycatalyzed reactions have actually been isolated from the reaction mixtures. Forexample Nitrosyl sulphuric acid (H2SO4.NO) in the manufacture of sulphuric acidby the lead chamber process.This theory also explains why the catalyst remains unchanged in mass andchemical composition at the end of the reaction and is effective even in smallquantity.

(b) Adsorption theory : This theory is applicable only for heterogeneous catalysis.The reactants are adsorbed at the surface of the catalyst and their highconcentration cause an increase in the frequency of collision per unit time. In thisprocess the atoms or molecules of one material becomes attached to the surface ofanother. For example, hydrogenation reaction of alkenes taking place at solid Nior Pt surfaces.

(ii) Negative Catalysis : When the rate of a reaction is retarded by the added substance, it issaid to be a negative catalyst or inhibitor and the phenomenon is termed as negativecatalysis. Negative catalysts acts in two ways :(a) In most cases –ve catalysts acts by poisoning or merely by destroying some catalyst

which already happens to be present in the reaction mixture. For example,decomposition of hydrogen peroxide (in glass tank) catalyzed by traces of alkali,but the addition of an acid would destroy the alkali catalyst.

(b) The –ve catalyst dislocates the mechanism by which the reaction would ordinarilyproceed. For example, a little sugar, alcohol or glycerin if added to a solution ofsodium sulphite, its oxidation by atmospheric oxygen is greatly retarded.

(iii) Induced catalysis : It is the phenomenon by which one reaction induces another reactionto take place rapidly. For example, sodium arsenite is not oxidized by air ordinarilywhile sodium sulphite is rapidly oxidized by the atmospheric oxygen. If air is passedthrough a mixture containing both the oxidation of sodium sulphite induces theoxidation of sodium arsenite as well.

(iv) Auto catalysis : A substance formed in the course of a reaction, sometimes acts as acatalyst. This phenomenon is known as auto catalysis. For example, when acidifiedpotassium permanganate is added to warm oxalic acid, the decolourisation of thepermanganate does not takes place readily in the beginning. But after the first portion ofthe permanganate is decolourised, the reaction becomes quite fast. The manganesesulphate formed in the reaction acts as an auto catalyst.


Chapter 5. Surface ChemistryQ. 1. Describe some features of catalysis by zeolites.Ans. Features of Catalysis by Zeolites : (i) Zeolites are hydrated alumino-silicates which have a

three-dimensional network structure containing water molecules in their pores.(ii) To use them as catalysts, they are heated so that water of hydration present in the pores islost and the pores become vacant.(iii) The size of the pores varies from 260 to 740 pm. Thus only those molecules can be adsorbedin these pores and catalysed whose size is small enough to enter these pores. Hence, they act asmolecular sieves or shape selective catalysts.An important catalyst used in petroleum industry in ZSM–5 (Zeolites sieve of molecularporosity 5). It converts alcohol into petrol by dehydrating them to form a mixture ofhydrocarbons.

Alcohols ZSM–5

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→→→→Dehydration

Hydrocarbons

Q. 2. Give four examples of heterogeneous catalysis.Ans. A catalyst in different phase than that of the reactants in a reaction is called heterogeneous

catalyst and this phenomenon is called heterogeneous catalysis. In heterogeneous catalysis,catalyst is generally a solid and reactants are usually gases (or some times liquids) someimportant industrial reactions which are heterogeneous catalytic reactions are given below :(i) Synthesis of ammonia from nitrogen and hydrogen in the presence of iron as a catalyst.

N2(g) + 3H2(g) Fe(s)

⎯⎯⎯⎯→Mo(s)

2NH3 (g)

(ii) Manufacture of sulphur trioxide (SO3) by the oxidation of sulphur dioxide (SO2) in thepresence of vanadium pentoxide (V2O5) as a catalyst.

SO2(g) + 12 O2(g)

V2O5(s)⎯⎯⎯⎯→ SO3(g)

(contact process)(iii) Dehydration of ethanol by Alumina (Al2O3).

C2H5OH (l) Al2O3(g)

⎯⎯⎯⎯⎯→ C2H4(g) + H2O (l)Ethanol Ethene

(iv) Reduction of nitrobenzene in the presence of nickel catalyst.

C6H5NO2(s) + 3H2 Ni(s)

⎯⎯⎯→ C6H5NH2(l) + 2H2O(l)Nitrobenzene Aniline

Q. 3. Physical and chemical adsorption responds differently with rise in temperature. What is thedifference and why is it so ?

Ans. Adsorption isobar for physical adsorption shows that the extent of adsorption decreases withthe increase in temperature. The adsorption isobar of chemical adsorption shows that theextent of adsorption first increases and then decreases with the increase in temperature. Theinitial unexpected increase in the extent of adsorption with temperature is due to the fact thatthe heat supplied acts as activation energy required for chemical adsorption which is muchmore than that of physical adsorption.


Q. 4. Write three distinct features of chemisorption which is not found in physisorption.Ans. Three distinct features of chemisorption which are not found in physisorption are :

(i) Enthalphy of adsorption : Enthalpy of adsorption is high (80-240 kJ mol–1) as it involveschemical bond formation.(ii) High specificity : Chemisorption is highly specific and it will only occur if there is somepossibility of chemical bonding between adsorbent or adsorbate. For example, hydrogenadsorbed by transition metal by virtue of hydride formation.(iii) Irreversibility : Chemisorption is usually irreversible in nature as it involves compoundformation. It is very slow at low temperature on account of high activation energy.

Q. 5. Discuss the effect of pressure and temperature on the adsorption of gases on solid.Ans. Effect of pressure on adsorption : At constant temperature,

the extent of adsorption of a gas (x/m) on a solid increaseswith pressure. A graph between x/m and the pressure P of agas at constant temperature is called adsorption isotherm.(i) At lower range of pressure x/m is directly proportional tothe applied pressure

xm

∝ P′ or xm

∝ Kp

(ii) At high pressure range, the extent of adsorption of a gas(x/m) is independent of the applied pressure

xm

∝ P° or xm

= K

(iii) At intermediate pressure range, the value of xm

is proportional to a fractional power of

pressure i.e.xm

∝ P1/n or xm

= KP1/n

where 1n

is a fraction. Its value may be between 0 and 1

log ⎝⎜⎛

⎠⎟⎞x

m= log K +

1n

log P

Effect of temperature on adsorption : Adsorption is generally temperature dependent. Mostlyadsorption process are exothermic and hence adsorption decreases with increasingtemperature. However, for an endothermic adsorption process, adsorption increases withincrease in temperature.

Q. 6. (a) With reference to Freundlich adsorption isotherm write the expression for adsorption ofgases on solid in the form of an equation.(b) Write an important characteristic of Lyophilic sols.(c) Based on type of particle of dispersed phase give an example each of associated colloid andmultimolecular colloid.

Ans. (a) Freundlich adsorption isotherms equation for adsorption of gases on solid.xm

= KP1/n (n > 1)

or log xm

= log K + 1n

log P

Where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure P. K and n areconstants which depends on the nature of the gas and adsorbent at a particular temperature.(b) Important characteristic of lyophilic sols.(i) They are reversible in nature i.e. once the dispersed phase is separated from dispersionmedium the sol can be made again by simply remixing with dispersion medium.(ii) They are quite stable and are not easily coagulated.


(c)Type of colloid Example

Associated colloid Surface active agents such as soap (CMC is1 0 – 4 to 10– 3 mol L–1) and syntheticdetergents.

Multimolecular colloid. Sulphur sol, gold sol.Q. 7. Consider the adsorption isotherms given along side and interpret, the variation in the extent of

adsorption (x/m) when(a) (i) Temperature increases at constant pressure.(ii) Pressure increases at constant temperature.

(b) Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia.

Ans. (a) (i) At constant pressure extent of adsorption ⎝⎜⎛

⎠⎟⎞x

m decreases with increase in temperature as

adsorption is an exothermic process.

(ii) At constant temperature, first adsorption ⎝⎜⎛

⎠⎟⎞x

m increases with increase in pressure upto a

particular presence and then it remains constant.

At low pressure xm

= kP

At intermediate range of pressure xm

= kP1/n

(n > 1).

At high pressure xm

= K (independent of pressure)

(b) Finely divided iron is used as a catalyst and molybdenum is used as promoter.


Q. 1. Copper can be extracted by hydrometallurgy but not zinc. Explain.

Ans. The E° of zinc (Zn2+/Zn= – 0·76V) is lower than that of copper (Cu2+/Cu = 0·34V). This meansthat zinc is a stronger reducing agent and can displace copper from solution of Cu2+ ions.

Zn(s) + Cu2+ (aq) ⎯⎯⎯→ Zn2+ (aq) + Cu(s)In order to extract zinc by hydrometallurgy, we need stronger reducing agents like

K (E°K

+/K = – 2·93 V), Mg. (E°

Mg2+/Mg = – 2·37V), Al (E°Al3+/Al = – 1·66 V), etc. However, all

these metals reduce water to hydrogen gas. Therefore, these metals, cannot be used to displaceZn from solution of Zn2+ions. Thus copper can be extracted by hydrometallurgy but not zinc.

Q. 2. Write chemical reactions taking place in the extraction of zinc from zinc blende.Ans. The various steps involved are :

(a) Concentration : The ore is crushed and then concentrated by Froth Floatation process.(b) Roasting : The concentrated ore is heated in the presence of excess of air at about 1200 K tofrom zinc oxide

2ZnS + 3O2 Heat

⎯⎯⎯→ 2ZnO + 2SO2

Zinc blend Zinc oxide


(c) Reduction : ZnO obtained above is mixed with powdered coke and heated to 1673 K in afire clay retort. It is reduced to zinc metal.

ZnO + C 1673 K

⎯⎯⎯⎯→ Zn + COAt 1673 K, zinc metal being volatile (b.p. 1180 K), distills over and is condensed.(d) Electrolytic refining : Impure zinc is made the anode and cathode consists of a sheet ofpure zinc. The electrolyte consists of ZnSO4 solution acidified with dil. H2SO4. On passingelectric current pure zinc gets deposited on the cathode.

Q. 3. Name the process by which chlorine is obtained as a by-product. What will happen if anaqueous solution of NaCl is subjected to electrolysis ?

Ans. In the electrolysis of NaCl by Down ’s process chlorine is obtained as by-product. This processinvolves the electrolysis of a fused mixture of NaCl and CaCl2 at 873 K. During electrolysissodium is liberated at the cathode and Cl2 is obtained at the anode.

NaCl (l) Electrolysis

⎯⎯⎯⎯⎯→ Na+ (melt) + Cl

– (melt)

Cathode : Na+ (melt) + e– ⎯⎯→ Na (s)

Anode : Cl– (melt) ⎯⎯→ Cl (g) + e–

2Cl (g) ⎯⎯→ Cl2(g)If an aqueous solution of NaCl is electrolysed H2 is evolved at the cathode and Cl2 is obtainedat the anode. The reason being that the E° of Na+/Na redox couple is much lower (E°

H2O/H2 =

– 0·83 V) and hence water in reduced to H2 in preference to Na+ ions. However NaOH isobtained in solution.

NaCl (aq) Electrolysis

⎯⎯⎯⎯⎯→ Na+(aq) + Cl

– (aq)

At cathode : 2H2O (l) + 2e – ⎯⎯⎯⎯→→→→ H2(g) + 2OH– (aq)

At anode : Cl– (aq) ⎯⎯⎯⎯→→→→ Cl(g) + e –

2Cl (g) ⎯⎯⎯⎯→→→→ Cl2(g)Q. 4. How is the concept of coupling reaction useful in explaining the occurrence of non-

spontaneous thermochemical reaction ? Explain giving an example.Ans. Couple reactions : Many reactions which are non-spontaneous (ΔG is +ve) can be made to

occur spontaneously if these are coupled with reaction having larger negative free energyCoupling means carrying out simultaneousy both non-spontaneous and spontaneous reactions.For example, decomposition of Fe2O3 into iron is a non-spontaneous reaction (ΔG° = + 1487 kJmol–1). However this decomposition can take place simultaneously (ΔG = – 514·4 kJ mol–1)

2Fe2O3(s) ⎯⎯→ 4Fe(s) + 3O2 (g) …(i)ΔG° = + 1487·0 kJ mol–1.

2CO(g) + O2(g) ⎯⎯→ 2CO2(g) …(ii)ΔG° = 514 Kj, mol– 1

Multiplying equation (ii) by and 3 then adding to equation (i), we get6CO(g) + 3O2(g) ⎯⎯→ 6CO2 (g) ΔG° = – 1543·2 kJ mol–1

2Fe2O3 (s) ⎯⎯→ 4Fe(s) + 3O2(s) ΔG° = + 1487·0 kJ mol–1 2Fe2O3 (s) + 6CO(g) ⎯⎯→ 4Fe (s) + 6CO2 (g) ΔG° = – 56·2 kJ mol–1

Since ΔG° in the reduction of Fe2O3 with CO is –ve therefore the reaction is feasible andspontaneous.

Q. 5. Write the chemical reaction which takes place in the following operations :(i) Electrolytic reduction of Al2O3 (ii) Isolation of zinc from zinc blende (iii) Mond’s process forrefining nickel.

Ans. (i) Electrolytic reduction of Al2O3

Cathode : Al3+(melt) + 3e– ⎯⎯→ AlAnode : C(s) +O2– (melt) ⎯⎯→ CO(g) + 2e –


(ii) Isolation of zinc from zinc blende.

Roasting : 2ZnS + 3O2Δ

⎯⎯→ 2ZnO + 2SO2

Zinc blende

Reduction : ZnO + C 1673 K

⎯⎯⎯→ Zn + CO(iii) Mond’s process for refining nickel

Ni + 4CO330-350K

⎯⎯⎯⎯→ Ni (CO)4

Ni (CO)4450–470K

⎯⎯⎯⎯→ Ni + 4COQ. 6. The value of ΔfG° for formation of Cr2O3 is – 540 kJ mol–1 and that of Al2O3 is – 827 kJ mol–1. Is

the reduction of Cr2O3 possible with aluminium.Ans. The two equations are :

43 Al (s) + O2 (g) ⎯⎯→

23 Al2O3 (s) ΔfG° = – 827 kJ mol–1 …(1)

43 Cr (s) + O2 (g) ⎯⎯→

23 Cr2O3 (s) ΔfG° = – 540 kJ mol–1 …(2)

Substracting eq. (2) from eq. (1), we get43 Al (s) +

23 Cr2O3 (s) ⎯⎯→

23 Al2O3 (s) +

43 Cr(s) ΔrG° = – 287 kJ mol–1

As ΔrG° of the combined redox reaction is negative, therefore, reduction of Cr2O3 by Al ispossible.

Q. 7. Free energies of formation (ΔfG) of MgO (s) and CO(g) at 1273 K and 2273 K are given below :Δf G [MgO (s)] = – 941 kJ/mol at 1273 K ΔfG [MgO (s) = – 314 kJ /mol at 2273 K

ΔfG [CO(g)] = – 439 kJ/mol at 1273 K ΔfG (CO(g) = – 628 kJ/mol at 2273 Kon the basis of above data, predict the temperature at which carbon can be used as a reducingagent for MgO (s).

Ans. At 1273 K

Mg(s) + 12 O2

(g) ⎯⎯→ MgO (s) ΔfG = – 941 kJ/mol …(i)

C(s) + 12 O2 (g) ⎯⎯→ CO(s) ΔfG = – 439 kJ/mol …(ii)

Substracting equation (i) from equation (ii), we get

MgO (s) + C(s) ⎯⎯→ Mg(s) + CO(g) ΔrG = 502 kJ/mol.

As ΔrG for the above reduction reaction is positive, therefore reduction of MgO by C is notfeasible at 1273 K.

At 2273 K Mg(s) + 12 O2(g) ⎯⎯→ MgO(s) ΔfG = – 314 kJ/mol … (iii)

C(s) + 12 O2 (g) ⎯⎯→ CO(g) ΔfG = – 628 kJ/mol. …(iv)

Substracting equation (iii) from equation (iv), we getMgO(s) + C(s) ⎯⎯→ Mg(s) + CO(g) ΔrG = – 314 KJ/mol

As ΔrG for the above reduction reaction is –ve, therefore reduction of MgO by carbon at 2273 Kis feasible and hence, carbon can be used as a reducing agent.

Q. 8. Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO ? What arethe conditions ?

Ans. Below 1683 K, i.e. the melting point of silicon, the ΔfG° curve for the formation of SiO2 liesabove the ΔfG° curve for MgO. So, at temperature below 1683 K, Mg can reduce SiO2 to Si. Onthe other hand, above 1683 K the ΔfG° curve for MgO lies above ΔfG° curve for SiO2, hence attemperature above 1683 K, Si can reduce MgO to Mg.


SiO2 + 2Mg< 1683K⎯⎯⎯→ 2MgO + Si ΔrG° = – ve

2MgO + Si > 1683K⎯⎯⎯→ SiO2 + 2Mg ΔrG° = – ve

Q. 9. Describe the process by which copper pyrites are concentrated.Ans. Copper pyrites are concentrated by Froth floatation process. The ore is crushed and put in water

in large tanks. To it pine oil is added along with ethyl xanthate. The mixture is agitated with acurrent of air when a froth is formed. The froth contains the grains of the ore whereas theimpurities settle down. The froth is collected and washed with water to get the concentrated ore.

Q. 10. Describe all the steps, with equations, to convert the concentrated ore to blister copper.Ans. The ore is roasted in the presence of air. Impurities like S, As, Sb forms volatile oxides.

S + O2 ⎯⎯→ SO24As + 3O2 ⎯⎯→ 2As2O34Sb + 3O2 ⎯⎯→ 2Sb2O3

Copper pyrites gets converted into cuprous and ferrous sulphide.2CuFeS2 + O2 ⎯⎯→ Cu2S + 2FeS + SO2

They are oxidized to ferrous and cuprous oxide.2Cu2S + 3O2 ⎯⎯→ 2Cu2O + 2SO2

2FeS + 3O2 ⎯⎯→ 2FeO + 2SO2The roasted ore is smelted in blast furnace by mixing with sand .Copper has greater affinity for sulphur than for oxygen, hence whatever Cu is formed duringroasting reacts with iron sulphide and gets reduced to copper sulphide.

Cu2O + FeS ⎯⎯→ Cu2S + FeOThe ferrous oxide combines with sand which acts as flux to form slag of ferrous silicate.

FeO + SiO2 ⎯⎯→ FeSiO3The slag floats over the molten layer of Cu2S with little FeS and is called matte.The matte is put in Bessemer converter with precalculated amount of sand and blast of air isblown when the following reactions occur :

2FeS + 3O2 ⎯⎯→ 2FeO + 2SO2FeO + SiO2 ⎯⎯→ FeSiO3

2Cu2S + 3O2 ⎯⎯→ 2Cu2O + 2SO2The cuprous oxide reacts with remaining copper sulphide to give blister copper.

Cu2S + 2Cu2O Auto-reduction⎯⎯⎯⎯⎯⎯→ 6Cu + SO2

Chapter 7. p-Block ElementsQ. 1. Give properties and structure of rhombic and monoclinic

sulphur.

Ans. (i) Rhombic sulphur : It is stable form, it is also called α-sulphur, yellow in colour, insoluble in H2O and CS2, densityis 2·07g cm– 3.

(ii) Monoclinic or ββββ-sulphur : It is stable only above 369 K,dull yellow in colour, soluble in CS2 and insoluble in H2O. Itslowly changes into rhombic sulphur. It exists as S8molecule which have puckered ring structure. It howeverdiffers in the symmetry of the crystals.

S

S

S S

S

S

S S

107°

204 pm

Fig. Puckered ring structure of sulphur

Q. 2. Describe the reactions of F2 and Cl2 with water.Ans. (i) Fluorine is highly reactive and decomposes water very readily even at low temperature and

in dark forming a mixture of O2 and O3

3F2 + 3H2O ⎯→ 6HF + O3

2F2 + 2H2O ⎯→ 4HF + O2

5F2 + 5H2O ⎯⎯⎯⎯→→→→ 10HF + O2 + O3

(ii) Chlorine decomposes water in the presence of sunlight forming halogen acid andoxoacid.

Cl2 + H2Ohv⎯→ HCl + HClO


Q. 3. How are Xenon fluorides XeF2, XeF4 and XeF6 prepared ? Deduce their structure applying VSEPRtheory.

Ans. Preparation of XeF2 , XeF4 and XeF6 : All three binary fluorides of Xe are formed by direct unionof elements under approximate experimental conditions. XeF2 can also be prepared by irradiatinga mixture of xenon and fluorine with sunlight or light from a pressure-mercury arc lamp.

Xe(g)1 Vol

+ Fe2(g)2 Vol

Δ at 675 K⎯⎯⎯⎯⎯⎯⎯→sealed Ni vessel

XeF2(g)(xenon difluoride)

Xe(g)1 Vol

+ 2F2(g)5 Vol

Δ at 75 K‚ 6 atm⎯⎯⎯⎯⎯⎯⎯→sealed Ni vessel

XeF4(s)xenon tetrafluoride

Xe(g)1 Vol

+ 3F2(g)20 Vol

Δ at 475 – 975 K⎯⎯⎯⎯⎯⎯⎯→sealed Ni vessel

XeF6(s)xenon hexafluoride

Structure of XeF2, XeF4 and XeF6. By applying VSEPR theory the shared and unshared pair ofelectrons around central atom tends to stay at maximum distance from each other as well as fromcentral atom. On this basis the structures are as follows :

Formulaof

fluoride

Centralatom

No. of shared + unshared pair of electronsaround central atom

Structure (or shape)of halide

1. XeF2 Xe 2 shared + 3 unshared pairs of e – = 5 pairs Linear molecule (3 lone pairs of eoccupy the equatorial position intrigonal bipyramid)

Xe

F

F

2. XeF4 Xe 4 shared + 2 unshared pairs of e– = 6 pairs Square Planar. .

. .

F F

FF

Xe

(octahedral with two positions ofoctahedron occupied with two lonepairs of e–)

3. XeF6 Xe 6 shared + 1 unshared pair of e– = 7 pairs Distorted octahedral shape

F

F F

F

Xe

.

.

F

F

(pentagonal bipyramidal with oneposition occupied by a lone pairresulting into distorted octahedralshape)

Q. 4. How XeO3 and XeO4 are prepared ? Describe their molecular shape.

Ans. (i) 6XeF4 + 12H2O ⎯⎯→ 4Xe + 2XeO3 + 24HF + 3O2

XeF6 + 3H2O ⎯⎯→ XeO3 + 6HF


(ii) Partial hydrolysis of XeF6 gives XeOF4

XeF6 + H2O ⎯⎯→ XeOF4 + 2HF

Molecular shape (or structure)

XeO3 has a trigonal pyramidal shape due to presence of one lone pair of electron at one corner oftetrahedron. Similarly XeOF4 has square pyramidal shape. Lone pair of electron lies opposite tooxygen atom.

Xe

O

O

O

. .

F

O

F

F F

Xe

Trigonal pyramidal Square pyramidalShape of XeO3 Shape of XeOF4

Q. 5. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms ofelectronic configuration, oxidation state and hydride formation.

Ans. (i) Electronic configuration :

8O = [He] 2s2, 2p4, 16S = [Ne]3s2, 3p4,

34Se = [Ar] 3d10,4s2,4p4

52Te = [Kr] 4d10 5s2, 5p4 and 84Po = [Xe] 4f19 5d10, 6s2, 6p4

All the elements have same ns2, np4 (n = 2 to 6) valence shell electronic configuration and henceare justified to be placed in group 16 of the periodic table.(ii) Oxidation state : They need two more electrons to form negative ions to acquire the nearestnoble gas configuration, so the minimum oxidation state of these elements should be – 2. Oxygenpredominantly and sulphur to some extent being electronegative shows an oxidation state of – 2.Since these elements have six electrons in the valence shell, therefore, at the maximum they canshow an oxidation state of +6. Other positive oxidation states shown by these elements are + 2and + 4. Although, oxygen due to the absence of d-orbitals does not show oxidation state of + 4and + 6. Thus on the basis of minimum and maximum oxidation states, these elements arejustified to be placed in the same group i.e. group 16 of the periodic table.(iii) Formation of hydrides : All the elements complete their respective octate by sharing two oftheir valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula EH2

i.e. H2O, H2S, H2Se, H2Te and H2Po. Therefore on the basis of the formation of hydrides of thegeneral formula EH2 these elements are justified to be placed in group 16 of the periodic table.

Q. 6. Illustrate how copper and zinc gives different products on reaction with HNO3.Ans. With conc. HNO3 :

2HNO3 ⎯→ H2O + 2NO2 + OCu + O ⎯→ CuO

CuO + 2HNO3 ⎯→ Cu(NO3)2 + H2O Cu + 4HNO3 ⎯⎯⎯⎯→→→→ Cu(NO3)2 + 2H2O + 2NO2 Zn + 3HNO3 ⎯→ Zn (NO3)2 + 2H

[HNO3 + H ⎯→ H2O + NO2] × 2 Zn + 4HNO3 ⎯⎯⎯⎯→→→→ Zn (NO3)2 + 2H2O + 2NO2

With dil HNO3 :2HNO3 ⎯→ H2O + 2NO + 3O[Cu + O ⎯→ CuO] × 3

[CuO + 2HNO3 ⎯→ Cu (NO3)2 + H2O ] × 3 3Cu + 8HNO3 ⎯⎯⎯⎯→→→→ 3Cu (NO3)2 + 4H2O + 2NO


Zn + 2HNO3 ⎯→ Zn (NO3)2 + 2H] × 4HNO3 + 8H ⎯→ 3H2O + NH3

NH3 + HNO3 ⎯→ NH4NO3

NH4NO3 ⎯→ N2O + 2H2O 4Zn + 10HNO3 ⎯⎯⎯⎯→→→→ 4Zn (NO3)2 + 5H2O + N2O

Q. 7. Knowing the electron gain enthalpy values of O → O– and O → O2– as – 141 and 702 kJ mol–1

respectively how can you account for the formation of a large number of oxides having O2–

species and not O– ?Ans. Let us consider the reaction of a divalent metal (M) with oxygen. The formation of M2O and MO

involves the following steps :

M(g) ΔiH1⎯⎯⎯→ M+ (g)

ΔiH2⎯⎯⎯→ M2+

(g)

O(g) ΔegH1⎯⎯⎯→ O– (g)

ΔegH2⎯⎯⎯→ O2–(g)

2M+(g) + O

–(g)

Lattice energy⎯⎯⎯⎯⎯→ M2O(s)

M2+

(g) + O2–

(g) Lattice energy⎯⎯⎯⎯⎯→ MO (s)

Although ΔiH2 is much more than ΔiH1 and ΔegH2 is much higher than ΔegH1, yet the latticeenergy of the formation of MO(s) due to higher charges is much more than that of M2O (s). Thusformation of MO(s) energetically more favourable than M2O. Due to this oxygen forms a largenumber of oxides having the O2– species and not O–.

Q. 8. Comment upon the nature of two S—O bonds found in SO2 molecule. Are the two S—O bondsin this molecule equal.

Ans.

Resonating structure of SO2– : In SO2, s is sp2 hybridised, two of the three sp2 orbitals forms two

σ-bonds with oxygen atom while the third contains the lone pair of electrons. Thus SO2 is nowleft with one half filled p-orbital and one half filled d orbital. These orbitals forms one pπ–pπ andone pπ-dπ double bond with oxygen atom. Due to resonance, the two S—O bonds have equalbond length = (143 pm).

Q. 9. Give uses of the following :(i) Hydrochloric acid (HCl) (ii) Sulphuric acid (H2SO4)

Ans. (i) Hydrochloric acid (HCl) :● It is used in the manufacture of glucose (from corn starch).● For extracting glues from bones and purifying bone black.● In the manufacture of chlorine and ammonium chloride (NH4Cl) used in dry cells.

● As a constituent of aqua regia, which is used for dissolving noble metals.● A saturated solution of zinc chloride in dil. hydrochloric acid is used for cleaning metals

before soldering or plating.● It is also used in medicine as a laboratory reagent.

(ii) Sulphuric acid (H2SO4)● It is used in the manufacture of fertilizers such as super phosphate of lime, ammonium

sulphate etc.● It is used in manufacture of pigments, paints, plastics, etc.● It is used in galvanising and metallurgical operations.● It is used in the manufacture of dye stuffs, explosives and drugs.● It is used in paper, textile and petroleum industry.● It is an important laboratory reagent.


Chapter 8. d – and f -Block ElementsQ. 1. Name the important ores of silver. Write all the steps and reactions involved in the cyanide

process for the extraction of silver from its ore.Ans. The important ore of silver are Argentite (Ag2S) and Horn silver (AgCl).

Extraction of Silver from ore by cyanide process : In this process the finely powdered ore isconcentrated by froth flotation process and then treated with dilute NaCN and a current of air ispassed into the solution so that Ag present in the ore is converted into soluble sodium argentocyanide complex.

Ag2S + 4 NaCN ⎯→ 2Na [Ag(CN)2] + Na2SSodium argento cyanidecomplex

The current of air passed into the solution oxidises Na2S formed in the above reaction intoNa2S2O3, Na2SO4 and S and thus enables the above reversible reaction to proceed in forwarddirection.

2Na2S + 2O2 + H2O ⎯→ Na2S2O3 + 2NaOH4Na2S + 5O2 + 2H2O ⎯→ 2Na2SO4 + 4NaOH + 2S

Then the solution containing Na [Ag(CN)2] is treated with Zn dust so that Ag gets precipitated.2Na[Ag(CN)2] + Zn dust ⎯→ Na2[Zn(CN)4] + 2Ag↓

ppt.Precipitated Ag is purified by electrolytic method. The cathode is pure silver, anode is block ofimpure silver and electrolyte is 6% AgNO3 with 1% HNO3. On passing current, impurities getdeposited at the anode and pure silver at the cathode. Zn and Cu remains in solution while Ausettles down as anode mud.

Q. 2. How is potassium dichromate prepared from a sample of chromite ore ? Give balanced equationsfor the chemical reactions inolved.

Ans. Preparation of potassium dichromate from chromite ore involves the following steps :(i) Conversion of chromite ore to sodium chromate by roasting finely powdered chromite ore

with sodium carbonate and quick lime in excess of air.

4FeCr2O4 + 8Na2CO3 + 7O2 CaO + air⎯⎯⎯→ 2Fe2O3 + 8Na2CrO4 + 8CO2

Chromite Sodium Ferric Sodium ore carbonate oxide chromate

(ii) Conversion of sodium chromate to sodium dichromate by the action of conc. H2SO4.2Na2CrO4 + H2SO4 ⎯→ Na2Cr2O7 + Na2SO4 + H2O

conc. Sodiumdichromate

(iii) Conversion of sodium dichromate to potassium dichromate by treating it with KCl solution.Na2Cr2O7 + 2KCl ⎯→ K2Cr2O7 + 2NaCl

Potassiumdichromate

From resulting solution potassium dichromate is crystallised.Q. 3. In the extraction of zinc from zinc blende :

(i) Give an equation to show how zinc oxide is converted to zinc.(ii) How is impure zinc finally electro-refined ?

Ans. (i) ZnOZinc oxide

+ Ccoke

heated

⎯⎯⎯→ Znzinc

+ CO ↑

(ii) Electrorefining of zinc :Electrolyte used : ZnSO4 solution containing a little H2SO4

Anode : Block of impure zinc.Cathode : Thin sheet of pure zinc.On passing current pure zinc is obtained at cathode.

Anode reaction : Znimpure

– 2e– ⎯→ Zn2+aq

Cathode reaction : Zn2+aq + 2e– ⎯→ Zn

pure metal


Q. 4. What are characteristics of the transition elements ?Ans. The important characteristics of transition metals are :

(i) All transition elements are metallic in nature.(ii) These metals exhibit variable oxidation states.(iii) Transition metal atoms or ions generally forms the complexes with neutral, negative and

positive ligands.(iv) Compounds of transition metals are usually coloured.(v) The compounds of these metals are usually paramagnetic in nature.(vi) Transition metals and their compounds act as good catalysts.(vii) These metals forms various alloys with other metals of the series.(viii) These metals form interstitial compounds with C, N, B and H.

Q. 5. Give preparation of KMnO4 from pyrolusite.Ans. Potassium permanganate is prepared by fusing MnO2 or pyrolusite with KOH or K2CO3 in

presence of atmospheric oxygen or any other oxidizing agent e.g., KNO3. The fused mass is greendue to the formation of potassium manganate.

2MnO2 + 4KOH + O2Δ

⎯⎯→ 2K2MnO4 + 2H2Oor 2MnO2 + 2K2CO3 + O2 ⎯⎯→ 2K2MnO4 + 2CO2

PotassiumManganate

The fused mass is extracted with water and treated with a current of ozone, chlorine or CO2when K2MnO4 is oxidized to potassium permanganate. The purple solution thus obtained isconcentrated when it deposits, dark purple, needle like crystals of KMnO4 are formed.

2K2MnO4 + O3 + H2O ⎯⎯→ 2KMnO4 + 2KOH + O2

or 2K2MnO4 + Cl2 ⎯⎯→ 2KMnO4 + 2KClCO2 neutralizes the alkali formed and facilitates the reaction

3K2MnO4 + 2CO2 ⎯⎯→ 2KMnO4 + 2K2CO3 + MnO2

Q. 6. Give oxidizing properties of KMnO4 in different mediums.Ans. KMnO4 acts as a very powerful oxidizing agent in acidic, neutral and alkaline media. The

equations representing oxidation in these media are given below :In acidic medium

2KMnO4 + 3H2SO4 ⎯⎯→ K2SO4 + 2MnSO4 + 3H2O + 5[O]

or MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O [E° = + 1.52 V]

In neutral or alkaline medium

2KMnO4 + H2O ⎯⎯→ 2KOH + 2MnO2 + 3[O]

or MnO4– + 2H2O + 3e– ⎯⎯→ MnO2

+ 4OH– [E °= + 1.23 V]In acidic medium potassium permanganate oxidizes :(a) Ferrous salts to ferric salts

MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O

[Fe2+ ⎯⎯→ Fe3+ + e–] × 5

MnO4– + 5Fe2+ + 8H+ ⎯⎯⎯⎯⎯⎯⎯⎯→→→→ Mn2+ + 5Fe3+ + 4H2O

(b) Oxalates to CO2

[MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O] × 2

[C2O42– ⎯⎯→ 2CO2 + 2e–] × 5

2MnO4– + 5C2O4

2– + 16H+ ⎯⎯⎯⎯⎯⎯⎯⎯→→→→ 2Mn2+ + 10CO2 + 8H2O

(c) Iodides to iodine

[MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O] × 2

[2I– ⎯⎯→ I2 + 2e–] × 5

10I– + 2MnO4– + 16H+ ⎯⎯⎯⎯⎯⎯⎯⎯→→→→ 2Mn2+ + 5I2 + 8H2O


(d) Sulphites to sulphates

[MnO4– + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O] × 2[SO3

2– + H2O ⎯⎯→ SO42– + 2H+ + 2e–] × 5

5SO32– + 2MnO4

– + 6H+ ⎯⎯⎯⎯⎯⎯⎯⎯→→→→ 2Mn2+ + 5SO42– + 3H2O

Q. 7. Describe the gradation in the following physio-chemical properties of transition metals :

(i) Ionic radii (ii) Metallic character (iii) Complex formation.

Ans. (i) Ionic radii : In general ionic radii decrease with increase of oxidation state. Since transitionmetals exhibit various oxidation states, the radii of ions are different in different oxidationstates. However, for the same oxidation state, ionic radii decreases with increase in thenuclear charge in a given transition series.

(ii) Metallic character : All transition metals are metallic in nature. They show gradual decreasein electropositive character on moving from left to right. The metallic bonds in transitionmetals are strong due to greater effective nuclear charge and large number of valenceelectrons.

(iii) Complex formation : The transition metals forms many coordination complexes which isattributed to the following reason :(a) Small size.(b) High charge density of the ions.(c) Presence of vacant d-orbitals of appropriate energy which can accept lone pair of

electrons donated by other groups (ligands).Q. 8. How can you separate alumina from silica in a bauxite ore associated with silica ? Give equations

if any.Ans. The bauxite ore containing SiO2 as an impurity is leached with sodium hydroxide at 473 – 523 K

and 35 – 36 bar pressure. Al2O3 forms sodium aluminate while SiO2 forms sodium silicate.Al2O3 + 2 NaOH + 3H2O ⎯⎯→ 2Na[Al(OH)3]

SiO2 + 2NaOH + 2H2O ⎯⎯→ Na2[Si(OH)6]The sodium aluminate in solution is neutralized by CO2 gas and hydrated Al2O3 is precipitate.

2Na[Al(OH)4] + CO2 ⎯⎯→ Al2O3.xH2O + 2NaHCO3

The sodium silicate remains in the solution.Al2O3.xH2O on heating gives pure Al2O3.

Al2O3.xH2O1470 K

⎯⎯⎯→ Al2O3(s) + xH2O↑

Q. 9. Compare the general characteristics of first series of Transition metals with those of second andthird series metals in the respective vertical columns on the basis of the following points :(i) Electronic configuration (ii) Oxidation state (iii) Ionisation enthalpies (iv) Atomic sizes.

Ans. (i) Electronic configuration : The elements in the same vertical column generally have similarelectronic configuration. Although the first series shows only two exceptions i.e. Cr = 3d54s1 andCu = 3d10 4s1 but the second series shows more exceptions of Mo(42) = 4d5,5s1, Tc (43) = 4d65s1, Ru(94) = 4d75s1, Rh = (4s), 4d185s1 Pd(46) = 4d105s1, Ag(47) = 4d105s1. Similarly, in the third series W(74) = 5d46s2, Pt (78) = 5d9 6s1 and Au (79) = 5d10 6s1. Hence in the same vertical column in anumber of cases, the electronic configuration of the three series are not similar.(ii) Oxidation states : The elements in the same vertical column generally shows similaroxidation state. The oxidation state shown by the elements in the middle of each series ismaximum and minimum at the extreme ends.(iii) Ionisation enthalpies : The first ionisation enthalpies in each series generally increasesgradually as we move from left to right though some exceptions are observed in each series. Thefirst ionisation enthalpies of some elements in the second (4d) series are higher while some ofthem have lower value than the element in 3d series in the same vertical column. However thefirst ionisation enthalpies of third (5d) series are higher than those of 3d and 4d series. This isbecuase of weak shielding of nucleus of 4f-electron in the 5d series.(iv) Atomic sizes : Generally ions of the same charge on atoms in given series shows progressivedecrease in radius with increasing atomic number though the decrease is quite small. But the size


of the atoms of the 4d series is larger than the corresponding elements of the 5d-series are nearlythe same as those of 4d-series due to lanthanoid contraction.

Q. 10. Compare the chemistry of actinoids with that of lanthanoids with special reference to :(i) Electronic configuration (ii) Oxidation states(iii) Atomic and ionic sizes (iv) Chemical reactivity

Ans. (i) Electronic configuration : The general electronic configuration of lanthanoids is [Xe]54 4f1–14

5d0–16s2 whereas that of actinoids is [Rn]86 5f1–14 6s0–1 7s2. Thus lanthanoids belongs to 4f-serieswhereas actinoids belongs to 5f-series.(ii) Oxidation states : Lanthanoid shows limited oxidation states (+ 2, + 3, + 4) out of which + 3,is most common this is because of large energy gap between 4f, 5d and 6s subshells. On the otherhand actinoids shows a large number of oxidation states because of small energy gap between 5f,6d and 7s subshells.(iii) Atomic and ionic sizes : Both shows decrease in size of their atoms or ions in + 3 oxidationstate in lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is calledactinoid contraction. However the contraction is greater from element to element in actinoidsdue to poorer shielding by 5f-electrons.(iv) Chemical reactivity :Lanthanoids : In general the earlier members of the series are quite reactive (similar to calcium)with increasing atomic number, they behave more like aluminium value for E° for the half-reaction

Ln3+ (aq) + 3e– ⎯→ Ln (s)are in the range of – 2·2 to – 2·4 V except for Eu for which the value is – 2·0V. This is of course asmall variation.The metals combine with hydrogen when gently heated in the gas. The carbides Zn3C, Ln2C3 andLnC2 are formed when the metals are heated with carbon which liberates hydrogen from diluteacid and burns in halogen to form halides. They form oxide and hydroxides M2O3 and M(OH)3.The hydroxides are definite compound not just hydrated oxides basic like alkaline earth metaloxides and hydroxide.Actinoids : The actinoids are highly reactive metals, especially when finely divided. The action ofboiling water on them. For example, gives a mixture of oxides and hydrides and combinationwith most non-metals takes place at moderate temperature. Hydrochloric acid attacks all metalbut most are slightly affected by nitric acid leading to the formation of protective oxide layers.Alkalies have no action.

Chapter 9. Coordination CompoundsQ. 1. For the complex ion of [Fe(CN)6]3– :

(i) Show the hybridization diagrammatically.(ii) Is it an inner orbital complex or an outer orbital complex ?

(iii) State its magnetic property.Ans. (i) In [Fe(CN)6]3– oxidation state of Fe is + 3

3d 4s 4pFe (Z = 26) atom inground state

↑↓ ↑ ↑ ↑ ↑ ↑↓

Fe3+ ion ↑ ↑ ↑ ↑ ↑

d2sp3 hybridization inpresence of CN– ligands

↑↓ ↑↓ ↑

Six d2sp3 hybrid orbitals(ii) It is an inner orbital complex.(iii) The ion is paramagnetic due to the presence of an unpaired electron.

Q. 2. Explain Werner’s postulates for bonding in coordination compounds.Ans. The main postulates of Werner’s theory of coordination compounds are as follows :

(i) Metal possess two types of valencies :(a) Primary valency which are ionisable.(b) Secondary valency which are non-ionisable.


(ii) Primary valency is satisfied by the negative ions and it is that which a metal exhibits in theformation of its simple salts. In terms of modern theories based on electronic configuration,the primary valency is now referred to as ‘oxidation state‘ of a metal.

(iii) Secondary valencies are satisfied by neutral molecules or anions and sometimes by cations,also. It is called unionisable valency. Molecules or ions satisfying secondary valency do notionise in complex.

For example, in CoCl3·6NH3, valency between Co and Cl is primary valency and valencybetween Co and NH3 is secondary valency. Here six ammonia molecules linked to Co bysecondary valencies are directed to six corners of a regular octahedron and thus accounts for itsstructure as follows :

NH3

H3N

NH3

3+

ClCo

H3N NH3

NH3

3–

Q. 3. Explain briefly the main features of the valence bond theory (VBT).

Ans. Valence bond theory is based on the following principles :

(i) In coordination compounds usually sp3, dsp2 or d2sp3 type of hybridisation of central metalatoms or ions are common due to which coordinate entities which shows tetrahedral,octahedral or square planar geometry.

(ii) Bonding between the ligands and the central metal atom or ion is a coordinate bonding.

(iii) Relationship between the observed magnetic behaviour of the coordinate complex and thebond type.

Q. 4. Using valence bond approach, explain shape and magnetic behaviour of [Ni (NH3)6]2+ (given AtNo. of Ni = 28).

Ans. Oxidation state of nickel is +2 in the given complex. The electronic configuration of Ni atom inground state and Ni2+ ion are :

3d 4s 4p 4d

Ni Groundstate

Ni2+ ion

Ni2+ in[Ni (NH3)6]2+ ×× ×× ×× ×× ×× ××

sp3d 2 hybridization 6 pairs of

electrons from six NH3 ligands

Nickel undergoes sp3d2 hybridization and its shape is octahedral. The complex is paramagneticdue to presence of two unpaired electrons.

Q. 5. Using valence bond approach, explain the shape and magnetic behaviour of [Fe (H2O)6]2+. [At.no. of Fe = 26]. Draw the structure.

Ans. Oxidation state of Iron (Fe) is + 2 in the given complex. The electronic configuration of Fe atom inground state and Fe2+ are :

3d 4s 4p 4d

Fe atom inGround state

↑↓ ↑ ↑ ↑ ↑ ↑↓


3d 4s 4p 4d

Fe2+ in excited state ↑↓ ↑ ↑ ↑ ↑

3d 4s 4p 4d

Fe2+ in [Fe(H2O)6]2+ ↑↓ ↑ ↑ ↑ ↑ ×× ×× ×× ×× ×× ××

sp3d2-Hybridization outerorbital complex

[Fe(H2O)6]2+ is sp3d2-hybridized and therefore, have octahedral geometry. The complex isparmagentic due to the presence of four unpaired electrons.

H2O

H2OH2O

H2O

Fe

H2OH2O

Octahedral Geometry

Q. 6. Draw structures of [Co(NH3)6]3+, [Ni (CN)4]2– and [Ni(CO)4]. Write the hybridization of atomicorbitals of the transition metal in each case.

Ans. In [Co(NH3)6]3+ cobalt is present as Co3+ and its coordination number is six.

27Co atom = 1s2, 2s2, 2p6, 3s2, 3p6, 3d7, 4s2.

Co3+ ion = 1s2, 2s2 2p6, 3s2, 3p6, 3d6

3d 4s 4p

Hence,

Co3+ ion is complex ion : ·· ·· ·· ·· ·· ··

d2sp3 hybridization

H3NCo 3+

NH3

NH3

H3N

NH3

NH3

CoNH3

NH3

3+

OrH3N

H3N NH3

NH3

Structure of [Co(NH3)]3+ is octahedral and it is an inner orbital complex.In [Ni (CN)4]2– nickel is presented as Ni2+ ion and its coordination number is 4.

28Ni atom = 1s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2.Ni2+ ion = 1s2, 2s2, 2p6, 3s2, 3p6, 3d8

3d 4s 4p

Ni2+ ion

Ni2+ ion in complex ion ·· ·· ·· ··

dsp2 hybridizationSo structure of [Ni (CN)4]2– ion is square planar.

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤N ≡ C

N ≡ C

Ni C ≡ N C ≡ N

2–

In [Ni (CO)4], nickel is present as Ni atom i.e., its oxidation number is zero and coordinationnumber is 4.


Ni atom = 1s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2

3d 4s 4p

Ni in complex ·· ·· ·· ··

sp3 hybridizationSo structure of [Ni(CO)4] is tetrahedral.

CO

CO

CO

CO

Q. 7. Discuss structural isomerism in coordination compounds.Ans. In structural isomerism, structural isomers are those isomers having different arrangement of

ligands around the central metal atom. Its various types are :(i) Ionisation isomer : When compounds having same composition but yield different ions in

solution : e.g.,[Co(NH3)5Br]2+ SO4

2– and [Co(NH3)5SO4]+ Br–

Pentaammine bromo cobalt (III) Pentaammine sulphatesulphate (violet) Cobalt (III) bromide (red)

(ii) Co-ordination isomerism : This type of isomerism is shown by compounds in which bothcation as well as anion are complexes . e.g.,

[Co(NH3)6] [Cr (CN)6] and [Cr (NH3)6] [Co (CN)6](iii) Linkage isomers : This type of isomerism results from two possible ways of attachment of a

ligand to the central atom. e.g.,[Co (NH3)5NO2]Cl2 and [Co (NH3)5ONO]Cl2

Pentaammine nitro Pentaammine nitritocobalt (III) chloride cobalt (III) chloride

(iv) Hydrate isomerism : This type of isomerism arises because of capability of water moleculesto appear in a variety of ways. e.g.,

[Cr (H2O6)]Cl3 [Cr(H2O)5Cl]Cl2·H2O and [Cr (H2O)4Cl2]Cl·2H2OHexaaquochromium Pentaaquochloro Tetra aquadichloro

(III) chloride chromium (III) chloride chloride dihydrate(violet) monohydrate (blue green) (green)

(v) Coordination position isomerism : This type of isomerism is exhibited by bridgedcomplexes and results from different attachment of ligands. e.g.,

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

(NH3)4 CoOH

Cl

Co (NH3)2 Cl2 SO4 and

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

Cl (NH3)3 CoOH

Cl

Co (NH3)3 Cl SO4

(vi) Polymerization isomerism : This is not a true isomerism. It is used to denote compoundhaving same empirical formula but different molecular formula. e.g. [Pt (NH3)2Cl2] and [Pt (NH3)4] [PtCl4].

Q. 8. What is crystal field splitting ? Draw figure to show splitting of degenerated d–orbitals in anoctahedral crystal field.

Ans. In an octahedral coordination entity due to repulsion between the electrons in metal d–orbitalsand the electrons (or negative charges) of the ligands, the degeneracy of d-orbitals is lost and theysplit up into two levels of different energy.


dx2 – y2, dz2

Average energy d -orbitals in

Δo

t2g

←⎯

⎯⎯

→

------------------------------------

spherical fieldof

eg

dxy, dyz, dzx

The splitting of degenerated levels is termed as crystal field splitting and the difference of energybetween these levels is known as CFSE (Crystal Field Splitting Energy). It is denoted by Δo.

Q. 9. A metal complex having composition Cr (NH3)4Cl2Br has been isolated in two forms (A) and (B).The form (A) reacts with AgNO3 to give a white precipitate readily soluble in dilute aq.ammonia, whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write theformula of (A) and (B) and state the hybridization of chromium in each. Calculate their magneticmoments (spin only value).

Sol. A metal complex having composition Cr(NH3)4Cl2Br have two forms A and B. The form ‘A’ giveswhite ppt. with AgNO3, hence it must have chloride ion in the form of non-complex ion i.e.,outside the complex sphere as [Cr (NH3)4ClBr]Cl.

[Cr (NH3)4ClBr]Cl + AgNO3 ⎯→ AgCl ↓ + [Cr(NH3)4ClBr] + NO3–

Form ‘A’ white ppt.

The precipitate of AgCl is soluble in NH4OH due to formation of complex saltAgCl + 2NH4OH ⎯→ [Ag (NH3)2Cl] + 2H2O

white ppt. Complex salt

Similarly form ‘B’ gives pale yellow precipitate of AgBr which are sparingly soluble in NH4OH.Hence form ‘B’ is [Cr(NH3)4Cl2]Br.

AgNO3 + [Cr(NH3)4Cl2]Br ⎯→ AgBr ↓ + [Cr(NH3)4Cl2]+ + NO3–

(Pale yellow ppt.)

AgBr + 2NH4OH ⎯→ [Ag(NH3)2Br] + 2H2OIn both complexes, chromium is present as central metal ion and its oxidation number is +3.Electronic configuration of Cr is as follows :

24Cr = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1

Cr3+ = 1s2, 2s2, 2p6, 3s2, 3p6 3d3.Number of ligands are six and Cr3+ shows d2sp3 hybridization in both complexes A and B.

3d 4s 4p

·· ·· ·· ·· ·· ··

d2sp3 hybridizationHence, in it number of unpaired electrons are 3. So magnetic moment

(μ) = √⎯⎯⎯⎯⎯⎯n (n + 2) (n = no. of unpaired electrons)(μ) = √⎯⎯⎯⎯⎯⎯3 (3 + 2) = √⎯⎯15 = 3·872 BM.

Chapter 10. Haloalkanes and HaloarenesQ. 1. A hydrocarbon C5H10 does not reacts with chlorine in dark but gives a single monochloro

compound C5H9Cl in bright sunlight. Identify the hydrocarbon.Ans. Since the hydrocarbon gives only one monochloro compound and all hydrogen atoms are

equivalent. Also the compound does not reacts with Cl2 in the dark therefore it cannot be analkene. Thus the compound is cyclopentane.

+ Cl2 ⎯⎯→hν

Cl

+ HCl

Cyclopentane 1-chloro cyclo pentane

Q. 2. What are ambident nucleophiles ? Explain with an example.

Ans. Nucleophiles such as cyanide (: –C ≡ N:) and nitrite ion ( )

˙˙..

: O — N = O :–

which can attack

nucleophilic centre from two sites are called ambident nucleophiles. These nucleophiles canattack from either of these sites depending upon the reaction conditions and the reagent used e.g;


CH3CH2Cl + KCN Alc.

⎯⎯→Δ

CH3CH2 — CN + KCl (propane nitrile)

CH3CH2Cl + AgCN Alc.

⎯⎯→Δ

CH2—CH2NC + AgCl (Ethyl isocyanide)

Q. 3. Write a note on Williamson’s synthesis.

Ans. When an alkyl halide (R—X) is heated with sodium alkoxide (R—O—Na), an ether is obtained. Inthis reaction halide (–X) of alkyl halide is replaced by an alkoxy group (—OR). This reaction isknown as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethersand mixed (or unsymmetrical) ethers.(i) Simple (symmetrical) ethers : When an alkyl halide and sodium alkoxide having similar

alkyl groups are heated symmetrical ether is obtained e.g., when ethyl bromide is heated withsodium ethoxide, diethyl ether is formed.

C2H5—O—Na + C2H5—Br Δ

⎯⎯→ C2H5—O—C2H5 + NaBr

Sodium ethoxide Bromoethane Diethyl ether

(ii) Mixed (unsymmetrical ether) : When an alkyl halide and sodium alkoxide having differentalkyl groups are heated, unsymmetrical ether is obtained e.g. when methyl bromide is heatedsodium ethoxide, ethyl methyl ether is formed.

C2H5 —O—Na 3 —Br Δ

⎯⎯→ C2H5—O—CH3 + NaBrSodium ethoxide

bromideMethyl Ethyl methyl ether

+ CH

Q. 4. How is Grignard reagent prepared ?Ans. Grignard reagent is an alkyl magnesium halide, R—Mg—X obtained by the reaction of alkyl

halide R—X with magnesium (Mg) in dry ether.

R—X + Mg dry ether⎯⎯⎯⎯→

δ–R—

2δ+Mg—

δ–X

Alkyl magnesium halide

When an alkyl halide like CH3I is added from a dropping funnel to a flask containing pieces ofpure Mg, dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH3—Mg—I isformed.

CH3—I + Mg dry ether⎯⎯⎯⎯→

δ–CH3—

2δ+Mg—

δ–I

Methyl Methyl magnesium iodide

iodide

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesiumiodide.

C2H5I + Mg dry ether⎯⎯⎯⎯→ C2H5—Mg—I

Ethyl magnesium

iodide

Q. 5. Illustrate Wurtz reaction.Ans. When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher

alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.

R– X + 2Na + X –Rether

⎯⎯⎯→alkaneR–R + 2NaX

In this reaction the alkyl radicals from two molecules of the reacting alkyl halides combines orcouple to form the higher alkane. Thus, methyl bromide reacts with sodium in ether to formethane (C2H6), while ethyl bromide under the same conditions forms butane (C4H10 ).

2CH3—Br + 2Naether

⎯⎯⎯→ CH3—CH3 + 2NaBrMethyl bromide Ethane

2CH3—CH2—I + 2Naether

⎯⎯⎯→ CH3—CH2—CH2—CH3 + 2Nal Ethyl iodide Butane

However, when a mixture of two different alkyl halides is treated with sodium in presence of dryether, a mixture of alkanes is formed i.e., self coupling products are formed.


CH3Br + Na + C2H5Br dry ether⎯⎯⎯⎯→ C2H6 + C4H10 + C3H8 + NaBr

Major Major

C2H6 and C4H10 are self coupling products.

Chapter 11. Alcohols, Phenols and EthersQ. 1. Why are phenols more acidic than alcohols ? Explain.Ans. Out of alcohol and phenol, phenols are more acidic because :

(i) Phenol is a resonance hybrid of the following contributing structures :

←⎯→ ←⎯→ ←⎯→ ←⎯→

[I]

++

––

–

— — — — — —— —

H — O H — O H — O H — O H — O

[II] [III] [IV] [V]

Due to resonance, oxygen acquires some positive charge (II to IV). It will strongly attract theshared electron of O—H bond towards itself and H+ ion is released.

H—O

+ H+

:: O ˙

Phenol

–

(ii) Phenoxide ion is a resonance stabilised ion.

––

←⎯→ ←⎯→ ←⎯→ ←⎯→

[VI]

+

––

–

— — — — — —— —

O O O O O

[VII] [VIII] [IX] [X]

Phenoxide ion is stabilised to a greater extent than phenol due to the dispersal of –ive charge.Thus phenoxide ion is readily formed due to the dissociation of phenol. On the other hand,neither alcohol nor alkoxide ion is resonance stabilised. Moreover alkyl group due to its + I effectincreases the electron density on oxygen in alkoxide ion and thus destabilises it .

R — O — H –

Not resonancestabilised (Not resonance stabilised

but destabilised)

. .. .> H R O + +

Thus alcohol do not dissociate to give H+ ion. Hence phenols are more acidic than alcohols.Q. 2. How can we distinguish between primary, secondary and tertiary alcohols ?

Ans. The primary, secondary and tertiary alcohols can be distinguished using the following tests :

Lucas test : This test is based upon relative reactivities of various alcohols towards HCl in thepresence of ZnCl2. In this test, the alcohol is treated with Lucas reagent which is equimolarmixture of HCl and ZnCl2. Alcohols are soluble in Lucas reagent and form a clear solution. Onreaction, alkyl chlorides are formed which being insoluble results in cloudiness in the solution.

R—O—H + H—Cl ZnCl2⎯⎯⎯→ R—Cl + H2OIf cloudiness appears immediately, tertiary alcohol is indicated.If cloudiness appears within five minutes, secondary alcohol is indicated.If cloudiness appears only upon heating, primary alcohol is indicated.Oxidation : The alcohol is treated with sodium dichromate in sulphuric acid at roomtemperature. Identification of the oxidation product gives information regarding the type ofalcohol.


Primary alcohols gives a carboxylic acid containing the same number of carbon atoms.Secondary alcohols gives a ketone containing the same number of carbons. However, on furtheroxidation carboxylic acids with less number of carbon atoms are obtained.In the above two cases the orange colour of sodium dichromate in conc. sulphuric acid changes togreen.Tertiary alcohols do not react under these conditions.Dehydrogenation (Reaction with Hot Reduced Copper)

Primary, secondary and tertiary alcohols gives different products when their vapours are passedthrough a tube packed with reduced copper at 575 K.(i) A primary alcohol is dehydrogenated to aldehyde :

R—

H|C|H

—O—H

Primary alcohol(1°)

Cu/575 K⎯⎯⎯⎯⎯→Aldehyde

R—C|H

= O + H2

(ii) A secondary alcohol is dehydrogenated to ketone :

R—

H|C|H

—OH

Secondary alcohol(2°)

Cu/575 K⎯⎯⎯⎯⎯→ R—C|H

= Ketone

O + H2

(iii) Tertiary alcohol does not dehydrogenates due to absence of α-hydrogen. However, it getsdehydrated to form an alkene.

CH3—

CH3|C—OH|CH3

tert-Butyl alcohol

(3°)

Cu/575 K⎯⎯⎯⎯⎯→

H H

C = C CH3 CH3

Isobutylene

+ H2

Q. 3. Starting from phenol prepare :(a) Salicylic acid(b) Salicylic acid-Aspirin, Salol, oil of winter green

Ans. (a) When sodium phenoxide is heated with CO2 at 400 K and at 4-7 atm pressure, sodiumsalicylate is formed as the major product, which on acidification yields salicylic acid. Thisreaction is known as Kolbe’s reaction.

OH|

NaOH⎯⎯⎯→

ONa|

400 K4 – 7 atm

+ CO2

⎯⎯⎯⎯⎯⎯→

OH| COONa

H+

⎯⎯⎯→

OH| COOH

Sod. Sodium 2–Hydroxybenzoic acidPhenoxide Salicylate (Salicylic acid)

⎝⎜⎛

⎠⎟⎞

MajorProduct

(b) Salicylic acid is the starting material for the manufacture of 2-acetoxybenzoic acid (aspirin),a well known analgesic.

OH| COOH

+ (CH3CO)2OEthanoic anhydride

Conc. H2SO4⎯⎯⎯⎯⎯→

OCOCH| COOH

3

+ CH3COOHAcetic acid

Salicylic 2-Acetoxy benzoicacid acid (Aspirin)

I


Salicylic acid can be employed to prepare salol, an intestinal antiseptic, by heating it with phenolin the presence of phosphoryl chloride.

Salicylic acid

OH| COOH

+ C6H5OH POCl3⎯⎯→ Phenyl salicylate

(Salol)II

OH| COOC H6 5

+ H2O

Methyl salicylate, a pleasant smelling liquid, also called oil of winter green, is used in perfumeryand as a flavouring agent can also be prepared from salicylic acid.

OH| COOH

+ CH3OH reflux

⎯⎯→HCl

OH| COOCH3

+ H2O

Salicylic Methyl salicylateacid (Oil of Winter green)

IIIQ. 4. How will you prepare :

(i) Phenol from benzene, chlorobenzene and benzene diazonium chloride.(ii) `Benzene from phenol.

Ans. (i) From Benzene : Phenol is prepared from benzene by the direct oxidation at 600 K inpresence of V2O5.

Benzene2C6H6 + O2

V2O5

600 K⎯⎯⎯→

Phenol2C6H5OH

From Chlorobenzene : In this method benzene is first converted into chlorobenzene byheating a mixture of benzene HCl and air at 50 K over a catalyst (CuCl2 + FeCl3). Thechlorobenzene, thus formed is then oxidized to phenol by heating with superheated steamat 750 K.

Benzene2C6H6 + 2HCl +

(air)O2

CuCl2 + FeCl3

500 K⎯⎯⎯⎯⎯⎯⎯→

Chlorobenzene2C 6H5Cl + 2H2O

C6H5Cl + H2O750 K

⎯⎯⎯⎯→ C6H5OH + HClChloro Super Phenolbenzene heated

steam

From Benzene diazonium chloride : When benzene diazonium chloride is heated with H2O,phenol is formed.

N+2Cl–

|H2OBoil

⎯⎯⎯→

OH|

+ N2 + HCl

Benzene diazonium PhenolChloride

(ii) Benzene from phenol : When phenol vapours are passed over heated zinc dust, benzene isformed.

OH|

+ ZnΔ

⎯⎯→ + ZnO

Phenol Benzene

Q. 5. Give reactions for the preparation of alcohols from :

(i) Alkenes

(ii) Grignard reagent


Ans. (i) Alcohols are prepared from alkenes in the following ways :(a) By hydration of alkenes :

CH3 —CH = CH 2 + H2 O ⎯⎯→ CH3 — CH—CH 3

( )Markonikovaddition

|OHPropan–2–ol

H+

(b) By hydroboration – Oxidation

CH3—CH = CH2 + BH3 H2O‚ 3H2O2

OH– ⎯⎯⎯⎯⎯→ (CH3CH2CH2)3 B H2O2

OH–⎯⎯⎯→ 1–Propanol

CH3CH2CH2OH

(ii) From Grignard reagent :

(a) HCHO + CH3CH2MgBr H2O

⎯⎯⎯→ H—

OH|C—|H

CH2CH3 + MgBr(OH)

1–Propanol(1° Alcohol)

Formaldehyde on reaction with Grignard’s reagent yields a 1° alcohol.

(b) CH3CHO + CH3MgBr H2O⎯⎯→ H3C

H

C

OH CH3

+ MgBr(OH)

2° Alcohol

2–Propanol

(Isopropyl alcohol)

Any other aldehyde on reaction with Grignard reagent yields a 2° alcohol.

(c)CH3 CH3

C = O + CH3MgBr H2O⎯⎯→

C

OH CH3

CH3

CH3

Tert butyl alcohol

(2–Methyl Propan–2–ol)

(3° alcohol)

Any ketone on reaction with Grignard’s reagent yields a 3° alcohol.


Q. 1. (a) Sodium bisulphite is used for the purification of aldehydes and ketones. Explain.

(b) Treatment of C6H5CHO with HCN gives a mixture of two isomers which cannot beseparated even by very careful fractional distillation. Explain :

Ans. (a)

C = O + NaHSO3 ⎯→ C

OH

SO3NaSod. bisulphite adduct

CH

C = O + NaHSO3 ⎯→H

CH3

C

OH

SO3NaH

3

Sod. bisulphite adduct


The addition products are crystalline solids. These can be decomposed by mineral acids oralkalies to give back the original aldehyde or ketone. Therefore this reaction is used forpurification of aldehydes or ketones.

(b)

C6H5C = O + HCN ⎯→ C6H5 — C*—OH + C H — C *—OH|H

|H

|CN

|CN(d–) (l–)

6 5|H

These two isomers are enantiomers and therefore cannot be separated by physical methods likefractional distillation.

Q. 2. An organic compound contains 69·77% carbon, 11·63% hydrogen and rest oxygen. The molecularmass of the compound is 86. It does not reduces Tollen’s reagent but forms an additioncompound with sodium hydrogen sulphite and gives positive iodoform test. On vigorousoxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Ans. The molecular formula of the compound is

Element Percentage Mole ratio Relative ratio Formula

1. C 69·77 69·7712 = 5·84

5·8141·16 = 5·0 C5H10O

2. H 11·63 11·631 = 11·63

11·631·16 = 10 M. Mass = 86

3. O 18·60 18·616 = 1·16

1·161·16 = 1

According to the given data, the compound is methylketone and its probable structure wouldbe :

CH3 —C

O||

—CH2 — CH2 — CH3, Pentan-2-one; Molecular mass = 86

The chemical reactions are :

CH3 —C

O||

—CH2 — CH2 — CH3 + NaHSO3 ⎯⎯→ CH3—C—CH2 — CH2 — CH3 |

OH

|OSO2Na

CH3 — C

O||

—CH2 — CH2—CH3 + 4I2 + 6NaOH ⎯⎯→ CHI3(Yellow)

ppt.

+ C2H5CH2COONa + 5H2O + 5NaI

Q. 3. Account for the following :(a) Chloroacetic acid has lower pka value than acetic acid.(b) Electrophilic substitution in benzoic acid takes place at meta position.(c) Carboxylic acid have higher boiling point than alcohol of comparable molecular masses.

Ans. (a) Chloroacetic acid is stronger acid than acetic acid, therefore chloroacetic acid has lower pka

(b) Because of resonance in benzoic acid, electron density is quite high at meta position

⊕

⊕ +

– – –—

— — — —— —

——

——

————

OHOHOHOH

←⎯→ ←⎯→ ←⎯→

O O O O

C C C C


⎯⎯⎯⎯⎯⎯→ —

—

—

COOHCOOH

NO2

+ H2OHNO3/H2SO4

(c) Both alcohol and carboxylic acid have hydrogen bonding. But hydrogen bonds amongcarboxylic acid molecules are more and stronger than among alcohols. Consequently carboxylicacids have higher boiling point than alcohols.

R — O — H |H

R R — COH – – – – O

O – – – – HOC —R

_ _ _ _ O —

Q. 4. What is the effect of substituent in the carbon chain on the acidic strength of carboxylic acid ?Ans. The substituent attached to the carboxylic acid, will increase its strength if -

(i) The substituent facilitates the release of H+ ions easier from acid.(ii) The substituent should be helpful in providing the resonance stabilization to the carboxylate

ion, which is formed after the release of H+ ion from the acid.The substituent attached to the carboxylic acid will decrease its strength if :(i) It makes the release of H+ ions difficult from the acid.(ii) It decreases the resonance stabilization of the carboxylate ion.

Q. 5. Give outline steps involved in the preparation of CH3COOH from acetylene.Ans. Acetic acid is manufactured by acetylene (Ethyne). Ethyne is firstly converted into acetaldehyde

by the action of 20% H2SO4 in presence of 1% HgSO4.

HC ≡ HCEthyne + H2O

20% H2SO4⎯⎯⎯⎯→1% HgSO4

CH3CHOAcetaldehyde

Acetaldehyde may be converted into acetic acid by the oxidation in presence of manganousacetate at 70°C.

2CH3CHO + O2 (CH3COO)2Mn

⎯⎯⎯⎯⎯→70°C

2CH3COOHAcetic acid

To convert acetaldehyde into acetic acid, the concentrated solution of acetaldehyde is taken in astainless vessel along with a small quantity of manganous acetate. A current of air is bulked inthe vessel. Manganous acetate prevents conversion of acetaldehyde into peracetic acid. With thehelp of steam coils, the temperature of vessel is maintained between 50–70°C. Under theseconditions acetaldehyde reacts with oxygen to form acetic acid. The 90–95% acetic acid obtainedfrom this method can be purified further by distillation.

Q. 6. Discuss the main properties of formic acid.Ans. Formic acid is a colourless liquid of irritating smell. Its boiling point is 100·8°C. It is miscible with

water, ether as well as ethanol. It causes burns and blisters on the skin because of its corrosivenature.Formic acid gives the reactions of aldehyde group as well as carboxyl group due to the followingunique structure.

H—

O||C—OH

Aldehyde group Carboxyl group

(i) Salt formation : When it is treated with alkalies carbonates or bicarbonates, it gives salt.HCOOH + NaOH ⎯→ HCOONa + H2O

Sodium formate

(ii) Reaction with PCl5 : On treating with PCl5, it gives formyl chloride, which decomposesimmediately into carbon monoxide and hydrogen chloride.

HCOOH + PCl5 ⎯→ HCOCl + POCl3 + HClFormyl chloride

HCOCl ⎯→ HCl + CO↑formyl chloride


(iii) Reaction with alcohols : Forms esters

HCOOH + CH3OHΔ

HCOOCH3 + H2O Methanoic acid Methanol Methylmethanoate

(iv) Reaction with sulphuric acid : It decomposes to carbon monoxide and water.

HCOOHMethanoic acid

H2SO4⎯⎯→

ΔH2O + CO ↑

(v) Reducing properties : It is a strong reducing agent. It reduces Tollen’s reagent and Fehling’ssolution.(a) HCOO– + 2Cu+2 + 5OH– ⎯→ CO3

2– + Cu2O↓ + 3H2O Formate ion From Fehling solution Red ppt.

(b) HCOO– + aq (NH3)2 OH ⎯→ CO32– + 2Ag↓ + 2H2O

From Tollen’s reagent Silver mirror

(c) With acidified KMnO4 Solution : It reduces purple coloured permanganate ion tocolourless manganese (II) ion.

2MnO4 + 11H+ + 5COO-⎯→ 2Mn+2 + 5CO2 ↑ + 8H2O

Q. 7. What happens when :(i) Formic acid is treated with ammonia and the product so formed is heated.(ii) Sodium formate is heated and the product is treated with dil. H2SO4.

Ans. (i) When formic acid is treated with ammonia, ammonium formate is formed, which on heatinggives formamide.

HCOOH + NH3 ⎯→ HCOONH4 Δ

⎯→ HCONH2 + H2OAmmonium formate Formamide

(ii) When sodium formate is heated upto 360°C, it gives sodium oxalate and hydrogen. Whensodium oxalate is treated with dil. H2SO4, oxalic acid is obtained.

HCOONa+

HCOONa

360° C⎯⎯→

–H2

COONa|COONa

dil. H2SO4⎯⎯⎯→

COOH|COOH

sodium formate Sodium oxalate Oxalic acid

Q. 8. Compare properties of formic acid and acetic acid.Ans. Formic Acid Acetic Acid

Points of Resemblance

1. Forms salts with bases and alkali. Forms salts with bases and alkali.

2. Forms ester with alcohol in presence ofconc. H2SO4.

This also forms ester with alcohols.

3. Forms formyl chloride with PCl5. Forms acetyl chloride with PCl5.

4. Ammonium salt on heating givesformide.

Ammonium salt on heating gives acetamide.

Points of difference1. Decomposes to CO andH2O in presence

of conc. H2SO4.Stable to heat and conc. H2SO4.

2. Reacts with chlorine to form CO2 + HCl. Chloroacetic acid is formed.3. Gives formaldehyde when calcium salt is

heated

Ca COOH COOH

⎯→ HCHO + CaCO3

Calcium acetate gives acetone

CH3COO CH3COO

Ca Δ

⎯→ CH3 CH3

CO + CaCO3

4. Gives sodium oxalate and hydrogen whensodium formate is heated.

No action of heat on sodium acetate.


5. Produces hydrogen when sodium formateis heated.

HCOONa + NaOHSodalime

⎯⎯⎯→Δ

Na2CO3 +H2

Produces methane with sodalime .

CH3COONa + NaOH Δ

⎯⎯⎯→Sodalime

CH4 + Na2CO3

6. Kolbe’s synthesis yields hydrogen. Kolbe’s synthesis yields ethane.

7. Reduces ammonical silver nitrate

solution.

No action with ammonical silver nitrate

solution.

Q. 9. What happens when oxalic acid reacts with :(i) NH3 (ii) PCl5

(iii) Acidic KMnO4 (iv) Glycerol at 110° C.

Ans. (i) Forms oxamide COOH|COOH

+2NH3⎯⎯⎯→

COONH4|COONH4

Heat

⎯⎯→– 2H2O

CONH2|CONH2

Oxalic acid Ammonium oxalate Oxamide

(ii) Forms phosgene with phosphorous pentachloride.COOH|COOH

PCl5⎯→

COCl|COOH

⎯→ COCl|COCl

⎯→ CO + COCl2

Oxalic acid Oxalylchloride Phosgene

(iii) Decolourises an acidic solution of KMnO4 (test)2KMnO4 + 3H2SO4 + 5(COOH)2 ⎯→ K2SO4 + 2MnSO4 + 8H2O + 10CO2 ↑

(iv) CH2OH + HOOC—COOH|CHOH|CH2OH

110°C

⎯⎯→

CH2O—OC—COOH|CHOH|CH2OH

+ H2O

Glycerol Glycerolmonoxalate

–CO2 ↓ 110°CCH2OH|CHOH + HCOOH|CH2OH

+H2O

←⎯⎯120° C

CH2OOCH|CHOH|CH2OH

Glycerol Glycerolmonoformate

Q. 10. What happens when-(i) Grignard’s reagent reacts with bubbling carbon dioxide ?(ii) Olefins reacts with carbon monoxide at 570K – 675K in presence of phosphoric acid ?

Ans. (i) When ethereal solution of Grignard’s reagent reacts with bubbling CO2, carboxylic acid isobtained.

CH3MgBr + CO2 ⎯→ ⎣⎢⎢⎡

⎦⎥⎥⎤

CH3—

O||C—OMgBr

H2O/H+⎯⎯⎯→ CH3—

O||C—OH + MgBr(OH)

Acetic Acid

(ii) When olefins reacts with CO and steam under pressure at 570 K-675 K in presence ofphosphoric acid, higher fatty acids are obtained.

CH2 = CH2 +Ethylene

CO + H2O 570–675 K

⎯⎯⎯⎯→H3PO4

CH3CH2COOHPropanoic acid

CH3CH = CH2Propene

+ CO + H2 570–675 K

⎯⎯⎯⎯→H3PO4

(CH3)2 CHCOOH iso-butyric acid

Q. 11. Arrange the following in increasing order of acidity and explain your order-Formic acid, Acetic acid, Monochloroacetic acid.

Ans. Acidity of any acid depends upon two factors :(i) The ease with which the H+ ion is released.(ii) After the release of H+, the stability of carboxylate ion.


In monochloroacetic acid – Cl atom has – I effect, i.e., it has a tendency to withdraw the electronstowards its own side. Therefore in this acid H+ ion is released easily as well as –ve charge onchloroacetate ion is reduced due to – I effect of –Cl atom which increases its stability. Thus itwould be strongest acid.

Cl ← CH2 ←⎯ CO

O–

In acetic acid +I group (methyl group) is present, therefore it would push the electrons away i.e.,towards the carbon of carboxyl group. Due to this effect H+ is released with difficulty. Moreoverthe –ve charge on acetate ion is further increased by – CH3 group, thus the stability of carboxylateion is decreased hence, it would be weakest acid.

The acidity of formic acid will be in between chloroacetic acid and acetic acid because it does nothave any +I group. Thus the order of increasing acidity will be-

CH3COOH < HCOOH < ClCH2COOH⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→Increasing acidity

Q. 12. What happens when:(i) Ammonium cyanate is heated ?(ii) Benzoic acid is treated with CaO.(iii) Benzoic acid is treated with a mixture of conc. nitric acid and conc. sulphuric acid.

Ans. (i) When ammonium cyanate is heated, it rearranges and forms urea.

NH4CNO Rearrangement⎯⎯⎯⎯⎯→ NH2.CO.NH2

Ammonium Ureacyanate

(ii) C6H5COOH + CaO ⎯→ C6H6 + CaCO3

Benzoic acid Benzene Calcium carbonate

(iii) Benzoic acid on treatment with conc. H2SO4 forms m-nitrobenzoic acid.COOH|

Benzoic acid +

Conc.HNO3

Conc. H2SO4⎯⎯⎯⎯⎯⎯⎯→

NO2m -nitrobenzoic acid

COOH|

Chapter 13. Organic Compounds Containing NitrogenQ. 1. (a) Give possible explanation for each of the following :

(i) The presence of a base is needed in the ammonolysis of alkyl halide.(ii) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

(b) Write the IUPAC name of CH3 — N — C — CH3|C2H5

||O

Ans. (a) (i) To remove the HX produced during the reaction.(ii) Aryl halides do not undergo nucleophilic substitution with potassium phthalimide.

(b) N—Ethyl — N—methyl ethanamide.Q. 2. Arrange the following :

(a) In decreasing order the pKb values.C2H5NH2, C6H5NHCH3. (C2H5) NH and C6H5NH2

(b) In decreasing order of basic strengthC6H5NH2, C6H5N(CH3), (C2H5) NH and CH3NH2.

Ans. (a) C6H5NH2 > C6H5 NHCH3 > C2H5NH2 > (C2H5)2 NH.(b) (C2H5)2NH > CH3NH2 > C6H5NH2 > C6H5 N (CH3)2


Q. 3. Describe amines and its classification.Or

What are amines ? How are they classified ?Ans. Amines may be regarded as amino derivatives of hydrocarbons. They may also be treated as the

derivatives of ammonia which are formed by the replacement of one or more hydrogens ofammonia by corresponding number of alkyl groups.

NH3↓

H + RR–NH2(Primary)

2H +2RR2NH

(Secondary)

3H +3RR3N

(Tertiary)

↓ ↓↓

The classification of amines is based upon the number of alkyl groups attached to the nitrogenatom. If one alkyl group is attached with nitrogen along with two hydrogens, it is known asprimary (1°) amine. In the same way, if two alkyl groups and one hydrogen is attached withnitrogen, it is called secondary (2°) amine and if only three alkyl groups are attached withnitrogen, it is called tertiary amine. Hence primary amine contains (–NH2) amino group,secondary amine contains (> NH) amino group and tertiary amine contains (> N–) tertiarynitrogen atom.

CH3—NH2, C2H5.NH 2

CH3 CH3

NH C2H5 CH3

NH

1° amine 2° aminesCH3 CH3

N—CH3,C2H5 C2H5

N—CH3

3° aminesQ. 4. Explain, oxidation in aniline at different conditions.Ans. Aliphatic amines slowly oxidized in air to form N-oxides

CH3CH2NH2(O)⎯→ CH3CH2—NH2

↓O

Ethylamine N-oxide

Aromatic amines are easily oxidizes on exposure to air or on treatment with oxidants to formcoloured complex compounds.

Aniline K2Cr2O7/H2SO4⎯⎯⎯⎯⎯⎯⎯→ Aniline black

(A black dye)

Controlled oxidation with K2Cr2O7, gives p-benzoquinone.

—NH2

Aniline

K2Cr2O7/H2SO4⎯⎯⎯⎯⎯⎯→Controlled oxidation

O = =O

p-benzoquinone

Q. 5. What happens when aniline undergoes bromination, nitration and sulphonation ?Ans. (i) Bromination : In aniline benzene ring is highly activated, with aqueous bromine all the

ortho and para positions gets substituted resulting in the formation of 2, 4, 6-tribromoaniline.

NH 2

Aniline

+ 3Br2 ⎯→ Br

NH2

|Br

+ 3 HBr

2, 4, 6-tribromoaniline

——Br

(ii) Nitration : Nitric acid is a strong oxidizing agent. Thus, direct treatment of aniline withHNO3 leads to oxidation of highly activated benzene ring resulting in the formation ofcomplex oxidation products. Before nitration, aniline is first acetylated to protect —NH2group. Acetanilide on treatment with nitric acid in the presence of sulphuric acid givesortho and para nitroacetanilide which on acidic hydrolysis gives ortho and para nitroaniline.


NH2

Aniline

+ CH3COCl Base

NHCOCH3

+ HCl

Acetanilide

288 K

HNO3/H2SO 4

NHCOCH3

NO 2p-nitroacetanilide

+

NHCOCH 3

—NO2

o-nitroacetanilide

H2O/H+ —CH 3 COOHNH2

|

NO2p-nitroaniline (major)

NH2|

—NO2

o-nitroaniline (Minor)

⎯→ ⎯⎯⎯⎯⎯→

⎯→

(iii) Sulphonation : Aniline reacts with conc. H2SO4 to form the salt, which on subsequently baking at455—475 K gives sulphanilic acid.

NH2

Aniline

+ H 2SO4 ⎯→

NH+HSO4

Anilinium hydogen sulphate

455 – 475 K⎯⎯⎯⎯⎯→Baking –H2O

NH 2

SOHp-aminobenzene-sulphonic acid

(sulphanilic acid)

NH3

SO3–

Zwitter ion structure

⊕4

–

Chapter 14. BiomoleculesQ. 1. Give one example of a fibrous protein. Name the final product of hydrolysis of proteins.

What is denaturation of proteins ?Ans. Myosin in muscles is a fibrous protein.

On hydrolysis of proteins, amino acid is obtained as the final product.Denaturation of proteins : When proteins are heated or subjected to the action of alkali etc. theirphysical and biological properties changes drastically without any change in their chemicalnature. This process is called denaturation of proteins.

Q. 2. Despite having an aldehyde group :(a) Glucose does not give 2, 4-DNP test. What does this indicates ?(b) Draw the Haworth structure of α-D-(+) – Gluco-pyranose.(c) What is the significance of D and (+) here ?

Ans. (a) Despite having —CHO group glucose does not undergoes certain characteristic reactions ofaldehyde due to the formation of the ring structure there is no free —CHO group.(b)

H

HH

H

H

———

——

O

OH

OH

α – D (+) – Glucopyranose

H — C1 — OH

—

H — C2 — OH

—

HO — C3 — H O

—

H — C4 — OH

—

H — C5

OH

OH

C

CH2OH

CH2OH(c) D means it has configuration like D-Glyceralydehyde (+) means it is dextro rotatory(optically) active.


Q. 3. What are the products obtained on complete hydrolysis of DNA ? Write down the names andstructures of pyrimidine and purine bases present in DNA.

Ans. Complete hydrolysis of DNA yields the following products :(i) Pentose sugar-deoxyribose.(ii) Pyrimidine and purine bases—cytosine, thymine, adenine and guanine.(iii) Phosphoric acid.Pyrimidine bases present in DNA are :

O

N

HThymine

CH3HN

O

N

HCytosine

2HN

O

N

Purine bases present in DNA are :

N|H

NAdenine

2NH

N N N

H

N

Guanine

O

HN N

2H N

Q. 4. Why DNA and RNA are called as nucleic acids ? Enumerate the differences between DNA andRNA.

Ans. Nucleic acids are long chain polymers and they are present in the nuclei of cells. RNA and DNAare two types of nucleic acids.

DNA RNA

1. It is deoxyribose nucleic acid. 1. It is ribose nucleic acid.

2. It contains deoxyribose sugar. 2. It contains ribose sugar.

3. It has a double helix structure. 3. It has single helix structure.

4. It contains thymine as pyrimidine base. 4. It contains uracil as pyrimidine base.

5. It can replicate. 5. It cannot replicate.Q. 5. Write three such reactions of glucose which cannot be explained by an open chain structure of

glucose molecule, What alternative structures has been proposed for glucose molecule ?Ans. The following reactions cannot be explained by the open chain structure of glucose.

(i) Despite of having aldehyde group, glucose does not gives 2, 4-DNP test Schiff’s test and itdoes not forms the hydrogen sulphite addition product with NaHSO3.(ii) The pentacetate of glucose does not reacts with hydroxyl amine indicating the absence of free–CHO group.(iii) D-glucose on treatment with methyl alcohol in the presence of dry HCl gas gives twoisomers, methyl α–D glucoside and methyl β-D glucoside. These glucosides do not reducesFehling’s solution and also do not react with hydrogen cyanide indicating the absence of free-CHO group. A ring structure called pyranose structure (α-and β) is proposed for glucosemolecule.

Q. 6. Comment on the specificity of enzyme action. What is the most important reason for theirspecificity ?

Ans. In case of enzymatic reaction, the enzyme is so built that it binds to the substrate in a specificmanner. Enzymatic reaction involves the following steps :Step I : Binding of substrate (S) to enzyme (E) to form a complexE + S ⎯⎯→ [ES] (Enzyme–Substrate complex).Step II : Product formation in the complex.[ES] ⎯⎯→ EP (Enzyme–Product complex


Step III : The dissociation of enzyme product complex, leaving the enzyme unchanged.EP ⎯→ E + P

The specificity of the enzyme is due to the presence of some specific regions called the activesites, on their surface. The shape of active sites is such that only a specific substrate can fit into,in the same way as one key can open a particular lock. This specific binding leads to theformation of an enzyme substrate complex which accounts for the high specificity of enzymecatalysed reaction. This most accepted model is popularly known as lock and key model . Forexample urease catalyses only the hydrolysis of urea and none of the several thousand otherenzymes present in the cell catalyses the reaction.


Q. 1. What are thermoplastics and thermosetting plastics ? Give one example of each kind.Ans. Thermoplastics : These are linear polymers. They soften on heating and harden on cooling.

Intermolecular forces in them are intermediate between those of elastomers and fibrouspolymers. Example : Polypropylene.Thermosetting plastics : These plastics are made by heating relatively low molecular mass semi-fluid polymers. These polymers become infusible and forms a hard mass on heating in a moulddue to cross linking. Once moulded, thermosetting plastics cannot be remoulded or reshaped byheating. Example : Bakelite made from urea - formaldehyde resins.

Q. 2. (a) What is the role of benzoyl peroxide in polymerization of ethene ?(b) What is LDPE and HDPE ? How are they prepared ?

Ans. (a) It acts as a chain initiator.(b) LDPE stands for low density polythene. It is prepared as follows :

nCH2 = CH2High Temp. High pres.

0·01% O2 CH2—CH2 n

LDPE⎯⎯⎯⎯⎯⎯⎯⎯⎯→ [ ]

HDPE stands for high density polythene. It is prepared as :

nCH2 = CH2 CH2—CH2 nHDPE

⎯⎯⎯⎯⎯⎯⎯⎯⎯→ [ ]

Ziegler Natta catalyst(C2H5)3 + TiCl4

Benzene 330 K, 5-6 atm

Q. 3. Give one example each of addition and condensation polymer. Name the monomers in eachcase.

Ans. Addition polymer : e.g., —[ CF2—CF2—] n

Teflon

Monomer : CF2 = CF2 TetrafluoroethyleneCondensation polymer : e.g.,

O CH2 CH2 O C

O

C

O

nDacron (Terylene)

Monomer :HO—CH2—CH2—OHEthane–1, 2–diol(Ethylene glycol)

CO

HO C OHO

+

Terephthalic Acidor (Benzene 1, 4-dicarboxylic acid)

Q. 4. Give the classification of polymers :(i) On the basis of their mode of synthesis(ii) On the basis of nature of forces between the macromolecules give suitable examples of each

case.


Ans. (i) There are two modes of synthesis for polymers :(a) Addition polymers(b) Condensation polymers

(ii) The mechanical properties of a polymer depends upon the intermolecular forces among thepolymer chain. On the basis of magnitude of intermolecular forces polymers may beclassified into four categories :(a) Elastomers e.g., rubber-synthetic or natural(b) Fibers e.g., nylon, dacron.(c) Thermoplastics e.g., polyethene, polystyrene(d) Thermosetting polymers e.g., bakelite

Q. 5. What are polyolefins ? Give the reaction for the preparation of polyethene, a polyolefin.Ans. Polymers obtained from unsaturated hydrocarbons (olefins) are called polyolefins.

Polythene is prepared by the addition polymerisation of ethylene by two methods :(i) Polymerisation at high pressure :

nCH2 = CH21000 –�5000 atm

520 K,O2Ethylene⎯⎯⎯⎯⎯⎯→

Low density polythene( CH2 = CH2 )

(ii) Polymerisation at low pressure :

nCH2 = CH26-7 atm and 333 –�443 K

Ziegler-Natta catalystEthylene High density polythene( CH2 = CH2 )⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Q. 6. What is a polyamide ? How is Nylon-6 synthesized ?

Ans. Polymers which have amide linkage (—

O||C—N

|H

—) in the chain are called polyamide.

Nylon-6 is obtained from the monomer caprolactum. Caprolactum is obtained from cyclohexaneaccording to the reaction sequence as follows—

Oxidation

O

⎯⎯⎯⎯→NH2OHH2O

⎯⎯⎯⎯→

NOH

Cyclohexane Cyclohexanone Cyclohexanoneoxime

Beckmannconc. H2SO4⎯⎯⎯⎯⎯→

Caprolactum

rearrangement

HON

Caprolactum on heating with traces of water hydrolysis to 6-aminocaproic acid which oncontinued heating undergoes self-condensation and polymerizes to give nylon-6.

|O

Caprolacum6-Amino caproic acid Nylon-6

n

N

H

H2O—HN—(CH2)5—C—Δ

H2N (CH2)5COOH⎯⎯→Polymerise

Δ⎯⎯⎯→ (

O

)||

Q. 7. What type of polymerisation takes place when a polyester is formed ? Give one example of apolyester and name the monomers from which it is formed.

Ans. Condensation polymerisation, having ester linkage. e.g., Terylene.It is an example of polyester

nHO—CH2—CH2—OH + nHO—

O||C— —

O||C—OH

Ethylene glycol Terephthalic Acid

425 – 475 K↓

– nH2O

(O—CH2—CH2—O—

O||C— —

O||C) n

Terylene


Q. 8. Write notes on :(i) Elastomers (ii) Fibres(iii) Thermoplastic polymers (iv) Thermosetting polymers

Ans. (i) Elastomers : In elastomers the intermolecular forces are very weak thus the elastomericcharacter is high. Hence the molecule can be stretched under stress and take its originalform when the stress is removed. Weak van der Waal’s forces are present among thepolymer chain. Such polymers can be stretched over ten times of their normal length.Elastomers have the randomly coiled molecular chains of irregular shape which have fewcross links. The examples of elastomers are natural rubber, vulcanized rubber and buna-S.When gummy rubber (poor elasticity) is heated with sulphur (3–5%) in vacuum, some crosslinks among the chains develops which would increase the elasticity. This process is knownas vulcanization.

Natural rubber Vulcanized rubber

Fig. Structure of Natural and Vulcanized rubber

(ii) Fibres : In this type of polymers the inter molecular forces are strongest. These forces arelike H-bonds or dipole–dipole interactions. Due to these strong forces, fibers show hightensile strength and low elasticity. They are thin, long and thread like and can be woven intofabrics. They have high melting points and low solubility. e. g., nylon (H—bonding ispresent), polyester (dipole-dipole interaction is present).

C

O

H

H

N N

N

C

O

H

C

O

H

NC

O

Fig. Bond between the chain of nylon-6 6.

(iii) Thermoplastic Polymers : In thermoplastic the inter–molecular force of attraction is in between those ofelastomers and fibers. These are linear polymers and canbe moulded into desired shape by heating and thencooling. This softening on heating and rigidness oncooling can be repeated many times without any changein mechanical properties and chemical composition ofplastic. Example—polyethene, polystyrene, polyvinylchloride, teflon etc.

(iv) Thermosetting Polymers : These are the low molecularweight, semifluid type polymers, which becomes veryhard, infusible and insoluble on heating. Excessive crosslinking among the polymer chains takes place on heatingand a three dimensional solid network is formed.Thermosetting polymers can not be remelted again likethermoplastic polymers. Examples—bakelite, urea-formaldehyde resin, melamine formaldehyde resin, etc.

Q. 9. How is bakelite synthesized ? Give its uses.

Ans. Bakelite is a condensation polymer of phenol and formaldehyde. When phenol is treated withformaldehyde in presence of a base, it gives linear or cross linked phenol formaldehyde resin orbakelite. During the reaction a methylene bridge is formed either at o-or p-position or both at o-orp-positions with respect to the phenolic group.

OH|

+ HCHOOH–⎯→

OH|

—CH2OH

-Hydroxy methyl phenol

+

OH|

CH2OHp-Hydroxymethyl phenol

n

OH|

CH2OH OH–

⎯⎯⎯⎯⎯→Polymerization

|

(i)

—OH|

OH|

—CH2 — —CH2 — + ( + 1)H2O

Linear polymer (Soft bakelite)

Phenol o

n


(ii) n

OH|

CH2OH—+

|

CH2OH|

OH

n OH–

⎯⎯⎯⎯⎯→Polymerization

OH|

CH2—

—

CH2

CH2

OH|

CH2

CH|

2 CH|

2|

|OH |

OH

CH2

|OH

CH2—

—

|

Cross-linked polymer (hard bakelite)

OH|

—

—

CH2

|

|

n

Uses of Bakelite :1. It is used for making plugs, switches, telephone cases and other electrical fittings.2. It is used for making handles of tea kettles, sauce pans and for making artificial leather.3. Radio and television casings are made of bakelite.

Chapter 16. Chemistry in Everyday Life

Q. 1. List two major classes of antibiotics with examples of each class.Ans. Two major classes of antibiotics are bactericidal and bacteriostatic. Examples are :

(a) Bactericidal (b) Bacteriostatic(i) Penicillin (i) Erythromycin(ii) Aminoglycosides (ii) Tetracycline(iii) Ofloxacin (iii) Chloramphenicol

The full range of micro-organisms attacked by an antibiotic is called its spectrum. Based onspectrum antibiotics can be divided into two classes (a) Broad spectrum antibiotics e.g.chloramphenicol (b) Narrow spectrum antibiotics e.g. pencillin G.

Q. 2. Label the hydrophilic and hydrophobic part in the following compounds.

(a) CH3(CH2)10 CH2OSO3– Na+

(b) CH3(CH2)15 —N+ (CH3) Br–

(c) CH3(CH2)16 COO (CH2CH2O)4 CH2CH2OH

Ans. (a) CH3—(CH2)10 — SO3– Na+

Hydrophobic part Hydrophilic part

(b) CH3(CH2)15 — N+ (CH3) Br–


(c) CH3(CH2)16 — COO (CH2CH2O)4 CH2CH2OH


Q. 3. Following types of non-ionic detergent are present in liquid detergents, emulsifying agents andweeting agents label the molecule. Identify the functional groups (s) present in the molecule.

C9H19 — — O (CH2CH2O)x CH2CH2OH (x = 5 to 10)

Ans. C9H19— — O (CH2CH2O)x—CH2CH2—OH

Hydrophobic Hydrophilic or polar part

or non-polar part

Functional group present in the molecule are

(i) ether (ii) alcohol


Q. 4. Answer the following questions :

(i) Dish washing soap are synthetic detergents. What is their chemical nature ?

(ii) What is the basic difference between antiseptics and disinfectants ?

(iii) Which category of the synthetic detergent is used in toothpaste ?

Ans. (i) They are non-ionic detergents.

(ii) Antiseptics are applied to living tissues whereas disinfectants are applied to non-livingobjects.

(iii) Anionic detergents.

Q. 5. Answer the following questions :

(i) Which class of drugs is used in sleeping pills ?

(ii) What is the commonality between the antibiotic Arsphenamine and azodyes ?

(iii) Pickles have a long shelf-life and do not get spoiled for months why ?

Ans. (i) Tranquilizers.

(ii) Arsphenamine possesses—As = As—Linkage that resembles —N = N— linkage in azodyes.

(iii) Plenty of salt and cover of oil acts as preservatives. These do not allow bacteria to grow onthem.

Q. 6. Write the chemical equation for preparing sodium soap from glyceryl oleate and glycerylpalmitate. Structural formula of these compounds are given below.

(i) (C15H35COO)3C3H5—Glyceryl palmitate

(ii) (C17H32COO)3 C3H5—Glyceryl oleate.Ans.

(i)CH2—O—C—C15—H35

CH—O—C—C15H35 + 3NaOH Heat⎯⎯→

CH2OH

CHOH

CH2OHGlycerol

+ 3C15H35COONa

CH2—O—C—C15H35

(C15H31COO)3C3H5

(Glyceryl Palmitate)

O

O

O

Sodium palmitate

(ii) CH2—O—C—C17–H32

CH—O—C—C17H32 + 3NaOH Sodium

⎯⎯→

CH2OH

CHOH

CH2OHGlycerol

+ 3C17H32COONa

CH2—O—C—C17H32

(C17H32COO)3C3H5

(Glyceryl oleate)

O

O

O

oleate

❐

Good book for practice. Category-wise solved questions with proper explanations. Liked it.- Tushar

Very helpful. Completes every topic with numerous questions and answers solved - Dhrubesh

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