SOLVED Model Specimen Paper 2 · 2019-11-28 · (ii) When NaCl is added to hydrated ferric oxide sol, it dissociates to Na+ and Cl-ions. Particles of ferric oxide sol are positively

Solution 1.(a) (i) Four, two

(ii) Glucose, sucrose(iii) basic, + I(iv) sodium chloride, copper.

(b) (i) (1) (ii) (1) (iii) (2) (iv) 2)(c) (i) (d), (ii) (c), (iii) (a), (iv) (b).

(d) (i) 3 2 2 6 3 2 2 3Propene

6CH CH CH B H 2(CH CH CH ) B− = + →+ 6CH O and NaOH (aq)2

6CH OH + 2H BO3CH CH2 2 3 3Propanol-1

(ii) The order is :Neoprene < Polyvinyl chloride < Nylon-6

(iii) Bactericidalmedicine—OfloxacinBacteriostatic medicine—Chloramphenicol

(iv) The lyophobic sols can be precipitated or coagulated by addition of small amounts of electrolytes,by heating or by shaking, hence they are not stable.

Solution 2.

We know that π = CRT or = n w

mVRT

VRT=

∴ 50076

= 81001000

0 0821 300m ×

× ×.

or m = 8 0 0821 300 1000

500 76 100× × ×

×.

/

m = 299.5 g/mol. Ans.OR

(i) The boiling point increases.(ii) ∆Tb = Kb × m

∆Tb = 0 526180

10001000

0 0173. .× × = °C

∴ B.P. = 100 + 0.0173°C = 100.0173°C. Ans.Solution 3.

(i) Density ρ =Z M

NA

××a3

Given : ρ = 4.4 g/cm3, M = 213, Z = 2, NA = 6.023 × 1023

∴ a3 = 2 213

4 4 6 023 1023×

× ×. .

= 160.74 × 10–24

Model Specimen Paper Chemistry

2SOLVED

| 2 | ISC Model Specimen Papers, XII

a = 160 74 10 243 . × −

= 5.438 × 10–8 cm = 544 pm Ans.(ii) 12.

Solution 4.The two cells may be represented as :Zn | Zn2+

(conc. = C) || Cu2+ (C = ?) | Cu emf = E1 (say)Zn | Zn2+

(conc. = C) || Cu2+ (C = 0.5 M) | Cu emf = E2 (say) E2 – E1 = 0.03 V

The cell reaction is :Zn (s) + Cu2+ (aq) � ⇀�↽ �� Zn2+ (aq) + Cu (s)

Ecell = E°Cell –2 303

2

2

2.

log[ ][ ]

RTF

ZnCu

+

+

= E°Cell −+

+0 06

2

2

2.

log[ ][ ]ZnCu

E1 = E°Cell − +0 06

2 2.

log[ ]

CCu

E2 = E°Cell −0 06

2 0 5.

log.C

∴ E2 – E1 = 0 06

2 0 52.

log log.

CCu

C+

−

0.03 = 0.03 log 0 52.

[ ]Cu +

log.

[ ]0 5

2Cu + = 0 5102

.[ ]Cu + =

[Cu2+] = 0 510

0 05.

.= M

Solution 5.(i) The role of desorption in the process of catalyst is to make the surface of the solid catalyst free for the

fresh adsorption of the reactants on the surface.(ii) (a) Aerosol : dispersed phase—solid; dispersion medium—gas

Example—Smoke.(b) Solid foam : dispersed phase—gas; dispersion medium—solid

Example—Pumicestone.OR

(i) When a beam of light is passed through a colloidal solution, scattering of light is observed. This isknown as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidalsolution.

(ii) WhenNaClisaddedtohydratedferricoxidesol,itdissociatestoNa+ and Cl- ions. Particles of ferricoxidesolarepositivelycharged.Thus,theygetcoagulatedinthepresenceofnegativelychargedCl- ions.

Solution 6.If K2SO4 is completely dissociated

K2SO4 � ⇀�↽ �� 2K++SO42–

i = 3Molecular mass of K2SO4 = 2 × 39 + 32 + 4 × 16

= 174p = i.CRT

Chemistry | 3 |

= i × ×

×W RTM V

B

B

= 3 25 10 0 082 298

174 2

3× × × ××

− .

π = 5.27 × 10–3 atm.Solution 7.

(i) Nitrobenzene is formed.

conc. HNO3

NO|

2

+ H2O

Benzene Nitrobenzene

conc. H SO42

(ii) Acetanilide is formed.

+ CH COCl3

NaOH

NH|

2

Aniline

NHCOCH|

3

+ HCl

Acetanilide

Solution 8. (i) Disinfectants when used in low concentrations work as antiseptics as well, e.g., Phenol.(ii) The medicines which inhibit or hinder the normal working mechanism of enzymes in a body by

blocking the binding site where the biological substrate binds or can also alter the activity of enzymeby binding to it and changing its shape hence function.

Solution 9.(i) Ferromagnetic : Thesearethesubstanceswhicharestronglyattractedbymagneticfieldandshows

permanentmagnetismeven in absenceofmagneticfield. In these substances a largernumberofunpaired electrons are present which gets spontaneously aligned in the same direction.

(ii) Ferrimagnetic : These are the substances which possess only small net magnetic moments inspite ofa large number of unpaired electrons. In these substances, the moments are aligned in parallel andantiparallel directions in unequal numbers.

(iii) Antiferromagnetic : These are the substances which possess zero net magnetic moment inspite of thepresence of unpaired electrons. In these substances alignment of magnetic moments takes place in acompensatory way so as to give zero net magnetic moment.

OR(i) The number of tetrahedral voids formed is equal to twice the number of atoms of element Y and only

2/3rd of these are occupied by the atoms of element X. Hence, the ratio of the number of atoms of Xand Y is 2 X (2/3) : 1 or 4 : 3 and the formula of the compound is X4Y3.

(ii) (a)Frenkeldefect,(b)Schottkydefect.Solution 10.

Suppose‘x’ is order with respect to A andSuppose‘y’ is order with respect to B thenIn1stexperiment r = k [0.01]x [0.01]y = 0.005 …(i)2ndexperiment r = k [0.02]x [0.01]y = 0.010 …(ii)3rdexperiment r = k [0.01]x [0.02]y = 0.005 …(iii)Dividing (ii) by (i)

k

k

x y

x y[ . ] [ . ]

[ . ] [ . ]

0 02 0 010 01 0 01

××

= 0 0100 005..

2x = 2∴ x = 1Now by dividing (iii) by (i)

k

k

x y

x y[ . ] [ . ][ . ] [ . ]0 01 0 020 01 0 01

= 0 0050 005..


2y = 1∴ y = 0∴ Order with respect to A = 1

Order with respect to B = 0.Solution 11.

(i) Potassium dicyanoargentate(I).(ii) Diamminesilver(I) chloride.(iii) Potassiumhexacyanoferrate(II)trihydrate.

Solution 12.Lab preparation of SO2 : Sulphurdioxidecanbepreparedbytheactionofdil.H2SO4 on sodium sulphite.

Na2SO3 + H2SO4 → Na2SO4 + H2O+SO2 The gas dried by passing it through conc. H2SO4 and is collected by upward displacement of air.

Dil. H 2 4

3

SO

Na SOConc.

2 H SOSO gas

2 42

Lab. preparation of sulphur dioxideSolution 13.

SulphurinSO2 involves sp2 hybridization two of these sp2 orbitals form two σbondswithoxygenatomby overlap of sp2 of s with 2p of O atoms. The remaining two unhybridised orbitals of s forms pπ–pπ and pπ –dπdoublebondswithOatoms.So,SO2hasthetwoS–Obondas:

OO

S

p

d�

�–

pp

��

–

ThoughthetwoS—Obondsappeartobedifferentbecauseoneisformedbypπ-pπ overlap and other is formed by pπ – dπoverlap.However,bothS—Obondsareequalinlengthbecauseofresonancebetweentwo structures.

OR(i) 3Cl2 + 8NH3 → N2 + 6NH4Cl(ii) 4O3+PbS→PbSO4 + 4O2(iii) Carbonmonoxideactsasareducingagentandreduceszincoxidetozinc.

ZnO + CO → Zn + CO2Solution 14.

(i) By heating ethanol with acetic acid + conc. H2SO4

CH COOH HOC H CH CEthanol

SOH O3 2 3 3

2 4

2Acetic acid

Conc. H+ →−

OOOC H2 5Ethyl acetate

(ii) 6 23 2 2 6 3 2 2 3

6 2

CH CH CH B H CH CH CH Bopene

CH O

Pr

( )= + → + and

NaOH (aq) →→ +6 23 2 2 3 3CH CH CH OH H BOPropanol-1

Chemistry | 5 |

(iii) NaOH, H2O

350°C

O Na|

+–

Cl2

FeCl3

H+

Cl|

OH|

(Benzene) (Chlorobenzene) (Sodium Phenoxide) (Phenol)

Solution 15.(i) Reimer-Tiemann reaction :

Conc. HNO3

NO2|

(Benzene)

+ H2O

(Nitrobenzene)

Conc. H SO42

(ii) Clemmenson’s reduction :

CH COOH HOC H CH COAcetic acid

SOH O3 2 3 3

2 4

2+ →

−Ethanol

Conc. H OOC H2 5Ethyl acetate

Solution 16.(i) Potassiumpermanganateisagoodoxidisingagenthence,itfindsusesinindustriesbasedonchemical

processes.(ii) As a consequence of contraction, the radii of all the elements hence their atomic sizes are similar, which

leadstosimilarchemicalpropertiesoftheelements.So,itbecomesdifficulttoseparatethemfromeachother.

(iii) The standard electrode potential for reduction of Zn2+ toZnishighest(–0.76)amongthefirstrowtransitionelementsduetostabilityofcompletelyfilledd10configurationofZn2+.

(iv) The high melting and boiling points of d-block elements are high and are attributed to involvement of(n–1) d electrons resulting into strong inter atomic bonding.

(v) TheorbitalsinHgarecompletelyfilledintheirgroundstate(5d106s2)aswellasoxidationstates.OR

CrO3 + 2KOH → K2CrO4 + H2O.(A) Yellow (B)

2K2CrO4 + H2SO4 → K2Cr2O7 + K2SO4 + H2O(B) (Conc.) (C)

(Yellow) (Orange)

K2Cr2O7 + 2NH4Cl → (NH4)2Cr2O7 + 2KCl

(Orange) (D)

(NH4)2Cr2O7 → CrO3 + N2+ 4H2O(A)

(Green)

A:ChromiumtrioxideB : Potassium chromateC : Potassium dichromateD : Ammonium chromate

Solution 17.(i) Hydraulic washing is based upon the differences in gravities of the ore and the gangue. It is therefore

a type of gravity separation.(ii) Frothfloatationprocessistheprocessusedforconcentrationofcopperfromitssulphideores.(iii) Inelectrolyticrefiningtheimpuremetalactsasanode,andastripofsamemetalinpureformisused

as cathode.(iv) Aluminium is stronger reducing agent than carbon and therefore, cannot be reduced by it.(v) Wrought iron is used in making anchors, chains and agricultural implements.

ORBauxite,aprincipaloreofaluminiumcontainssilica,ironoxideandtitaniumoxideasimpurities.Theoreisdigested with a concentrated solution of NaOH at 473- 523 K and 35-36 bar pressure. Al2O3 is leached out as sodium aluminate and silica as sodium silicate while impurities are left behind.


Al2O3(s) + 2NaOH(aq) + 3H2O → 2Na[Al(OH)4](aq)The aluminate in the solution is neutralized by passing CO2 gas and hydrated Al2O3 is precipitated.

2Na[Al(OH)4] (aq) + CO2(g) → Al2O3.xH2O + 2NaHCO3(aq)Thesodiumbicarbonateremainsinthesolutionwhilehydratedaluminaisfilteredanddried.Thehydratedaluminathusprecipitatedisfiltered,driedandheatedat1473Ktogivebackpurealumina.

Al2O3.xH2O → Al2O3(s) + xH2O Hydrated alumina Alumina

Solution 18.(i) Grignard reagents (RMgX) are a potential sources of carbanions because of the polar nature of C—Mg

bond.(ii) The reaction of haloalkanes with aqueous sodium or potassium hydroxide or moist silver oxide

produces alcohols by hydrolysis reaction mechanism.(iii) Ethanol is added to chloroform on prolonged storage to destroy traces of phosgene formed by aerial

oxidation.(iv) SN1 reaction mechanism leads to Walden inversion.(v) For the same alkyl group, the boiling point of haloalkanes decreases in the order RI > RBr > RCl > RF. This

is due to the increase in van der Waals forces when the size and mass of the halogen atom increases.OR

(i) CH3CH2Br + AgF → CH3CH2—F + AgBr

(ii) + NaCl

CH3Cl|

+ Na + CH CH Cl3 2

Ether

(iii) H C3

CH2

+ H—BrPeroxide

H

BrH C

3

(iv) CH3CH=CH2 + HI → CH3CH2CH2I + CH3CH(I)CH3(Minor) (Major)

(v)

Cl| O

CH3 +

Cl|

H C3 O

2-Chloroacetophenone(Minor)

4-Chloroacetophenone(Major)

O

Anhyd. AlCl3

Cl|

+ H C3 Cl

qq

Solution 1.(a) (i) monoclinic, rhombic

(ii) outer, inner(iii) adenine, guanine, thiamine(iv) formic acid

(b) (i) (3) (ii) (2) (iii) (1) (iv) (1)(c) (i) (b) (ii) (c) (iii) (d) (iv) (a)(d) (i) Hexagonal closed packed structure.

(ii) De-icing is a process of removal of snow, ice and frost from surfaces such as roads using de-icingagents. De-icing agents such as common salt decreases the freezing point of ice and eventuallymelts the ice deposited and clears the surface.

(iii) Inordertobeeffective,collisionofreactivespeciesneedsthemtocollidewithsufficientkineticenergy (threshold energy) as well as proper orientation. So, in the above reaction rate is slow dueto improper orientation of colliding molecules to give effective reaction.

(iv) The d-blockelementsshowvariableoxidationstatesduetoincompletelyfilledd-orbitals whichare used for different arrangements of electrons to give different oxidation states.

Solution 2.Faraday’s first law of electrolysis : The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). Electrochemical equivalent of a substance (ECE) is the amount of a substance in grams produced or consumed by the passage of one coulomb of electricity in an electrochemical reaction. A voltmeter is used to measure the electrochemical equivalent of an element.

OR(i) UnitofspecificconductanceisSm–1.(ii) Itisdefinedastheconductivityofanelectrolytesolutiondividedbythemolarconcentrationofthe

electrolyte.

λmkc

=

Solution 3.(i) Natural food preservative—Salt

Synthetic food preservative—Sodium benzoate(ii) ‘Tincture of iodine’ is the common name of 2-3 percent solution of iodine in alcohol water mixture. It is

used as antiseptic and applied on wounds.Solution 4.

(i) Members of the 16th group are oxygen, sulphur, selenium, tellurium and pollonium. The elements ofgroup 16 have six electrons in the outermost shell and have ns2np4generalelectronicconfiguration.

(ii) The elements of group 16 exhibit a number of oxidation states such as – 2, – 1, 1, 2, 4, 6 where – 2 is themost common oxidation state.

OR(i) 3Cl2(g) + 6NaOH (conc.) → 5NaCl(aq) + NaClO3(aq) + 3H2O(ii) 5SO2 + 2KMnO4 + 2H2O → 2MnSO4 + K2SO4 + 2H2SO4

Solution 5.Nylon-6 is prepared by heating caprolactum (a seven membered cyclic amide) with water at high temperature. The notation ‘6’ is to denote presence of 6 carbon atoms in caprolactum.


5SOLVED


O

NH533-543 K/H O2 [ CO—(CH ) —CO ]

Nylon-62 5 ——

Caprolactum

Nylon-6, 6 is prepared by the condensation of hexamethylene diamine with adipic acid, where a molecule of water is given out as a by-product.

n nH N CH NH HOOC CH COOHAdipic ac

2 2 6 2 2 4( ) ( )Hexamethylene diamine

+iid Nylon water

H CH NHCO CH H O→ + [ ( ) ( ) ],

2 6 2 46 6

2n n-

The notation ‘6, 6’ is to denote presence of 6 carbon atoms in the two monomers hexamethylene diamine and adipic acid both.

Solution 6.(i) Zwitter ion structure of glycine can be shown as follows :

H CH COO

NH

−

+|

3

(ii) Proteins have an unique three dimensional structure in their native form. If the native form of proteinis subjected to any physical change (such as temperature change) or any chemical change (such aschange in pH), the hydrogen bonds are disturbed. Due to this globules unfold and helix get uncoiledand protein loses its biological activity. This is called denaturation of protein. During denaturation 2°and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg whiteis an example of denaturation of protein.

Solution 7.

(i) Ni

CO

COCOCO

Ni

CO

COCOCO

CO CO

Ni(CO)Tetrahedral

4 Cr(CO)Octahedral

6

(ii) The magnetic moment is determined by the number of unpaired electrons and is calculated by usingthe ‘spin-only’ formula :

µ = +n n( )2

Solution 8.Forafirstorderreaction:

Rate constant k = 2.303/(t2 – t1) log[ ][ ]RR

1

2

Putting the values given :k = 2.303/5 min × log (0.6 mol L–1/0.2 mol L–1) = 0.4606 × 0.477 = 0.22 min–1

Solution 9.(i) As depression of freezing point (∆Tf) is directly proportional to molality of solution :

∆Tf = Kf m Now putting the values :

m = ∆Tf /Kfm = 273.0 K – 272.4 K/1.86 K kg mol–1

= 0.323 mol kg–1

(ii) Similarly,∆Tb = Kbm

= 0.323 mol kg–1 × 0.512 K kg mol–1

= 0.165 K

Chemistry | 3 |As, boiling point of water is 373.15 K, therefore boiling point of solution would be

373.15 K + 0.165 K = 373.31 K(iii) Relative lowering of vapour pressure can be given the below formula :

P1 – p1 /P1 = w2 × M1 / M2 × w1

or P1 – p1 /P1 = (m/1000) × 18Now the values know here,

m = 0.323 mol kg–1, P1 = 23.756 mm HgPutting the values :

(23.756 – p1)/23.756 = (0.323 mol kg–1/1000) × 18 g mol–1

p1 = 23.62 mm HgSo, lowering of vapour pressure would be by 0.138 mm Hg.

Solution 10.(i) NaCl has a face centered cubic unit cell, so number of Na+ and Cl– ions present per unit cell are 4 each.

Na+

Cl–

(ii) Coordination number of Na+ ions is 6 here.(iii) In sodium chloride crystal the constituent particles are held by strong electrostatic (coulombic) forces.

Solution 11.(i) CaF2 + H2SO4 → CaSO4 + 2HF(ii) C + 2 H2SO4 (conc.) → CO2 + 2SO2 + 2H2O(iii) 2NaHCO3 (s) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l) + 2CO2 (g)

OR

(i) NH4NO3(s) Heat → N2O + 2H2O

(ii) N2O + 2NaNH2 → NaN3 + NaOH + NH3(iii) Cu2+ (aq) + 4NH3 (aq) → [Cu(NH3)4]2+ (aq)

Solution 12.(i) Propanoic acid to acetic acid conversion :

H C3COOH

H C3 H C3

COCl CONH2Propanoic acid Propanoyl chloride Propanamide

Hofmann bromamidedegradation

Br , NaOH2

Acetic acid Ethanol Ethanamine

SOCl2 NH3

(ii) Formic acid contains both aldehyde (–CHO) and acidic group (–COOH). Due to presence of aldehydicgroup. It acts as reducing agent and reduces Tollen’s reagent to metallic silver (silver mirror test)according to the following reaction :

HCOOH + 2[Ag(NH3)2]2 + 2OH– → 2Ag + CO2 + 2H2O + 4NH3

OR

(i) Sodium hydrogen sulphite adds to aldehydes and ketones to form the addition products.


(ii) CHO+ NaHSO3

H3C

ONa

SO3HProton transfer

H3C

OH

SO3Na

Acetaldehyde Sodium bisulphiteBisulphite addition

product

H C3

Acetone reacts with phenyl hydrazine to give corresponding hydrazone.C6H5NHNH2 + CH3COCH3 → C6H5NHN = C(CH3)2 + H2O

(iii) 2-pentanone and 3-pentanone2-Pentanone can be differentiated from 3-pentanone with iodoform reaction. As methyl ketones reactwith I2 and NaOH to give CHI3 (iodoform) and corresponding sodium carboxylate, so 2-pentanone would undergo the reaction whereas 3-pentanone won’t give the reaction.

H C CH3 3

O||

NaOI

I , NaOH2

NaO CH3

O||

+ CHI32-Pentanone Sodium butanoate Iodoform

Solution 13.(i) Persistent dialysis leads to coagulation of sols.(ii) Delta formation takes place due to coagulation of river water as there is difference in concentration of

river and sea water.(iii) Clot is formed due to coagulation of blood which stops further bleeding.

Solution 14.(i) TheprocesswherepurifiedAl2O3 is mixed with cryolite (Na3AlF6) would yield aluminium better.

This is because cryolite lowers the melting point of the mix and brings conductivity which enhancesthe electrolysis process.

(ii) The electrolysis of the molten mass is carried out in an electrolytic cell using carbon electrodes(graphite). The oxygen liberated at anode reacts with the carbon of anode producing CO and CO2.

Solution 15.(i) Lanthanoid elements have similar sizes, that is atomic radii of atoms are nearly same due to the

lanthanoidcontraction,whichmakestheirseparationdifficult.(ii) Transition metal contain large number of unpaired electrons, and they have strong interatomic

attractions, that’s why they have higher atomization enthalpies.(iii) Transition metals have high enthalpy of hydration due to their small size and large nuclear charge.

Solution 16.(i) Alkyl halides are considered to be very reactive compounds towards nucleophile because they have

an electrophilic carbon and a good leaving group. As we go down the periodic table, halides that arelarger in size are able to distribute their charge over a larger volume, making them less reactive (lessbasic) hence a good leaving group.

(ii) 1 mole of butane would be formed.(iii) Carbocations are stabilized by electron donating groups (i.e., alkyl groups). So, the order would be :

2 < 1 < 3(iv) Butane < 1-chlorobutane < 1-bromobutane < 1-iodobutane

Higher the surface area, higher will be the intermolecular forces of attraction and thus boiling pointtoo. Boiling point increases with increase in molecular mass of halogen atom for the similar type ofalkyl halide. Butane has no halogen atom and rests of all three compounds are halo derivatives ofbutane. Atomic mass of iodine is highest so, the boiling point of 1-iodobutane is maximum among allthe given compounds.

OR(i) (d) Hydration of alkene is not the method of preparation of alkyl halide.(ii) Primary alkyl halide gives best SN2 reaction.(iii) The best method of preparation of alkyl halides is a reaction of alcohol with SOCl2/ Pyridine because

by-products formed in the reaction are SO2 and HCl which are in gaseous form and escape into theatmosphere leaving behind pure alkyl chlorides.

CH3CH2 – OH + SOCl2 → CH3CH2 – Cl + SO2 ↑ + HCl ↑(iv) Decreasing reactivity order for the alkyl halides towards SN2 reaction is :

R–I > R–Br > R–Cl > R–FThis can be explained by comparative ability of halogen atom to behave as a better leaving groupcompared to the other.

Chemistry | 5 |Solution 17.

(i) Rectified spirit : It is obtained from the distillation of the fermentation liquor wash containing about95% of ethanol.

(ii) Isopropyl alcohol and methyl iodideWhen ethers are treated with strong acid in the presence of a nucleophile, they can be cleaved to givealcohols and alkyl halides. If the ether is on a primary carbon this may occur through an SN2 pathway.

(iii) When the vapours of a secondary alcohol are passed over copper-nickel couple, heated at 573 K, thecorresponding ketones are formed.

(iv) o-cresol is least acidic because presence of electron withdrawing groups at ortho and para-positionincreases the acidic strength, whereas electron releasing or donating groups at ortho and para-positiondecreases acidic strength. In cresol (ortho or para cresol), methyl group is +I group, so it destabilizesthe anion formed after ionisation, hence cresol is less acidic than carbolic acid (phenol).

OR(i) SOBr2. Thionyl bromide will react with secondary alcohols and give a reasonable percent yield of the

bromide.(ii) Phenol is considered as a better Bronsted acid than cyclohexanol because phenol is able to stabilize the

anion, formed after proton loss, by resonance. In this case there is delocalization of the electrons intothe benzene ring increasing the overall stability of the phenoxy anion.

(iii) Reaction of HX with alcohol depends on the degree of carbocation generated due to heterocycliccleavage of HX bond. If alcohol is primary alcohol then reaction will proceed through SN2, for tertiaryalcohol it will proceed with SN1 mechanism. For secondary alcohol it can go through both path (SN2or SN1). SN1 being ionic in nature takes place very fast while SN1 is molecular in nature and reactsslowly. Hence, we can sum up above, rate of reaction is 3° alcohol > 2° alcohol > 1° alcohol.

(iv) When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along withwater).Thisreactioniscalledtheesterification.

Solution 18.(i) Aniline is less basic than methylamine because the –NH2 group in aniline is directly attached to the

benzene ring, which results in the unshared electron pair on nitrogen atom to be in conjugation withthe benzene ring and thus, making it less available for protonation, whereas there is no such case withmethylamine.

(ii) Activation of benzene ring by amino group reduced by acetylating aniline. The lone pair of electronson nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below :

N—C—CH3

N — C— CH3

O

||O

||: :

..:: –

..

+

Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH3 group is less than that of amino group.

(iii) Sulphanilic acid is prepared from aniline by sulphonation of aniline. Aniline reacts with concentratedsulphuric acid to form anilinium hydrogen sulphate which on heating with sulphuric acid at453-473 K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid.

NH2453-473K

HO S3AnilineAnilinium hydrogen

sulphateSulphanilic acid

NH HSO3+

4–

NH2H SO2 4

ORNO2|

Sn/HClBr2

Excess

NH2|

NaNO , HCl2

5°C

N2+ –

Cl| BrBr

H PO , H23 2 O

NH2| BrBr BrBr

|Br

|Br

|Br

Nitrobenzene Aniline(A)

2, 4, 6-Tribromoaniline(B)

1, 3, 5-Tribromobenzene(D)

(C)

qq

Solution 1.(a) (i) less, same (ii) bidentate, monodentate

(iii) below (iv) HCl(b) (i) (1), (ii) (3), (iii) (4), (iv) (4).(c) (i) (d) (ii) (a) (iii) (b) (iv) (c)(d) (i) Glycerol is a trihydric alcohol.

(ii) Cannizzaro’s reaction is given by the aldehydes which does not have α-hydrogen.(iii) Aromatic compounds undergo electrophilic substitution reaction on the ring.(iv) The oxidising power of halogen decreases down the group.

Solution 2.Given, Weight of glucose = 9 g, (Mol. wt. of glucose = 180 g mol–1)Hence, Number of moles of glucose 9/180 = 0.05 mol

Volume of solution, V = 500 cm3 or 0.5 LTemperature = 27 °C or 300 K

Now, Osmotic pressure, π = (n/V) RTPutting the values :π = (0.05 mol/0.5 L ) 300 K × 0.0821 L atm K–1 mol–1

= 2.46 atmOR

More the number of molecules more is the elevation in boiling point, now 0.1 M solution of K4[Fe(CN)6] would give 0.1 × 4 of K+ and 0.1 × 1 of [Fe(CN)6]4–, i.e., total 0.5 moles of ions. Similarly, 0.2 M solution ofNaCl would give 0.4 moles of ions and 0.3 M solution of sucrose would give 0.3 moles of sucrose molecules. Hence, in increasing order of boiling point :0.3 M solution of sucrose > 0.2 M solution of NaCl > 0.1 M solution of K4[Fe(CN)6].

Solution 3.(i) In diamond carbon atoms are bonded through covalent bonds which are strong and directional, there

are no loose electrons present in the structure and atoms are held very strongly at their positions. Ingraphite carbon atoms are arranged in different layers and each carbon atom is covalently bonded tothree of its neighbouring atoms in the same layer. The layers can slide over each other making graphitea soft and slippery solid to touch.

(ii) Crystalline compounds are usually anisotropic, that is some of their physical properties like electricalresistance or refractive index show different values when measured along different directions in thesame crystal. This arises from different arrangement of particles in different directions. Since, thearrangement of particles is different along different directions, the value of same physical property isfound to be different along each direction.

Solution 4.The plot of ln k vs. 1/T gives a straight line according to the equation given below :

ln k = – Ea/RT + ln AActivation energy can be calculated by the value of slope (= – Ea/R) of the graph.Given, Slope = – 5841 K and R = 8.314 JK–1 mol–1

Putting the values :Slope = – Ea/R

– Ea = – 5841 K × 8.314 JK–1 mol–1

= 48562 J mol–1


8SOLVED

Chemistry | 2 |OR

(i) If reaction is 20% completed then the remaining reactant is 80%,Hence, R1 = 100; R2 = 80R1Forafirstorderreaction,rateconstantisgivenby:Rate constant k = 2.303 / (t2 – t1) log ([R]1/[R]2)Putting the values given :

k = 2.303 / 10 min × log [100 / (100 – 20)]= 0.2303 × 0.1= 0.023= 2.3 × 10-2 min–1

(ii) Time, t = 2.303 / k log ([R]1/[R]2)= 2.303/2.3 × 10-2 min–1 log [100 / (100 – 75)] = 0.6 × 102 min= 60 min.

Solution 5. The Tyndall effect is the effect of light scattering in colloidal dispersion, when a light beam is passed through. Theindividualsuspensionparticlesscatterandreflectlight,thisscatteringoflightilluminatesthepathofbeam in the colloidal dispersion.

Scattered Light

Microscope

Colloidal Solution

Flashlight

Solution 6.

(i)

CONH|

2

(i) SOCl2

(ii) NH3

Br /NaOH2

Hofmann bromamidedegradation

COOH|

NH|

2

Benzoic acid Benzamide Aniline

(ii) CH CHO CH COOHK Cr O i NHii3 3

2 2 7 3

Acetaldehyde Acetic acid → ( )

( ) ∆∆ ,− →

H OAcetamide

CH CONH2

3 2

Solution 7.Benzaldehyde and acetaldehyde can be distinguished with help of iodoform reaction. Acetaldehyde bearing methyl carbonyl group would give a positive iodoform test whereas benzaldehyde would not give this test.

H C3

O

H

NaOH I–

2

NaOIH C3

O

ONa–+ CHI3

Acetaldehyde Sodium ethanoate

Iodoform

Solution 8.Chlorobenzene reacts with Cl2 gas in presence of AlCl3 to give 1, 2-dichlorobenzene (A) and 1, 4-dichlorobenzene (B).Chlorobenzene reacts with CH3Cl gas in presence of AlCl3 to give 2-methyl chlorobenzene (C) and 4-methylchlorobenzene.


Cl|

Anhyd. AlCl3

Anhyd. AlCl3

CH Cl3

Cl2

Cl| Cl

Cl|

+

|Cl

Cl|

Cl|

+

|CH3

CH3

(A)

(B)

(C)

(D)Solution 9.

A binary compound of oxygen with another element is called oxide. Those oxides which combine with water to give an acid are known as acidic oxides and those which combine with water to give a base are known as basic oxide.Oxygen reacts with non metals to give acidic oxides in general, such as SO2, Cl2O7, CO2, N2O5.For example, SO2 combines with water to give H2SO4.

SO2 + H2O → H2SO4Oxygen reacts with metals to give basic oxides in general, such as Na2O, CaO, BaO. For example, CaO combines with water to give Ca(OH)2.

CaO + H2O → Ca(OH)2However, there are some exceptions where some metal in high oxidation states also have acidic character, such as Mn2O7, CrO3, V2O5 etc. Some metallic oxides show dual behaviour, they show characteristic of both acidic as well as basic oxides. For example, Al2O3.

OR(i) 4NaCl + MnO2 + 4H2SO4 → MnCl2 + 4NaHSO4 + 2H2O + Cl2(ii) 2NaCl + H2SO4 → 2HCl + Na2SO4(iii) Cu + 2H2SO4 (conc.) → CuSO4 + SO2 + 2H2O

Solution 10.(i) In acidic medium :Permanganateasoxidizingagentworksmostefficientlyinacidicsolution,because

it is reduced to the greatest extent in this medium, from oxidation state +VII in MnO4 to +II in Mn2+.

8H++ MnO4– + 5e– → Mn2++ 4H2O

e.g., 1. Iodine is liberated from potassium iodide :10I– + 2MnO4

– + 16H+ →2Mn2+ + 8H2O + 5I22. Fe2+ ion (green) is converted to Fe3+ (brownish yellow) :

5Fe2+ + MnO4- + 8H+ → Mn2+ + 4H2O + 5Fe3+

(ii) In neutral medium : Theoxidation in neutralmedium is less efficient, asmanganese (VII) is onlyreduced to manganese (iv), yielding insoluble MnO2 as a byproduct which needs to be removed fromthereactionmixtureduringpurificationofthedesiredproduct(carboxylicacid).

2H2O + MnO4– + 3e– → MnO2↓ + 4OH–

e.g., Oxidation of iodide to iodate :2MnO4

– + H2O + I– → 2MnO2 + 2OH- + IO3–

(iii) In basic medium :Oxidation in strongly alkaline medium is least efficient in terms of transferredelectrons per mole permanganate, as the latter is only reduced to manganate(VI).

MnO4– + e– → MnO4

2–

ORThe actinoids show a greater range of oxidation states, which is due to the following reasons. The oxidation state shown in general by actinoids is +3.1. This is due to the very small energy gap between 5f, 6d and 7s sub-shells.2. The principal oxidation states are +3 and + 4 and +3 oxidation state is the most stable.3. The +4 oxidation state is the most stable in Th and Pu.4. +5 in Pa and Np and +6 is seen in U.

Chemistry | 4 |5. The distributions of oxidation states in actinides are uneven.6. The oxidation state shown in general by actinoids is +3.

Solution 11.Alcohols are compounds in which one or more hydrogen atoms in an alkane have been replaced by an –OH group. Alcohols fall into different classes depending on how the –OH group is positioned on the chain of carbon atoms.1. Primary alcohols : In a primary (1°) alcohol, the carbon which carries the –OH group is only attached to

one alkyl group.For example : Ethanol, CH3CH2OH

2. Secondary alcohols : In a secondary (2°) alcohol, the carbon with the –OH group attached is joineddirectly to two alkyl groups, which may be the same or different.For example : Propan-2-ol, CH3CH(OH)CH3.

3. Tertiary alcohols : In a tertiary (3°) alcohol, the carbon atom holding the –OH group is attached directlyto three alkyl groups, which may be any combination of same or different.For example : 2-Methyl-propan-2-ol, CH3(CH3)CH(OH)CH3.

The primary, secondary and teriary alcohols can be distinguished by two tests.1. Jones reagent test :Jonesreagentorasolutionofsodiumorpotassiumdichromate(VI)acidifiedwith

dilute sulphuric acid is used to oxidise the alcohols to corresponding aldehydes or ketones. If oxidationoccurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containingchromium(III) ions.

2. Lucas reagent test : Lucas reagent is a mixture of ZnCl2 and HCl. On reaction alcohols are turned to thecorresponding alkyl halide giving turbidity to reaction solution due to precipitation of the insoluble alkyl chloride generated.

Alcohol type Jones reagent Lucas Reagent

Primary Reacts readily at room temperature and turns the reaction solution green.

Takes a lot of time to react and show turbidity.

Secondary Does not react readily and when heated turns the reaction solution green.

Reacts to give turbidity in some time like in 5 minutes.

Tertiary No reaction. Reacts immediately to give turbidity.

Solution 12.Structure of sodium lauryl sulphate or CH3(CH2)10CH2 SO3

– Na+ can be shown as :

H C3 O—S—O

O

O

||

||Na

Hydrophobic Region Hydrophilic Region

Now the micelle formed (where the hydrophobic part, that is long hydrocarbon chains are pulled into the bulk of solution and aggregate to from a spherical shape, whereas the polar group SO4

– remains on thesurface of sphere formed facing the water molecules) can be shown as :

O

O

O

O

O

O

O

O

OO

O

O

O

O

O

O

SO3–

SO3–

SO3–

SO3–

SO3–

SO3–

SO3–

SO3–– SO3

– SO3

– SO3

– SO3

– SO3

– SO3

– SO3

– SO3

OSO3

–

Monomer

Micelle


Solution 13.A protein is a complex, high-molecular-weight organic compound and are polymers of a-amino acids connected to each other through peptide bonds (also called polypeptides). Proteins are essential to the structure and function of all living cells. Protein play structural and mechanical roles and they also work as enzymes or form subunits of enzymes. Theycanbeclassifiedintotwotypesbasedontheirmolecularshapes:1. Fibrous proteins : In this polypeptide chains run parallel and are held together by hydrogen and sulphidebonds.2. Globular proteins : This structure results when the chains of polypeptides coil around to give a sphericalshape.

Solution 14.(i) For the reaction, MgO (s) + C (s) → Mg (s) + CO (g)

At 1273 K, ∆Gr = ∆Gf [CO (g)] – ∆Gf [MgO (s)] = – 439 – (– 941) kJ mol–1 = 502 kJ mol–1

And at 2273 K, ∆Gr = –628 – (– 314) kJ mol–1 = – 314 kJ mol–1

So, the temperature is 2273 K at which carbon can be used for reduction of MgO.(ii) Pyrometallurgy is reducing a metal oxide by heating with either coke or some other reducing agent.

e.g., Zn, Al, Mg etc.

ZnO C Zn COK+ → +975

(iii) Entropy of the reaction is more positive when the metal is in liquid state as compared with solid state,so ∆G becomes more positive.

Solution 15.(i) We will use detergent because it will not form insoluble precipitate with Ca2+ ions.(ii) Drugs which bind to the receptor site and inhibits its natural function are known as antagonists. They

are useful when blocking of message is required for action of drug.(iii) Chloramphenicol is a broad spectrum antibiotic. Other three have antiseptic properties.

Solution 16.(i) Given :

Time (t) = 1.5 hrs = 90 min. = 5400 sCharge = Current × Time = 4 A × 5400 s = 21600 CAccording to the chemical reaction :

Cu2+ (aq) + 2e– → Cu (s)We need 2 F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.For 21600 C, the mass of Cu deposited

= (63 g mol × 21600 C)

(2 × 96487 C mol

−

−

1

1) = 7.05 g

So, theoretically, 7 g of Cu should have been deposited, but the amount deposited was 3.2 g.Hence,currentefficiencyis:

= 3 2

7 05.

. = 0.45

Chemistry | 6 |(ii) Standard electrode potential for Ag+ + e– → Ag (s) is 0.80 V, which is higher than standard electrode

potential for 2H+ + e– → H2 (g). Which means Ag+/Ag is a weaker reducing couple than H+/H2,hence silver could not reduce H+ ions present in solution and hence can’t displace hydrogen fromsolution.

OR(i) Equivalent conductance L = k (1000 /N), where k isspecificconductance.

Now, Given :conductivityorspecificconductance=2.6×10–2 ohm–1 cm–1

N = M × Basicity = 1.0 × 2 = 2 N

So, putting the values in equation :L = k (1000 /N)

= 2.6 × 10–2 ohm–1 cm–1 (1000/2 equivalents cm–3)= 14 ohm–1 cm2 equivalent–1

(ii) Passage of direct current through ionic solution over a prolonged period can lead to change in itscomposition due to electrochemical reactions.

(iii) A > D > B > C.Solution 17.

S. No. Polymer Name of monomers Use

(i) Nylon-6 Caprolactum Manufacture of fabrics

(ii) Teflon Tetrafluoroethene Making of non-stick surface for utensils.

(iii) PVC Vinyl chloride Manufacture of rain coats.

(iv) Nylon-6, 6 Hexamethylene diamine and adipic acid Making bristles for brushes.

(v) Buna-S 1, 3-Butadiene and styrene Used as synthetic rubber.

OR(i) Structurally, the natural rubber is a linear cis-1, 4-polyisoprene. In this polymer the double bonds are

locatedbetweenC2andC3ofisopreneunits.Thiscisconfigurationaboutdoublebondsdonotallowthe chains to come closer for effective attraction due to the weak intermolecular attractions. Hence, thenatural rubber has a coiled structure and shows elasticity.

(ii) (a) Copolymerisation is a process in which a mixture of more than one monomeric species is allowedto polymerise. The copolymer contains multiple units of each monomer in the chain. One of the examples is copolymers of 1, 3-butadiene and styrene giving butadiene-styrene copolymer :

nCH = CH—CH = CH +2 2

1, 3-Butadiene

CH2

—CH —CH=CH—CH —CH—CH —2 2 2

|

n

Butadiene-styrene polymerStyrene

(b) Vulcanisation of rubber is a process where a mixture of raw rubber is heated with sulphur and anappropriate additive at a temperature range between 373 K to 415 K. Sulphur forms cross-links atthe reactive sites of double bonds present in rubber molecules and thus the rubber gets stiffened.This is done to improve the physical properties of rubber, because natural rubber becomes soft athigh temperature, brittle at low temperatures, shows high water absorption capacity etc.

Solution 18.Crystal Field Theory describes important properties of complexes (magnetism, absorption spectra, oxidation states, coordination,) considering that the ligands are point charges and d-orbitals of the central atom interacts with ligands. According to CFT, the attraction between the central metal and ligands in a complex is purely electrostatic.Theenergyofthefivedegenerated-orbitals ( , , , )d d d d dxy zx yz x y z2 2 2−

and changes due to the surrounding

ligands (considered as point charges). As a ligand approaches the metal ion, the electrons from the ligand


will be closer to some of the d-orbitals and farther away from others, causing a loss of degeneracy in the d- orbitals. The electrons in the d-orbitals and those in the ligand repel each other due to repulsion betweenlike charges. Thus, the d-electrons closer to the ligands will have a higher energy than those further away,which results in the d-orbitals splitting in energy. So, two sets of orbitals are thus formed. This splitting isaffected by nature and oxidation state of ligands, size of d-orbitals and geometry of the complex. Crystalfieldsplittingenergy(∆) is the energy difference between two sets of d- orbitals (of metal atom) formed aftersplitting in presence of ligands.For example, in tetrahedral complexes, d-orbitals split in two energy levels t2 (it has dxy, dyz and dxz orbitals, higher than degenerate level) and e (it has dx2-y2 and dz2 lower than degenerate level). The splitting is shown In below diagram :

dxz dyz dxy

dx –y22 dz2e

Degenerate -orbitalsd

∆t

t2

OR(i) (a) COhashighvalueofcrystalfieldsplittingenergythanNH3.

(b) Ni has d8configurationwhichisnotaffectedbyfieldstrengthofligandhence,lowspincompoundsare not formed.

(c) In this case, Ti3+ has t2g1 eg0configuration.Itcanperformd-d transition and hence is coloured.(ii) EDTA or Ethylene diamine tetra acetate is used for treatment of lead poisoning.(iii) As coordination number of Co is 6 and complex should be A3B or AB3 type (as the conductivity

experiment indicates) the best derived formula is : [Co(NH3)4(H2O)2]Cl3Name : Tetraamminediaquacobalt (iii) chloride.

qq

Solution 1.(a) (i) Activation energy

(ii) Molarity, Molality(iii) Formaldehyde(iv) Formaldehyde and benzaldehyde

(b) (i) (2), (ii) (2), (iii) (3), (iv) (1).(c) (i) (b), (ii) (e), (iii) (a), (iv) (d).(d) (i) Fe2+ is a better reducing agent as it acquires d6 (Fe3+) configuration upon oxidation from d5 (Fe2+)

configuration, whereas there is no such stability gain takes place in case of Cr2+ (d4) getting oxidized to Cr3+ (d3).

(ii) Allosteric site is place on an enzyme where a molecule that is not a substrate may bind and changethe structure of enzyme rendering it inactive for further purposes.

(iii) It's homopolymer, since the unit —(NH—CHR—CO)n— is obtained from a single monomer unit.(iv) It has been found experimentally that osmotic pressure is proportional to the molarity, C of the

solution at a given temperature T. Thus,π = CRT, Here π = Osmotic pressure, R = Gas constant.π = (n2/V) RTHere, V is the volume of a solution in litres containing n2 moles of solute. If w2 grams of solute, ofmolar mass, M2 is present in the solution, then n2 = w2/M2 and we can write,

π V = w2RT/M2M2 = w2RT/ π V

Thus, knowing the quantities w2, T, π and V we can calculate the molar mass of the solute.Solution 2.

Osmotic pressure : π = CRT

or π = nV

RT

= wmV

RT

Given : Molecular mass of glucose = 180, w = 0.45, V = 250/1000, R = 0.821l atm K–1 mol–1, T = 273 K.

∴ π = 0 45 1000180 250

0 821 273.

.××

× × = 2.24 atm Ans.

Solution 3.(i) Glucose on treatment with nitric acid gets oxidized to give saccharic acid.

HNO3

CHO|

(CHOH)|

CH OH

4

2

COOH|

(CHOH)|

COOH

4

Glucose Saccharic acid

(ii) The main characteristics of enzymes are :1. They speed up the biological reaction upto million times.2. They are highly specific in nature.3. They are highly sensitive to pH and temperature.


11SOLVED


OR(i) (a) Xerophthalmia is caused by deficiency of Vitamin A.

(b) Scurvy is caused by deficiency of Vitamin C (Ascorbic acid).(c) Beri-beri is caused by deficiency of Vitamin B1 (thiamine).(d) Rickets is caused by deficiency of Vitamin D.

(ii) Amylose and Amylopectin are the two components of starch.Solution 4.

(i) By the reduction of acetyl chloride with hydrogen over palladium supported over BaSO4.

CH COCl H CH CHO HPdBaSO S3 2 3

4Acetylchloride

Acetaldehyde+ → +

/CCl

(ii) 2

2 2 7 2 4

3Ca(OH)[O] Heat3 3 3 3 3K Cr O H SO

Acetone3

CH COOCH CHO CH COOH Ca CH COCH CaCO

CH COO+

→ → → +Acetaldehyde

Solution 5.Structures :

Resonance structures XeF (square planar)4

OR(i) 2NaOH + Cl2 → NaClO + NaCl + H2O(ii) 3F2 + 3H2O → 6HF + O3

Solution 6.(i) Phenol is more acidic than aliphatic alcohols because the phenoxide ion formed after protonation can

be stabilized by resonance.C6H5OH → C6H5O– + H+

O|

: :..

–..

:

..–

:

–..

..

..–

O||

O||

O||

:..

In aliphatic alcohols the alkoxide ion so formed cannot be stabilized by resonance.(ii) Due to the + M effect of the –OH group the benzene ring becomes electron rich, therefore, the nitro

group which is an electrophile can easily attack the ring at these positions and shows substitution.Solution 7.

(i) Lanthanoids usually show three different oxidation states. +3 is the common oxidation states oflanthanoids, in addition to + 3, oxidation states +2 and + 4 are also exhibited by some of the lanthanoids.

(ii) Copper shows +1 oxidation state frequently because it attains d10 configuration with +1 oxidation state.Solution 8.

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 different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer. Due to orderly arrangement of carbon atoms in graphite, the magnitude of physical properties is different in different direction. Therefore graphite is anisotropic.

Solution 9.(i) Standard electrode potential is the electrode potential of a given half cell when the concentration of

ions is 1 M, temperature is 25°C and pressure is maintained at 1 atm.


(ii) For the construction of a standard hydrogen electrode which is used as a reference electrode, 1 Msolution of HCl is taken and maintained at 25°C. The electrode consisting of a platinum wire is sealedin a glass tube attached with a platinum foil at the end and is dipped in a 1 M HCl solution. Hydrogengas at one atmosphere is bubbled continuously. The gas is adsorbed on the surface of the platinum

electrode and the following equilibrium is set up 12

H2 (g) = H+ + e–.

OR(i) The half cells in a cell must be separated only if the oxidizing and reducing agent can migrate to the

other half cell. In lead storage cell, the oxidising agent, PbO2 and the reducing agent, Pb as well as theiroxidation and reduction products of PbSO4 are solids. Therefore there is no need to separate half cells.

(ii) Cu2+ + e– → Cu+ E1° = 0.15 V\ DG1° = – 1 × F × 0.15 V = – 0.15 F

Cu+ + e– → Cu E2° = 0.50 V\ DG1° = – 1 × F × 0.50 V = – 0.50 VNow Cu2+ + 2e– → Cu

DG3° = – 2F × E3° = – 2FE3°\ DG3° = DG1° × DG2°

E° = ?Solution 10.

The value of K corresponding to t1/2 = 10 minutes = 600 s is given by

K = 0 693 0 693

6001 155 10

1 2

3 1. ..

/t= = × − − second

Given : A = 4 × 1013 s – 1, Ea = 98.6 kJ mol–1

According to Arrhenius equation,K = Ae – Ea/RT

Taking log, we get log10 K = log10 A – E

RTa

2 303.

Substituting the values

log 1.155 × 10 – 3 = log 4 × 1013 – 98 6

2 303 8 314 10 3.

. .× × ×− T

– 2.9374 = 13.6021 – 5149 6.T

5149 6.T

= 13.6021 + 2.9374 = 16.5395

T = 5149 6

16 5395.

.= 311.35 K. Ans.


OR(i) For a zero order reaction k = 0.0030 mol L– 1 s– 1

[A]0 = 0.10 M. [A] = 0.075 M

K0 =

[ ] [ ]A A0 −t

t = [ ] [ ] . .

..

A As0 0 10 0 075

0 00308 33

−=

−=

t

(ii) Order of the reaction is zero because the rate of reaction is independent of the concentration of reactants.Solution 11.

(i) (a) Hydrate isomerism.(b) Ionization isomerism.

(ii) (a) In analytical chemistry : Formation of coordination compounds is largely used in analyticalchemistry. For example, in the first group of qualitative analysis white ppt. of AgCl, can be differentiatedfrom Hg2Cl2 and PbCl2 by dissolving it in NH3 solution. AgCl forms a soluble complex with ammoniawhile Hg2Cl2 and PbCl2 do not.

AgCl + 2NH3 → [Ag(NH3)2]Cl(b) Biological importance : Several coordination compounds play significant role in a number ofbiological processes. For example, chlorophyll present in plants is a complex of magnesium whichplays a central role in photosynthesis.

Solution 12.(i) Both oxygen and flourine being highly electronegative element can easily cause promotions of electrons

from filled orbitals of noble gases to vacant d-orbitals.(ii) Bleaching by Cl2 is by oxidation while SO2 bleaches by reduction. Hence product bleached by SO2 is

re-oxidised by air to its original state.(iii) XeF2 — sp3d

XeF4 — sp3d2

Solution 13.

(i) + O2Air

CH —COOH|

3

CH|

3

Dil. H SO2 4

CH(CH )23|

OH|

+ CH COCH3 3

Cumene Cumene hydroperoxide Phenol

Acetone

(ii) C H I C H ONa C H OC H2 5 2 5 2 5 2 5Ethyliodide

Sodiumethoxide

Diethyl + →

eether+ NaI

(iii) C H O C H O C H OHOHInvertase

Fr12 22 11 6 12 6 6 12 6

Sucrose Glucose

+ → +uuctose

C H O C H OH COZymase6 12 6 2 5 22 2Glucose Ethanol

→ +

Solution 14.(i) Iproniazid and Phenelzine.(ii) Tranquilizers are class of compounds used for treatment of stress and mental diseases. These relieve

anxiety, stress, irritability or excitement by inducing a sense of well being. e.g., Equanil.(iii) If taken in excess the neurologically active drugs (or for that matter any drug) can be harmful. So, a

doctor should always be consulted for appropriate usage of such drugs.Solution 15.

(i) The complete reactions are as follows :Fe3O4 + CO → 3FeO + CO2

FeO + C → Fe + CO


(ii) Leaching is a process for concentration of ores, this is used when the ore is soluble in some suitablesolvent. Leaching of alumina is done from bauxite to concentrate aluminium metal. In the metallurgyof silver and gold, the respective metal is leached with a dilute solution of NaCN or KCN.

(iii) ‘Mond Process’ is used for refining nickel.

Solution 16.(i) CH3CH2CH2Cl + KOH(aq)

→ CH3CH = CH2 + KCl + H2O

(ii)

Cl|

+ HOHSO3

Cl| HSO3

+ H O2

o-chloro benzenesulphonic acid

(iii) CH3OH + PCl5 → CH3Cl + HCl + POCl3

(iv) ArN Cl2

+ –

OH–

OH|

N=N—|

|OH

(v) KMnO

H SO Heat

4

2 4

COOH

COOH

Hexane-1, 6, dioic acid

OR(i) (a) CH3CH2—Cl + AgNO3

→ AgCl (white ppt.)CH2 = CHCl + AgNO3

→ No ppt.

(b) —NH + CHCl + 3KOH (alc.)2 3Heat

—N C + 3KCl + 3H O≡ 2

Phenyl isocyanide(foul smelling)

Carbon tetrachloride does not gives this test.(ii) Bromination reaction of phenol in solvents of low polarity (Such as chloroform or carbon disulphide)

give monobromophenols :

Br

Br

+

Phenol 2-Bromophenol(Minor)

4-Bromophenol(Major)

OH OH OH

Br in CS2 2

273 K

(iii) Aniline can be converted to aryl halide through formation of aryl diazonium salt :

NaNO + HX2

273-278 K

Aniline

NH2

Aniline

NH2Cu X22

N X2+ –

Benzene diazoniumhalide

Aryl halideX=Cl, Br

X

Solution 17.(i) A = Tin + HCl

B = NaNO2 + HCl at 0°C to 5°CC = CuCl


(ii) Amines are alkyl derivatives of ammonia. They are classified as primary, secondary and tertiaryamines, depending on the number of hydrogen atoms of ammonia replaced by the alkyl group.

R N

H

H R N

R

H R N

R

R | | |

1 2 3° ° °

CH3CH2CH2NH2 is a primary amine and on reacting with alcoholic KOH and chloroform, it will give offensive odour of isocyanide while CH3CH2NHCH3 which is a secondary amine, will not give this reaction.

OR

(i) (a) H C CN CH CH NHLiAlH3 3 2 2

4Ethane nitrile Ethanamine

→ −

(b) Ethanamine

Benzene sulphonyl chloride

N-ethylbenzenesulphonamide

—SO Cl + CH –CH NH2 3 2 2

O

S

O

||

||— — N

|H

— C H + HCl2 5

(c)

2-Bromo-4-nitrotoluene 3-Bromo-4-methylaniline

Sn/HCl

CH|

3

Br

|NO2

CH|

3

Br

|NH2

(ii) Ammonia is basic due to the presence of the lone pair of electrons on nitrogen atom. In aniline the lonepair of electron gets involved in the resonance structure of phenyl ring so becomes less available, henceless basic than ammonia whereas in methyl amine the + I inductive effect of –CH3 group renders thelone pair on nitrogen more available so its more basic than ammonia.

(3) Hinsberg’s reagent is benzenesulphonyl chloride, C6H5SO2Cl.Solution 18.

(i) Adsorption arises due to the fact that the surface particles of the adsorbent are not in the sameenvironment as the particles inside the bulk. Inside the adsorbent all the forces acting between theparticles are mutually balanced but on the surface the particles are not surrounded by atoms ormolecules of their kind on all the sides, and hence they possess unbalanced or residual attractive forces.These forces of the adsorbent are responsible for attracting the adsorbate particles on its surface.

(ii) Surfaces of certain precipitates such as silver halides have the property of adsorbing some dyes likeeosin etc., thereby producing a characteristic colour at end point. Such precipitates can be consideredas adsorption indicators.

(iii) Urease is the enzyme which convert Urea into ammonia and carbon dioxide, it can be found fromsoybean.

(iv) When FeCl3 is added to excess of hot water, a positively charged sol of hydrated ferric oxide is formeddue to adsorption of Fe3+ ions.

ORThe variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve termed as adsorption isotherm. Freundlich adsorption isotherm gives a relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed by the following equation :

x / m = k.p1/n (n > 1)where, x is the mass of the gas adsorbed on m of the adsorbent at pressure P, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.Plotting log x/m against log Ca straight line is obtained which shows the validity of Freundlich isotherm.

qq

Slope = 1/n

log p

log k (intercept)

log

x/m

Solution 1.(a) (i) Butadiene

(ii) Galena, malachite(iii) 273 K, one(iv) Endosmosis

(b) (i) (3), (ii) (3), (iii) (4), (iv) (2)(c) (i) (d), (ii) (a), (iii) (b), (iv) (c)(d) (i) Order of increasing boiling point :

0.1 M Urea < 0.2 M KCl < 0.2 M BaCl2(ii) Bases present in DNA are : Adenine, Guanine, Cytosine and Thymine.(iii) Increase in temperature has a marked effect on the rate of chemical reaction. On increasing the

temperature by 10°C, the rate of reaction nearly doubles in most of the cases. This can be explained on the basis of collision theory. The above graph shows energy distribution in molecules at twodifferent temperatures, wheretemperature T2 has high value than temperature T1. It can be inferred from the graph that on increasingtemperature more number of moleculegain high kinetic energy. E.g., in this graph, area of molecules having a particular kinetic energy range is given by the area abdc, whereas on increasing energy the area increases and becomes abfe. Which in turn refers to that on increasing temperature rate of reaction increases as more number of molecules with required energy would collide to give the reaction in positive direction.

(iv) When phenol is treated with chloroform in the presence of sodium hydroxide, a CHO group isintroduced at ortho position of benzene ring. This reaction is known as Reimer Tiemann reaction.The complete sequence of reaction is as follows :

OH|

Phenol Intermediate

aq. NaOH

CHCl3CHCl2

O Na|

+–

NaOH H+CHO

O Na|

+–

CHO

OH|

Salicylaldehyde

Solution 2.The half cell reactions can be written as :

Mg2+ + 2e– → Mg; Eel = – 2.36 + 0.0591/2 (log [Mg2+]) = – 2.27 V ...(i)

Cu2+ + 2e– → Cu; Eel = 0.34 + 0.0591/2 (log [Cu2+]) = 0.428 V ...(ii)

Considering : Cu2+ (aq) + Mg(s) → Mg2+ (aq) + Cu(s), as the cell reaction

So, Ecell will be :Ecell = 0.428 V – (– 2.27) V

= 2.69 VOR

Writing the equations :H+(aq) + e– → 1/2 H2 (g)

No. of Particleswith a givenkinetic energy

Kinetic energya b

cd

ef

T1

T2


14SOLVED


Cu2+ + 2e– → Cu (s)Zn2+ + 2e– → Zn (s)

Now, as the cells are in series, amount of current is same throughout the cells. So, in case of hydrogen, half mole of hydrogen is evolved by 1 F charge so, 0.5 F would release 0.25 mol, i.e., 0.25 × 1 = 0.25 g H2Similarly, 2 F charge is required to deposit 63.5 g of copper,

0.5 F would give = 63.5/2 × 0.5 g = 15.87 g of copper

Similarly, 2 F charge is required to deposit 63.5 g of zinc, 0.5 F would give = 65/2 × 0.5 g

= 16.25 g of zinc.Solution 3.

Given : Weight of solute, w2 = 2.175 gWeight of solvent, w1 = 39.0 g

Molar mass of solvent, benzene = 78 g mol–1

Initial pressure, P1, of pure benzene = 640 mm HgPressure of solution, p1 = 600 mm Hg

Relative lowering of vapour pressure can be given the below formula : P1 – p1 /P1 = w2 ×M1/M2× w1

Putting the given values :(640 – 600)/600 = (2.175 g × 78 g mol–1)/ M2 × 39.0 g

0.067 = 4.35/ M2M2 = 65 g mol–1

Solution 4.(i) Glasspanesfixedtowindowsordoorsofveryoldbuildingareinvariablyfoundtobeslightlythicker

atbottomthanonthetop.Thisisbecausetheglassisamorphousinnatureandhasatendencytoflowlikeliquids,thusitflowsdownveryslowlyandmakesthebottomportionthickerinlongrun.

(ii) There are only 14 possible three dimensional lattices. These are called Bravais lattices.OR

(i) The three possible variations of a cubic crystal system are 1. primitive, 2. body centered and 3. facecentered cubic systems.

(ii) Various combinations of n-type and p-type semiconductors are used to prepare electronic componentstransistors made by sandwiching a layer of one type of semiconductor between two layers of the othertype of semiconductor, there could be npn and pnp, two types of transistors.

Solution 5.(i) HCl is not recommended as a co-reagent with KMnO4 due to the fact that HCl is oxidized by KMnO4

to Cl2, hence depleting both the reagent in reaction.(ii) CrO < Cr2O3 < CrO3.

Solution 6.(i) Polyacrylonitrileisusedforpreparingsyntheticfibresandsyntheticwool,itsmonomerhasmolecular

formula CH2 = CHCN (Acrylonitrile).(ii) Based on molecular forces neoprene is elastomer.(iii) Benzoyl peroxide acts as initiator in polymerization of ethene to polythene.

Solution 7.

S. No. Physisorption Chemisorption

1. It takes place due to van der Waals’ forces andnotspecificinnature.

2. Enthalpy of adsorption is low (20 – 40 kJ mol–1) in this case.

3. Low temperature is favourable for adsorption. It decreases with increase in temperature and is reversible in nature.

It is caused by chemical bond formation and highlyspecificinnature.Enthalpy of adsorption is high (80 – 240 kJ mol–1) in this case.High temperature is favourable for adsorption. It increases with the increase in temperature and is irreversible in nature.

Solution 8.(i) The alcohols in which the –OH group is attached to a sp3 hybridized carbon next to the carbon-carbon

double bond, that is to an allylic carbon, are known as allylic alcohols.Example : CH2= CH–CH–OH (Prop-2-ene-1-ol)

Chemistry | 3 |(ii) IUPAC names are :

(a)

CH3

H C3

O – CH3

2-Methoxy-1, 1-dimethylcyclohexane

(b)

Cl

NO2|

|OH

H3C

5-Chloro-2-methyl-4-nitrophenolSolution 9.

(i) There are some substances which at low concentrations behave as normal strong electrolyte, but athigher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregatedparticles thus formed are called micelles. Micelles turn into normal electrolytic solution on dilutionhence, they differ from normal colloidal solution which is a heterogeneous system of one substancebeingdispersedasveryfineparticlesinanothersubstancecalleddispersionmedium.

(ii) Soap is sodium or potassium salt of higher fatty acid and may be represented as RCOO–Na+. Whendissolved in water it dissociates into RCOO– and Na+ ions, so at lower concentrations soap behavesas a normal electrolyte. However, at higher concentrations, an aggregate is formed known as ‘ionicmicelle’, which can be described as follows. The RCOO- ions have a hydrocarbon chain (R) which isnon-polar (hydrophobic) and the COO– group which is polar (hydrophilic). At higher concentrations(critical micelle concentration), the anions are pulled in the bulk of the solutions and aggregate to forma sphere where the hydrocarbon part points towards the centre of sphere, these aggregates are knownas ‘ionic micelle’.

Solution 10.(i) Addition of bromine in CCl4 to an alkene gives vicinal dihalide as product.

1, 2-dibromoethane

Ethene

H

H

H

H

+ Br2

CCl4BrCH –CH Br2 2

(ii) Propene would add HI along the double bond, to give 2-iodopropane as major product according toMarkovnikov’s rule. But if a peroxide is added to the reaction mixture, the propene would undergoanti-Markovnikov’s addition of HI along the double bond to give 1-iodopropane as the major product.

CH2

+ HIPeroxide

H C3 H C3

H

I

Propene1-Iodopropene

Solution 11.(i) Crystal Field Theory describes important properties of complexes (magnetism, absorption spectra,

oxidation states, coordination,) considering that the ligands are point charges and d-orbitals of thecentral atom interacts with ligands. According to CFT, the attraction between the central metal andligands in a complex is purely electrostatic.

(ii) Theenergyofthefivedegenerated-orbitals (dxy, dzx, dyz, dx2-y2 and dz2) changes due to the surroundingligands (considered as point charges). As a ligand approaches the metal ion, the electrons from theligand will be closer to some of the d-orbitals and farther away from others, causing a loss of degeneracy in the d-orbitals. The electrons in the d-orbitals and those in the ligand repel each other due to repulsionbetween like charges. Thus, the d-electrons closer to the ligands will have a higher energy than thosefurther away, which results in the d-orbitals splitting in energy. So, two sets of orbitals are thus formed.This splitting is affected by nature and oxidation state of ligands, size of d-orbital and geometry of thecomplex.

(iii) Crystalfieldsplittingenergy(D) is the energy difference between two sets of d-orbitals (of metal atom)formed after splitting in presence of ligands. For example, in octahedral complexes, d-orbitals split in


two energy levels eg (higher than degenerate level) and t2g (lower than degenerate level). The eg level has d d

x y z2 2 2− and (pointing towards the axes along the direction of the ligand) and t2g level has dxy,

dyz and dxz orbitals (directed between the axis).dxz dyz dxy

dx –y22 dz2eg

Degenerate -orbitald

∆

t2g

ORCoordination number of a metal ion in a complex is the number of ligand donor atoms to which the metal atom is directly bonded (through sigma bond).

(i) Coordination number of [Pt(NH3)2Cl2] is 2(ii) Coordination number of [Ag(NH3)2Cl] is 2(iii) Coordination number of K4 [Fe(CN)6] is 6

Solution 12.(i) D-Glucose is an aldohexose.

On heating with HI, it forms n-hexane :

HI4 3

2

CHO|

(CHOH) H C|CH OH

∆→

CH3

n-Hexane

Glucose

(ii) Glucose gets oxidized to six carbon carboxylic acid (Gluconic acid) on reaction with bromine waterwhich is a mild oxidizing agent.

CHO

CHOH

CH OH

COOH

CHOH

CH OH

Br Water|

( )|

|( )|

4

2

4

2

2

Glucose Gl

→

uuconic acid

(iii) Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid (HNO3) :

CHO

CHOH

CH OH

COOH

CHOH

COOH

HNO|

( )|

|( )|

4

2

43

Glucose Sacchari

→

cc acid

Solution 13.

Sulphuric acid is manufactured industrially by the “Contact Process” which involves three steps :

1. Burning of sulphur or sulphide ores in air to generate SO2. The SO2producedispurifiedbyremovingdust and other impurities such as arsenic compounds.

2. Conversion of SO2 to SO3 by the reaction with oxygen in presence of a catalyst (V2O5), this is the keystep of the process. The reaction is exothermic, reversible and the forward reaction leads to a decreasein volume. Therefore, low temperature (should not be very low) and high pressure are the favourableconditions for maximum yield.

2 22 2 32 5SO g O g SO gV O( ) ( ) ( )+ →

3. Absorption of SO3 in H2SO4 to give oleum which is diluted with water to give desired concentration ofsulphuric acid.

SO3 + H2SO4 → H2S2O7 (Oleum)

The sulphuric acid obtained by contact process is 96-98% pure.

Chemistry | 5 |OR

(i) Upper atmosphere has layer of ozone molecules (O3) which saves earth from the harmful ultravioletrays coming from sun. Experiments have shown that nitrogen oxides (particularly nitric oxide) combine very rapidly with ozone and thus, the nitrogen oxides emitted from the exhaust systems of supersonicjet aeroplanes slowly deplete the concentration of the ozone layer in upper atmosphere.

NO(g) + O3(g) → NO2(g) + O2(g)Similarly,Freons(chlorofluorocarbons)usedinaerosolspraysandrefrigerantsalsodepletetheozonelayer.

(ii) All the halogens exhibit –1 oxidation state. However, chlorine, bromine and iodine exhibit + 1, + 3,+ 5and+7oxidationstatesalso.Belowarethedifferentexcitedstateelectronicconfigurationtoexplainthe different oxidation states.

Ground state, –1 or + 1

oxidation state

3 unpaired electrons, + 3

oxidation state


oxidation state


oxidation state

ns np nd

ns np nd

ns np nd

ns np nd

Thehigheroxidationstatesofhalogens(exceptfluorine)arerealizedmainlywhenthehalogensarein combinationwith the small andhighly electronegativefluorine andoxygenatoms.Thefluorineatom has no d-orbitals in its valence shell and therefore, cannot expand its octet. Being the most electronegative, it exhibits only –1 oxidation state.

Solution 14.(i) Uracil is the nitrogenous base which is present in RNA but not in DNA. Its structure is :

NH

ON

H

O

Uracil

(ii) (a) XerophthalmiaiscausedbydeficiencyofvitaminA.(b) ScurvyiscausedbydeficiencyofvitaminC(ascorbicacid).(c) Beri-beriiscausedbydeficiencyofvitaminB1 (thiamine).(d) RicketsiscausedbydeficiencyofvitaminD.

(iii) Amylose and amylopectin are the two components of starch.Solution 15.

Let us write the available information :Radius (r) = 248 pm = 2.48 × 10–8 cm

As it is a bcc structure, no. of atoms per unit cell, z = 2 Atomic mass of Pd = 106 Now, Atomic radius (r) = 3 4a/

Volume, V = a3

Density (d) = (z × M)/(NA × V)So, applying Atomic radius (r) = 3 4a/

2.48 × 10–8 cm = 1.732 a/4a = 2.48×10–8 cm × 4/1.732


= 5.727 × 10–8 cmNow, Volume, V = (5.727 × 10–8 cm)3

= 18.78 × 10–23 cm3

Now, Density, d = 2 × 106/(6.023 × 1023 × 18.78 × 10–23)d = 1.87 g/cm3

Solution 16.(i) Ethyl bromide to ethyl amine :

Ethanolic NH3

Ethyl bromide Ethyl amine

H3C Br NH2H3C

(ii) Phenol to picric acid :

Phenol

Phenol-2, 4-disulphonic acid 2, 4, 6-Trinitrophenol

(Picric acid)

Conc. H SO42

OH

|

OH

| NO2

|

NO2

O N2

OH

| SO H3

|

SO H3

Conc. HNO3

(iii) Benzene to aniline

H SO42

HNO3

H /Pd2

Ethanol

NO2 NH2

Benzene Nitrobenzene Aniline

(iv) Aniline to benzene diazonium chloride

Benzene diazonium chloride

NH2

Aniline

NaNO2

HCl

N Cl2+ –

OR(i) Ammonia is basic due to the presence of the lone pair of electrons on nitrogen atom. In aniline the lone

pair of electron gets involved in the resonance structure of phenyl ring so becomes less available, henceless basic than ammonia whereas in methyl amine the +I inductive effect of –CH3 group renders thelone pair on nitrogen more available so its more basic than ammonia.

(ii) Hinsberg’s reagent is benzenesulphonyl chloride, C6H5SO2Cl.

(iii) Ethyl chloride

H3C ClNaCN

Ethanol

Reduction

Propane nitrile

H3C CN1-Aminopropane

H3CNH2

(iv) Gabriel phthalimide synthesis gives aliphatic primary amine only, it can’t be used for the preparationof 2º and 3º amines.

(v) During direct nitration of aniline in presence of nitric acid and sulphuric acid (highly acidic medium),aniline is protonated to form anilinium ion which is meta directing, that’s why, besides o- and p- derivativessignificantamountof m- derivative is also formed.

HNO /H SO43 2

288 K

Benzene

NH2|

NH2|

|

NO2p-Nitroaniline

m-Nitroaniline

+

NH2|

NO2

+

NH2| NO2

o-Nitroaniline

Solution 17.(i) The aldehydes and ketones having at least one a-hydrogen undergo a reaction in the presence of dilute

alkali as catalyst to form a-hydroxy aldehydes (aldol) or b-hydroxy ketones (ketol), respectively. Thisis known as aldol reaction.

Chemistry | 7 |

dil. NaOH

Ethanal

OH

O3-Hydroxybutanal

H C3 OH C3

(ii) The aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction(disproportionation) reaction on heating with concentrated alkali, to give a molecule of carboxylicacid and a molecule of alcohol respectively. Formaldehyde and benzaldehyde undergo Cannizzaroreaction.

Methanol

Formaldehyde

= O + + Conc. KOHH

H

H

H

H C OH + HCOOK3Potassium acetate

O

(iii) Aldehydes are more reactive than ketones due to steric and electronic reasons. Sterically, the presenceof two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbonthan in aldehydes having only one such substituent. Electronically, aldehydes are more reactive thanketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectivelythan in former.

(iv) Chloroacetic acid is stronger acid than acetic acid. The chlorine atom attached to a-carbon (to carbonylcarbon) imparts a –I inductive effect, that is it withdraws the bond pair of electrons towards it (dueto its high electronegativity) and thus facilitates the removal of proton from chloroacetic acid. It alsostabilizes the carboxylate anion by dispersing the negative charge developed on it after loss of proton.

OR(i) Rosenmund reaction.(ii) Stephen’s reduction : Stephen aldehyde synthesis reaction involves the preparation of aldehydes

(R–CHO) from nitriles (R–CN) using tin (II) chloride (SnCl2), hydrochloric acid (HCl) and quenchingthe resulting iminium salt ([R–CH = NH2]+Cl–) with water (H2O).

(iii) X is benzaldehyde, C6H5CHO. When benzene is reacted with HCl and carbon monoxide in presence ofanhydrous aluminium chloride benzaldehyde is formed.

(iv) The descending order of all four of the provided compounds is :Butanoic acid > 2-hexanone > 2-pentanone > 3-methyl-2-butanone.

(v) Isopropyl acetone is a commonly used solvent and is the active ingredient in nail polish remover andsome paint thinners.

Solution 18.(i) Considering the homogeneous gaseous reaction :

Consider the moles of HI formed as n HI = xMoles of H2, nH2 = 0.2 – (x/2)

and Moles of I2, nI2 = 0.2 – (x/2)Kp = x2 / [0.2 – (x/2)]2

49.5 = {x/ [0.2 – (x/2)]}2

x = 0.31So, no. of moles at equilibrium :

Moles of H2, nH2 = moles of I2, nI2 = 0.2 – (0.31/2) = 0.045

Molar concentration is given by c= nV (considering this an ideal gas mixture)Putting the values, the molar concentration of hydrogen and iodine is :

= 0.045 × 10 = 0.45 mol L–1

And, molar concentration of hydrogen iodidec = 0.31 × 10

= 3.1 mol L–1


(ii) Relation between Kp and Kc can be given by the following equation :Kp = Kc (RT)Dn

Where, ∆n is the difference between the number of moles of reactant and product.(iii) The plot of ln k vs. 1/T gives a straight line according to the equation given below :

ln k = – Ea/RT + ln A

ln k

1/T

Slope = – E /Ra

Intercept = Aln

A plot between k and 1/Tln

Activation energy can be calculated by the value of slope of the graph.OR

(i) The energy distribution in molecules at two different temperatures can be shown through the belowgraph. Where temperature T2 has high value than temperature T1.It can be inferred from the graph that on increasing temperature more number of molecule gain highkinetic energy. Example : in this graph, area of molecules having a particular kinetic energy range isgiven by the area abdc, whereas on increasing energy the area increases and becomes abfe. Which inturn refers to that on increasing temperature rate of reaction increases as more number of moleculeswith required energy would collide to give the reaction in positive direction.

No. of particleswith a given

kinetic energy

Kinetic energya b

c

d

ef

T1

T2

(ii) Order of the reaction is sum of powers to which the concentration terms are raised in a rate lawexpression, hence in this case :

Order = + −32

1( )

=12

Hence, the reaction is half order reaction.qq

Solution 1.(a) (i) Tetra ethyl lead

(ii) Coulometer, electrochemical cell(iii) CN, Fe(iv) Chloroxylenol and α-terpineol

(b) (i) (3), (ii) (1), (iii) (2), (iv) (4).(c) (i) (c), (ii) (d), (iii) (a), (iv) (b)(d) (i) Preparation of glycerol from oils :

CH OOCR

CH OOCR

CH OOCR

CH OH

CH OH

CH O

H O

2

2

2

2

2

2

2|

|

|

|

(Oil or Fat)

→

HH

RCOOH

Glycerol( )

+ 3Fatty acid

(ii) Lucas test : Lucas reagent is a mixture of ZnCl2 and HCl. On reaction alcohols are turned tothe corresponding alkyl halide giving turbidity to reaction solution due to precipitation of theinsoluble alkyl chloride.

Alcohol type Treatment with Lucas Reagent

Primary Takes a lot of time to react and show turbidity.

Secondary Reacts to give turbidity in some time like in 5 minutes.

Tertiary Reacts immediately to give turbidity.

(iii) The polymer A—B—B—A—A—A—B—A— is a copolymer.(iv) Mechanism of esterification : Esterification is the formation of esters from carboxylic acids and

alcohols in the presence of an acid acting as the catalyst. Let’s consider the following example :

R—C

O

OH

H+

R—C

OH+

OH

R –OH′..

R—

OH|C—O—R|OH

′.

|H: ..

Tetrahedral intermediate

Protontransfer

Carboxylic acid

Ester

R—C

O

O–R′

– H+

R—C

OH+

O–R′

–HOH

Protonated ester

R—

OH|C—O—R|OH:

H

:

Solution 2.Differences between crystalline and amorphous solids are as follows :

Crystalline Amorphous

1. A crystalline solid has a definite and regulargeometry, due to definite and orderlyarrangement of molecules or atoms in threedimensional space.

An amorphous solid does not have any pattern of arrangement of molecules or atoms and thus does not have any definite shape.


17SOLVED


2. A crystalline substance has a sharp meltingpoint.

An amorphous substance does not have a sharp melting point.

3. Crystalline solids are anisotropic in nature and their physical properties such as electricalresistance, refractive index etc., show differentvalues when measured along different directions in the same crystal.

Amorphous solids are isotropic in nature and their physical properties such as electrical resistance, refractive index etc., have same values when measured along different directions in the same substance.

OR(i) The primitive cublic unit cell consists of one atom at each of the 8 corners. Each atom is thus shared

by 8 unit cells. Hence, n = 8x (1/8) = 1.(ii) Vacancy defect the interstitial defect combine together to form Frenkel defect.

Solution 3.(i) Condensation polymerization : This type of polymerization involves a repetitive condensation reaction

between two bi-functional monomers. These polycondensation reactions may result in the loss of somesimple molecules as water, alcohol etc.Copolymerization : It is a polymerization reaction in which a mixture of more than one monomericspecies is allowed to polymerize and form a copolymer. The copolymer can be made not only bychain growth polymerization but by step growth polymerization also. It contains multiple units of eachmonomer used in the same polymeric chain.

(ii) The formation of Terylene or Dacron by the interaction of ethylene glycol and terephthlic acid is anexample of condensation polymerization.

n nHOH C–CH OH +2 2

O

HO— —

O

OHOCH –CH –O–C– –C—2 2

O||

O||

—n

Terelyne or DacronEthylene glycolTerephthalic acid

Solution 4.(i) Concentrated H2SO4 is diluted by adding the acid in small portions to water. When acid is mixed with

water, the acid ionizes to H+ and SO42– ions which are hydrated in presence of water and release a

large amount of heat energy. When water is in excess, this heat is absorbed by the water (acting as sinkfor the heat evolved) preventing boiling and splattering.

(ii) Xenon has large radius compared to other noble gas atoms, therefore, the electron attraction to thenucleus is weaker in comparison to the smaller noble gases. When it gets close to other atoms, itselectron cloud gets distorted easily (due to large distance from the nucleus and screening of the innershells) and achieves a dipole moment which makes it suitable to create polar covalent bonds.

ORAqua regia is a mixture of concentrated nitric acid (1 part by volume) and concentrated hydrochloric acid (3 parts by volume). This is a yellow brown fuming liquid. Aqua regia is used for dissolving noble metals like gold and platinum.

Au + 4H+ + NO3– + 4Cl– → AuCl4– + NO + 2H2O3Pt + 16H+ + 4NO3– + 18Cl– → 3PtCl6

2– + 4NO + 8H2OAqua regia is also used in etching and in specific analytic procedures. It is also used in laboratories to clean glassware of organic compounds and metal particles.

Solution 5.(i) The molecules of Butan-1-ol form hydrogen bonds with each other and need higher energy for boiling

whereas the molecules of ethers have no such association within the molecules, that’s why boilingpoint of ethoxy ethane is less than that of butanal.

(ii) The chemical reagent which can distinguish between an aldehyde and a ketone qualitatively is2, 4-Dinitrophenylhydrazine or 2, 4-DNP. Formation of orange coloured arylsulphonamides take place with aldehydic group whereas there is no reaction with ketones due to the absence of α-hydrogen atomin ketones.


(i) The two requirements are :1. The metal to be refined should form a volatile compound with an available reagent.2. The volatile compound should be easily decomposable, to have an easy recovery.

(ii) The term benefaction means removal of unwanted materials from the ores.Solution 7.

The reasons are as follows :(i) Silica gel packets absorb water from nearby air and thus keep the photographs dry and safe.(ii) Whipped cream is made by dispersing air (gas) into liquid (cream), so it’s a colloid.

Solution 8.According to Henry’s law, the pressure of a gas over a solution in which gas is dissolved is proportional to the mole fraction of the gas dissolved in the solution :i.e., p2 = kx2Given, Standard pressure or p2 = 760 torrSolubility of H2S at STP or n2 = 0.195 m, i.e., 0.195 mol per litre of solution.Now x2 = n2/(n1 + n2), now considering the number of moles of gas H2S dissolved (n2 is negligible compared to the moles of solvent n1)

x2 = n2/n1For water, n1 = 1000 g/18 g mol–1 = 55.5 mol

x2 = 55.5/0.195 = 284.6

So, substituting the values :760 torr = k × 284.6

So, Henry’s law constant is :k = 2.67 torr.

Solution 9.Corrosion of iron is its oxidation by loss of electron to atmospheric oxygen and formation of iron oxides. This is also known as rusting and occurs in presence of water and air. The electrochemical phenomena of rusting or corrosion can be shown as :Anode : 2Fe (s) → 2Fe2+ (aq) + 4e–

Cathode : O2 (g) + 4H+ (aq) + 4e– → 2H2O (l)The overall reaction being :

2Fe (s) + O2 (g) + 4H+ (aq) → 2Fe2+ (aq) + 2H2O (l)The ferrous ions thus obtained are oxidized further by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3.xH2O).

OR(i) Given :

T = 298 K, PH2 = 2 atm, H+ = 0.02 M, E° for H-electrode at 298 K is 0.0 V.The Nernst equation can be written as :

EE H+/ 2

= ERT

FHPE H

H+

°+

+

/

ln[ ]

22

2

n

Putting the values :E = 0 + 0.059 ln [0.02 /2]

= 0.059 - 4.6 V = – 4.54 V

(ii) Given :Current = 4.0 A

Reaction equation :Fe2+ + e– → Fe3+, so here 1 F or 96487 C of charge is required

So, putting the given values in equationCharge = Current × Time96487 C = 4.0 A × time

Time = 24121 s = 6.7 hrs


Solution 10.(i) Acetone reacts with alkali and iodine solution to give iodoform and sodium acetate.

O

H C3 CH3

I /NaOH

Iodoform reaction2 CHI + CH COONa33

Iodoform

Acetone

(ii) 2 2

2 4 4

3 3

H O H /H OHCN3dil. H SO HgSOAcetylene Acetaldehyde(A)

CyanohydrinIntermediate(C)(B)

H C H C OHO

HC CH CH CHO

CNH HCN

+

−

+

≡ → → →

Solution 11.Given, A = 4.4 × 10-12

Ea = 99.1 kJ mol–1 = 99100 J mol–1

t/1/2 = 7 min = 420 sFor first order reactions :

Rate constant, k = 0.693/t1/2The Arrhenius equation is given as :

k = A e –Ea/RT

or, ln k = ln A – Ea/RTSubstituting the given information :

ln 0.693 /420 s = ln 4.4 × 1012 – (99100 J mol–1 / 8.314 JK–1 mol × T)– 6.4 = 29.11 – 11919 /T

– 35.51 = – 11919 / TT = 335 K

So, at 335 K its half life would be 7 minutes.Solution 12.

(i) Co

NH

NH3

3 Cl

NH3

Cl

Cl

mer-triamminetri-chlorido cobalt(III)

Co

N

OH2 OH2

OH2

||O

O||

O—N

O = N|O

fac-triaquatrinitro-N-cobalt (III)

O

(ii) Geometric isomers are two or more coordination compounds which contain the same number andtypes of atoms, and bonds (i.e., the connectivity between atoms is the same), but which have differentspatial arrangements of the atoms.For example,


(i) (a) Sodium chloride, water and sulphur dioxide are formed when dil. HCl is added to sodium sulphite :Na2SO3 + 2HCl → 2NaCl + H2O + SO2

(b) Fluorine reacts with hydrogen sulphide to give hydrogen fluoride and HS radical.F + H2S → HF + HS

(ii) High electronegativity of 17th group elements : Electronegativity is the ability of an atom toattract other electrons. Atoms that have high electronegativities will attract electrons strongly. Theelectronegativity of a bonded atom is defined as its relative tendency (or ability) to attract the sharedelectron pair towards itself.Electronegativity increases as we move down from left to right across a period. This is because theatomic size decreases across the period and nuclear attraction over electrons increases. So, elementsof group 17 have highest electronegativity compared to the elements of corresponding periods. Also,fluorine being the smallest in the group 17 elements has the highest electronegativity.

Solution 14.(i) Lanthanoid contraction : With the successive filling of the inner orbitals 4f, there is a gradual decrease in

the atomic and the ionic sizes of inner transition elements along the series. This is known as ‘lanthanoidcontraction’. This phenomenon can be attributed to the imperfect shielding of one electron by anotherin the same sub-shell. The net result of the lanthanoid contraction is that the second and the third dseries exhibit similar radii and have very similar physical and chemical properties.

(ii) Shielding effect in transition elements : ‘When the attraction between the outermost electron and thenucleus decreases due to the presence of inner electrons, the effect is called the shielding effect.” Or theelectrons in the outer or remote shells are not attracted to the nucleus with a force of attraction that isequal to the electrons in the inner shells. Hence, these electrons are more loosely bound to the nucleus,simply, the inner electrons ‘shield’ the nucleus from the outer electrons. In the d-block transitionelements, ions of the same charge in a given series show progressive decrease in radius with increasingatomic number, this is because with increasing atomic number the electron is added to a d-orbital hereand shielding effect of a d-orbital is not that effective because of its shape. The net electrostatic chargebetween the nucleus and the outermost electron increases and size decreases. Same pattern is observedin atomic radii of a given series as well.

Solution 15.Given :

Osmotic pressure = 2.0 atmosphereTemperature = 300 K

Kf = 1.86 K kg mol–1

R = 0.0821 L atm K–1 mol–1

Osmotic pressure can be given by :p = i M R T

(where i is Van’t Hoff factor and is 1 for urea, there is no association or dissociation) and M is the molar concentration of solution in mol/L.So, putting the values :

2.0 atm = M × 0.0821 L atm K–1 mol–1 × 300 KM = 0.08 mol L–1

Now, let’s assume that the molarity and molality (m) of solution is same.So, the depression in freezing point ∆Tb = Kf m

= 1.86 K kg mol–1 × 0.08 mol kg–1

= 0.148 KNow, freezing point of water is 273 K hence the freezing point of aqueous solution would be :

273 – 0.148 = 272.85 KOR

Given, Vapour pressure of urea solution (p1) at 100°C= 736.2 mmVapour pressure of water (p1°) at 100 °C = 760 mmNow, osmotic pressure, p = i MRT (where i is Van’t Hoff factor and is 1 for urea, there is no association or dissociation)


The vapour pressure of water (pure solvent) is 760 mm at 100 °CMolarity = (p1° – p1) / p1° × 1000/M2Molarity = (760 – 736.2)/760 × 1000/60 (Molecular mass of Urea)

= 0.5As, density of water is 1, molarity is equal to molality here :

p = CRT (where C is molarity) = 0.5 mol L–1 × 0.0821 L atm K–1 mol–1 × 288 K = 11.8 atm

So, Osmotic pressure is 11.8 atm at 15 °C (288 K). Solution 16.

(i) (a) Acetamide (CH3CONH2) undergoes Hofmann degradation in presence of bromine and NaOH togive methanamine.

CH3CONH2 + Br2 + 4NaOH → CH3NH2 + Na2CO3 + 2NaBr + 2H2O(b) Ethylamine being a primary amine on heating with chloroform and alcoholic potassium hydroxide

form foul-smelling isocyanides or carbylamines.

CH CH NH CH NCCHCl

Methyl3 2 2 3

3Ethyl Amine

Alc. KOH

Isocyan →

iide

(ii) Formaldehyde, acetaldehyde and acetone being carbonyl compounds undergo nucleophilic additionreaction with ammonia, to give the corresponding imines.

+ NH H3

O

HH

O

H

+ H O2

Formaldehyde Methanimine

+ NH H3 3CO

HH C3

NH

H

+ H O2

Acetaldehyde Ethanimine

+ NH H3 3CO

CH3

H C3

NH

CH3

+ H O2

Acetone Propan-2-imine

OR

(i) (a)

NH

|2

Aniline

(CH CO) O3 2

Pyridine

NHCOCH

|3

N-Acetylaniline

(b) + (CH ) NH2 →3

SO Cl|

2

Benzene sulphonylchloride

S|

O O

— N

CH3

CH3

N-Dimethylbenzenesulphonamide

(c)

N Cl|

2+ –

+ CH CH OH3 2 + N + CH CHO + HCl2 3

Benzene diazoniumchloride

Benzene

Chemistry | 7 |(ii) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would

react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’tundergo this reaction.

N Cl|

2

+ –

+ —NH + Cl + H O–

2 2

Benzene diazoniumchloride

H+

—N = N —

p-Aminoazobenzene

(yellow dye)

NH|

2

Aniline

(iii) In increasing order of pKb values : C2H5NH2 < C6H5NH2 < C6H5NH2CH3

Solution 17.(i) (a) Polysaccharides : Polysaccharides are food storage materials and most commonly found

carbohydrates in nature. These are the compound which are formed of large number of monosaccharide units joined together by glycosidic linkages. Example : Starch, main storage polysaccharide of plants.

(b) Denatured protein : Proteins have an unique three dimensional structure in their native form. Ifthe native form of protein is subjected to any physical change (such as temperature change) or anychemical change (such as change in pH), the hydrogen bonds are disturbed. Due to this globuleunfold and helix get uncoiled and protein loses its biological activity. This is called denaturationof protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structureremains intact. Coagulation of egg white is an example of denaturation of protein.

(c) Essential amino acids : The amino acids which are not synthesized in our body and have to beobtained through diet are known as essential amino acids. Example : Tryptophan.

(ii) Glucose gets oxidized to six carbon carboxylic acid (Gluconic acid) on reaction with bromine water which is a mild oxidizing agent.

CHO

CHOH

CH OH

COOH

CHOH

CH OH

|( )|

|( )|

4

2

4

2

2

Glucose

Br Water

G

→

lluconic acid

OR(i) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid :

CHO

CHOH

CH OH

COOH

CHOH

COOH

|( )|

|( )|

4

2

43

Glucose

Conc. HNO →

SSaccharic acid

(ii) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) andbasic (amino group) in the same molecule. So, in basic medium the carboxyl group can lose a protonand in acidic medium amino group can accept a proton.

(iii) Difference between α-helix and β-pleated sheet :

In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a righthanded screw (helix) with the –NH group of each amino acid residue hydrogen bonded to the –C=Oof an adjacent turn of the helix (intra molecular bonding), whereas in β-structure all peptide chainsare stretched out to nearly maximum extension and then laid side by side which are held together byintermolecular hydrogen bonds (intermolecular bonding).

(iv) Glucose is also known as dextrose and is an aldohexose.Solution 18.

(i) (a) H C3

Br

+ KCNC H OH

2 5H C

3

N


(b) + Cl — ClFe

CH|

3

+

CH|

3

Cl

CH|

3

|Cl

(ii)

H C3

H C3CH3

Cl

(iii) H C3

CH3

Br

OH–

H C3

CH2CH

3 H C3

+

2-BromopentanePent-2-ene 81% Pent-1-ene 19%

(iv) There are four different types of hydrogen atoms are present in the substrate, replacing which following four monochlorinated products are formed.(CH3)2CHCH2CH2Cl, (CH3)2CHCH(Cl)CH3, (CH3)2C(Cl)CH2CH3, CH3CH(CH2Cl)CH2CH3

OR(i) (a) 2, 2-Dimethylpropane has all the hydrogen atoms equivalent and replacement of any one to yield

a monobrominated product would be same product structurally.CH3

CH3

H3C CH3

2, 2-Dimethylpropane

(b) n-Pentane has three type of carbon atoms, marked as a, b and c types below :CaH3CbH2CcH2CbH2CaH3

Where, there are 6 equivalent ‘a’ type hydrogen atoms, 4 equivalent ‘b’ type hydrogen atoms and 2 equivalent ‘c’ type hydrogen atoms are there, hence there would be three isomeric monobrominated products are possible if n-Pentane undergoes monobromination reaction.

(ii) (a) CH3CH2Br + AgF → CH3CH2—F + AgBr

(b) + Na + CH CH Cl3 2

Ether

Cl|

CH3

+ NaCl

(c) H C3

CH2

+ H — BrPeroxide

H C3 Br|

H

qq

Documents

SOLVED Model Specimen Paper 2 · 2019-11-28 · (ii) When NaCl is added to hydrated ferric oxide sol, it dissociates to Na+ and Cl-ions. Particles of ferric oxide sol are positively