156/1/3 P.T.O.

Series : OSR/1

Roll No.

Code No. 56/1/3Candidates must write the Code onthe title page of the answer-book.

Please check that this question paper contains 15 printed pages. Code number given on the right hand side of the question paper should be written on the title page of the answer-

book by the candidate. Please check that this question paper contains 29 questions. Please write down the Serial Number of the questions before attempting it. 15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m.

From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not write any answer on theanswer script during this period.

Chemistry (Theory)[Time allowed : 3 hours] [Maximum marks : 70]

General Instructions:

(i) All questions are compulsory.

(ii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each.

(iii) Question numbers 9 to 18 are short-answer questions arid carry 2 marks each.

(iv) Question numbers 19 to 27 are also short-answer questions and carry 3 marks each.

(v) Question numbers 28 to 30 are long-answer questions and carry 5 marks each.

(vi) Use Log Tables, if necessary. Use of calculators is not allowed.

Studymate Solutions to CBSE Board Examination 2013-2014

DISCLAIMER : All model answers in this Solution to Board paper are written by Studymate Subject Matter Experts.This is not intended to be the official model solution to the question paper provided by CBSE.The purpose of this solution is to provide a guidance to students.

256/1/3 P.T.O.

1. Give one example each of lyophobic sol and lyophilic sol.

Sol. Lyohilic Sol gelatin

Lyophobic sol Metal sulphides (AS2S

3)

2. Write the IUPAC name of the compound.

3 2 3| ||

CH CH CH C CH

OH O

Sol. 4-hydroxypentan-2-one

3. What type of intermolecular attractive interaction exists in the pair of methanol and acetone ?

Sol. H-bounding

H C – O3

H O = CCH3

CH3

4. Which of the following is more stable complex and why ?

[Co(NH3)6]3+and [Co(en)

3]3+

Sol. [(Co(en)3]3+ in more stable than [Co(NH

3)6]3+ because ‘en’ i.e. ethan-1, 2-diamine is a bidentate

chelating ligand which leads to stability.

Co

NH3

NH3

NH3

NH3

NH3

NH3

Co

en

enen

3+3+

5. Arrange the following in increasing order of basic strength :

C6H

5NH

2, C

6H

5NHCH

3, C

6H

5N(CH

3)2

Sol. In vapour phase order of basic strength will be :

C6H

5NH

2 < C

6H

5NHLH

3 < C

6H

5N(CH

3)2

Because 3°amines are more basic in aquesous phase order of basic strength will be

C6H

5N(CH

3)2 < C

6H

5NH

2 < C

6H

5NH(CH

3)

becuase of polar solvent solubility increases in this order.

H C – N – CH3 3

<

NH2

<

H– N – CH3

6. Name the products of hydrolysis of sucrose.

Sol. Sucrose on hydrolysis give glucose and fructose.+H

12 22 11 6 12 6 6 12 6Glucose Fructose

C H O C H O C H O

7. Which of the following isomers is more volatile :

o-nitrophenol or p-nitrophenol ?

Sol.

O – H

N O||

O- nitrophenol

O

OH

N = 0O

p-nitrophenol

356/1/3 P.T.O.

O-nitrophenol is more volatile due to intramolecular H-bonding b/w H of –OH and 0 of – NO2 of

the same nmolecule.

8. Which reducing agent is employed to get copper from the leached low grade copper ore ?

Sol. Scrap iron or H2 gas is used to get copper from leached low grade copper ore.

9. State Raoult’s law for the solution containing volatile components. What is the similaritybetween Raoult’s law and Henry’s law ?

Sol. Accoridng to Raoul’ts law, “In a solution, the vapor pressure of a component at a giventempreature is equal to the product of vapour pressure of the pure component and its molefraction.

i.e., 0A A AP P H

According to Henry’s law, “The mass of a gas dissolved per unit volume of the solvent at a

constant temperature of the gas in equilibrium with the solution.

P = KHH

In both Raoult’s law and Henry’s law partial pressure of the volatile component or gas is directly

propotional to the mole fraction in solution. Raoult’s Law become special case of Henry’s law in

which KH become equal to vapour pressure of the pure component 0

AP i.e., 0A HP K

10. Explain the following terms:

(i) Rate constant (k)

(ii) Half life period of a reaction (t1/2

) .

Sol. (i) Rate = K [A]n

[A] = I

Rate constant in that rate of reaction in which concentration of reactant become equal to1.

(ii) Half life of the reaciton is defined as the time during which the concentration of reactantis reduced to half of its initial concentration

1/2

0.693t

k

11. Write the principles of the following methods :

(i) Froth floatation method

(ii) Electrolytic refining

Sol. (i) Froth floatation. This method is widely used for the concentration of sulphide ores suchas zinc blende (ZnS), copper pyrites (CuFeS

2), galena (PbS), etc. This method is based upon

the fact that the surface of sulphide ores ios preferentially wetted by oils while that ofgangue is preferentially wetted by water.

The ore is crushed inot a fine powder and mixedc with water to form a suspension in atank. To this suspension are added collectors (e.g. pine oil, xanthates, i.e., ethyl xanthateand potassium ehtyl xanthate, and fatty acids) which enthance the non-wettability of theore particles and froth-stabilizers (e.g., cresols and aniline) which stabilize the froth.

During this process, the ore particles which are preferentially wetted by the oil becomeslighter and thus rise to the surface along with the froth while the gangue particles whichare preferentially wetted by water become heavier and thus settle down at the bottom ofthe tank. The froth is skimmed off. It is allowed to collapse and finally dired to get theconcentrated ore.

If the mineral to be concentrated consists of suphides of two metals, then by adjusting theproportion of oil to water, it is often possible to seoparate one sulphide from the other.Sometimes additional reagents called froth depressants are used to prevent one type ofsulphide ore particles from forming the froth with air bubbles. For example, sodium cyanide

456/1/3 P.T.O.

is used as a depressant to separate lead sulphide (PbS) ore form surface of ZnS ore. This isdue to the reason that NaCN forms a zinc complex, Na

2[Zn(CN)

4] one the surface of ZnS

thereby preventing it from the formation of froth Under these conditions, only PbS formsfroth and hence can be separated from ZnS ore.

2 4 2Sod. tetracyanozincate (II)

4NaCN ZnS Na [Zn(CN) ] Na S

(ii) Electrolytic Refining : In this method, the impure metal is converted into a block whichforms the anode while cathode is made up of a pure strip of the same metal.Thse electrodesare suspended in an electrolyte which is the solution of a soluble salt of the metal usuallya double salt of the metal. When electric current is passed, metal ions from the electrtolyteare deposited at the cathode in the form of pure metal while an equivalent amount ofmetal dissolves from the anode and goes into the electrolyte solution as metal ions, i.e.,

Anode : M(s) ( )

Cathode : M ( ) ( )

n

n

M aq ne

aq ne M s

12. An element with density 11.2 g cm–3 forms a f.c.c. lattice with edge length of 4 × 10–8 cm. Calculatethe atomic mass of the element. (Given : N

A = 6.022 × 1023 mol–1)

Sol. Density = 11.2g cm–3

Z = 4

a = 4 × 10–8 cm

NA = 6.022 × 1023 mol–1

3

Z Mdensity

Naa

38 13

4 M11.2

4 10 6.022 10

24 2316 10 11.26.022 10M

4

M = 107.91 g mol–1

13. Examine the given defective crystal

A+ B– A+ B– A+

B+ 0 B– A+ B+

A+ B– A+ 0 A+

B– A+ B– A+ B–

Answer the following questions :

(i) What type of stoichiometric defect is shown by the crystal ?

(ii) How is the density of the crystal affected by this defect ?

(iii) What type of ionic substances show such defect ?

Sol. (i) Schottky defect

(ii) Density of the crystal will decrease by this effect

(iii) Ionic substance having high coordination number i.e., r

r

ratio high show such defects.

14. Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 g of benzeneto lower its freezing point by 0.48 K (Kf = 5.12 K kg mol-1).

Sol. TF = K

f × m

mass of compoundm

56 75

kF = 5.12 k Kg mol–1

556/1/3 P.T.O.

TF

= 0.48 k

5.12 mass of compound0.48 1000

256 75

0.48 256 75mass of compound

1000 5.12

mass of compound = 1.8 g

15. (i) Which alkyl halide from the following pair is chiral and undergoes faster SN

2 reaction ?

(a)Br

(b)Br

(ii) Out of SN

1 and SN

2, which reaction occurs with

(a) Inversion of configuration

(b) Racemisation

SOL. (i) (a) compound (b) is chiral (b) Br

and compound (a) will undergo SN

2 reaction faster due to

less steric hinderance

(ii) (a) SN

2 reaction occur with inversion of configuration.

(b) SN

1 reaction occur with racemisation.

16. Draw the structure of major monohalo product in each of the following reactions :

(i) OH 2SOCl

(ii) CH2 – CH = CH

2 + HBr Peroxide

Sol. (i) OH 2SOCl Cl

(ii) CH2 – CH = CH

2 + HBr Peroxide CH – CH – CH22 2

Br

17. Complete the following chemical equations :

(i) Ca3P

2 + H

2O

(ii) Cu + H2SO

4(conc.)

Sol. Complete the following chemical equations :

(i) Ca3P

2 + 6H

2O 3 Ca (OH)

2 + 2PH

3

(ii) Cu + H2SO

4(conc.) CuSO

4 + H

2

OR

17. Arrange the following in the order of property indicated against each set

(i) HF, HCl, HBr, HI - increasing bond dissociation enthalpy.

(ii) H2O, H

2S, H

2Se; H

2Te - increasing acidic character.

Sol. (i) HI < HBr < HCl < HF

Due to increasing bond strength and decreasing bond length bond dissociation enthalpyincreases in this order.

(ii) H2O < H

2S < H

2Se < H

2Te

Due to increase in size from O to Te, loss of H+ become easier.

656/1/3 P.T.O.

18. Write the IUPAC name of the complex [Cr(NH3)4Cl

2]+. What type of isomerism does it exhibit?

Sol. Tetraamminedichloridochromium (III). It will exhibit geometrical isomerism.

H N3

Cl

CrNH3

NH3H N3

Cl

+

(–trans)

H N3

Cl

CrCl

NH3H N3

Nh3

+

(–cis)

19. (a) Write the mechanism of the following reaction:

CH3CH

2OH HBr CH

3CH

2Br + H

2O

(b) Write the equation involved in Reimer-Tiemann reaction.

Sol. (a) H C – CH – OH3 2 H+

CH CH — O – H3 2

H

+ Br–

CH – CH – Br3 2

(b)

OH

+ CHCl3 + 3KOH

OHCHO

20. (a) Draw the structures of the following compounds :

(i) XeF4

(ii) N2O

5

(b) Write the structural difference between white phosphorus and red phosphorus.

Sol. (a) (i) XeF

F

F

F

(ii) N – O – NO

O

O

O(b) Write phosphorus has reqular tetrahedral with < PPP = 60

P

P

P

P

Red phosphorus exist as P4 tetrahedral but have polymeril structure consisting of P

4 tetrahedral

linked together.

P

P

P

P

P P

P P

P P –

P P21. Give the structures of A, B and C in the following reactions :

(i) 24 HNOLiAlHKCN3 273K

CH Br A B C

(ii) 23 3Br KOHNH CHCl NaOH3CH COOH A B C

756/1/3 P.T.O.

Sol. (i) 4

2

LiAlHKCN 273K3 3 3 2 2 3 2HNO

CH Br CH CN CH CH NH CH CH OH

(ii) 23 3Br +KOHNH CHCl NaOH3 3 2 3 2 3CH COOH CH CONH CH NH CH NC

OR

How will you convert the following:

(i) Nitrobenzene into aniline

(ii) Ethanoic acid into methanamine

(iii) Aniline into N-phenylethanamide

(Write the chemical equations involved.)

Sol. (i)

NO2

LiAlH4

NH2

(ii)

H C – COOH3

NH3 CH – CONH3 2

Br + KOH2 CH – NH3 2

(iii)

NH2

CH – C –Cl3

H – N – C – CH 3

O

O

NaOH

22. Account for the following :

(i) Sulphur in vapour form exhibits paramagnetic behaviour.

(ii) SnCl4 is more covalent than SnCl

2.

(iii) H3PO

2 is a stronger reducing agent than H

3PO

3.

Sol. (i) In vapour state sulphur exist as S2 which has two unpaired electron in * orbital due to

which it exhibhit paramagnetic behaviour.

(ii) SnCl4 is more covalent than SnCl

2 because of more polarising nature of Sn4+.

(iii) H3PO

2 is a stronger reducing agent than H

3PO

3 becuase it has two hydride ions to donate.

23. (i) What are disinfectants ? Give an example.

(ii) Give two examples of macromolecules that are chosen as drug targets.

(iii) What are anionic detergents ? Give an example.

Sol. (i) Disinfectant are chemical substances which kill microorganisms but are not safe to beapplied to the living tissues. Ex. Phenol.

(ii) Carbohydrates and proteins.

(iii) Anionic detergents are detergents having large part of their molecules as anions and it isthe anionic part of the molecule which involve in cleaning action.

Ex. C11

H23

CH2SO

3

– Ma+

24. (i) Deficiency of which vitamin causes scurvy ?

(ii) What type of linkage is responsible for the formation of proteins ?

(iii) Write the product formed when glucose is treated with HI.

Sol. (i) Vitamin C

(ii) Peptide linkage is responsible for formation of proteins.

(iii)

CHO

(CHOH)

CH OH

2

Glucose

4

HI CH ( CH ) CH

n-hexane

4 4 3

856/1/3 P.T.O.

25. (i) In reference to Freundlich adsorption isotherm write the expression for absorption ofgases on solids in the form of an equation.

(ii) Write an important characteristic of lyophilic sols.

(iii) Based on type of particles of dispersed phase, give one example each of associated colloidand multimolecular colloid.

Sol. (i)1

log log logx

k Pm n

(ii) Prepared easily by directly mixign with the liquid dispersion medium. They are quite stableand are not easily precipitated or coagulated.

(iii) Multimolecular colloid = gold sol

Associated colloid micelles.

26. The following data were obtained during the first order thermal decomposition of S02Cl

2 at a

constant volume :

SO2Cl

2(g) SO

2(g) + Cl

2(g)

Experiment Time/s–1 Total pressure/atm

1 0 0.4

2 100 0.7

Calculate the rate constant.

(Given : log 4 = 0.6021, log 2 = 0.3010)

Sol. 2 2 2 2SO Cl SO Cl

Intial pressume 0.4 – –

Pressume decompound x – –

Presume at 0.4–x x x

and 0.4 – x + x + x = 0.7

x = 0.3

2.313 0.4log

0.4k

t x

2.313 0.4

log100 0.1

k = 1.3 × 10–2 sec–1.

27. After the ban on plastic bags, students of one school decided to make the people aware of theharmful effects of plastic bags on environment and Yamuna River. To make the awarenessmore impactful, they organized rally by joining hands with other schools and distributed paperbags to vegetable vendors, shopkeepers and departmental stores. All students pledged not touse polythene bags in future to save Yamuna River.

After reading the above passage, answer the following questions :

(i) What values are shown by the students ?

(ii) What are biodegradable polymers ? Give one example.

(iii) Is polythene a condensation or an addition polymer?.

Sol. (i) Save environment

(ii) Biodegradable polymers are the polymers which can be degrade easily and eco-friendly.

(iii) Polythene is an addition polymer.

28. (a) How do you prepare :

(i) K2MnO

4 from MnO

2 ?

(ii) Na2Cr

2O

7 from Na

2CrO

4 ?

(b) Account for the following :

956/1/3 P.T.O.

(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.

(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.

(iii) Actinoid elements show wide range of oxidation states.

Sol. (a) (i) Pyrolusite on treatment with potassium hydroxide in presence of air given potassiummagnanate.

2 2 2 4 22MnO 4KOH+O 2K MnO 2H O

(ii) Sodium chromate solution on treatement with concentrated sulphuric acid gives

sodium dichromate.

2 4 2 4 2 2 7 2 4 2Sod.Chromate Connc. Sod.dichromate

2Na CrO H SO Na Cr O Na SO H O

(b) (i) Mn2+ is more stable than Fe2+ towards oxidation to + 3 state as Mn2+ do have half filled3d5 configuration whereas Fe2+ has partially filled 3d6 configuration.

(ii) The enthalpy of atomization is lowest for Zn as there are no unpaired electrons. So,metallic bonding is the weakest.

Zn = 3d10 4s2

(iii) Actionoids show wide range of oxidation states as 5f, 6d, 7s levels are comparable

energies. So removing electrons from these energy levels are equally energydemanding.

OR

28. (i) Name the element of 3d transition series which shows maximum number of oxidationstates. Why does it show so ?

(ii) Which transition metal of 3d series has positive E°(M2+/M) value and why ?

(iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why ?

(iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidationstate.

(v) Complete the following equation : -

MnO–

4 + 8H+ + 5e–

Sol. (i) Mn (Z = 25) exhibit the largest number of oxidation states because it has the maximumnumber unpaired electrons hence, it shows oxidation states from + 2 to + 7.

(ii) Cu; Cu has high enthalpy of atomisation and enthalpy of ionization. Therefore, the high

energy required to transform Cu(s) to 2+aqCu is not balanced by its low enthalpy of hydration.

(iii) Mn3+ is stronger oxidising Mn3+/Mn2+ has larger standard reduction potential (+1.5 volt)

with respect to Cr3+/Cr2+ couple (SRP = – 0.4 volt). Cr3+ has t2g

3 whereas Mn3+ t2g

4

configuration.

(iv) Eu2+, Yb2+, Sm2+

(v) - + - 24 2MnO 8H 5e Mn 4H O

29. (a) Write the products of the following reactions

(i) O + H2N – OH H

(ii) 2C6H

5CHO + conc. NaOH

(iii) CH3COOH 2Cl /P

(b) Give simple chemical tests to distinguish between the following pairs of compounds :

(i) Benzaldehyde and Benzoic acid

(ii) Propanal and Propanone

1056/1/3 P.T.O.

Sol. (a) (i) N – OH

:

(ii)

CH OH2

+

COO Na– +

(iii) CH – C – OH2

O

Cl(b) (i) Tollen’s reagent or NaHCO

3 test.

C H CHO+ 2 [Ag(NH ) ] + 3OH6 5 3 2+ C H – C – O + 2 Ag 6 5 + 4NH + 2H O3 2

O

(ii) Iodoform test is given by propanone.

H C – C – CH + 3NaOI3 3 CH – C – O Na + CHI + 2NaOH3 3

OO

OR

29. (a) Account for the following :

(i) CH3CHO is more reactive than CH

3COCH

3 towards reaction with HCN.

(ii) Carboxylic acid is a stronger acid than phenol.

(b) Write the chemical equations to illustrate the following name reactions :

(i) Wolff-Kishner reduction

(ii) Aldol condensation

(iii) Cannizzaro reaction

Sol. (a) (i) CH3CHO is more reactive than CH – C – CH3 3

O

due to smaller +I effect of one methyl

group. CH3CHO is less hindered than CH

3 – CO – CH

3 towards Nu .As a result

nucleophilic addition reactions occur more in CH3CHO.

(ii) Carboxylic acid is more acidic than phenol as carboxylate ion is much more resonancestabilized than phenoxide ion.

So, release of a proton from carboxylic acid is much easier than form phenol.

(b) (i) 2 2

2

NH NH KOH,glycol2 3 2H O 453 473KAldehyde AlkaneHydrazone

R CH O R CH NNH R CH N

In this reaction aldehyde reduce to alkane directly in the presence of hydrazine.

(ii) C = O + H – CH CHO2

Dil. NaOHCH – C – CH – CHO3 2

14 23OH

HEthanal or Acetaldehyde

3-Hydroxybutanal ( )Aldol

CH3

H

CH – CH = CH – CHO3

Aldehydes containing alpha hydrogen undergo aldol condensation in the presence ofalkali.

(iii) 2H – C – H + NaOH CH – OH + H + C ONa3

OO

Formaldehyde (50%) Methyl alcohol Sod. Formate

Aldehydes which do not contain alpha hydrogen undergo Cannizaro reaction in whichaldehyde undergo self oxidation and reduction in the presence of alkali.

30. (a) Define the following terms :

(i) Limiting molar conductivity

(ii) Fuel cell

1156/1/3 P.T.O.

(b) Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 . If the resistanceof the same cell when filled with 0.02 mol L–1 KCl solution is 520 , calculate theconductivity and molar conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L_1

KCl solution is 1.29 × 10–2 cm–1.

Sol. (a) (i) The molar conductvity at infinite dilution is called limiting molar conductivity.

(ii) Fuel cells are the devices which convert the energy produced during the combustionof fuels like hydrogen methane, methanol etc. directly into electrical energy.

(b) For 0.1 M KCl solution, R = 100, k = 1.29 s m–1

Cell constant = Conductivity × Resistance = 1.29 Sm–1 × 100= 129 m–1

Conductivity of 0.02 M KCl solution

1

1 1Cell constant 129m0.248 m

Resistance 520k

Concentration of the solution = 0.02 M = 0.02 M = 0.02 mol L–1 = 0.02 × 103 mol m–3 = 20 molm–3

Molar conductivity 1

2 13

0.248 m0.0124 S m mol

20mol mm

k S

c

OR

30. (a) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is requiredfor the reduction of 1 mol of Cu2+ to Cu.

(b) Calculate emf of the following cell at 298 K :

Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)

[Given E°cell

= + 2.71 V, 1 F = 96500 C mol–1]

Sol. (a) The amount of chemical reaction and hence the mass of any substance deposited or

liberated at any electrode is directly proportional to the quantity of electricity passed through

the electrolyte (solution or melt).

2 2Cu e Cu So, two Faraday of charge is required.

(b)0 0.0591

log cE E kn

0.0591 0.12.71 log

2 0.01

= 2.71 – 0.02955

= 2.68 volts.

× × × × ×