CHEMISTRY (Theory) SOLUTION · Please check that this question paper contains 19 ... on the answer-book during this period. CHEMISTRY ... are also short answer questions and carry

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CHEMISTRY (Theory) & SOLUTION

Time allowed : 3 hours Maximum Marks : 70

General Instructions :

(i) All questions are compulsory.

(ii) Questions number 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Questions number 6 to 10 are short answer questions and carry 2 marks each.

(iv) Questions number 11 to 22 are also short answer questions and carry 3 marks each.

(v) Question number 23 is a value based question and carries 4 marks.

(vi) Questions number 24 to 26 are long answer questions and carry 5 marks each.

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SECTION A

Q 1. Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.

Ans. Due to variable valency of metal ion of FeO crystal, some Fe2+ cations are missing and loss of positive

charge is made up by the presence of required number of Fe3+ ions.

Q 2. CO(g) and H2(g) react to give different products in the presence of different catalysts. Which ability of thecatalyst is shown by these reactions ?

Ans. Selectivity, This is the ability of catalyst to direct reaction towards a particular product.

Q 3. Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2].

Ans. In [Pt (en)2Cl2],(i) the co-ordination number of Platinum is 6(ii) the oxidation state of Platinum is +2

Q 4. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH andwhy?

Ans. Chlorobenzene is

Benzyl chloride is

Benzyl chloride will get hydrolysed easily with aq NaOH because in chlorobenzene the C of the C�Cl

bond is sp2 hybridised so the bond is a very strong one which can�t be broken easily.

More over due to resonance the C�Cl bond in chlorobenzene acquires a double bond character, which is

difficult to be broken.

Q 5. Write the IUPAC name of the following :

OHHC||

CHCHCCH

|CH

52

33

3

Ans.

53

42

13

233

3

CHCH|

CHCHCCH||

OHCH

3,3-dimethyl pentan-2-ol

SECTION B

Q 6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol�1) in 250 gof water.(Kf of water = 1.86 K kg mol�1)

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CLASS-XII / (CBSE)

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P.T.O

Ans. Given, 60g of glucose in 250g of water, Molality of solution, m

100025018060

=

34

mol kg�1

Depression of freezing point, Tf = kf × m

= 1.86 × 34

= 2.48 Kelvin

As. we know, freezing point of pure water, fT = 273.15 K

So, freezing point of solution can be given as

Tf= Tf0� Tf

= 273.15 � 2.48 = 270.67 Kelvin

Q 7. For the reaction (g),O(g)4NO(g)O2N 2252 the rate of formation of NO2 (g) is 2.8 × 10�3 M s�1,

Calculate the rate of disappearance of N2O5(g).

Ans. 2(g)2(g)5(g)2 O4NOO2N

Rate Expression can be given as,

r =

t

ON ][

21 52

=

t

O

t

NO ][][

41 22

given, Rate of formation of NO2 is 2.8 ×10�3Ms�1

i.e.

t

NO ][ 2= 2.8 × 10�3 Ms�1

From Rate Expression,

Rate of disappearance of N2O5

t

NO

t

ON ][

4

12

)( 252

)1082(41

2 3

= 1.4 × 10�3 Ms�1

Q 8. Among the hydrides of Group-15 elements, which have the(a) lowest boiling point ?(b) maximum basic character ?(c) highest bond angle ?(d) maximum reducing character ?

Ans . (a) Lowest boiling point � PH3(b) Maximum basic character � NH3

(c) Highest bond angle � NH3

(d) Maximum Reducing character � BiH3







CLASS-XII / (CBSE)

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P.T.O

Q 9. How do you convert the following ?(a) Ethanal to Propanone(b) Toluene to Benzoic acid

ORAccount for the following :(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid

Ans. (a) Ethanal to propanone :-

(b) Toulene to Benzoic acid

OR(a) Aromatic benzoic acids one �M groups that is electron with drawing groups and thus the alkyl

or acyl group get attached to meta position instead of ortho & para position, so not considerdas friedal craft reaction.

(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid because Nitro group is anelectron with drawing group so it stabilizes the conjugate base (carboxylate an ion) throughdelocalisation of the negative charge by resonance effect. Hence it increases the acidity ofbenzoic acid. We conclude that 4-nitrobenzoic acid is stronger acid than benzoic acid so itspKa value is lower.

Q 10. Complete and balance the following chemical equations :

(a) HMnOFe 42

(b) IO24 HMnOAns. Complete and balance :-

(a) 5Fe+2 + MnO4� + 8H+ Mn+2 + 4H

2O + 5Fe3+

(b) 2MnO4� + H

2O + 2MnO

2 + 2 OH� + IO

3�

SECTION-CQ 11. Give reasons for the following :

(a) Measurement of osmotic pressure method is preferred for the determination of molar masses ofmacromolecules such as proteins and polymers.

(b) Aquatic animals are more comfortable in cold water than in warm water.(c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

Ans. (a) Osmotic pressure = CRT

= RTvn

= RT

vMm







CLASS-XII / (CBSE)

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P.T.O

= Mv

mRT

M = v

mRT

knowing the value of m, R, T, & V, M can be calculated.

(b) All aquatic species use dissolved oxygen in water for respiration.Oxygen is in dissolved state in water As temperature rises solubility of oxygen decreases .It meanssolubility of dissolved oxygen in warm water is less than that of cold water. This makes aquaticspecies respirate more comfortably in cold water than in warm.

(c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution becausethe van�t hoff factor for 1M KCl is 2 i.e. i = 2 and that of sugar is 1.

Q 12. An element �X� (At. mass = 40 g mol�1) having f.c.c. structure, has unit cell edge length of 400 pm.

Calculate the density of �X� and the number of unit cells in 4g of �X�. (NA = 6.022 × 1023 mol�1)

Ans. Give, edge length of unit cell, a = 400 pm= 4 × 10�8 cm

Molar mass of element �x�, M = 40 g mol�1

for f.c.c. structure, Z = 4

density of unit cell, d = 3A aN

MZ

d = 3823 )104(10022.6

404

= 4.15 g cm�3

Given,Mass of �X� = 4 gram

No. of moless of element x =

40

4

= 0.1 molNo. of atoms of element x = 0.1 × NA atomsAs we know, In f.c.c. structure4 atoms occupy, 1unit cell

So, 1atom will occupy

4

1unit cell

therefore, 0.1 NA atoms will occupy.

41

× 0.1 NA unit cell

= 41

× 0.1 × 6.022 × 1023

= 1.5055 × 1022 unit cells

Q 13. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate theactivation energy of the reaction. (Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK�1 mol�1)

Ans. Let, at temp 300 k (T1), rate constant is k1 and at temp. 320 k (T2), rate constant is k2According to question.at 300 k, time required for 50% completion

t1/2 = 40 minutes







CLASS-XII / (CBSE)

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P.T.O

for 1st order rate constant, k1 = 2/1t693.0

= 1min

406930

at 320 K, time required for 50% completion, t1/2 = 20 minutes

rate constant, k2 = 1

2/1

min206930

t6930

So, 2

406930206930

k

k

1

2

According to arrhenius Equation,

21

12a

1

210 TT

TT

R3032

E

k

klog

300320300320

30483032

Elog a

2

0.3010 =

30032020

31483032

Ea

Ea = 20

3003203148303230100

= 27663.8 Joule/mol or 27.66 kJ/mol

Q 14. What happens when(a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution ?(b) persistent dialysis of a colloidal solution is carried out ?(c) an emulsion is centrifuged ?

Ans. (i) Fe(OH)3 Sol is formed, due to peptization.(ii) On persistent dialysis, traces of the electrolyte present in the sol are removed almost completely

and colloid become unstable and ultimately coagulate.(iii) Demulsification takes place i.e. constituents of an emulsion are separated.

Q 15. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCNand Zn in this process.

Ans. Extraction of gold involves leaching the metal with NaCN. This is also an oxidation reaction. The metal islater recovered by displacement method.4Au + 8NaCN + 2H

2O + O

2 4Na[Au(CN)

2] + 4NaOH

2Na[Au(CN)2] + Zn 2Au + Na

2 [Zn(CN)

4]

Dilute NaCN acts as a leaching agent and Zn acts as a reducing agent.

Q 16. Give reasons :(a) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.(b) Iron has higher enthalpy of atomization than that of copper.(c) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.

Ans. Give reasons :(a) Much larger third ionisation energy of Mn where the required change is d5 to d4 mainly responsible

for this. Mn+2 has 3d5 configuration which is extra stable but for Fe+3 / Fe+2, low value of E° shows the

stability of Fe+3(d5).







CLASS-XII / (CBSE)

PAGE # 7

(b) Iron has 4 unpaired electrons in its 3d while copper has only 1 unpaired electron in its 4s. Hence Fehas stronger interatomic interactions which result in stronger bonding between its atoms, so its hashigher enthalpy of atomization than Cu.

(c) In aqueous solution, the configuration of Sc+3 is 3d° 4s°, so it does not have any unpaired electron

which makes it colourless, but, Ti+3 in aqueous solution has outer configuration 3d1 4s°, so it has

unpaired electron which makes it coloured.

Q 17. (a) Identify the chiral molecule in the following pair :

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in thepresence of sodium metal and dry ether.

(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexanewith alcoholic KOH.

Ans. (a) is the chiral molecule

(b) Cl + 2Na + Cl � CH3

etherdry CH3 + 2NaCl

Product obtained is Toluene

(c) KOH.alc or + KBr + H

2O

Q 18. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formulaC

4H

8O. Isomers (A) and (C) give positive Tollens' test whereas isomer (B) does not give Tollens' test but

gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl give the sameproduct (D).(a) Write the structures of (A), (B), (C) and (D).(b) Out of (A), (B) and (C) isomers, which one is least reactive towards addition of HCN ?

Ans. (i) A, B, C are non-cyclic functional isomers of carbonyl compound with molecular formula C4H

8O.

(ii) A and C give positive Tollen�s test so they are aldehyde

(iii) B does not give Tollen�s test but gives positive Iodoform test, So it must be a ketone.

(a) A is CH3 � CH

2 � CH

2 � HC == O ; B is CH

3�CH

2� �CH

3

C is

3

3

CH|

CHO�CH�CH ; D is CH3�CH

2�CH

2�CH

3







CLASS-XII / (CBSE)

PAGE # 8

(b) B is least reactive towards addition of HCN as it is ketone.

Q 19. Write the structures of the main products in the following reactions.

(i) O||

OCHCCH 4NaBH32

(ii) H

2OH

(iii) IH

Ans. (i) O||

OCHCCH 4NaBH32

(ii) H

2OH

(iii)

Q 20. (a) Why is bithional added to Soap ?(b) What is tincture of iodine ? Write its one use.(c) Among the following, which one acts as a food preservative ?

Aspartame, Aspirin, Sodium Benzoate, Paracetamol

Ans. (a) Bithional is added to soaps to impart antiseptic properties.(b) A 2�3% solution of Iodine in alcohol-water mixture is called tincture of iodine., It is a powerful

antiseptic applied on wounds.(c) Sodium Benzoate acts as a food preservative.

Q 21. Define the following with an example of each :(a) Polysaccharides(b) Denatured protein(c) Essential amino acids

OR(a) Write the product when D-glucose reacts with conc. HNO

3.

(b) Amino acids show amphoteric behaviour. Why ?(c) Write one difference between -helix and -pleated structures of proteins.

Ans 21. (a) Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysisare called polysaccharides. Some common examples are starch, cellulose. glycogen, gums, etc.Polysaccharides are not sweet in taste, hence they are also called non-sugars.

(b) As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and tertiarystructures of protein are destroyed, but the primary structures remain unaltered. It can be said that







CLASS-XII / (CBSE)

PAGE # 9

during denaturation, secondary and tertiary structured proteins get converted into primary structuredproteins.

(c) Essential amino acids are required by the human body, but they cannot be synthesised in the body.They must be taken through food. For example: valine and leucine

OR

(a) On being treated with HNO3, D-glucose get oxidised to give saccharic acid.

(b) Amino acid have both NH2 and COOH group in their structure, so amphoteric in nature.

(c) The alpha helix is joined by intramolecular hydrogen bonding between amino and carbonyl of otherunit and is insoluble in water but soluble in acid and alkali.beta pleated structure is joined by intermolecular hydrogen bonding between carbonyl and aminogroup of neighboring unit and is soluble acid, alkali and water.

Q 22. (a) Write the formula of the following coordination compound Iron(III) hexacyanoferrate(II)(b) What type of isomerism is exhibited by the complex [Co(NH

3)

5Cl]SO

4 ?

(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3�.

(Atomic No. of Co = 27)

Ans. (a) Fe4[Fe(CN)

6]3

(b) Ionisation isomerism.(c) [CoF

6]3� Co+3 = 3d6 4s°

As F � is a weak field ligand so electrons will not pair up.

Co+3 in [CoF6]3�

Hybridisation is sp3d2

Number of unpaired electrons is 4.

SECTION D

Q 23. Shyam went to a geocery shop to purchase some food items. The shopkeeper packed all the items inpolythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and askedthe shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penaltyimposed by the government for using polythene bags. The shopkeeper promised that he would use paperbags in future in place of polythene bags.Answer the following :(a) Write the values (at least two) shown by Shyam.(b) Write one structural difference between low-density polythene and high-density polythene.







CLASS-XII / (CBSE)

PAGE # 10

(c) Why did Shyam refuse to accept the items in polythene bags ?(d) What is biodegradable polymer ? Given an example.

Ans. (a) Shyam has shown his maturity towards the environment.� As polythene is a non-biodegradable polymer it will produce solid waste and thus he had refused

to take it.(b) Low density polythene is prepared by the polymerisation of ethene under high pressure of 1000 to

2000 atm and temp of 350 k to 570 k.It has a highly branched structure High density plythene is prepared at a pressure of 6-7atm by theuse of zieglar-Natta Catalyst.

(c) Shyam refused to take polythene bags as it produces non-biodegradable solid waste.(d) Biodegradable polymers are those polymers which get degrade by the action of micro organism

present in the environment.Eg. PHBV Poly -hydroxybutyrate-to--hydroxy valerate

SECTION EQ 24. (a) Given reasons :

(i) H3PO

3 undergoes disproportionation reaction but H

3PO

4 does not.

(ii) When Cl2 reacts with excess of F

2, ClF

3 is formed and not FCl

3.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature.(b) Draw the structures of the following :

(i) XeF4

(ii) HClO3

OR(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas

(A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling,the gas(A) changed into a colourless solid (B).(i) Identify (A) and (B)(ii) Write the structures of (A) and (B).(iii) Why does gas (A) change to solid on cooling ?

(b) Arrange the following in the decreasing order of their reducing character.HF, HCl, HBr, HI

(c) Complete the following reaction :XeF

4 + SbF

5

Ans. (a) Give reasons :-

(i) H3PO3 has P in +3 oxidation state, so it tends to disproportionate to higher and lower oxidation

state .POH3POH4 4333

H3PO4 has P in + 5 oxidation state which cannot go to any higher oxidation state. it does not

disproportion ate.

(ii) Because Flourine is more electronegative than Cl.(iii) Oxygen is very small in size and highly electronegative. It does not have d orbitals and cannot

extend its covalency beyond 4, but sulphur has d-orbitals so it can extend its covalency beyond







CLASS-XII / (CBSE)

PAGE # 11

4 and exist as S8. O2, thus is a gas but S8 is a solid

(b) (i) XeF4 (ii) HClO3

OR

(a) (i) A is NO2 and B is N2O4

(ii) A structure

B structure

(iii) A which is NO2 undergoes dimerisation on cooling so it converts into N2O4 which is a solid.

(b) HI > HBr > HCI > HF

(c) XeF4 + SbF

5 XeF3 + SbF6

Q 25. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :Sn(s) | Sn2+(0.004 M) || H+(0.020M) | H

2(g) (1 bar) | Pt(s)

(Given : 2

0

Sn /SnE 0.14V )

(b) Given reasons :(i) On the basis of E° values, O

2 gas should be liberated at anode but it is Cl

2 gas which is

liberated in the electrolysis of aqueous NaCl.(ii) Conductivity of CH

3COOH decreases on dilution.

OR(a) For the reaction

2AgCl(s) + H2(g) (1 atm) 2Ag(s) + 2H+ (0.1 M) + 2Cl� (0.1 M), G° = � 43600 J at 25°C.

Calculate the e.m.f. of the cell [log 10�n = �n]

(b) Define fuel cell and write its two advantages.

Ans 25. (a) At anode : 2e(aq)SnSn(s) 2 (oxidation)

At cathode (g)H2e2H 2(aq) (Reduction)

Cell reaction : (g)H(aq)Sn2HSn(s) 22

(aq)

cellE = + 0.14 V =

SnSn

HH 2

2

EE

= 0 � (�0.14)

= 0.14 volt

Ecell =

2

2

10cell][H

][Snlog

n0.059

E

= 0.14

210(0.02)

004.0log

20.059







CLASS-XII / (CBSE)

PAGE # 12

= 0.14 10log2

0.05910

= 0.1105 V

(b) (i) The reaction at anode with lower value of E° is prefferred and therefore, water should get oxidised

in preference to Cl� i.e. on account of overpotential of oxygen oxidation of Cl� is preffed.

(ii) On dilution, the number of ions per unit volume that carry current in solution decreases, thereforeconductivity also decrease.

OR

(a)(0.1M)

(aq)(0.1M)

(aq)(s)atm)(12(g)(s) 2Cl2H2AgH2AgCl

Given, G° = � 43600 J

also, G° = � nF cellE

cellE =

nFG

cellE =

965002)43600(

cellE = 0.2259 volt

From nernst Equation, at 25°C E.M.E of cell

Ecell = cellE �

2H

22

10 P][Cl][H

logn

0.0591

=

1)1.0()1.0(

log2

0591.00.2259

22

= )410(log2

0591.00.2259

= 10log42

0591.00.2259

= 0.2259 + 0.118= 0.3439 volt

Q 26. (a) Write the reactions involved in the following :(i) Hofmann bromamide degradation reaction(ii) Diazotisation(iii) Gabriel phthalimide synthesis

(b) Give reasons :(i) (CH

3)

2NH is more basic than (CH

3)

3N is an aqueous solution.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.OR

(a) Write the structures of the main products of the following reactions :

(i) Pyridine

OCO)(CH 23







CLASS-XII / (CBSE)

PAGE # 13

(ii) NH)(CH 23

(iii) OHCHCH 23

(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.

(c) Arrange the following in the increasing order of their pKb values :

C6H

5NH

2, C

2H

5NH

2, C

6H

5NHCH

3

Ans. (a) (i) Hofmann bromamide degradation reaction.It is a method for preparation of primary amines by treating an amide with bromide in an aqueous

or ethanolic soln.

O2H2NaBrCONaNHR4NaOHBrNHCR||O

232222

(Mechm:-)2

OHNaOH

Br2 NHROCHRNHCR

||O

2

2

(ii) DiazotitationThe conversion of primary aromatic amines into diazonium salts is known as diazotization.It is very unstable and thus decompose further and this is the reason that is not stored and isused immediately after its preparation.

O2HNaClClNHCHClNaNONHHC 2

È

256278K273

2256

(iii) Gabriel pthalimide synthesisGabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment withethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkylhalide followed by alkaline hydrolysis produces the corresponding primary amine. Aromaticprimary amines cannot be prepared by this method because aryl halides do not undergonucleophilic substitution with the anion formed by phthalimide.







CLASS-XII / (CBSE)

PAGE # 14

(b) (i) (CH3)2 NH in aqueous solution forms 2 H-bond with H2 and (CH3)3N forms only 1 Hydrogen

bond. So the former is more basic.

223223

2

OHHN)CH(OHHN)CH(|

OHH

233233 OHHN)CH(OHHN)CH(

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because of resonancein aromatic diazonium

OR

(a) (i)Pyridine

OCO)(CH 23

(ii) NH)(CH 23

(iii) OHCHCH 23

(b) Reagent Aniline N, N�dimethyl aniline

CHCl3 + alc. KOH Gives a foul Does not give the foul smelling(carbylamine reaction) smelling compound compound on warming.

which is isocyanide, onwarming

(c) C2H5NH2 < C6H5NHCH3 < C6H5NH2



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