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Solved Question Paper 2011ChemistryClass – XII

Set - II

1. 'Crystalline solids are anisotropic in nature.’ What does this statement mean?

Answer

It means that some of the physical properties of crystalline solids such as refractiveindex show different values when measured along different directions in the samecrystals.

2. Express the relation between conductivity and molar conductivity of a solutionheld in a cell.

Answer

The molar conductivity of a solution is related to conductivity of that solution.

m

Al

��

3. Define Electrophoresis.

Answer

Electrophoresis is the phenomenon of movement of colloidal particles under theapplied electric potential.

4. Draw the structure of XeF2 molecule.

Answer



5. Write the IUPAC name of the following compound:

� �3 2CH CCH Br

Answer

2, 2 -dimethylbromopropane

6. Draw the structure of 3-methylbutanal.

Answer

7. Arrange the following compounds in an increasing order of their solubility inwater :

� �6 5 2 2 5 2 5 22C H NH , C H NH,C H NH

Answer

� �6 5 2 2 5 2 5 22C H NH C H NH C H NH� �

8. What are biodegradable polymers?

Answer

Biodegradable polymers are those polymers which can be decomposed by bacteria.

9. The chemistry of corrosion of iron is essentially an electrochemical phenomenon.Explain the reactions occurring during the corrosion of iron in the atmosphere.



Answer

Corrosion is a redox reaction. In this reaction, simultaneous oxidation and reductionreactions takes place at cathode & anode.

Because of presence of air and moisture, oxidation takes place anode. The pointwhere oxidation takes place, it behaves as the anode.

2( ) ( )

Anode:

2s aqFe Fe e� ��

Electrons released at the anodic position move through the metallic object and go toanother position of the object. Presence of H ions helps the electrons to reducemolecular oxygen. This point behaves as the cathode. These H+ ions come either fromH2CO3, which are formed due to the dissolution of carbon dioxide from air into wateror from the dissolution of other acidic oxides from the atmosphere in water.

� � ( ) 2 ( )2

Cathode:

4 4 2aq lgO H e H O� ��

The overall reaction is:

� �2

( ) ( ) ( ) 2 ( )22 4 2 2s aq aq lgFe O H Fe H O� ��

Ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferricions combine with moisture & forms hydrated ferric oxide i.e., rust.

10. Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction:

�+ 2+ 0(s) (aq) (aq) (s)Ni + 2Ag Ni + 2Ag , E = 1.05 V

(1F = 96500 C / mol)

Answer



11. Distinguish between 'rate expression' and 'rate constant’ of a reaction.

Answer

Rate expression: It is defined as the stoichiometric coefficients of reactants and products. In this expression, the rate of reaction is in terms of the molar concentrationof the reactants, with each term raised to some power, which may or may not be thestoichiometric coefficient of the reacting species in a balanced chemical equation.

Rate constant: It is defined as the rate of the reaction when the concentration of eachof the reactant is taken as unity.

12. State reasons for each of the following:(i) The N —O bond in NO2

- is shorter than the N -O bond in NO3-

(ii) SF6 is kinetically an inert substance.

OR

State reasons for each of the following:(i) All the P-Cl bonds in PCI5 molecule are not equivalent.(ii) Sulphur has greater tendency for catenation than oxygen.



Answer

(i) In NO2- the lone pair is delocalised between the two oxygen atoms. Hence bond

order equal to 1 + 1/2 = 3/2

Whereas in NO3- lone pair shared between three O atoms hence bond order = 1 + 1/3

= 4/3We know that, greater the bond ordershorter the bond. Thus, bond length ofNO2

- less than that of NO3-

(ii) In SF6, the six fluoride atoms areformed because of which it is verydifficult for a reagent to attack it. Thatis why SF6 is kinetically an inertsubstance.

OR

(i) In PCl5, there is trigonal bipyramidal geometry. In this structure, the two axial P—Cl bonds are longer than the three equatorial P — Cl bonds. Thus, axial bonds areless stable. This is because of the greater bond pair - bond pair repulsion in theaxial bonds. Hence, all the bonds in PCl5 are not equivalent.

(ii) S — S bonds are stronger as compared to O—O bonds. Thus, sulphur has agreater tendency for catenation than oxygen.

13. Assign reasons for the following:(i) Copper (I) ion is not known in aqueous solution.(ii) Actinoids exhibit greater range of oxidation states than lanthanoids.

Answer

(i) In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion.

0( ) ( )

2 0( ) ( )

, 0.52

2 , 0.34

aq s red

aq s red

Cu e Cu E V

Cu e Cu E V

� �

� �

� � �

� � �

Cu+ has more reduction electrode potential value. Thus, it undergoes oxidationreaction quite feasibly. Thus, Cu+ ion is not known in aqueous solution.



(ii) The actinoids exhibits greater range of oxidation states because there is very smallenergy gap between the 5f, 6d and 7s sub shells. Thus, all their electrons can takepart in bond formation.

14. Explain the following giving one example for each:(i) Reimer-Tiemann reaction.(ii) Friedel Craft's acetylation of anisole

Answer

(i) Reimer-Tiemann reaction:

In the presence of sodium hydroxide, phenol is treated with chloroform &introduces

−CHO group at the ortho position of the benzene ring.

This reaction is known as the Reimer-Tiemann reaction.

The intermediate is hydrolyzed in the presence of alkalis to produce salicyclaldehyde.

(ii) Friedel-Crafts acetvlation of anisole: It involves the reaction of anisole witheither acetyl chloride or acetic anhydride to give 2-methoxyacetophenone and 4-methoxyacetophenone.



15. How would you obtain(i) Picric acid (2, 4, 6-trinitrophenol) from phenol,(ii) 2-Methylpropene from 2-methylpropanol?

Answer

(i) Picric acid (2, 4, 6-trinitrophenol) from phenol

(ii) 2-Methylpropene from 2-methylpropanol



16. What is essentially the difference between α-form of glucose and β-form ofglucose Explain?

Answer

α -form of glucose and β -form of glucose can be distinguished by the position ofhydroxyl group on the first carbon atom.

In open chain β -glucose, the hydroxyl group on the first carbon atom is presenttowards the left side whereas in the closed ring β –glucose, the hydroxyl group on thefirst carbon atom is above the plane of the ring.

In open chain α –glucose, the hydroxyl group on the first carbon atom is towards theright whereas, in the closed ring α -glucose, the hydroxyl group on the first carbonatom is below the plane of the ring.



17. Describe what you understand by primary structure & secondary structure ofproteins.

Answer

Primary structure of proteins: In this structure, each polypeptide chain of a proteinhas amino acids. These amino acids are linked with each other in a specific sequence.

Secondary structure of proteins: This structure refers to the shape in which a long polypeptide can exist.

There are two different secondary structures possible:

(a) Helical structure: In this, a polypeptide chain forms all possible hydrogen bondsby twisting into a helix with—NH group of each amino acid residue and hydrogen bonded to >C=O of an adjacent turn of helix.

(b) b -pleated structure: In this, all peptide chains are stretched out to theirmaximum extensions and then laid side by side which are held together byintermolecular hydrogen bonds.

18. Mention two important uses of each of the following:(i) Bakelite(ii) Nylon 6

Answer

(i) Uses of Bakelite(a) for making combs(b) for manufacturing electrical switches

(ii) Uses of Nylon 6 (a) for making tyre cords(b) for making fabrics and mountaineering ropes

19. Silver crystallizes in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms justtouch each other on the diagonal across the face of the unit cell. That is each faceatom is touching the four corner atoms.)

Answer



10

10

For a fcc structure,

2 2400 400 10

400 102 1.414

141.4

ar

a pm cm

r

r pm

�

�

�

� � �

��

��

Thus, the radius of the silver atom is 141.4 pm

20. Nitrogen pentoxide decomposes according to equation

�2 5(g) 2(g) 2(g)2N O 4NO + O

This first order reaction was allowed to proceed at 40 °C and the data below were collected:

� �2 5 ( )N O M Time (min)

0.400 0.00

0.289 20.0

0.209 40.0

0.151 60.0

0.109 80.0

(a) Calculate the rate constant. Include units with your answer.(b) What wil1 he the concentration of N2O5 after 100 minutes? (c) Calculate the initial rate of reaction.

Answer



(a)

� �2 5 ( )N O M Time (min) � �2 5log N O

0.400 0.00 -0.3979

0.289 20.0 -0.5391

0.209 40.0 -0.6798

0.151 60.0 -0.8210

0.109 80.0 -0.9625



� ��

� �

2 5 0

2 5

1

2 5

3

4 1

( )

2.303log

After 100 min,2.303 0.4

log100 0.098

0.1406 min

( )

1.15 10 0.4

4.6 10 s

t

b

N Ok

t N O

k

k

c

r k N O

r

r

�

�

� �

�

�

�

�

� � �

� �

21. Explain how the phenomenon of adsorption finds application in each of thefollowing process:(i) Production of vacuum(ii) Heterogeneous catalysis(iii) Froth Floatation process

OR

Define each of the following terms:(i) Micelles(ii) Peptization(iii) Desorption

Answer

(i) Production of high vacuum: Charcoal can adsorb traces of air from a vessel,evacuated by a vacuum pump & gives a very high vacuum.

(ii) Heterogeneous catalysis: In this catalysis, the gaseous reactants are adsorbed onthe surface of the solid catalysts. Because of which, the concentration of the reactants increases on the surface. Thus, the rate of reaction increases.



(iii) Froth floatation process: The basic principle involved in this process isadsorption

This process is used to remove gangue from sulphide ores. In this process, a mixtureof water and pine oil is taken in tank. The impure powdered sulphide ore is dropped inthrough hopper and the compressed air is blown in through the agitator. The agitatoris rotated several times. As a result, froth is formed and the sulphide ores get adsorbed in the froth. The impurities settled down and are let out through an outlet at thebottom. The froth formed is collected in froth collector tank. After sometime, the oreparticles in the froth collecting

OR

i. Micelle: When, soaps and detergents dissolved in water, they form micelles. Themolecules of such substances contain a hydrophobic and a hydrophilic part. Whenpresent in water, these substances arrange themselves in spherical structures insuch a manner that their hydrophobic parts are present towards the centre, whilethe hydrophilic parts are pointing towards the outside. This structure is known asmicelle formation

ii. Peptization: It is the process of conversion of a precipitate into a colloidal sol byshaking it with the dispersion medium in the presence of an electrolyte. Theelectrolyte used is known as a peptizing agent.

iii. Desorption: It is the process of removing an adsorbed substance from the surfacethrough which it was adsorbed.

22. Describe the principle behind each of the following processes:(i) Vapour phase refining of a metal.(ii) Electrolytic refining of a metal.(iii) Recovery of silver after silver ore was leached with NaCN.

Answer

(i) Vapour phase refining of a metal:

It is the process of refining metal by converting it into its volatile compound and then,decomposing it to obtain a pure metal. To carry out this process, metal should form avolatile compound with an available reagent, and the volatile compound should beeasily decomposable so that the metal can be easily recovered.

(ii) Electrolytic refining: In this process, impure metals are refined through theuse of electricity.

In this process, impure metal is made the anode and a strip of pure metal is made thecathode. A solution of a soluble salt of the same metal is taken as the electrolyte.



When an electric current is passed, metal ions from the electrolyte are deposited at thecathode as pure metal and the impure metal from the anode dissolves into theelectrolyte in the form of ions. The impurities present in the impure metal getscollected below the anode. This is known as anode mud.

Anode:

Cathode:

n

n

M M ne

M ne M

� �

� �

� �

� �

(iii) Leaching: The powdered ore is digested with a dilute sodium cyanide while acurrent of air is continuously passed. As a result. silver pass into the solution forming sodium dicyanoargenate (I) while the impurities remain unaffected which are filtered off.

� �2 224 2

Sod. dicyanoargentate(I)

Ag S NaCN Na Ag CN Na S� ��

23. Complete the following chemical equations:

�

��

�

- 2- +4 2 4

Heated4

2- +2 7 2

(i) MnO + C O + H

(ii) KMnO

(iii) Cr O + H S + H

Answer

2 24 2 4 2 2

4 2 4 2 2

2 32 7 2 2

( ) 2 5 16 2 8 10

( )

( ) 2 7 3

Heated

i MnO C O H Mn H O CO

ii KMnO K MnO MnO O

iii Cr O H S H Cr H O S

� � � �

� � �

� � � � �

��

� � � � �

24. Write the name, stereochemistry and magnetic behaviour of the following:(At.nos. Mn = 25, Co = 27, Ni = 28)



� ��

� �

� ��

� ��

4 6

3 25

2 4

(i)K Mn CN

(ii) Co NH Cl Cl

(iii)K Ni CN

Answer

(i)

� �4 6K Mn CN� ��

Name: - Potassium hexacyanomanganate (II)Stereochemistry— does not show geometric or optical isomerismMagnetic behaviour — Paramagnetic

(ii)

� �3 25Co NH Cl Cl� ��

Name-Pentaamminechloridocobalt (III) ChlorideStereochemistry- Does not show geometric isomerism but is optically activeMagnetic behaviour- Paramagnetic

(iii)

� �2 4K Ni CN� ��

Name - Potassium tetracyanonickelate (II)Stereochemistry - Does not show geometric or optical isomerismMagnetic behaviour- Diamagnetic

25. Answer the following:

1) Haloalkanes easily dissolve in organic solvents, why?

2) What is known as a racemic mixture? Give an example.

3) Of the two bromoderivatives, C6H5CH (CH3) Br and C6H5CH (C6H5) Br, whichone is more reactive in SN1 substitution reaction and why?



Answer

1. Polarity of Haloalkanes & organic solvents is quite less. Thus, they can easilydissolve in organic solvents. The new forces of attraction set up between haloalkanesand the solvent molecules are of same strength as the forces of attraction being broken.

2. Racemic mixture: A mixture of equal amounts of two enantiomers is knownas racemic mixture.

For example: When a 3° halide undergoessubstitution, the reactionproceeds through SN1 mechanism forming theracemic mixture in whichone of the products has thesame configuration as areactant while the otherproduct has an inverted configuration.

3. C6H5CH (C6H5) Br will be more reactive towards SN1 substitution reactionbecause SN1 substitution reaction involves the formation of carbocation which is not affected by the presence of bulky groups

26.

A. Explain why an alkylamine is more basic than ammonia.

B. How would you convert(i) Aniline to nitrobenzene(ii) Aniline to iodobenzene?

Answer



A. The basicity of amines depends on the +I effect of the alkyl groups. Inalkylamine, -CH3 groups increase the electron density on the nitrogen atom and thusincreases the basicity whereas in ammonia - CH3 groups are absent. Thus, alkylamineis more basic than ammonia

B.

(i)

(ii)

27. Describe the following giving one example for each(i) Detergents(ii) Food preservatives(iii) Antacids



Answer

a) Detergents: A detergent is a surfactant or a mixture of surfactants having cleaning properties in dilute solutions. Commonly, detergents arealkylbenzenesulphonates. For example: Sodium dodecylbeuzene sulphonate.

b) Food preservatives: The chemicals which are added into food in order toprevent their spoilage by inhibiting microbial growth are known as food preservatives.For example- Table salt, sugar, sodium benzoate (C6H3COONa), vegetable oil, and salts of propanoic acid.

c) Antacids: Any drug that is used to counteract the effects of excess acid in thestomach and raise the pH to an appropriate level is called an antacid. For example: Omeprazole.

28. (a) Differentiate between molarity and molality for a solution. How does achange in temperature influence their values? b) Calculate the freezing point of an aqueous solution containing 10.50 g ofMgBr2 in 200 g of water. (Molar mass of MgBr2= 184 g), (Kf for water = 1.86 K kg mol-1)

OR

(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (Kb for water = 0.512 K kg mol-1. (Molar mass of NaCl=58.44 g)

Answer

(a)

Molality: - it can be defined as the number of moles of the solute per kilogram of thesolvent.

moles of solute,

mass of solvent in KgMolality m �

Molarity: - it can be defined as the number of moles of the solute dissolved in oneLitre of the solution.



moles of solute,

volume of solution in LitreMolarity M �

Molarity decreases with an increase in temperature & molality is independent oftemperature. Because molality involves mass, which does not change with a change intemperature, while molarity involves volume, which is temperature dependent.

(b)

2

1

2

2

1 2

0

10.50

200

184

Now,1000

1000 1.86 10.50200 184

0.53

273 0.53

272.47

ff

f

f

f f

f

f

w g

w g

M g

k wT

w M

T

T

T T T

T

T K

��

� ��

��

� ��

� �

� � �

� �

�

OR

(a)

Osmosis: The process of flow of solvent molecules from pure solvent to solution orfrom solution of lower concentration to solution of higher concentration through asemi-permeable membrane is called osmosis.

Osmotic Pressure: The pressure required to just stop the flow of solvent due to osmosis is called osmotic pressure of the solution.

Yes, the osmotic pressure of a solution is a colligative property.



nRT

V� �

Where

� = osmotic pressuren = number of moles of soluteV= volume of solutionT= temperature

(b)

2

1

2

2

1 2

0.512 k kg/mol

w 15

250

58.44

Now,1000

b

bb

K

g

w g

M g

k wT

w M

�

�

��

� ��

�

0

1000 0.512 15250 58.44

0.52

373 0.52

373.53

f

f

b b

f

f

T

T

T T T

T

T K

� ��

��

� � �

� �

�

29. (a) Give chemical tests to distinguish between(i) Propanal and propanone,(ii) Benzaldehyde and acetophenone.

(b) How would you obtain(i) But-2-enal from ethanal,(ii) Butanoic acid from butanol,(iii) Benzoic acid from ethylbenzene?

OR



(a) Describe the following giving linked chemical equations:(i) Cannizzaro reaction(ii) Decarboxylation

(b) Complete the following chemical equations:

Answer

(a)

(i) Propanal can be distinguished from Propanone by iodoform test.

Propanone is a methyl ketone. On treatment with I2/NaOH, it undergoes iodoformreaction to give a yellow ppt. of iodoform.

3 3 3 33 2

Propanone Iodoform

CH COCH NaOI CH COONa CHI NaOH� � � � �

Whereas, Propanal doesn’t form yellow ppt

3 2 No yellow ppt of Iodoform

Propanal

CH CH CHO NaOI� �

(ii) Benzaldehyde and acetophenone can be distinguished by iodoform test.

Acetophenone is a methyl ketone. On treatment with I2/NaOH, it undergoes iodoformreaction to give a yellow ppt. of iodoform.

6 5 3 6 5 33 2

Iodoform

C H COCH NaOI C H COONa CHI NaOH� � � � �

On the other hand, Benzaldehyde does not give this test.



6 5 No yellow ppt of IodoformC H CHO NaOI� �

(b)

(i)

3.3 3 2 32 ( )

But-2-enal

H ODil NaOHCH CHO CH CH OH CH CHO CH CH CH CHO�

��

(ii)

2 2 7

2 43 2 2 2 3 2 2.

Butanol Butanoic acid

K Cr ODil H SOCH CH CH CH OH CH CH CH COOH��

(iii)

OR

(a)

(i) Cannizzaro reaction:



Aldehydes which have don’t have α-hydrogen, they undergo this reaction. In this,

aldehydes undergo self oxidation-reduction on treatment with concentrated alkalis. In

this reaction, two molecules of aldehydes participate where one is reduced to alcohol

and the other is oxidized to carboxylic acid.

For example: -

3 3 2 32 .

Ethanal Ethanol Potassium Ethanoate

CH CHO Conc KOH CH CH OH CH COOK� � �

(ii) Decarboxylation:

In this reaction, carboxylic acids lose carbon dioxide to form hydrocarbons when their

sodium salts are heated with soda-lime.

� �Soda-lime : 3:13 4 2 3

Sodium ethanoate Methane

NaOH CaOCH COONa CH Na CO��

(b)

(i)

(ii)



(iii)

36 5 2 6 5

H OheatC H CONH C H COOH

�

��

30. (a) Explain the following:(i) NF3 is an exothermic compound whereas NCl3 is not.(ii) F2 is more reactive of all the four common halogens.


�

��

2 4(conc)

4 2

2 2(excess)

(i)C + H SO

(ii)P + NaOH + H O(iii)Cl + F

OR

(a) Account for the following:(i) the acidic strength decreases in the order HCI> H2S> PH3 (ii) Tendency to form pentahalides decreases down the group in group 15 of theperiodic table.


��

4 2 2

2 2

2 3(conc)

P + SO ClXeF + H OI + HNO

Answer

(a)

(i) We know that, size of element increases on moving down the group. There is largedifference in the size of N and Cl. It results in the weakness of strength of N — Clbond. But this is not so in case of N & F. The difference in size of N and F is small.Thus, the N - F bond is quite strong. As a result, NF3 is an exothermic compound.

(ii) Fluorine is smaller in size. The three lone pair present on F atom in F - F moleculerepels the bond pair. Thus, F - F is most reactive of all the four common halogens.



(b)

2 4( ) 2 2 2

4 2 3 2 2

2 2( ) 3

( ) 2 2 2

( ) 3

Phosphine( ) 3 2

Chlorine trifluoride

conc

excess

i C H SO SO CO H O

ii P NaOH H O PH NaH PO

iii Cl F ClF

� � � �

� � � �

� �

OR

(a)

1) We know that, electronegativity increase across the periods. Thus, the electronegativity will be in order:

Cl > S > P

Thus, the ability to lose H+ ions decreases. Hence, the acidic strength of the hydridesdecreases in the following order:

HCl > H2S > PH3

2) The tendency to form pentahalides decreases down the group. This happensbecause of inert pair effect.

(b)

4 2 2 5 2

2 2 2

2 3( ) 3 2 2

( ) 10 4 10

( ) 2 2 2 4

( ) 2 10 4conc

i P SO Cl PCl SO

ii XeF H O Xe HF O

iii I HNO HIO NO H O

� � �

� � � �

� � � �

Copyright ©Jagranjosh.comAll rights reserved. No part or the whole of this eBook may be copied, reproduced, stored in retrieval system or transmitted and/or citedanywhere in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the written permission ofthe copyright owner. If any misconduct comes in knowledge or brought in notice, strict action will be taken.

cbse cbse class 12th chemistry solved question paper 2011 set ii ebook

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