CBSE CBSE Class 12th Chemistry Question Paper Set I 2014 2

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Solved Question Paper 2014Subject: Chemistry

Class: XIISet: I

1. What is the effect of temperature on chemisorption?

Ans.1

Chemisorption increases with increase in temperature.

2. What is the role of zinc metal in the extraction of silver?

Ans.2

Zn acts as a reducing agent in the extraction of silver.

2( ) ( )2 4( ) ( )2 2s saq aqAg CN Zn Ag Zn CN

3. What is the basicity of H3PO3?

Ans.3

23 3 32H PO H HPO

It is dibasic in nature. Thus, its basicity is 2

4. Identify the chiral molecule in the following pair:



Ans.4

5. Which of the following is a natural polymer?

Buna-S, protein, PVC

Ans.5

Protein is a natural polymer.

6. The conversion of primary aromatic amines into diazonium salts is known as _______?

Ans.6

The conversion of primary aromatic amines into diazonium salts is known as diazotisation reaction.

7. What are the products of hydrolysis of sucrose?

Ans.7

Hydrolysis of sucrose gives glucose and fructose.

8. Write the structure of p-methyIbenzaldehyde.



Ans.8

9. An element with density 2.8g/cm3 forms of f.c.c. unit cell with edge length 4 108 cm. Calculate the molar mass of the element. [Given: NA = 6.022 1023 mol1]

Ans.9

8

3

Given

4 10

2.8 /Lattice is fcc type. Thus,

4

a cm

d g cm

z

3

38 23

,

42.8

4 10 6.022 10

A

Now

ZMd

a N

M

24 232.8 64 10 6.022 104

27 /

M

M g mol

Thus, molar mass of the element is 27g/mol.



10.

(i) What type of non-stoichiometric point defect is responsible for the pink colour ofLiCl?

(ii) What type of stoichiometric defect is shown by NaCl?

Ans.10

(i) The metal excess defect caused by anionic vacancies is responsible for the pink colour of LiCl.

(ii) The Schottky defect is shown by NaCl.

OR

How will you distinguish between the following pairs of terms?

(i) Tetrahedral and octahedral voids

(ii) Crystal lattice and unit cell

Ans.10

(i) Tetrahedral void: It is surrounded by 4 spheres.

Octahedral void: It is surrounded by 6 spheres.



(ii) Crystal lattice: Arrangement of the constituent particles of a crystal in a regular three-dimensional network is called crystal lattice.

Unit cell: It is the smallest three-dimensional part of a crystal lattice. When these cells joined each other they make whole crystal lattice.

11. State Kohlrausch's law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?

Ans.11

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum individual contributions of its cations & anions.

Conductivity of a solution decreases with dilution as with dilution number of ions per unit volume decreases.

12. For a chemical reaction R P , the variation in the concentration (R) vs. time (t) plot is given as

(1) Predict the order of the reaction.

(2) What is the slope of the curve?

Ans.12

(i) Graph shows a zero order reaction.



0Rate R(ii) For a zero order reaction, Slope = K

13. Explain the principle of the method of electrolytic refining of metals. Give one example.

Ans.13

Electrolytic refining:

It is the process of refining impure metals by using electricity. In this process, anode is made up of impure metal and cathode is made up of a strip of pure metal. A solution of asoluble 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 the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.

Example: Pure aluminium is obtained from aluminium ore by the electrolytic refiningmethod

14 . Complete the following equations:

4 2

4 2 2

(i) P + H O(ii) XeF + O F



Ans.14

4 2

4 2 2 6 2

( ) No reaction

( )

i P H O

ii XeF O F XeF O

15. Draw the structures of the following:

(i) XeF2

(ii) BrF3

Ans.15

(i) 2XeF

(ii) 3BrF

16. Write the equations involved in the following reactions:

(i) Reimer - Tiemann reaction



(ii) Williamson synthesis

Ans.16

(i) Reimer-Tiemann reaction:

In the presence of sodium hydroxide, phenol is treated with chloroform &introduces CHOgroup 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) Williamson ether synthesis:

It is a laboratory method to prepare symmetrical and unsymmetrical ethers.

Alkyl halide is reacted with sodium alkoxides & forms ether.

This reaction involves SN2 attack of the alkoxide ion on the alkyl halide.

' 'Sodium alkoxide Alkyl Ether halide

R O Na R X R O R NaX



If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

17. Write the mechanism of the following reaction:

HBr3 2 3 2 2CH CH OH CH CH Br + H O Ans.17

18. Write the names of monomers used for getting the following polymers:

(i) Bakelite

(ii) Neoprene

Ans.18

(i) Monomers of Bakelite are Phenol and formaldehyde

(ii) Monomer of Neoprene is Chloroprene



19.

0r

2+ 2+s aq aq s

(a) Calculate G for the following reaction : Mg + Cu Mg + Cu

Given : +2.71 V,1F = 96500 C / mol

(b) Name the type of cell which was used in Apollo space programme for providingelectrical power.

Ans.19

2 2

0

0 0

0

0

(a)Given cell reaction is:

2.71

2 96500 2.71

523030 J/mol

s aq aq s

cell

r cell

r

r

Mg Cu Mg Cu

E V

G nFE

G

G

(b) The fuel which uses the reaction of hydrogen with oxygen to form water was used in Apollo programme for providing electrical power.

20. The following data were obtained during the first order thermal decomposition of SO2Cl2at a constant volume:

Experiment Time/sec Total pressure/atm

1 0 0.4

2 100 0.7

Calculate the rate constant.

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



Ans.20

2 2( ) 2( ) 2( )

0

0

0

0

0

0

0

At 0 0 0

At

Now,

for a first order reaction,

2.303log

2.

g g g

t

t

t

SO Cl SO Cl

t P

t t P p p p

P P p p p

P P p

p P P

Pk

t P p

k

0

0

2 1

303log

2

At 1002.303 0.4

log100 0.8 0.7

1.386 10 sec

t

Pt P P

t s

k

k

21. What are emulsions? What e their different types? Give one example of each type.

Ans.21

Emulsion: Emulsions are the colloidal solution in which both the dispersed phase and dispersion medium are liquids.

There are two types of emulsions:

(a) Oil in water type: In this, oil is the dispersed phase while water is the dispersion medium. For example- vanishing cream, milk, etc.



(b) Water in oil type: In this, water is the dispersed phase while oil is the dispersion medium. For example- butter, cold cream, etc.

22. Give reasons for the following:

(i) (CH3)3 P = O exists but (CH3)3 N = O does not.

(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.

(iii) H3PO2 is a stronger reducing agent than H3PO3.

Ans.22

(i) The d-orbitals are absent in nitrogen atom. This restricts nitrogen to expand its coordination number beyond four. Hence, (CH 3)3N=O does not exist.

(ii) Oxygen is the smallest element of group 16. Due to its small size, the incoming electron is not accommodated with ease As a result; oxygen has less electron gain enthalpy than sulphur.

(iii) H3PO3 has one PH bond & H3PO2 has two P-H bonds. Thus, H3PO2 can release 2 hydrogen atoms while H3PO3 can release one hydrogen atom. Hence, H3PO2 is a stronger reducing agent than H3P03.

23. Write the IUPAC name of the complex 3 24Cr NH Cl Cl

(i) What type of isomerism is exhibited by the complex 3+3Co en ? (en = ethane-1,2-diamine)

(ii) Why is [NiCI4]2- paramagnetic but [Ni (CO)4] is diamagnetic?

(Atomic no.: Cr = 24, Co = 27, Ni = 28)



Ans.23

(i) 3 24Cr NH Cl ClIUPAC name = Tetraamminedichlorochromium(IlI) chloride

(ii) The complex 3+3Co en exhibits optical isomerism. Its optical isomers are:

(iii) Chloride is a weak field legend & carbonyl is a strong weak field legend. In [NiCl4]2-, the oxidation state of Ni is +2. Chloride does not pair up d-orbital electrons. Thus, the two unpaired electrons present in the valence d-orbitals are responsible of paramagneticnature of the complex. On the other hand, carbonyl pair up the d-orbital electrons. As aresult no electrons are left unpaired and the complex becomes diamagnetic in nature.

Q.24

(a) Draw the structures of major mono halo products in each of the following reactions:



(b) Which halogen compound in each of the following pairs will react faster in SN2 reaction?

3 3

3 33

(i) CH Br or CH I(ii) CH C - Cl or CH - Cl

Ans.24

(a)

(b)

In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in thefollowing order.

RF



In case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered becauseof the presence of bulky substituents on that carbon atom. In case of, there are no bulkysubstituents on the carbon in CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH.

25. Account for the following:

(i) Primary amines (R-NH2) have higher boiling point than tertiary amines (R3N).

(ii) Aniline does not undergo Friedel- Crafts reactions:

(iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.

Ans.25

(i) In primary amines, two hydrogen atoms are present whereas in tertiary amine, there are no Hatoms. Due to the presence of Hatoms, primary amines undergo intermolecularHbonding.

(ii) As a result, extra energy is required to break the hydrogen bonds of primary amines. Thus, primary amines have higher boiling points than tertiary amines.

(i) AlCl3 is acidic in nature & aniline is a strong base. Hence, aniline reacts with AlCl3 to form a salt.



(ii) But it does not undergo the Friedel-Crafts reaction. It happens because of the positivecharge on the N-atom. This leads to deactivation of electrophilic substitution in the benzenering takes place.

(i) Basic strength in aqueous solution depends on 2 factors: -

Steric factor- alkyl group is larger than hydrogen atom which causes sterichindrances. As the number of alkyl group increases, the steric hindrances increase in the same direction.

Amines form H-bonds with water molecules & release hydration energy & forms protonated amines. Protonated amines are more stable. Greater thehydration enthalpy more is the stability of protonated amine. Thus greater is the tendency of the amine to form cation leading to greater base strength of theamine.

Hence, the order of the basic strength in aqueous phase is:

Secondary > Tertiary > Primary

OR

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

2 2

+32 2

NaNO +HCl H OSn+HCl6 5 2 273K

NHH O/ H Br +KOH3

(i) C H NO A B

(ii) CH CN A B C

Ans.25

(i)



(ii)

26. Define the following terms as related to proteins:

(i) Peptide linkage

(ii) Primary structure

(iii) Denaturation

Ans.26

(i) Peptide linkage:

When COOH group of one molecule of an amino acid and NH2 group of anothermolecule of the amino acid is reacted, amide is formed. In this reaction, elimination of awater molecule takes place. This linkage is known as a peptide linkage.



(ii) Primary structure:

It is a specific sequence of linkages between amino acids in a polypeptide chain. Different sequences are present in different proteins. Any change in the sequence creates a different protein.

(iii)Denaturation:

Protein has is a unique 3-dimensional structure and a unique biological activity. In such condition, it is active & takes part in reactions, & is known as native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as its H-bonds are disturbed, change in pH. These activities disturb the structure of protein. As a result, unfolding of the globules takes place which uncoils the helix. In this form, the protein loses its biological activity. This loss of biological activityby the protein is known as denaturation.

Denaturation destroys the secondary and the tertiary structures of the protein, but it does not alter the primary structure.

For example: Coagulation of egg white when an egg is boiled

27. On the occasion of world Health Day, Dr. Satpal organized a 'health camp for the poorfarmers living in a nearby village. After check-up, he was shocked to see that most of thefarmers suffered from cancer due to regular exposure to pesticides and many were diabetic. They distributed free medicines to them. Dr. Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care, financial assistance, setting up of super-speciality hospitals for treatment and prevention of the deadly disease in the affected villages all over India.

(i) Write the values shown by (a) Dr. Satpal (b) NHRC

(ii) What types of analgesics are chiefly used for the relief of pains of terminal cancer?

(iii) Give an example of artificial sweetener that could have been recommended to diabetic patients.



Ans.27

(i) (a) Dr. Satpal is an altruist

(b) Responsibility

(ii) Narcotic analgesics are used for the relief of pains of terminal cancer.

(iii)Saccharin

28.

(a) Define the following terms: (i) Molarity (ii) Molal elevation constant (Kb)

(b) A solution containing 15 g urea (molar mass = 60 g/mol) per litre of solution in waterhas the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g/mol) in water. Calculate the mass of glucose present in one litre of its solution.

Ans.28

(a)

(i) Molarity: It can be defined as the number of moles of the solute dissolved in one Litre of the solution.

moles of solute,

volume of solution in LitreMolarity M

(ii) Molal elevation constant (Kb): It is defined as the elevation in boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent.

(b)

we know that urea solution is isotonic with glucose solution.

CRT

nRT

V



cos

cos

cos

Thus,

156014

1180 45

4

glu e urea

glu e

glu e

n nRT RT

V Vn

V V

n

m g

Thus, mass of glucose present is 45 g.

OR

(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.

(b) A solution of glucose (molar mass = 180 g/mol) in water is labelled as 10% (by mass) whatwould be the molality and molarity of the solution? (Density of solution = 1.2 g/ml)

Ans.28

(a) Mixture of ethanol & acetone shows positive deviation from Raoults law. Pure ethanol possesses hydrogen bonding. On addition of acetone, some hydrogen bonds break &mixture shows positive deviation.

(b)

10 g of glucose in present in = 100 g of the solution

10 g of glucose is present in = (100 10) g = 90 g of water

6 12 6

6 12 6

180 /

100.056

180

C H O

C H O

M g mol

n mol

Thus,



2

0.056Molality of solution 0.62

0.0990

5 18H O

m

n mol

3

1.2 /100

83.33 101.2

density g mL

V L

Now,

3

0.056Molarity of solution 0.67

83.33 10M

29.

(a) Complete the following equation:

2- -2 7

- + -4

(i) Cr O + 2OH

(ii)MnO + 4H + 3e

(b) Account for the following:

(i) Zn is not considered as a transition element.

(ii) Transition metals form a large number of complexes.

(iii) The E0 value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+

couple.

Ans.29

(a)

2 22 7 2 4 2

4 2 2

(i) 2 2

( ) 4 3 2

Cr O OH Cr O H O

ii MnO H e MnO H O

(b)



(i) Zinc has the stable filled valence shell d10 electronic configuration in its ground state as well as in its most common oxidation state of +2. Thus, it is not considered as a transition element.

(ii) Transition elements possess vacant d-orbitals to accommodate electrons for bond formation. Also, ions formed by transition metals have small sizes and high ioniccharges. Thus, they form a large number of complexes.

(iii) The E value for Mn3+ / Mn2+ is much more positive than that for Cr3+/Cr2+.

This is because Mn2+ is more stable because of half filled orbitals than Mn3+

So, the reduction of Cr3+ /Cr2+ is easier than the reduction of Mn3+/ Mn2+.

OR

(i) With reference to structural variability & chemical reactivity, write thedifferences between lanthanoids & actinoids.

(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

(iii) Complete the following equations:

- + -4MnO + 8H + 5e(iv) Out of Mn3+ & Cr3+, which is more paramagnetic & why?

Ans.29

(i) Lanthanoids & actinoids:a) Actinoids has far greater irregularities in metallic radii than lanthanoids. Thus,actinoids shows a variety of structures. b) The ionisation enthalpies of the early actinoids are lower than those of the earlylanthanoids. c) Actinoids are more reactive and have more complex magnetic properties than lanthanoids.

(ii) Cerium exhibits +4 oxidation state

(iii) 24 28 5 4MnO H e Mn H O



(iv) Mn3+ has four unpaired electrons in its valence shell whereas Cr3+ has three unpaired electrons. Thus, Mn3+ is more paramagnetic than Cr3+

30.

(a) Write the products formed when CH3CHO reacts with the following reagents:

(i) HCN

(ii) H2N-OH

(iii) CH3CHO in the presence of dilute NaOH

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

(i) Benzoic acid and Phenol

(ii) Propanal and Propanone

Ans.30

(a)

i. 3 3( )

2-Hydroxypropanenitrile

CH CHO HCN CH CH CN OH

ii. 3 2 3 2 Acetaldoxime

CH CHO NH OH CH CH NOH H O

iii.



(b)

(i) Propanal and propanone can be distinguished by tollens test, fehlings test & Iodoform test.

Tollens testPropanal can be reduces Tollens reagent. But, propanone does not reduce Tollens reagent as it is a ketone.

3 2 3 3 2 3 222 3 4 2Propanal Tollen's reagent Pranoate ion silver mirror

CH CH CHO Ag NH OH CH CH COO Ag NH H O

Fehlings testPropanal being an aldehyde reduces Fehlings solution to a red-brown precipitate of Cu2O, but propanone being a ketone does not.

23 2 3 2 2 22 5 3

Propanal Pranoate ion brown ppt

CH CH CHO Cu OH CH CH COO Cu O H O

(ii) Phenol and benzoic acid can be distinguished by ferric chloride test.

Ferric chloride test:

Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.

36 5 3 6 5 66 3 3Phenol Iron-Phenol complex

C H OH FeCl Fe OC H H Cl

But benzoic acid reacts with neutral FeCl3 to give a buff coloured ppt. of ferric benzoate.

6 5 3 6 5 33 3

Benzoic acid ferric benzoate

Buff coloured ppt

C H COOH FeCl C H COO Fe HCl

OR

(a) Account for the following:

(i) CI-CH2COOH is a stronger acid than CH3COOH.



(ii) Carboxylic acids do not give reactions of carbonyl group.

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

(i) Rosenmund reduction

(ii) Cannizzaro's reaction

(c) Out of CH3CH2-CO-CH2-CH3 and CH3CH2-CH2-CO-CH3, which gives iodoform test?

Ans.30

(a)(i) Chlorine is an electron withdrawing group & it increases the acidity of carboxylic

acid by stabilizing the conjugate base due to delocalisation of the negative chargeby resonance effects.

(ii) Lone pairs present in carboxylic acid are involved in resonance. Thus, it doesnt show the reactions of carbonyl group.

(b)(i) Rosenmund reduction:

(ii) Cannizzaro reaction:

Aldehydes which have dont have -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

(c)

3 2 2 3 3 3 23 2

yellow ppt

CH CH CH CO CH NaOI CHI CH CH COONa NaOH

3 2 2 3 No yellow pptCH CH CO CH CH NaOI

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CBSE CBSE Class 12th Chemistry Question Paper Set I 2014 2