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Class XII

Code 56/1

Chemistry (Theory)

Time allowed: 3 hours Maximum Marks: 70

General Instructions:

1. All questions are compulsory.2. Question nos. 1 to 8 are very short answer questions and carry 1 markeach.3. Question nos. 9 to 18 are short answer questions and carry 2 marks each.4. Question nos. 19 to 27 are also short answer questions and carry 3 markseach.5. Question nos. 28 to 30 are long answer questions and carry 5 marks each.6. Use log tables if necessary, use of calculators is not allowed.

Q.1 How do metallic and ionic substances differ in conducting electricity?

(1 mark)

Q.2 What is the coagulation process? (1 mark)

Q.3 What is meant by the term pyrometallurgy? (1 mark)

Q.4 Why is red phosphorus is less reactive that white phosphorus? (1

mark)

Q.5 Give the IUPAC name of the following compound: (1 mark)

2 2 2 3|

OH

H C CH CH CH CH CH=

Q.6 Write the structural formula of 1 phenylpentan 1 one. (1 mark)

Q.7 Arrange the following compounds in an increasing order of basic

strengths in their aqueous solutions: (1 marks)

NH3, CH3NH2, (CH3)2NH, (CH3)3N

Q.8 What does 6, 6 indicate in the name nylon 6,6?

Q.9 What type of cell is a lead storage battery? Write the anode and thecathode reactions and the overall cell reaction occurring in the use of a

lead storage battery? (2 marks)


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Or

The half cell reactions of an electrochemical cell are given below:

( ) ( ) ( ) ( )

( ) ( )

24 2

2 4

MnO aq 8H aq 5e Mn aq 4H O ,E 1.51 V

Sn aq Sn aq 2e ,E 0.15 V

+ +

+ +

+ + + = +

+ = +

Construct the redox equation from the two half cell reactions and

predict if this reaction favours formation if reactants or product shown

in the equation. (2 marks)

Q.10 Define the following: (2 marks)

i. Elementary step in a reaction

ii. Rate of a reaction

Q.11 Describe the underlying principle of each of the following metal

refining methods: (2 marks)

i. Electrolytic refining of metals

ii. Vapour phase refining of metals

Q.12 Complete the following chemical reaction equations: (2 marks)

i. XeF2 + H2O

ii. PH3 + HgCl2

Q.13 Complete the following chemical reaction equations: (2 marks)

i. ( ) ( ) ( )24 2 4MnO aq C O aq H aq +

+ +

ii. ( ) ( ) ( )2 22 7Cr O aq Fe aq H aq + +

+ +

Q.14 Which one in the following pairs undergoes SN1 substitution reaction

faster and why? (2 marks)

i.

ii.

ClCl

Or

Or


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Q.15 Complete the following reaction equations: (2 marks)

i.

+ HI

ii. 3 2 2CH CH CH CH HBr= +

Q.16 Name the four bases present in DNA. Which one of these is not

present in RNA? (2 marks)

Q.17 Name two fat soluble vitamins, their sources and the diseases caused

due to their deficiency in diet. (2 marks)

Q.18 Differentiate between molecular structures and behaviours of

thermoplastic and thermosetting polymers. Give one example of each

type. (2 marks)

Q.19 A first order reaction has a rate constant of 0.0051 min-1. If we begin

with 0.10 M concentration of the reactant, what concentration of the

reactant will be left after 3 hours? (3 marks)

Q.20 Silver crystallises with face- centred cubic unit cells. Each side of the

unit cell has a length of 409 pm. What is the radius of an atom of

silver? (Assume that each face atom is touching the four corner

atoms.) (3 marks)

Q.21 A copper- silver cell is set up. The copper ion concentration in it is 0.10

M. The concentration of silver ion is not known. The cell potential

measured 0.422 V. Determine the concentration of silver ion in the

cell. (3 marks)

Given:o o

2Ag / Ag Cu /CuE 0.80 V, E 0.34 V+ += + = +

Q.22 What happens in the following activities and why? (3 marks)

i. An electrolyte is added to a hydrated ferric oxide sol in water.

ii. A beam of light is passed through a colloidal solution.

iii. An electric current is passed through a colloidal solution.

CH3CH3


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Q.23 Giving a suitable example for each, explain the following: (3 marks)

i. Crystal field splitting

ii. Linkage isomerism

iii. Ambidentate ligand

Or

Compare the following complexes with respect to structural shapes of

units, magnetic behaviour and hybrid orbitals involved in units?

( ) ( ) ( )3 3

3 36 6 4Co NH , Cr NH , Ni CO

+ +

(At. Nos.: Co =27, Cr=24, Ni = 28) (3 marks)

Q.24 Explain the following observations: (3 marks)

i. The boiling point of ethanol is higher than that of methoxymethane.

ii. Phenol is more acidic than ethanol.

iii. o and p nitrophenols are more acidic than phenol.

Q.25 How would you account for the following: (3 marks)

i. Many of the transition elements and their compounds can act as

good catalysts.

ii. The metallic radii of the third (5d) series of transition elements are

virtually the same as those of the corresponding members of the

second series.

iii. There is a greater range of oxidation states among the actinoids

than among the lanthanoids.

Q.26 Complete the following reaction equations: (3 marks)

i.

O||

LiAlH42 H O2

R C NH

ii. 6 5 2 3 2 2C H N Cl H PO H O+ +

iii. ( )6 5 2 2C H NH Br aq+

Q.27 Describe the following substances with one suitable example of each

type: (3 marks)


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i. Non ionic detergents

ii. Food preservatives

iii. Disinfectants

Q.28 a. Define the following terms:

i. Mole fraction

ii. Vant Hoff factor

b. 100 mg of a protein is dissolved in enough water to make 10.0 mL

of a solution. If this solution has an osmotic pressure of 13.3 mmHg at

25 C, what is the molar mass of protein? (R=0.0821 L atm mol-1 K-1

and 760 mmHg = 1 atm) (5 marks)

Ora. What is meant by:

i. Colligative properties

ii. Molality of a solution

b. What concentration of nitrogen should be present in a glass of water

at room temperature? Assume a temperature of 25C, a total pressure

of 1 atmosphere and mole fraction of nitrogen in air of 0.78. [KH for

nitrogen = 8.42 x 10-7 M/mm Hg] (5 marks)

Q.29 a. Draw the structures of the following: (5 marks)

i. H2S2O8

ii. HClO4

b. How would you account for the following:

i. NH3 is a stronger base than PH3.

ii. Sulphur has a greater tendency for catenation than oxygen.

iii. F2 is a stronger oxidising agent than Cl2.

OR

a. Draw the structures of the following:

i. H2S2O7

ii. HClO3

b. Explain the following observations:


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i. In the structure of HNO3, the NO bond (121 pm) is shorter than the

NOH bond (140 pm).

ii. All the PCl bonds in PCl5 are not equivalent.

iii. ICl is more reactive than I2. (5 marks)

Q.30 a. Write chemical equations of illustrate the following name bearing

reactions: (5 marks)

i. Cannizzaros reaction

ii. Hell- Volhard Zelinsky reaction

b. Give chemical tests to distinguish between the following pairs of

compounds:i. Propanal and Propanone

ii. Acetophenone and Benzophenone

iii. Phenol and Benzoic acid

Or

a. How will you bring about the following conversions?

i. Ethanol to 3 hydroxybutanal

ii. Benzaldehyde to Benzophenone

b. An organic compound A has the molecular formula C8H16O2. It gets

hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and

an alcohol C. Oxidation of C with chromic acid also produced B. C on

dehydration reaction gives but-1ene. Write equations for the

reactions involved. (5 marks)


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Class XII

Code 56/1

Chemistry (Theory)

1. Metallic substances are conductors in solid state as well as in molten stateIonic substances are insulators in solid state but conductors in molten stateand in aqueous solution. (1)

2. The process of settling down of colloidal particles as insoluble precipitateby addition of some suitable electrolytes is known as coagulation process.

(1)

3. Pyrometallurgy is the branch of metallurgy which consists of the thermaltreatment of minerals and metallurgical ores and their concentrates to bring

about physical and chemical transformations in the materials to enablerecovery of valuable metals. (1)

4. Red phosphorus is polymeric which consists of chains of P4 tetrahedralinked together whereas white phosphorus consists of a single tetrahedron.(1mark)

5. Hex-1-en-3-ol (1)

6.

(1)7. In aqueous solutions, the increasing order of basic strengths is:NH3 < (CH3)3N < CH3NH2


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9. Lead storage battery is a secondary cell. (1/2)

Anode: Pb(s) + SO42- (aq) PbSO4 (s) + 2e- (1/2)

Cathode: PbO2(s) + SO42- (aq) + 4H+(aq) + 2e-PbSO4 (s) + 2H2O (l)

(1/2)

Overall cell reaction:

Pb (s) + PbO2 (s) + 2H2SO4 (aq) 2PbSO4 (s) + 2H2O (l) (1/2)

Or

MnO4-(aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O(l) Eo= +1.51V

Sn4+ (aq) + 2e- Sn2+(aq) Eo= -0.15V

Cathode: Reduction reaction:2

4 2MnO (aq) 8H (aq) 5 e Mn (aq) 4H O(l) + +

+ + +

Anode: Oxidation reaction:Sn2+(aq) Sn4+ (aq) + 2e-

Redox equation:2 MnO4

-(aq) + 16 H+(aq) + 5 Sn2+(aq) 2 Mn2+(aq) + 8 H2O(l) + 5 Sn4+(aq)(1)

Cell potential = Eoreduction - Eo

oxidation (1/2)

= 1.51 - (-0.15)= 1.66 V (1/2)

10. (i) An elementary step in a reaction is a chemical reaction in which oneor more of the reactants react directly to form products in a single reactionstep. (1)

(ii) Rate of a reaction is defined as the rate of decrease in concentration ofany on of the reactants or the rate of increase in concentration of any one ofthe products. (1)

11. (i) In this method ,the impure metal acts as anode which undergoesoxidation and pure metal acts as cathode which undergoes reduction. The

metal ions deposit on the cathode as metal. (1)

(ii) Vapour phase refining is based on the principle that metal is converted toits volatile compound and collected which decomposes to give pure metal.

(1)


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12.(i) 2XeF2 + 2H2O 2Xe + 4HF + O2 (1)

(ii) 2PH3 + 3HgCl2

Hg3P2 + 6HCl (1)

13.(i) 2MnO4

-(aq) + 5C2O42- (aq) + 16H+(aq) 2Mn2+ (aq) + 8H2O + 10CO2

(1)

(ii) Cr2O72- (aq) + 6Fe2+(aq) + 14H+(aq) 2Cr3+ + 6Fe3+(aq) + 7H2O

(1)

14.

(i) will undergo NS 1 reaction faster than

because is a tertiary alkyl halide and is asecondary alkyl halide. Tertiary carbocation is more stable than secondarycarbocation. (1 )

(ii) will undergo NS 1 reaction faster than

because is a

secondary alkyl halide and is a primary alkylhalide. Secondary carbocation is more stable than primary carbocation.


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(1)

15. (i)

+ HI

(1)

(ii) CH3CH2CH=CH2 + HBr CH3CH2CH-CH3Br

(1)

16. Four bases in DNA are Adenine, Guanine, Thymine and Cytosine. (1)Thymine is not present in RNA. (1)

17. Vitamin A and D are fat soluble vitamins.

Sources Disease

A Carrots, butter Xerophthalmia

(1)

D Exposure to sunlight,fish

Rickets(1)

18.

Thermoplastic polymers Thermosetting polymers

Molecular Structure Linear or slightlybranched long chainmolecules

Cross linked or heavilybranched molecules(1)

Behaviour Can undergo softeningon heating andhardening on cooling

Do not soften onheating(1)


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19.

k = 0.0051 min-1

[A0] = 0.10 Mt = 3 hours = 3 60 min

k = 2.303 log [A0] (1)t [A]

0.0051 = 2.303 log 0.1 (1)3 60 [A]

log [A] = -1.39[A] = 4.074 10-2 M (1)

20. a = 409 pm, r = ?


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21. Ecell = 0.422 V

Since EAg+/Ag > E

Cu2+

/Cu

So, +Ag / Ag is cathode and Cu2+/Cu is the anode.

cell Reduction Oxidation

2

22

cell cell 2

cell 2

E E E

0.80 0.34

0.46V (1 / 2)

Cell r eaction : 2Ag Cu Cu 2Ag

Cu Ag0.059E E log (1 / 2)

nAg Cu

0.059 [0.1]E 0.46 log

2 [Ag ]

0.422 0.46 0.029{log[0.1] log Ag

+ +

+

+

+

+

=

=

=

+ +

=

=

=

2

1

}

log[.1] 2 log Ag 1.310

(log10 ) 2 log[Ag ] 1.310

1 2 log[Ag ] 1.310

2 log Ag 2.310 (1)

log Ag 1.155

Ag Antilog( 1.155)

Ag 0.0699M (1)

+

+

+

+

+

+

+

=

=

=

=

=

=

=


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22. (i) If an electrolyte is added to a hydrated ferric oxide sol, the colloidalparticles get precipitated due to neutralization of charges. (1)

(ii) Scattering of light takes place and the the path of the beam isilluminated. (1)

(iii) The colloidal particles move towards oppositely charged electrodes, getdischarged and precipitated. (1)

23. (i) The splitting of the degenerate d orbitals into eg and t2g orbitals dueto the presence of ligands in a definite geometry is called crystal fieldsplitting.It is denoted by o. (1)

(ii) Linkage isomerism arises in coordination compounds containingambidentate ligands in which two different atoms of the same ligand canform coordinate bond with metal ion.

Example: 3 5 2 2Co(NH ) NO Cl and 3 5 2Co(NH ) ONO Cl (1)

(iii) Ligands which can bind the central metal atom through two differentatoms are called ambidentate ligands.

Example: SCN ,NCS (1)

OR

[Co(NH3)6]3+

+3Co (Ground state): 3d6 4s0 4p0

[Co(NH3)6]3+

Structural shape Octahedral


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Hybrid Orbitals-2 3d sp

Magnetic behaviour- Diamagnetic (1)

3

3 6[Cr(NH ) ]+

+3Cr (Ground state): 3d3 4s0 4p0

3

3 6[Cr(NH ) ]

+

:

Structural shape Octahedral

Hybrid Orbitals-2 3d sp

Magnetic behaviour- Paramagnetic (1)

4Ni(CO)

Ni - 3d8 4s2


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4Ni(CO) :

Structural shape- tetrahedral

Hybrid orbitals- sp3

Magnetic behaviour- Diamagnetic (1)

24.(i) Hydrogen bonding in ethanol causes the boiling point of ethanol to behigher. (1)

(ii) Phenol on releasing a proton forms phenoxide ion which is resonancestabilised .So, phenol is more acidic than ethanol. (1)

(iii)

2NO group has -I effect or electron withdrawing inductive effect. (1)

25 (i) Ability to show variable oxidation state and to form complexes. (1)(ii) Due to Lanthanoid contraction. (1)(iii) 5f, 6d and 7s orbitals are of comparable energies. (1)

26 (i)

(1)

(ii)

(1)


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(iii)

(1)

27. (i) Non ionic detergents do not contain any ion in their constitution.(1/2)

Example: Non- ionic detergent is formed when stearic acid reacts withpolyethyleneglycol.

(1/2)

(ii) Food preservatives are substances which present spoilage of food due tomicrobial growth. (1/2)For example: sodium benzoate (1/2)

(iii) Disinfectants are chemicals which either kill or prevent the growth ofmicroorganisms .They cannot be applied on a living tissue. (1/2)For example: 1% solution of phenol (1/2)

28. (a) (i)Mole fraction of a component is the ratio of number of moles ofthe component to the total number of moles of all the components.

(1)(ii) Vant Hoff factor is the ratio of normal molar mass to the abnormal molarmass. (1)Vant Hoff factor is the ratio of observed value of colligative property tocalculated value of colligative property assuming no association or

dissociation.(b) Mass of protein = 100mg = 0.1 g

V= 10 mL = 0.01L

(1)


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( 1 mark for calculations)

13 3 0 10 01 0 0821 298

760

13980

V nR Tw

V R TM

. .x . x . x

M

M g

=

=

=

=

Molar mass of protein=13980 g mol-1 (1)

Or(a)

(i) All the properties which depend on the number of solute particlesirrespective of their nature relative to the total number of particles presentin the solution are known as colligative properties. (1)(ii) Molality of solution is the number of moles of solute present in 1kilogram of solvent. (1)

(b)

2 2

2

2 2

2

2

2

N N total

7

H

N

N H N H

7

N

7

N

4

H O

p x xP

0.78x760mmHg

592.8 mmHg

K 8.42 x10 M /mmHgx ?

x K x p [Since K is given in M/mmHg therefore this formula is being used]

(1/2)

x 8.42 x10 M /mmHg x 592.8 mmHg

x 4991.376x10 M

4.99x10 M (1)

1000n

=

=

=

=

=

=

=

=

=

=

2 2

2

2

2

2

N H O

N

N

H O

4 4N

55.5mol18

Since n n

nx (1 /2)

n

n 4.99x10 x55.5 276.9x10 M (1)

=


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(1)

(ii) 4HClO

(1)

(b) (i) N is smaller in size and more electronegative than P. So lone pair of

electrons is concentrated more on nitrogen. (1)

(ii) Oxygen has a small size as compared to sulphur. So, the lone pairs onoxygen repel each other. Thus, the O-O bond is weak whereas in case ofsulphur, the lone pair do not repel to the same extent.As a result, S-S bond is stronger than O-O bond. So, S has a greatertendency for catenation than oxygen. (1)

(iii) The net energy release for 21

X (g) e X (aq)2

+ is more negative

for fluorine as compared to chlorine i.e. sum of dissociation energy, electron

gain enthalpy and hydration energy is more negative for fluorine. So, it astronger oxidising agent than chlorine. (1)

Or

(a)(i) 2 2 7H S O


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(1)

(ii) 3HClO

(1)

(b)(i) Because of partial double bond character of (N-O) bond. (1)

(ii) P in PCl5 is sp3d hybridized. The axial bonds due to more repulsion

are longer than equatorial bonds. (1)

(iii) ICl is an interhalogen compound having electronegativity

difference. The I-Cl bond is more reactive than 2I . (1)

30. (a)

(i) Cannizaros reaction: Aldehydes which do not have an -hydrogenatom undergoes self oxidation and reduction on treatment withconcentrated alkali.

(1)(ii) Hell-Volhard-Zelinsky reaction: Carboxylic acids having an -

hydrogen are halogenated at the -position on treatment with chlorineor bromine in the presence of small amount of red phosphorus to give- halocarboxylic acids.


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(1)

(b) (i) Propanal and propanoneFehlings test

2

3 3 3 2 2 2CH CH CHO 2Cu 5OH CH CH COO Cu O 3H O

Red ppt

+

+ + + +

Propanone being a ketone will not give this test. (1)

(ii) Acetophenone and benzophenone

Iodoform test

Benzophenone does not give this test. (1)

(iii) Phenol and Benzoic acidSodium bicarbonate test:Add NaHCO3 solution to each of them. Phenol will not react. Benzoicacid will give brisk effervescence due to CO2.

(1)

OR(a)(i)


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(1)(ii)

(1)


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(b)

(1)

3

2

CrO3 2 2 2 3 2 2

dehydration

3 2 2 2 3 2 2H O

CH CH CH CH OH CH CH CH COOH (1)

C B

CH CH CH CH OH CH CH CH CH (1)

C But 1 ene

=

last yrr sol

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