HBSE Class 12 Sample Paper 2025 Answers Chemistry

Board Of School Education Haryana

मॉडल पेपर
उत्तर
2025

BSEH MARKING SCHEME

CLASS- XII Chemistry (March-2024) Code: B

The answer points given in the marking scheme are not final. These are
suggestive and indicative. If the examinee has given different, but
appropriate answers, then he should be given appropriate marks.
Q. Answers Marks
No.

1. d) Molality 1

2. c) No reaction 1
3. c) mol L-1s-1 1

4. a) La 1

5. b) cis-platin 1
6. b) Racemization 1

7. c) 4-Nitroanisole 1
8. b) β-D-Glucose 1
9. b) Vitamin C 1
10. Ideal solution 1

11. Rare earth 1

12. Cobalt 1
13. 51 1
14. Tert-butyl Alcohols 1

15. Carbonyl Chloride 1

16. a) Both A and R are true, and R is the correct explanation of A 1

17. d) A is false but R is true 1

18. b) Both A and R are true, and R is not the correct explanation of
A

19. The properties which depend on the number of solute particles 2
irrespective of their nature relative to the total number of
particles present in the solution are called colligative properties.
Examples: (1) relative lowering of vapour pressure of the
solvent
(2) depression of freezing point of the solvent
(3) elevation of boiling point of the solvent
(4) osmotic pressure
(Any two, ½ mark each)

20. a) Fuel cell 2

(½ mark)

b) Lead storage

(½ mark)

c) Mercury cell

(½ mark)

d) Dry cell

(½ mark)

Or
Given
Production of Al from Al2O3 has a reaction as following:
Al3+ + 3e- → Al

(½ mark)

i.e. production of 1 mole of Al (27 g) from Al2O3 requires
electricity = 3 F
or production of 1 g of Al from Al2O3 requires electricity = 3/27 F

(½ mark)
So, production of 40 g of Al from Al2O3 requires electricity =
40/9 F
= 4.44 F
(½ mark
for answer, ½ mark for unit)

21. concentration of reactants & pressure in case of gases,
temperature, and catalyst. 2
(½ mark each)

22. In the first transition series, Cu exhibits +1 oxidation state very
frequently.
(1 mark) 2
2K2Mno4+2H2O
2Cr3+7H2O+3T2
23. tert-butyl bromide < sec-butyl bromide < isobutyl
2
bromide < n-butyl bromide

24. The difference in the relative acidic strength if we compare the
resonance hybrids of carboxylate ion and phenoxide ion

The electron charge is more dispersed in compression to the
phenol ion the release of H+ ion from carboxylic acid is easier 2
than phenol.
Or
The nucleophile which has two different electron donor atoms
and can attack through two different sites are called as ambident
nucleophiles.
For examples cyanide ion and nitrite ion represent ambident
nucleophiles.
25. i) p-nitroaniline, Aniline, p-toluidine
2
ii) NH3, C2H5NH2, (C2H5)2NH, (C2H5)3N

26. Positive Deviation NonIdeal Negative Deviation Nonideal
Solutions solutions

1. Those liquid-liquid 1. Those liquid-liquid
solutions which has vapour solutions which has vapour
pressure more than pressure less than 3
expectations from Raoults’ expectations from Raoults’
law. law.

2. The molecular interactions 2. The molecular interactions
of solution is weaker than of solution is stronger than
that of solute and solvent. that of solute and solvent.

3. ∆𝑉 >0 3. ∆𝑉 <0

4. ∆𝐻 >0 4. ∆𝐻 <0
5. They form minimum 5. They form maximum
boiling azeotrops. boiling azeotrops.

(Any three, 1 mark each)

27. For a first order reaction:

2.303 [𝑅]
𝑡= 𝑙𝑜𝑔
𝑘 [𝑅]
3
(½ mark) Using this we get:
2.303 100
𝑡 = 𝑙𝑜𝑔
𝑘 1

(½ mark)

2.303 × 2
𝑡 =
𝑘

(½ mark)

Also
2.303 100
𝑡 = 𝑙𝑜𝑔
𝑘 10

(½ mark)

2.303
𝑡 =
𝑘

(½ mark)

. ×

Now

𝑡
=2
𝑡

(½ mark)
Or
Consider the reaction, R P is zero order reaction.
𝑑[𝑅]
𝑅𝑎𝑡𝑒 = − = 𝑘[𝑅]
𝑑𝑡
(½ mark)
𝑑[𝑅]
⇒ 𝑅𝑎𝑡𝑒 = − =𝑘
𝑑𝑡

⇒ 𝑑[𝑅] = −𝑘𝑑𝑡

Integrating both sides
[𝑅] = −𝑘𝑡 + 𝐼 ………Eq. 1
Where I is the constant of integration
(½ mark)
At t = 0, the concentration of the reactant R = [R]0 , where [R]0 is
initial concentration of the reactant.
(½ mark)
Substituting in above equation 1
[𝑅] = −𝑘 × 0 + 𝐼
[𝑅] = 𝐼
(½ mark)
Substituting the value of I in the equation 1 [𝑅] = −𝑘𝑡 + [𝑅]
(½ mark)
[𝑅] − [𝑅]
⇒ 𝑘=
𝑡
This is the integrated rate equation for a zero-order reaction.
(½ mark)

28. i) ability to adopt multiple oxidation states ii) ability to form
complexes. iii) transition metals utilise outer d and s electrons
for bonding. This has the effect of increasing the concentration
3
of the reactants at the catalyst surface and also weakening of the
bonds in the reacting molecules.
(1 mark each)
29. i) Freon-12 is used for aerosol propellants, refrigeration and
air conditioning purposes.
ii) Carbon tetrachloride is used in the synthesis of
chlorofluorocarbons and other chemicals, pharmaceutical
3
manufacturing, and general solvent use.
iii) Iodoform can be used as antiseptic.
(1 mark each)

30. A: CH3CH2CN

B: CH3CH2CH2NH2
C: CH3CH2CH2OH
(½ mark each)

A: C6H5NH2
B: C6H5N+2Cl-

C: C6H5OH 3

(½ mark each)
Or
i) Ethylamine is capable of forming hydrogen bonds with water
as it is soluble but in aniline the bulk carbon prevents the
formation of effective hydrogen bonding and is not soluble.

ii) A Friedel-Crafts reaction is carried out in the presence of
AlCl3. But AlCl3 is acidic in nature, while aniline is a strong base.
Thus, aniline reacts with AlCl3 to form a salt and benzene ring is
deactivated. Hence, aniline does not undergo the Friedel-Crafts
reaction.
iii) Gabriel phthalimide reaction gives pure primary amines
without any contamination of secondary and tertiary amines.
Therefore, it is preferred for synthesising primary amines.

31. (i) Dicholorocarbene, CCl2
mark)
(ii) Salicylic acid

Or

4
(iii)

(iv)

32. (i) β-D-2-Deoxyribose (1 mark)
(ii) Cytosine, uracil (1 mark)
(iii) Hydrogen bonds (1 mark) 4
(iv) RNA (1 mark)

33. 2Cr(a) + 3Fe3+ (aq) === 2Cr3+ 3Fe(s)
0.059 (0.01)2
E = E° - log (1 mark)
6 (0.01)3
E° = 0.261 V
0.059
E = 0.261 - log 10−2 (1 mark)
6
0.059
= 0.261 - × (−2)
6
= 0.261 + 0.0197 = 0.2807 V (1 mark)
(Deduct ½ mark for no or incorrect unit)
‘A’ will prevent iron from corrosion.
So, we can cost the iron surface with metal A because it has
more negative Eo value.
Or
𝑘 × 1000
Λ𝑚 =
𝐶
𝐶 = 0.001 𝑀, 𝑘 = 3.905 × 10−5 𝑆 𝑐𝑚−1 5
3.905×10−5 ×1000
⸫ Λ𝑚 = (1mark)
0.001
= 39.05 S cm2 mol-1
CH2COOH === CH3COO- + H+
Λ°𝑚 = 𝜆° 𝐶𝐻3 𝐶𝑂𝑂− + 𝜆° 𝐻+
= 40.9+349.6=390.5 S cm2 mol-1 (1mark)
Λ𝑚 39.05
Degree of dissocistion α = = = 0.1 (1 mark)
Λ°𝑚 390.5
(Deduct 1 mark for no or incorrect unit)
Electrochemical cell is a device used for the production of
electricity from energy released during spontaneous chemical
reaction. Electrochemical cell converts chemical energy into
electrical energy. (1mark)
If E°cell (external) > E°cell the cell starts acting as an electrolytic cell. In this
case, electrical energy is used to carry out non-spontaneous chemical
reaction.

34. (i) [Fe(CN)6]3- : Oxidation state of Fe = +3 (1 mark)

Fe(III) 3d5 4so (1 mark)

Hybridisation:

3d 4s 4p

↑ ↑ ↑ ↑ ↑

[Fe(CN)6]3-

↑↓ ↑↓ ↑ xx xx xx xx xx xx

d2sp3 hybridisation (1 mark)

Hybridisation: - d2sp3 : Magnetic character: - Paramagnetic. Spin type: -
Low spin complex.

5

Dextru laevo
Or
(i) In [CoF6]3- the Co(III) ion has 3d6 electronic
configuration. In the formation of the complex, it
involves sp3d2 hybridisation using outer d-orbitals. It is
therefore paramagnetic having 4 unpaired electrons.
(1+1 mark)
↑ ↑ ↑ ↑ ↑ xx xx xx xx xx xx

(ii) Dibromidobis (ethylenediamins) cobalt(III) ion.
(iii) It ionizes as : [Co(NH3)6]Cl3 [Co(NH3)6]3+ + 3Cl
⸫ 4 ions are produced. (1 mark)
(Deduct 1 mark for no or incorrect Hybridisation)

35.

(1+1+1 mark)

(b) (i) CH3COOH < HCOOH < FCH2COOH < NO2 – CH2COOH

(II) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde
5
(1+1 mark)

Or

Organic compound A is an ester as on acid hydrolysis it gives a
mixture of an acid and an alcohol.
(½ mark)
Oxidation of alcohol (C) gives acid (B). Hence, the number of
carbon atoms in (B) and (C) are the same.
(½ mark)

Ester (compound A) has eight C atoms. Hence, both carboxylic
acid (B) and alcohol (C) must contain 4 C atoms each.
(½ mark)
Dehydration of alcohol C gives but-1-ene. Hence, C must be a
straight chain alcohol, i.e butan-1-ol.
(½ mark)

Reactions:
𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝑂𝑂𝐶𝐻2𝐶𝐻2𝐶𝐻2𝐶𝐻3 + dil. H2SO4

𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝑂𝑂𝐻 + 𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝐻2𝑂𝐻

𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝐻2𝑂𝐻 Dehydratio 𝐶𝐻3𝐶𝐻2𝐶𝐻 = 𝐶𝐻2

CrO3/CH3COOH
𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝐻2𝑂𝐻 𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝑂𝑂𝐻

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