HBSE Class 12 Sample Paper 2026 Answers Chemistry

BSEH MARKING SCHEME

CLASS- XII Chemistry (March-2025-26) CHE-856

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) Benzene and Toluene 1

2. b) Gibbs energy 1
3. c) 0 1

4. c) 𝑀𝑛𝑂 1
5. c) Cr2+ 1
6. a) 0 1

7. b) Bromoform 1
8. a) tert-Butyl bromide 1
9. c) Both of the above 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 1

19. 2
The partial pressure of the gas in vapour phase (p) is proportional
to the mole fraction of the gas (x) in the solution.

(i) To increase the solubility of CO2 in soft drinks and soda water,
the bottle is sealed under high pressure.

(ii) To avoid bends the tanks used by scuba divers are filled
with air diluted with helium.

iii) At high altitudes the partial pressure of oxygen is less than that
at the ground level. This leads to low concentrations of oxygen in the
blood and tissues of people living at high altitudes or climbers.
(Any two, ½ mark each)
(OR)
Mass of solute (w1) = 1g
Mass of solvent (w2) = 50g
Kf = 5.12 K kg mol-1
Tf = 0.40
×
Since Tf = Kf × ( ½ mark)
×
×
So M2 = Kf ×
 ×
×
= 5.12 x 𝑔𝑚𝑜𝑙 ( ½ mark)
. ×
= 256 𝑔𝑚𝑜𝑙
(½ mark for answer, ½ mark for unit)
20. The law states that limiting molar conductivity of an electrolyle can 2
be represented as the sum of the individual contributions of the
anion and cation of the electrolyte.

i) to calculate limiting molar conductivity of any electrolyte.

ii) to calculate the degree of dissociation of weak electrolyte.

(iii) to calculate dissociation constant of weak electrolyte.

(Any two,½ mark each)

21.

2

22. (i)
N – OH

2
(ii)
H OH
C
H OCH3
23. (i) In 2,2,6 trimethyl cyclohexanone, three methyl groups are presents
at  - Position with respect to the ketonic (>C=O) group. Therefore,
these groups cause steric hindrance during the nucleophilic attack of
CN-ion so cyanohydrins is not formed. Due to the absence of methyl
groups in cyclohexanone, there is no steric hindrance and
cyanohydrins is formed.
2
(ii) Semicarbazide has two amino (NH2) groups, out of which one is
involved in resonance. Electron – density on this (NH2) decreases and
it does not act as a nucleophile. But the other (NH2) group (attached O
NH) has a lone pair of electrons which are not involved in resonance.
So, this pair is available for the nucleophilic attack on the carbonyl
group (>C=O) of aldehydes and/or ketones.

24. Aldehydes and having at least one -hydrogen undergo a reaction in
the presence of dilute alkali as catalyst to form -hydroxy aldehydes,
this is known as Aldol.
.
2CH3 – CHO ⎯⎯⎯⎯⎯ CH3 – CH – CH2 – CHO –
. Ethanal OH
3 – Hydroxybutanal (Aldol)

(1 mark) 2
OR
(i) Tollen’s test / Fehling’s test;
(½ mark)
Ethanal gives the test while propanone does not.
(½ mark)
(ii) Fehling’s test;
(½ mark)
Propanal gives the test while benzaldehyde does not.
(½ mark)
25. Quaternary structure of proteins: Some of the proteins are composed
of two or more polypeptide chains referred to as sub-units. The
spatial arrangement of these subunits with respect to each other is
known as quaternary structure.

Example:- Haemoglobin 2
(½ mark)
Which has four polypeptide chains around a haeme group.
(½ mark)

26. Molar conductivity of a solution at a given concentration is the
conductance of volume V of a solution containing one mole of
electrolyte kept between two electrodes with an area of cross-section A
and distance of unit length.

Molar conductivity increases with a decrease in concentration. This is
because the total volume V of the solution containing mole of the
electrolyte increases on dilution.
(½ mark)
For weak electrolytes, molar conductivity increases steeply on 3
dilution, especially near lower concentrations as shown in graph.
(½ mark)

(½ mark for proper diagram, ½ mark for each of 3 labelling)

27. Given :
𝐸 = 0.236 V

Number of electrons involved in cell reaction (n) = 2
(½ mark)
0
G = -nF 𝐸
(½ mark)
0
G = -2 x 96500 x 0.236 3
G0 = -45548 Jmol-1
(½ mark for answer ½ mark for unit)
0
𝑛 × 𝐸𝑐𝑒𝑙𝑙
𝑙𝑜𝑔𝐾 =
0.059
(½ mark)
𝐾 = 10
(½ mark)

28. i) Diamminechloridonitrito-N-platinum (II)
ii) Potassium trioxalatochromate (III)
iii) Pentaamminecarbonatocobalt (III) chloride
(1 mark each)
3

29. Step 1: Formation of protonated alcohol.

Step 2: Formation of carbocation: It is the slowest step and hence, the
rate determining step of the reaction.

3
Step 3: Formation of ethane by elimination of a proton.

Or
Lucas test can be used to distinguish between primary, secondary and
tertiary alcohols.
(½ mark)

For this Lucas reagent (conc. HCl and ZnCl2) is used.
(½ mark)
Alcohols are soluble in Lucas reagent while their halides are
immiscible and produce turbidity in solution.
(½mark)
In case of tertiary alcohols, turbidity is produced immediately as they
form the halides easily.
(½mark)
Secondary alcohols produce turbidity after some time.
(½mark)
Primary alcohols do not produce turbidity at room temperature.
(½mark)
30. (i) CHO
,∆
(CHOH)4 ⎯ CH3 – CH2 – CH2 – CH2 – CH2 – CH
CH2OH (n-Hexane)

(ii) CHO COOH

(CHOH)4 ⎯⎯⎯⎯⎯⎯ (CHOH)4
CH2OH CH2OH
3
Gluconic acid

(iii) CHO COOH

(CHOH)4 ⎯⎯⎯⎯⎯⎯ (CHOH)4
CH2OH COOH
Saccharic acid

31. (i) if a pressure larger than the osmotic pressure is applied to the
solution side, the pure solvent flows out of the solution through the
semi permeable membrane. This phenomenon is called reverse
osmosis.
(ii) Red blood corpuscles will swell up
OR
Two solutions having the same osmotic pressure at a given 4

temperature are called isotonic solutions.
(iii) The preservation of meat by salting or the preservation of fruits
by adding sugar
(iv) 1 M KCl

32.
(i) Position isomerism
(1mark)
(ii) 2, 5 – Dimethylhexane
(1mark)
(iii) (CH3)2C=CH2 4
(1mark)
(iv) (CH3)2C=CH2 + HBr ⎯⎯⎯⎯⎯ (CH3)2CHCH2Br
(1mark)
OR
Wurtz reaction

2Cr(a) + 3Fe3+ (aq) === 2Cr3+ 3Fe(s)
33. . ( . )
E = E° - log
( . )
(1mark)
E° = 0.261 V
.
E = 0.261 - log 10
(1mark)
.
= 0.261 - × (−2)
= 0.261 + 0.0197 = 0.2807 V
(1mark)
(Deduct ½ mark for no or incorrect unit)
‘A’ will prevent iron from corrosion.
(1mark)
So, we can cost the iron surface with metal A because it has more 5
negative Eo value.
Or
Give for same time period:
T1 = 298 K, reaction completed is 10%
T2 = 208 K, reaction completed is 25%
A = 4 x 1010s-1
For a fist order reaction:
2.303 100
𝑘 = 𝑙𝑜𝑔
𝑡 90

0.1055
𝑘 =
𝑡

(½mark)
And
2.303 100
𝑘 = 𝑙𝑜𝑔
𝑡 75

0.2879
𝑘 =
𝑡
(½mark)
According to Arrhenius equation:
𝑘 𝐸 𝑇 −𝑇
𝑙𝑜𝑔 = [ ]
𝑘 2.303 × 𝑅 𝑇 × 𝑇
(½mark)
.
𝐸 308 − 298
𝑙𝑜𝑔 . = [ ]
2.303 × 8.314 298 × 308
(½mark)
-1
𝐸 =76622.7 Jmol
(½mark for answer, ½ mark for unit)
For k at 318K:
According to Arrhenius equation
𝐸
𝑙𝑜𝑔𝑘 = 𝑙𝑜𝑔𝐴 −
2.303 × 𝑅𝑇
(½mark)

76622.7
𝑙𝑜𝑔𝑘 = log(4 × 10 ) −
2.303 × 8.314 × 318

𝑘 = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔(−1.9822)
(½mark)
𝑘 = 1.042 × 10 𝑠
(½mark for answer, ½ mark for unit)

34. The elements of the first series of the inner transition metals; 4f (Ce to
Lu) are known as lanthanoids.
General electronic configuration for lanthanoids is:
(𝑛 − 2)𝑓 (𝑛 − 1)𝑑 𝑛𝑠 where n = 6
The steady decrease in the atomic and ionic radii of lanthanide
elements with increasing atomic number is called Lanthanide
contraction.
Consequence:
(i) The size of the atom of the third transition series is closely the
same as that of the atom of the second transition series.
(ii) As there is only a minute change in the iconic radii of lanthanides,
their chemical properties are the same. This makes the separation very
difficult.
OR 5
Potassium permanganate is prepared by fusion of MnO2 with an alkali
metal hydroxide and an oxidizing agent like KNO3.
(½ mark)
2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O
(½ mark)
This produces the dark green K2MnO4 which disproportionate in a
neutral or acidic solution to give permanganate.
(½ mark)
3MnO42- + 4H+  2MnO4- + MnO2+2H2O
(½ mark)
(a) 5 Fe2+ + MnO4- + 8H+  Mn2+ + 4H2O + 5Fe3+
(b) 5SO2 + 2 MnO4- + 2H2O  2Mn2+ + 4H+ + 5SO42-
(b) 5C2O4 2-+ 2 MnO4- + 16H+  2Mn2+ + 8H2O + 10CO2

35. It is given that compound ‘C’ having the molecular formula, C6H7N is
formed by heating compound ‘B’ with Br2 and KOH. This is a
Hoffmann bromamide degradation reaction. Therefore, compound ‘B’
is an amide and compound ‘C’ is an amine.
(½ mark)

Compound A is Benzoic acid.

(½ mark)

5

(1Mark)

Or

(i) C6H6
(ii) C6H5NHCOCH3
(iii) C6H5SO2NHCH3
(iv) C6H5NC
(iii) C6H5N+H3HSO-4