HBSE Class 12 Sample Paper 2024 Answers Chemistry

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2024
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Marking Scheme
Sample Paper (2023-24) CHEM-856 Class: 12th
SECTION-A
1. (d) Vitamin B12 (1)
2. (c) About three times (1)
3. (b) (1)
ECu2+/Cu

log[Cu2+]

4. (d) 40 min (1)
5. (b) Sorbitol (1)
6. (a) CH3 NC (1)
7. (d) [Cr (H2O)6]Cl3 (1)
8. (d) Benzyl alcohol (1)
9. (b) CrO42- (1)
10. (b) Diethyl ether (1)
11. (a) CH3 NH2 (1)
12. (d) Aspirin (1)
13. (d) P-Benzoquinone (1)
14. (c) ix = iy = iz (1)
15. (c) Assertion (A) True, Reason (R) False (1)
16. (b) Assertion (A) True, Reason(R) True (1)
But Reason(R) not true explanation
17. (c) Assertion (A) True, Reason (R) False (1)
18. (d) Assertion (A) False, Reason (R) True (1)
SECTION-B
19. (a) Aniline being lewis base react with Anhydrous AlCl3 which is lewis acid
to form salt. (1)

(b) Methylamine accept proton from water and liberate OH- ion which
combine with Fe3+ ion to form hydrated ferric oxide Fe(OH)3 or (1)
Fe2O3.3H2O
CH3
|
20. (a) CH3-C-CH3 or 2,2-dimethyl propane (1)
|
CH3
(b) CH3-CH-CH2-CH3 or 2-methyl butane (1)
|
CH3
OR
(a) Because Grignard reagent reacts with moisture and form Alkane. (1)
(b) C-Cl bond in chloro benzene acquire some double bond character due to
delocalization of ions pair on chlorine so bond length decreases
OR
any other relevant answer. (1)
21. (a) amylose is water soluble linear polymer of -D glucose whereas
amylopectin is water insoluble branched (C1-C6) glycosidic linkage carrying
branched polymer. (1)
(b) Intra molecular H-Bonding (1)
22. Geometrical Isomers (1)

Cis Trans
Optical isomers (1)

dextro(+) laevo(-)

23. Q =I x t
= 0.5 x 4 x 60 x 60
= 20x360
= 7200C (1)
96500 corresponds to 6.02 x 1023 e-
6.02𝑥1023
7200 C gives = x 7200
96500

= 4.49 x 1022 e- (1)
24. (a)(i) Azeotropic mixture is type of liquid mixture having definite
composition and boiling like a pure liquid (½)
eg. 95.37% C2H5OH + 4.63% H2O (½)
OR
Any other relevant example
(ii) Solutions which have the same osmotic pressure at same temperature (½)
eg. 0.9% solution of pure NaCl is isotonic with RBC (½)
OR
Any other relevant example
OR
(b) If we have two completely miscible volatile liquid A and B having mole
fraction xA and xB Then at certain temperature partial pressures P A and PB
and vapour pressure in pure state PA° and PB° are expressed as
PA=PA°.xA
PB=PB°.xB
PT=PA+PB (½)
PT=PA°.xA+ PB°.xB
PT=PA°(1-xB)+PB°xB
when xA=1 PT= PA°.xA
when xB=1 PT=PB°xB (½)

PT=PA+PB PB°
PA° III
°x B
=PB
PB
PA =
PA °
II I xA

xA=1 xA=0
xB=0 xB=1 (½)
P
yA= A yB=1-yA (½)
PT

25. (i) Ea decrease (1)
(ii) No effect on G (1)
SECTION-C
26. (a) It is the amide linkage present between – COOH group of one  amino
acid and NH2 group of other amino acid. (1)
(b) When protein in native form is subjected to physical changes like change
in temperature or pH then hydrogen bonds are broken, it looses its
biological activity and all structures are destroyed and only primary
structure remain intact. (1)
(c) It is the sequence in which various -amino acids present in a protein are
linked to one another. (1)
OR
Amino acids contain acidic and basic group within same molecule. In
aqueous solution they neutralize each other, carboxyl group loses a
proton and amino group accept it. (1)
NH2-CH-COOHNH3+-CH-COO-
| |
R R
(Zwitter ion)
H+ OH-
NH3+-CH-COOH NH3+-CH-COO- NH2-CH-COO- (1)
| OH- | +
H |
R R R
NH3+ group Amphoteric COO- group (1)
act as acid react with act as base
acid and base

27. (a) CH3CH2COOH (1)
(b) OH OH (1)
Br
+

Br

O OMgBr
|| |
(c) CH3-C-CH3+CH3 Mg Br  CH3-C-CH3 (1)
|
CH3
H+ / H2O
OH
|
CH3-C-CH3
|
CH3

28. (a) 1st order (1)
(b) min-1 (1)
0.693
(c) t½ = (1)
𝐾
29. For AB2
𝐾𝑓.𝑊𝐵.1000
MAB2 = (½)
𝑊𝐴.𝑇𝑓

5.1 𝑥 1 𝑥 1000
=
20 𝑥 2.3

= 110.87u (½)

5.1 𝑥 1 𝑥 1000
MAB4 =
20 𝑥 1.3

= 196.5u

Atomic mass of A=a and Atomic mass of B is b
 a+2b = 110.87 (i) (1)
a+4b = 19.65 (ii)
(ii) – (i)
196.5 – 110.87 = a+46 – a-2b
85.28 = 2b (1)
b = 42.64u

a+2b=110.87
a+2x42.64=110.87
 a = 110.8 - 8528
= 25.29u
i.e. atomic mass of A = 25.59 u
atomic mass of B = 42.64 u

30. (a) C6H5CH=N NHCONH2 (1)

(b) O 4- Oxocyclohexane (1)
Carboxylate anion (1)

COO-

(c)
O

OR
OH-
(a) 2CH3CHO CH3-CH-CH2-CHO (1)
|
OH 

CH3-CH=CH-CHO
CHO dil CHOH-CH2CHO CH=CH-CHO
(b) +CH3CHO (1)
NaOH 

H2 / Ni

CH2-CH2-CH2OH

COOH COCl CHO

(c) SOCl2 Pd / BaSO4/S (1)
H2

31. (a) The difference of energy between the two sets of a orbitals is called as
crystal field splitting energy. (1)
(b)

(2)

OR
Mn2+=4s° 3d5
H2O being weak ligand, don't cause pairing 5 unpaired e (1)
t2g3 eg2
CN- strong ligand, cause pairing so there is 1 unpaired e- (1)
i.e. t2g5 eg°
(c)  O> P pairing occurs (1)
 O< P No pairing occurs
32. (a) C6H5CHClC6H5 (1)
(b) 1-Bromo pentane>2-Bromopentane>2-Bromo-2-methylbutane (1)
(c) Allylic carbocation is stable (1)
- -
(d) I is better leaving group than Cl (1)
OR
Br
(d) Allylic substitution (1)

33. (a) (i) 4 FeOCr2O3 + 8 Na2CO3 + 7O2  8Na2CrO4+2Fe2O3+8CO2 (1)
(ii) 2Na2CrO4+H2SO4  Na2Cr2O7+Na2SO4+H2O (1)
(Conc)

(iii) Na2Cr2O7+2KCl K2Cr2O7 + 2NaCl (1)
(b) (i) Cr2O2-7 +14H+ + 6I-  2Cr3++7H2O+3I2 (1)
(ii) Cr2O2-7 +6Fe2+ + 14H+  2Cr3++6Fe3++7H2O (1)
OR
(a)(i) 2MnO2+4KOH+O22K2MnO4+2H2O (1)
(ii) 2K2MnO4+Cl22KMnO4+2KCl (1)
OR
any other relevant answer.
(b) Lanthanoids Actinoids
(i) Electronic Configuration
[xe]4f1-145d0-1 6s2 [Rn] 5f1-14 6d0-17s2 (1)
(ii) Regular decrease in Regular decrease in (1)
size from left to size from left to
right known as right is known as
lanthanoid contraction Actinoid contraction

(iii) Lanthanoids react with  Actinoids are
dilute acid to liberate highly reactive in
H2 gas divided state
 Form oxide and hydroxides  React with boiling water
of type M2O3 / M(OH)3 to give mixture of oxide and
hydride
 With C form carbides  Attacked by HCl but the effect of
HNO3 is very small.
 With halogen form halides  No action of alkalies
OR
any other relevant difference
34. (a) E°cell=E°cathode-E°Anode
= E°Cu2+/ Cu - E°Mg2+/Mg
= 0.34 – (-2.36) (½)
= 2.70V
2+
Mg(s)+Cu Mg2+ + Cu(s)
(0.0001M) (0.001M)
0.0591 [Mg2+]
Ecell=E°cell – log
2 [Cu2+]

0.0591 0.001]
= 2.70 - log
2 0.0001

= 2.70 – 0.0295 log 10 (1)
= 2.70 – 0.0295 x 1
= 2.6705V

(b) Because the number of ions per unit volume decreases. (2)
OR
(a) (i) During recharging, cell is operated like electrolytic cell.
(ii) Electrical energy is supplied to it from external source.
(iii) Electrode reactions are reverse of that of discharging. (1)
(iv) At cathode (Reduction) (1)
PbSO4(s) + 2e- Pb(s)+SO42-(aq)

At Anode (oxidation)
PbSO4(s) + 2H2O PbO2(s)+SO42-(aq) + 4H++2e-

Overall reaction
2PbSO4 + 2H2O Pb(s)+PbO2(s)+4H(aq)++2SO42-(aq) (1)

(b) E° Zn2+/Zn = -0.76V
E° Cu2+/Cu=0.34V
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
Red
Zn(s) + Cu2+ Zn2+ + Cu(s)
oxid
E°cell=0.34 – (-0.76)
= 1.10V
(c) E°cell +ve means reaction is spontaneous and in this reaction zinc is
oxidised  we can't store CuSO4 in zinc pot. (2)

35. (a) CHO (1)

NO2

COOH
(b) (1)

(c) NOH (1)

conc
(d) HCHO+HCHO CH3OH + HCOONa (1)
NaOH

(e) CH3 CHOH-CH-CHO CH3CH= C-CHO (1)
| -H2O |
CH3 CH3
OR
(a) Phenol gives violet colouration with neutral FeCl3 solution but benzoic
acid does not. (1)
OR
any other relevant test
(b) Acetaldehyde is more reactive towards nucleophillic addition reaction
because of stearic hindrance in acetone. (1)
(c)

(1)
CHO

C2H5

2-Ethyl benzaldehyde

CHO COOH
[O]

C2H5 COOH
+ -
1,2 Benzene dicarboxylic acid
[Ag(NH3)2] OH

COO-

+ Ag
C2H5
2,4 DNP NO2

CH = N-NH- NO2 + H2O (1)
+ Ag
C2H5