CBSE Class 12 Marking Scheme 2021 for Chemistry

MARKING SCHEME

SAMPLE PAPER 1

SECTION A

Q.No. Value Point Marks
1(i) D 1
(ii) B OR A 1
(iii) B 1
(iv) C 1
2(i) B 1
(ii) A 1
(iii) A 1
(iv) A or B 1
3 C 1
4 D or C 1
5 C 1
6 B OR B 1
7 B OR D 1
8 A OR A 1
9 C 1
10 A 1
11 A 1
12 A 1
13 D 1
14 B OR B 1
15 B 1
16 A 1
SECTION B, C, D

Q.No. VALUE POINTS MARKS
SECTION B
17 Nitro group at ortho position withdraws the electron density from the benzene ring 2
and thus facilitates the attack of the nucleophile on haloarene.

OR

(i) NH2 N2Cl Cl

NaNO2 + HCl Cu2Cl2 1
273 – 278 K

(ii) CH3CH(Br)CH3 alc KOH CH3CH=CH2 HBr, organic peroxide
CH3CH2CH3Br 1

18 ΔTb = Kf m ΔTb = 101.04-100 = 1.04 oC
or m= 1.04 /0.52 = 2 1
Relative lowering of VP = x2 1/2
Relative lowering of VP = n2/n1+n2 1/2
= 2/ 2+55.5 = 2/57.5 = 0.034 atm

19 (i)t2g4eg2 Paramagentic ½, ½
(ii)Dichloridobis(ethane-1,2-diamine)cobalt(III)nitrate 1
OR
(i)Square planar 1
(ii)Cu2+ = 3d9 1 unpaired electron so √ 1(3) = 1.73BM 1
20 Reaction is a complex reaction. 1/2
Order of reaction is 1.5.
Molecularity cannot be 1.5, it has no meaning for this reaction. The reaction occurs
in steps, so it is a complex reaction. 1/2
(ii)units of k are mol-1/2L1/2s-1 1
OR
Ans : let the rate law expression be Rate = k [P]x[Q]y
from the table we know that
Rate 1 = 3.0 x 10-4 = k (0.10)x (0.10)y
Rate 2 = 9.0 x 10-4 = k (0.30)x (0.30)y
Rate 3 = 3.0 x 10-4 = k (0.10)x (0.30)y

Rate 1/ Rate 3 = (1/3)y or 1 = (1/3)y
So y = 0 ½
Rate 2/ Rate 3 = (3)x or 3 = (3)x
So x = 1 ½
Rate = k [P] 1

21 k = 0.693/t1/2
k = 0.693/5730 years-1 ½
t = 2.303 log Co
k Ct ½
let Co = 1 Ct = 3/10 so Co/Ct = 1/ (3/10) = 10/3
t = 2.303 x 5730 log 10 ½
0.693 3
t = 19042 x (1-0.4771) = 9957 years 1

22
½

½

½

½

23 XeF6 1
. Central atom Xe has 8 valence electrons, it forms 6 bonds with F and has
1 lone pair. According to VSEPR theory, presence of 6 bp and 1 lp results in 1
distorted octahedral geometry

24. (a)inverted product will be given by 1 Chlorobutane as it undergoes SN2
reaction. ½+1/2
(b)racemic mixture will be given by 2 chloro-2-methylpropane as it undergoes ½+1/2
SN1 reaction
25 Let no. of Atoms of element P be x
1/2
No. of tetrahedral voids = 2x
No. Of octahedral voids = x

½
Atoms of Q = 1/3 (2x) + x = 5x/3
PxQ5x/3
P3Q5 1

SECTION C
26
(i)Due to large surface area and ability to show variable oxidation states 1
(ii)Due to high value of third ionisation enthalpy 1
(iii) Oxidation state of Cr in Cr2O3 is +3 and of CrO is +2. When oxidation number 1
of a metal increases, ionic character decreases so CrO is basic while Cr 2O3 is
amphoteric.
OR
(i) The general trend towards less negative E V values across the series is 1
related to the general increase in the sum of the first and second
ionisation enthalpies.
(ii) The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its 1
hydration
enthalpy.
(iii) The stability of the half-filled d sub-shell in Mn2+ and the completely 1
filled d10 configuration in Zn2+ are related to their more negative Eo V
values

27 (i) Aniline, N-ethylethanamine Etanamine 1
(ii)Ethanamine,ethanol, ethanoic acid 1
(iii) N, N dimethylmethanamine, methanamine, N-methylmethanamine 1
OR
(i) N-methyletahnamine is a secondary amine. When it reacts with 1
benzenesulphonyl chloride, it forms N- Ethyl -N methyl sulphonamide while and
N,N-dimethyl etahnanmine is a tertiary amine it does not react with
benzenesulphonyl chloride.

1

½
(iii)Butan-1-ol
Alcohol forms stronger hydrogen bonds with water than formed by amine due to ½
higher electronegativity of O in alcohol than N in amine

28 We know that d = zM/ Na a3 1/2
For fcc, z=4 therefore d = 4 x M / Na (3.5 x 10 -8)3 g/cm3 1
For bcc, z=2 therefore d’ = 2 x M / Na (3.0 x 10 -8)3 g/cm3 1
d/d’ = 4/(3.5 x 10-8)3 / 2/(3.0 x 10-8)3 = 3.17:1 1/2

29 (i)
CH3 CH2 –COOH (CH2)4- NH2 1

HOOC CH NHOC CH NHOC CH NH2

CH2COOH CH3 (CH2)4- NH2
1
HOOC CH NHOC CH NHOC CH NH2

(ii)
1

30 i. Arrange the following in decreasing order of bond dissociation enthalpy 1
I2 < F2 < Br2 < Cl2 ,
ii. Bi does not form pπ-pπ bonds as its atomic orbitals are large and diffuse so
1
effective overlapping is not possible
iii.Due to small size of oxygen, it has greater elecrton electron repulsions 1
SECTION D
31. (i)
(a) 3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O 1
(b)

1

1
(ii)‘X’ is Helium
It is used as a diluent for oxygen in modern diving apparatus because of its very 1
low solubility in blood.
It monoatomic having no interatomic forces except weak dispersion forces and has 1
second lowest mass therefore bp is lowest.

OR
(a) H2Te, H2Se, H2S, H2O 1
(b) [Fe (H2O)5 (NO)]2+ 1
(ii) A is chlorine gas 1
Its bleaching action is due to oxidation.
Cl2 + H2O → 2HCl + O , Coloured substance + O → Colourless substance 1
6 NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
1

36

½ each
A: C CH B: COCH3 C: COOK D: CHI3

1
C CH COCH 3

HgSO4, H2SO4

1
COCH3

CH3COCl, anhy AlCl3

1

COCH3 COOK

KOH , I2 + CHI3

OR

1

1

1

1

1

37
(i) limiting molar conductivity of an electrolyte can be represented as the sum of 1
the individual contributions of the anion and cation of the electrolyte.
(ii) Eocell = Eocathode -Eoanode = 0.34-(-1.66) = 2.00 V
Ecell = Eocell – 0.059 log [Al3+]2 1/2
n [Cu2+]3 1
Here n = 6 ½
Ecell = 2 – 0.059 log [0.15]2
6 [0.025]3 1
= 2 – 0.059/6 ( 2log 0.15 – 3 log 0.025)
= 2 – 0.059/6 (-1.6478 +4.8062) = 2- 0.0311 = 1.9689V
1

OR
(i) MnO4– 1

(ii)(a) Molar conductivity of a solution at a given concentration is the conductance 1
of the volume V of solution containing one mole of electrolyte kept between two
electrodes with area of cross section A and distance of unit length.
(b)Strong electrolyte, For strong electrolytes, Λm increases slowly with dilution 1

( c) Λm = Λm° – A c ½
Therefore Λm° = 150 S cm2 mol–1 1

(d)
1

A = – slope = - (149 – 147.8/ 0.010-0.022) = 100 S cm2 mol–1/(mol/L–1)1/2.