MARKING SCHEME 2023

CHEMISTRY (Theory) - 043

QP CODE 56/4/1

Q.No Value points Mark

SECTION A
1 (D) 1
2 (A) 1
3 (C) 1
4 (A) 1
5 (A) 1
6 (B) 1
7 (C) 1
8 (A) 1
9 (C) 1
10 (B) 1
11 (A) 1
12 (A) 1
13 (D) 1
14
(A)
1
15 1
(C)
16 (C) 1
SECTION B
(a) The sum of powers of the concentration of the reactants in the rate law expression. 1
17
(b) The energy required to form activated complex / The minimum amount of extra energy
required by reactingmolecules to get converted into a product. 1
18
Δ Tf = K f m
K × w B × 1000 ½
MB = f
w A ×  Tf
o
 Tf = T f – Tf
 Tf = 273  15 – 272  07 = 1  08 K
1  86 × 18 × 1000
MB = 1
200 × 1  08
= 155 g mol –1
½
19 a) 𝐶𝐻 – 𝐶𝐻 – 𝐼; Iodide is a better leaving group/ due to larger size of iodine. ½,½
b) Butane < 1 – Chlorobutane < 1– Bromobutane < 1 – Iodobuatane. 1

20

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 a)

OR
20 (b)
(i)

(ii)

1
(Or by any other suitable method)
21 (a)

(b) Thymine – DNA, Uracil – RNA ½,½

SECTION C
22
(a)

½,½

(b) t 32g e1g 1

(c) When a ligand is bound to a metal atom or ion through a single donor atom.
Example: Cl- / H2O / NH3(or any other one correct example) ½
½

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23 Sn 2 + 
0  059
log  
o ½
E cell = E cell –
2 + 2
H 
o ½
E cell = 0 – (– 0  14 V) = 0  14 V
⋅ ( ⋅ )
E cell= 0 ⋅ 14 – 𝑙𝑜𝑔
( ⋅ )
1
0  059
= 0  14 – log 10
2
= 0 ⋅ 14 – 0 ⋅ 0295 = 0 ⋅ 1105 𝑉 or 0.11 V 1
(Deduct ½ marks for incorrect or no unit)
(Or by any other suitable method)
24
(a)

(b) 1×3

(or any other correct equation)

(c)

(d)

(Any three)
25 (a)
(i) On adding neutral FeCl3, phenol gives violet colouration whereasbenzoic acid does not give 1
violet colour.
(ii) On adding Tollens reagent, propanal gives silver mirror whereas propanone does not. 1
(or any other suitable chemical test).
(b) CH3CHF CH2 COOH ; due to stronger – I effect or electron withdrawing nature of F, as F is closer ½,½
to the carboxyl group.
26. (a) Amino acids that cannot be synthesised in the body and must be obtained through diet. 1
(b) The amide linkage between – COOH group and -NH2 group. / The amide linkage (-CONH-)which 1
joins two amino acids.
(c) When a protein in its native form, is subjected to physical change like change in pH, 1
temperature etc it loses itsbiological activity (Or destruction of secondary and tertiary structure.)

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27 (a) 3 – Bromo– 1 – chlorocyclohexene 1
(b) Nitro group is electron withdrawing group, it withdraws electron density from the benzene 1
ring and facilitates the attack of nucleophile on haloarene.
(c) CH 3 – CH 2 – Cl + KOH(a lc) 
 CH 2 = CH 2 + KCl + H 2 O / Ethene is formed. 1

28.
2  303  R 0
k= log ½
t R 
For 99·9 % completion

Let  R 0 = 100,

R  = 100 – 99·9 % = 0·1

2  303 100 ½
t 99·9 % = log
k 0 1

2  303
= log 1000
k
2  303
= × 3 …………………… (i) ½
k
Let  R 0 = 100,  R  = 100 – 50 = 50

2  303 100
t 50 % = log ½
k 50

2  303
= log 2
k

2  303 ½
= × 0  3010 ………………………… (ii)
k
Divide (i) by (ii)

2  303
t 99·9 % × 3
k
=
t 50 % 2  303
× 0  3010 ½
k
t 99·9 %
= 10
t 50 %

or t 99·9 % = 10 t 50 %
(or by any other suitable method)
SECTION D

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29.
(a) Paramagnetic, F – does oes not cause pairing of electrons and hence unpaired electrons are
½,½
left.
(b) 6 1
(c) (i) diamminedichloridoplatinum(IV) ion 1
(ii) It uses inner d orbitals because NH3 causes pairing of electrons 1
OR
c)

Shape: Octahedral ; Hybridization : sp3d2 ½,½

30 (a) It allows flow of ions
ns and the circuit is completed / itmaintains
it the electrical neutrality.
neutrality 1
(or any other correct reason).
(b) When E ext > E cell 1
o o o
(c) E cell = E –E
Cu 2+ /Cu Zn 2+ //Zn
Zn
= 0  34 – (– 0  76) = 1  10 V 1
o
As E cell = +ve, the reaction takes place,
place so copper sulphate cannot be stored ina zinc pot. 1
OR
(c) (i) 5F 1
(ii) 2 F 1
SECTION E
31 (a) Zn hasfully filled d-orbital
orbital configuration in ground state and in its oxidized state
state.
(b) The filling of 4f orbital before 5d orbital results in steady decrease in atomic radii and ionic
radii. / The steady decrease in the atomic ra radii or ionic radii of the elements with increase in
atomic number.
(c) In chromium an electron is removed from 4s1 while in Zn it is from fully filled 4s2orbital.
(d) Due to variable
ariable oxidation stateandcomplex
stat formation /provide large surface
urface area. 1x5
(e) Due to d–d d transition of electrons in d
d– orbitals / unpaired electrons in d-orbital
orbital.
(f) K2MnO4, due to the presence of one unpaired electron.
(g) Cr 2 O 7 2 – + 14 H + + 6 e –   2 Cr 3 + + 7 H 2 O (Any five)
32. (a) (i) If a pressure larger than the osmotic pressure is applied to the solution side,
side, resulting in the 1
movement of solvent particles from solution to solvent.
(ii) Solubility of gases in water decreases with rise in temperature
temperature.. More oxygen will be 1
available in the cold water.
(iii)
1

. × 1
.
= ×
 32.8- 𝑝 = 0.0656
𝑝 = 32.734 mm Hg (Deduct ½ mark for no unit or incorrect unit) 1
OR
32 (a) (i) i will be less than 1. 1
(ii) Solution which obeys Raoult's law over the entire range of concentration. 1
(iii)
i=3
𝛥𝑇 = 𝑖 ×𝐾 × 𝑚 1

𝑖 × 𝐾 × 𝑤 × 1000
𝛥𝑇 =
𝑀 ×𝑤

3 × 1.86 × 𝑤 × 1000 1
2𝐾 =
111 × 500

2 × 111 × 500
wB = 1
3 × 1  86 × 1000
= 19  89 g (Deduct ½ mark for no unit or incorrect unit)
33

1x5

(1 mark for identification of A, ½ + ½ each for identification and reaction of formation of B, C,

D, E).
OR
33 (b) (i) (1) Benzene Sulphonyl Chloride (C6H5SO2Cl) (Name or formula) 1
(2)C2H5NH2< (C2H5)2 NH < (C2H5)3 N 1

(ii) (1) In methylamine, electron donating effect of – CH3 group increases the availability of 1
lone pair of electrons on nitrogen of the amino group. / In aniline, benzene withdraws electrons
due to resonance therefore electron pair is less easily available for protonation.
(2) Due to strong activating effect of amino group. 1
(3) Due to intermolecular hydrogen bonding in primary amines. 1

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