CBSE Class 12 Chemistry Question Paper 2020 Set 56-5 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – CHEMISTRY
(SUBJECT CODE -043) (PAPER CODE – 56/5/1,2,3)

General Instructions: -

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assessment of the candidates. A small mistake in evaluation may lead to serious
problems which may affect the future of the candidates, education system and
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Evaluation is a 10-12 days mission for all of us. Hence, it is necessary that you
put in your best efforts in this process.
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should not be done according to one’s own interpretation or any other consideration.
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and/or are innovative, they may be assessed for their correctness otherwise
and marks be awarded to them.
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marks should be retained and the other answer scored out.
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only once.
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 Leaving answer or part thereof unassessed in an answer book.
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mark is correctly and clearly indicated. It should merely be a line. Same is with
the X for incorrect answer.)
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awarded.

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should be marked as cross (X) and awarded zero (0)Marks.

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14. The Examiners should acquaint themselves with the guidelines given in the
Guidelines for spot Evaluation before starting the actual evaluation.

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over to the title page, correctly totaled and written in figures and words.

16. The Board permits candidates to obtain photocopy of the Answer Book on request in
an RTI application and also separately as a part of the re-evaluation process on
payment of the processing charges.

Marking scheme – 2020

CHEMISTRY (043) / CLASS XII

56/5/1
Q.No Expected Answer / Value Points Marks
SECTION A
1 By gaining one electron they acquire noble gas configuration/ smallest size and high effective 1
nuclear charge in their respective period.
2 Extremely small size/ absence of d orbital/highest electronegativity / low bond dissociation 1
enthalpy of F-F bond.
3 HI>HBr>HCl>HF 1
4 Low bond dissociation enthalpy and high hydration enthalpy. 1
5 X >X’ /X is bigger in size and X’ is smaller. 1
6 Mercury cell 1
7 5F 1
8 k/2.303 1
9 Saccharine/Sucralose / alitame (any other except Aspartame) 1
10 Bakelite 1
11 (c) 1
12 (b) 1
13 (c) 1
14 (a) 1
15 One mark may be awarded to any option 1
16 (D) 1
17 (D) 1
18 (A) 1

19 (C) 1
20 (D) 1
SECTION B
21
 For a solution of volatile liquids, the partial vapour pressure of each component of the 1
solution is directly proportional to its mole fraction present in solution.

 If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the
partial pressure of the volatile component or gas is directly proportional to its mole 1
fraction in solution.
22 (i) NaCN acts as a leaching agent / it forms complex with gold/ [Ag(CN)2]- 1
4Au + 8CN– + 2H2O + O2 4 [Au(CN2)] – + 4OH– (Balancing may be ignored)
(ii) CO acts as a reducing agent 1
OR
22  It is leached out using acid or bacteria 1
 Electrolytic refining 1

23  The accumulation of molecular species at the surface rather than in the bulk of a solid or 1+½
liquid. Example: adsorption of gases on surface of active charcoal (or any other suitable
example) ½
 Adsorption of reactants occurs on surface of catalyst and reaction takes place.

OR
23  A state of continuous zig-zag motion of particles. 1
 Unbalanced bombardment of the particles by the molecules of the dispersion medium. ½
 The Brownian movement has a stirring effect which does not permit the particles to settle. ½

24 (a) Hexacyanidoferrate(III) / Hexacyanoferrate(III) ½
d2sp3 ½
(b) Ligand which can ligate through two different atoms is called ambidentate ligand whereas
di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion. / a 1
chelating ligand forms a more stable complex as compared to an ambidentate ligand. / chelating
ligand forms a cyclic complex while ambidentate ligand forms a non-cyclic complex.
25  Antiseptic is applied on living tissue, to kill or stop growth of microbes while disinfectant is 1
applied on inanimate/ non -living objects

 0.2 per cent solution of phenol is an antiseptic while its one percent solution is disinfectant. 1
26.

1
(i)HOCH2CH2OH and / ethylene glycol and phthalic acid / Ethane-1,2-diol and Benzene-1,
2 -dicarboxylic acid
1
(ii)CH2=CHCN / Acrylonitrile / Propene nitrile
27

(i) 1

(ii)

1

SECTION C
28 ΔTf = iKf m ½
0.068 = i x 1.86 x 0.01 ½
i = 3.65 or 3.656 ½
α = i-1/n-1 ½
α = 0.883 or 0.885
88.3% or 88.5% (or by any other correct method) 1

29 m=Z I t ½
2 = 63.5 x 2 x t/2x96500
t = 3039.4 s 1
m1/m2 = eq wt 1/eq wt 2 ½
2 / m2 = 63.5/2 / 65/2
m2 = 2.05 g (or by any other correct method) 1
(deduct ½ mark for incorrect or no unit)
30 (i) Amylose is water soluble component of starch while amylopectin is insoluble in water 1
(ii) Globular proteins are spherical in shape while fibrous are linear. 1
(iii) Nucleoside consists of a sugar and a base
1
When nucleoside is linked to phosphate group, it forms a nucleotide

(or any other suitable difference in each case)
31 A: ( CH3 )2 C=CH2 B : ( CH3 )2 CBrCH3 C : ( CH3 )3 C - C(CH3 ) 3 ½ X6
D: ( CH3 )2 CHCH2MgBr E : ( CH3 )2 CHCH3 F: ( CH3 )2 CHCH2OC2H5 =3
32 (i) CH3 CH2CH2OH 1
(ii) (CH3 )2C=CH2 1
(iii)

1

OR
32
1

(i)
(Intermediate compound in above equation may be ignored)
1
1. CH3MgBr
(ii)HCHO CH3CH2OH
2. H2O
(iii) C6H5OH +CH3COCl C6H5OCOCH3 1
(or by any other correct method)
33 (i) Aniline forms salt with AlCl3, the Lewis acid. 1
(ii) Aryl halides do not undergo nucleophilic substitution with the anion formed by 1
phthalimide
(iii) Due to +I effect of alkyl group electron density on N atom increases. 1
34 Lyophobic sol Lyophilic sol
Interaction between dispersed phase Interaction between dispersed phase and
and dispersion medium are weak dispersion medium are strong 1x3=3
irreversible reversible
Can be easily coagulated Can’t be easily coagulated

(or any other suitable difference)

OR
34 (i) Lyophilic colloids have a unique property of protecting lyophobic colloids./ Lyophilic 1
colloids form a layer around the lyophobic colloids to protect them from the electrolyte or
coagulation. 1
(ii) Potential difference between the fixed layer and the diffused layer of opposite charges of a
colloid. 1
(iii) Substances used for stabilisation of an emulsion.

SECTION D
35 a) i) Variable or multiple oxidation states / ability to form complexes / they provide large 1
surface area for adsorption.
ii) Similar size/similar properties 1
iii)No unpaired electron/weak interatomic metallic bonding / completely or fully filled d 1
orbitals 1
+ +
b) i) 2Na2CrO4 + 2 H → Na2Cr2O7 + 2 Na + H2O 1
ii) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(Balancing may be ignored in both above reactions)
OR
a) i) Ti has an unpaired electron while there are no unpaired electrons in Sc3+.
3+ 1
35 ii) Stable t2g3 of Cr3+ ion 1

b) 1. Both show variable oxidation states 1
2. Both show f-f transitions 1
3. Electrons of f-orbital in both show poor shielding effect
4. both have common +3 oxidation state
5. both show contraction in atomic radii. (any two suitable differences)
2– + –
c) 3MnO4 + 4H → 2MnO4 + MnO2 + 2H2O 1

36 a) (i) 3-hydroxy-3-phenylpropanal /
1

/ C6H5CH(OH)CH2CHO
(ii) Phenyl hydrazone of benzaldehyde / 1
C6H5CH=N-NHC6H5

(iii)Sodium benzoate and benzyl alcohol /

½+½

and
b) (i) On heating with NaOH and I2 : CH3CH=CHCOCH3 will form yellow ppt of CHI3 while 1
other compound doesn’t .
(ii) On adding NaHCO3 : Benzoic acid produces brisk effervescence while other 1
compound doesn’t.
(or any other suitable chemical test)
OR
36 1
a) (i) CH3CH2CH3
1
(ii) C6H6
1
(iii) CH2=CH-CH2CHO

b) C6H5COCH3 < CH3COCH3 < CH3CHO < HCHO
1

c)
1

37 a) k = (2.303 / t) log ([A]o / [A]t) ½
k = (2.303 / 40) log (100 / 75)
1
= 0.007 min-1or 0.0071 min-1or 0.0072 min-1
t = (2.303 / k) log ([A]o / [A]t)
½
t = (2.303 / 0.0071) log (100/20)
t =230 min or 226.7min or 223.7 min. (deduct ½ mark if incorrect or no unit) 1
b) Sum of powers of the concentration of the reactants in the rate law expression. 1
When one of the reactant is present in large excess. 1
OR
37
a) k = 0.693/ t1/2
½
k1= 0.693/ t1/2= 0.693 / 30
k2= 0.693/ t1/2= 0.693 / 10
log k2/k1 = Ea /2.303 R (1/ T1 -1/ T2) 1
log 3 = Ea /2.303 x 8.314 (1/ 300 -1/ 320)

Ea = 2.303 X 8.314 x 0.4771 x ( 300 x 320/20) ½
= 43848.5 J/mol OR 43855 J/mol or 43.8 kJ/mol 1
b) Proper orientation ½
Energy of the colliding particles should be more than threshold energy ½
c) For a complex reaction, order of reaction is applicable while molecularity has no 1
meaning.

Marking scheme – 2020

CHEMISTRY (043)/ CLASS XII

56/5/2
Q.No Expected Answer / Value Points Marks
SECTION A
1 By gaining one electron they acquire noble gas configuration 1
2 Extremely small size/ absence of d orbital/highest electronegativity 1
3 HI>HBr>HCl>HF 1
4 Low bond dissociation enthalpy and high hydration enthalpy 1
5 X >X’ 1
6 Benzylchloride / C6H5CH2Cl 1
7 N,N-dimethylaniline OR N,N-dimethylbenzenamine 1
8 Glycosidic linkage 1
9 Aspartame 1
10 Teflon/PTFE 1
11 (c) 1
12 (c) 1
13 (d) 1
14 (c) 1
15 (d) 1
16 (A) 1
17 (D) 1
18 (C) 1
19 (D) 1
20 (A) 1
SECTION B
21 i) NaCN acts as a leaching agent / it forms complex with gold/ [Ag(CN)2]- 1
4Au + 8CN- + 2H2O + O2 4 [Au(CN2)]- + 4OH- (Balancing may be ignored) 1
ii) CO acts as a reducing agent
OR
21  It is leached out using acid or bacteria 1
 Electrolytic refining 1
22  For a solution of volatile liquids, the partial vapour pressure of each component of the 1
solution is directly proportional to its mole fraction present in solution.

 If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the 1
partial pressure of the volatile component or gas is directly proportional to its mole
fraction in solution.
23 (i)

1

(ii)

1

24  The accumulation of molecular species at the surface rather than in the bulk of a solid or 1+½
liquid e.g. adsorption of gases on surface of active charcoal (or any other suitable example)
 Adsorption of reactants occurs on surface of catalyst and reaction takes place. ½

OR
24  A state of continuous zig-zag motion of particles. 1
 Unbalanced bombardment of the particles by the molecules of the dispersion medium. ½
 The Brownian movement has a stirring effect which does not permit the particles to settle. ½

25 (i) Formaldehyde and phenol / HCHO and C6H5OH 1
(ii) Adipic acid and hexamethylenediamine / HOOC (CH2)4 COOH and H2N (CH2)6 NH2 1

26 2 marks to be given for attempting the question. 2
27  Antiseptic is applied on living tissue, to kill or stop growth of microbes while disinfectant is 1
applied on inanimate/ non -living objects

 0.2 per cent solution of phenol is an antiseptic while its one percent solution is disinfectant. 1
SECTION C
28 A: ( CH3 )2 C=CH2 B : ( CH3 )2 CBrCH3 C : ( CH3 )3 C - C(CH3 ) 3 ½ X6
D: ( CH3 )2 CHCH2MgBr E : ( CH3 )2 CHCH3 F: ( CH3 )2 CHCH2OC2H5
29 ΔTf = iKf m ½
0.068 = i x 1.86 x 0.01 ½
i = 3.65 or 3.656 ½
AlCl3  Al3+ + 3 Cl-
1 0 0
1-α α 3α
α = i-1/n-1 ½
α = .883 or 0.885
88.3% or 88.5% (or any other suitable/ correct method) 1
30 a) Polysaccharides contain a large number of monosaccharide units joined together by ½
glycosidic linkages./ carbohydrates which give a large number of monosachharides on
hydrolysis.
Example: Starch / Cellulose / Glycogen ½
b) Loss of biological activity of native form of protein when subjected to a change in ½
temperature or pH./During denaturation 2o and 3o structures are destroyed. ½
Example: Coagulation of egg white / Curdling of milk
c) When the polypeptide chains run parallel and are held together by hydrogen and ½
disulphide bonds, then fibre-like structure is formed. ½
Example: Keratin / Myosin
31 m=Z I t ½
2 = 63.5 x 2 x t/2x96500 ½
t = 3039.4 s ½
m1/m2 = eq wt 1/eq wt 2 ½
2 / m2 = 63.5/2 / 65/2 ½
m2 = 2.05 g (or by any other correct method) ½

32
Lyophobic sol Lyophilic sol
Interaction between dispersed phase Interaction between dispersed phase and 1
and dispersion medium are weak dispersion medium are strong

Unstable stable 1
Irreversible reversible 1
Can easily be coagulated Can’t easily be coagulated
(any three from above differences) (or any other suitable difference)

OR
32 i) Lyophilic colloids have a unique property of protecting lyophobic colloids./ Lyophilic colloids 1
form a layer around the lyophobic colloids to protect the lyophobic colloid from the electrolyte
in order to prevent coagulation.
ii) Potential difference between the fixed layer and the diffused layer of opposite charges of a 1
colloid.
iii) Substances used for stabilisation of an emulsion. 1

33 i) CH3 CH2CH2OH 1
ii) (CH3 )2C=CH2 1
iii)

1

OR

1
33
(i)
(Intermediate compound in above equation may be ignored)
1. CH3MgBr
(ii)HCHO CH3CH2OH 1
2. H2O

(iii) C6H5OH +CH3COOH H+ C6H5OCOCH3 1
(or any other suitable method)
34 i) Aniline is a Lewis base and anhydrous AlCl3 the catalyst is a Lewis acid which form a salt 1
ii) Aryl halides do not undergo nucleophilic substitution with the anion formed by 1
phthalimide.
iii) Due to +I effect of alkyl group electron density on N atom increases. 1
35 a) k = (2.303 / t) log (Ao / At) ½
k = (2.303 / 40) log (100 / 75)
= 0.007 min-1or 0.0071 min-1or 0.0072 min-1 1
t = (2.303 / k) log (Ao / At)
t = (2.303 / 0.0071) log (100/20) ½
t =230 min or 226.7min or 223.7 min. (deduct ½ mark if incorrect or no unit) 1
b) Sum of powers of the concentration of the reactants in the rate law expression. 1
When one of the reactant is present in large excess. 1
OR
35 a) K1= 0.693/ t1/2= 0.693 / 30 =0.0231 min-1 ½
K2= 0.693/ t1/2= 0.693 / 10 =0.0693 min-1 ½
log K2/K1 = Ea /2.303 R (1/ T1 -1/ T2) 1
Ea = 2.303 R log K2/K1 ( T1T2/T2-T1)
= 2.303 X 8.314 log( 0.0693/0.0231) X ( 300X320/320-300) ½
= 43848.5 J/mol OR 43855 J/mol OR 43.8 kJ/mol ½
b) Proper orientation ½
Energy of the colliding particles should be more than threshold energy ½
c) For a complex reaction, order of reaction is applicable while molecularity has no meaning. 1

36 a) i) Variable or multiple oxidation states / ability to form complexes / they provide large 1
surface area for adsorption.
ii) Similar size/similar properties 1
iii)No unpaired electron/weak metallic bonding/ completely or fully filled d orbitals 1
b) i) 2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O 1
ii) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 1

(Balancing may be ignored in both above reactions)
OR
36 a) i) Ti3+ has an unpaired electron while there are no unpaired electrons in Sc3+.
ii) Stable t2g3 of Cr3+ ion 1
1
b) 1. Both show variable oxidation states
2. Both show f-f transitions 1
3. Electrons of f-orbital in both show poor shielding effect 1
4. both have common +3 oxidation state
5. both show contraction in atomic radii. (any two suitable differences)
c) 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O
1
37 a) (i) 3-hydroxy-3-phenylpropanal /
1

/ C6H5CH(OH)CH2CHO

(ii) Phenyl hydrazone of benzaldehyde / 1
C6H5CH=N-NHC6H5

(iii)Sodium benzoate and benzyl alcohol /

½+½

and

b) (i) On heating with NaOH and I2 : CH3CH=CHCOCH3 will form yellow ppt of CHI3 while 1
other compound doesn’t .
(ii) On adding NaHCO3 : Benzoic acid produces brisk effervescence while other 1
compound doesn’t.
(or any other suitable chemical test)
OR
1
a) (i) CH3CH2CH3
1
(ii) C6H6
37 1
(iii) CH2=CH-CH2CHO

b) C6H5COCH3 < CH3COCH3 < CH3CHO < HCHO
1

c)
1

Marking scheme – 2020

CHEMISTRY (043)/ CLASS XII

56/5/3
Q.No Expected Answer / Value Points Marks
SECTION A
1 By gaining one electron they acquire noble gas configuration 1
2 Extremely small size/ absence of d orbital/highest electronegativity 1
3 HI>HBr>HCl>HF 1
4 Low bond dissociation enthalpy and high hydration enthalpy 1
5 X >X’ 1
6 (CH3)4C 1
7 (CH3)2NH 1
8 Cis-[Pt(en)2Cl2]2+ 1
9 Zone refining 1
10 Copolymer 1
11 (b) 1
12 (b) 1
13 (c) 1
14 (a) 1
15 (d) 1
16 (D) 1
17 (D) 1
18 (C) 1
19 (A) 1
20 (D) 1
SECTION B
21 (a) Hexacyanidoferrate(III) / Hexacyanoferrate(III) ½
d2sp3 ½
(b) Ligand which can ligate through two different atoms is called ambidentate ligand whereas
di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion. / a 1
chelating ligand forms a more stable complex as compared to an ambidentate ligand. / chelating
ligand forms a cyclic complex while ambidentate ligand forms a non-cyclic complex.
22 (i)

1

(ii)

1

23 i) NaCN acts as a leaching agent / it forms complex with gold/ [Ag(CN)2]- 1
4Au + 8CN- + 2H2O + O2 4 [Au(CN2)]- + 4OH- (Balancing may be ignored)
ii) CO acts as a reducing agent 1
OR

23  It is leached out using acid or bacteria 1
 Electrolytic refining 1

24 (i) Glycol and terephthalic acid / ½+½

,
(ii) Melamine and formaldehyde/
½+½

,
25  The accumulation of molecular species at the surface rather than in the bulk of a solid or 1+½
liquid. Example: adsorption of gases on surface of active charcoal (or any other suitable
example)
 Adsorption of reactants occurs on surface of catalyst and reaction takes place. ½

OR
25  A state of continuous zig-zag motion of particles. 1
 Unbalanced bombardment of the particles by the molecules of the dispersion medium. ½
 The Brownian movement has a stirring effect which does not permit the particles to settle. ½

26
 For a solution of volatile liquids, the partial vapour pressure of each component of the 1
solution is directly proportional to its mole fraction present in solution.

 If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the
partial pressure of the volatile component or gas is directly proportional to its mole 1
fraction in solution.
27 (i) A chemical substance which in low concentrations inhibits the growth or ½
destroys microorganisms.
Example: Penicillin/Aminoglycosides/Ofloxacin ½
(ii) Prevent spoilage of food due to microbial growth. ½
Example: table salt/sugar/vegetable oils/sodium benzoate/salts of sorbic acid/salt of ½
propanoic acid.
SECTION C
28 i) CH3 CH2CH2OH 1
ii) (CH3 )2C=CH2 1
iii)
1

OR
28 (i)

1

(Intermediate compound in above equation may be ignored)
1. CH3MgBr
(ii)HCHO 2. H2O CH3CH2OH
1
+
(iii) C6H5OH +CH3COOH H C6H5OCOCH3
(or any other suitable method) 1
29 A: ( CH3 )2 C=CH2 B : ( CH3 )2 CBrCH3 C : ( CH3 )3 C - C(CH3 ) 3 ½ X6
D: ( CH3 )2 CHCH2MgBr E : ( CH3 )2 CHCH3 F: ( CH3 )2 CHCH2OC2H5

30 ΔTf = iKf m ½
0.068 = i x 1.86 x 0.01 ½
i = 3.65 or 3.656 ½
AlCl3  Al3+ + 3 Cl-
1 0 0
1-α α 3α
α = i-1/n-1 ½
α = .883 or 0.885
88.3% or 88.5% (or any other suitable/ correct method) 1

31 (i) Deoxyribose sugar , Nitrogenous base and phosphoric acid 1
(ii) Gluconic acid /
1

1
(iii) 2o and 3o structures are destroyed.
32 m=Z I t ½
2 = 63.5 x 2 x t/2x96500 ½
t = 3039.4 s ½
m1/m2 = eq wt 1/eq wt 2 ½
2 / m2 = 63.5/2 / 65/2 ½
m2 = 2.05 g (or any other suitable/ correct method) ½

33
Lyophobic sol Lyophilic sol
Interaction between dispersed phase Interaction between dispersed phase and 1
and dispersion medium are weak dispersion medium are strong

Unstable stable 1
irreversible reversible 1
Can easily be coagulated Can’t easily be coagulated
(any three from above differences) (or any other suitable difference)

OR
33 i) Lyophilic colloids have a unique property of protecting lyophobic colloids./ Lyophilic 1
colloids form a layer around the lyophobic colloids to protect the lyophobic colloid from
the electrolyte in order to prevent coagulation.
ii) Potential difference between the fixed layer and the diffused layer of opposite charges of a 1
colloid.
iii) Substances used for stabilisation of an emulsion. 1

34 i) Aniline is a Lewis base and anhydrous AlCl3 the catalyst is a Lewis acid which form a 1
salt
ii) Aryl halides do not undergo nucleophilic substitution with the anion formed by 1
phthalimide.
iii) Due to +I effect of alkyl group electron density on N increases. 1
SECTION D
35 a) (i) 3-hydroxy-3-phenylpropanal /
1

/ C6H5CH(OH)CH2CHO

(ii) Phenyl hydrazone of benzaldehyde / 1
C6H5CH=N-NHC6H5

(iii)Sodium benzoate and benzyl alcohol /

½+½
and
b) (i) On heating with NaOH and I2 : CH3CH=CHCOCH3 will form yellow ppt of CHI3 1
while other compound doesn’t .
(ii) On adding NaHCO3 : Benzoic acid produces brisk effervescence while other 1
compound doesn’t.
(or any other suitable chemical test)
35 OR
1
a) (i) CH3CH2CH3
1
(ii) C6H6
1
(iii) CH2=CH-CH2CHO

b) C6H5COCH3 < CH3COCH3 < CH3CHO < HCHO
1

c)
1

36 a) k = (2.303 / t) log ([A]o / [A]t) ½
k = (2.303 / 40) log (100 / 75)
= 0.007 min-1or 0.0071 min-1or 0.0072 min-1 1
t = (2.303 / k) log ([A]o / [A]t)
t = (2.303 / 0.0071) log (100/20) ½
t =230 min or 226.7min or 223.7 min. (1/2 mark deducted for incorrect or no unit) 1
b) Sum of powers of the concentration of the reactants in the rate law expression. 1
When one of the reactant is present in large excess. 1
OR
-1
a) K1= 0.693/ t1/2= 0.693 / 30 =0.0231 min ½
36 K2= 0.693/ t1/2= 0.693 / 10 =0.0693 min-1 ½
log K2/K1 = Ea /2.303 R (1/ T1 -1/ T2) 1
Ea = 2.303 R log K2/K1 ( T1T2/T2-T1)
= 2.303 X 8.314 log( 0.0693/0.0231) X ( 300X320/320-300) ½
= 43848.5 J/mol OR 43855 J/mol OR 43.8 kJ/mol ½
b) Proper orientation ½
Energy of the colliding particles should be more than threshold energy ½
c) For a complex reaction, order of reaction is applicable while molecularity has no meaning. 1
37 a) i) Variable or multiple oxidation states / ability to form complexes / they provide large 1
surface area for adsorption.
ii) Similar size/similar properties 1
iii)No unpaired electron/weak metallic bonding/ completely or fully filled d orbitals 1
b) i) 2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O 1
ii) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 1
(Balancing may be ignored in both above reactions)
37 OR
a) i) Ti has an unpaired electron while there are no unpaired electrons in Sc3+.
3+
1
ii) Stable t2g3 of Cr3+ ion 1

b) 1. Both show variable oxidation states 1

2. Both show f-f transitions 1
3. Electrons of f-orbital in both show poor shielding effect
4. both have common +3 oxidation state
5. both show contraction in atomic radii. (any two suitable differences)
c) 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O 1