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Jee Advance 2015
Chemistry PaperPART II - CHEMISTRY

C
I
SECTION - 1 (Maximum Marks : 32)
(One Integer Value Corret Type)
This section contains a group of 8 questions.
Each question, when worked out will result in one integer from 0 to 9 (both inclusive)
21. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride- ammonia complex
(which behaves as a strong electrolyte) is – 0.0558°C, the number of chloride (s) in the coordination
sphere of the complex is
(Kf of water = 1.86 K kg mol–1]
21. (1)
Tf = 0 – ( – 0.0558) = 0.0558
Tf = i Kf m
0.0558 = i × 1.86 × 0.01
i = {0.0558 / (1.86 × 0.01)} = 3
So n = 3
[Co(NH3 ) x Cl y ]ClZ  [Co(NH3 ) x Cl y ]n   zC l
1 + z= 3 z= 2
Hence y = 1
22. All the energy released from the reaction X  Y, r G° = – 193 kJ mol–1
is used for oxidizing M+ as M+  M3+ + 2e–, E° = – 0.25 V.
Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted
to Y is [F = 96500 C mol–1]
22.
(4)
G° = – nFE°
– 193 × 103 = – n × 96500 × (– 0.25)
n=8
for 1 mole of M+ oxidation 2 mole e– releases
for 4 mole of M+ oxidation 8 mole e– releases
23. For the octahedral complex of Fe3+ in SCN – (thiocyanato -S) and in CN– ligand environments, the
difference between the spin-only magnetic moments in Bohr magnetons (when approximated to
the nearest integer) is
[ Atomic number of Fe = 26]
23. (4)
Since SCN– is a weak field ligand
So the magnetic moment of Fe3+  5(5  2)  5.92
& CN– is a strong field ligand so magnetic moment =  1(1  2)  1.73
difference  4
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24.
The total number of lone pairs of electrons in N2O3 is
O
N
24. (8)
N
O
O
25. Among the triatomic molecules/ ions, BeCl2, N3–, N2O, NO2+, O3, SCl2, ICl2– , I3– and XeF2, the
total number of linear molecule (s)/ ion(s) where the hybridization of the central atom does not have
contribution from the d-orbital(s) is
[ Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54]

C
25. (4)
BeCl2, N3–, N2O, NO2+, all are sp hybridisation and linear
Although ICl2– , I3– and XeF2 are linear but due to sp3d hybridisation these are not involved.
26. Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom
is 9, while the degeneracy of the second excited state of H– is
26. (3)
In multielectronic system the energy depends on (n + l) so 2nd excited state is 2p i.e.
degenerency = 3.
27. The total number of stereoisomers that can exist for M is
H3C CH3
H3C
M
O
H3C CH3
27. (2)
H3C
M
O
(CH3) group must be in wedge position.
So, only two isomers.
28. The number of resonance structures for N is
OH
NaOH
N

O
28. (9)
O
O






O
O
O


O

O
O
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SECTION - 2 (Maximum Marks : 40)
(One or More Options Correct Type)
This section contains 10 multiple choice questions.
Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE are correct.
29. The major product of the reaction is
CO2H
H3C
,aqueous HCl
NaNO
 2
 
0C
CH3 NH2

C
NH2
H3C
(A)
29. (C)
CH3 OH
CO2H
H3C
(B)
NaNO2 / HCl

 
CH3 NH2
CO2H
H3C
CH3 OH
COOH
CH3
(C)
(D)
CH3 OH
COOH
CH3

CH3 N2Cl
NH2
H3C
CH3 OH
CO2H
H3C
H2O
+ N2
CH3 OH

Due to NGP reaction.
30. The correct statement(s) about Cr2+ and Mn3+ is (are)
[ Atomic numbers of Cr = 24 and Mn = 25]
(A) Cr2+ is a reducing agent
(B) Mn3+ is an oxidizing agent
(C) Both Cr2+ and Mn3+ exhibit d4 electronic configuration
(D) When Cr2+ is used as reducing agent, the chromium ion attains d5 electronic configuration.
30. (ABC)
Cr3+ also -d3 - configuration
In Cr2+ & Mn3+ both have d4 configuration & Mn3+ is oxidising agent because it reduces
to Mn2+ to attain the stable configuration.
31. Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this
process is (are)
(A) Impure Cu strip is used as cathode
(B) Acidified aqueous CuSO4 is used as electrolyte
(C) Pure Cu deposits at cathode
(D) Impurities settle as anode-mud
31. (BCD)
In electron refining pure metal strip is taken as cathode & impure metal is at anode
32. Fe3+ is reduced to Fe2+ by using
(A) H2O2 in presence of NaOH
(C) H2O2 in presence of H2SO4
32. (AB)
(B) Na2O2 in water
(D) Na2O2 in presence of H2SO4
In basic medium H2O2 act as reducing agent.
2K3Fe(CN)6 + 2KOH + H2O2  2K4Fe(CN)6 + 2H2O + O2
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33. The % yield of ammonia as a function of time in the reaction
N2 (g) + 3H2(g)
2NH3 (g), H < 0
at (P, T1) is given below :
T1

C
time
If this reaction is conducted at (P, T2), with T2 > T1 , the %yield of ammonia as a function of time is
represented by
T1
T2
T2
T1
(A)
(B)
time
time
T2
T1
T1
T2
(C)
(D)
time
33. (B)
time
It is exothermic reaction, so on increasing temperature the rate of reaction increases & so at
T2 the slope is more stipper.
34. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of
octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by
magnesium ions, m and n, respectively, are
(A)
34. (A)
1 1
,
2 8
(B)
1,
1
4
(C)
1 1
,
2 2
(D)
1 1
,
4 8
Its an example of spinel
Mg Al2O4
Where ‘ O’ undergoes ccp arrangement.
So 2 Al3+ occuppies octahedral void out of four octahedral void available. Hence
m = fraction of octahedral hole = (2 / 4) = (1 / 2)
Mg2+ occupies the tetrahedral void
n = fraction of tetrahedral hole = (1 / 8)
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35.
Compound (s) that on hydrogenation produce(s) optically inactive compound (s) is(are)
H Br
(A)
H3C
CH3
H
Br
H
(B)
H2C
(D)
H2C
CH3
Br
H2C
CH3
(C)
Br H
CH3
CH3

C
H
35. (BD)
Br
CH2
Br
H
Ni
2 /
H
 H3C
CH3
CH3
Plane of symmetry present
(optically inactive)
Br H
Br H
Ni
2 /
CH3 H
 CH3
CH2
CH3
Plane of symmetry
(optically inactive)
36. The major product of the following reaction is
O
CH3
.KOH ,H2O
i


ii .H  , heat
O
CH3
(A)
36. (A)
CH3
O
(B)
O
O
O
O
CH3
(C)
H
CH3
(D)
O

CH3
KOH

6

5
2 1
3
4
O
CH3
O
CH3
O
O
CH3
H+

CH3
O
O

H 

3
5
4
(major)
JEE Advanced 2015
(24-May-15) Question & Solutions Paper-I
(minor)
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37. In the following reaction, the major product is
CH3
CH2
H2C
equivalent HBr
1
 

CH3
CH3
CH3
(A)
(B)
H2C
H3C
CH3
CH3
(C)
Br
H2C
Br
(D)
CH2
CH2
37. (D)
CH3
CH2
1eq. H+Br
CH2

Br
CH2
CH3
H2C
Br
H3C

C
Br
CH3 Br
CH3
CH2 Br
CH3
(Major)
38. The structure of D-(+) -glucose is
CHO
H
OH
HO
H
H
OH
OH
H
CH2OH
The structure of L-(–)- glucose is
CHO
(A)
HO
H
H
HO
OH
H
HO
H
CH2OH
38. (A)
CHO
(B)
CHO
CHO
H
OH
HO
H
HO
H
HO
H
H
OH
HO
H
H
OH
HO
HO
H
H
HO
H
HO
H
(C)
CH2OH
CH2OH
(D)
H
OH
CH2OH
D(+) glucose & L ( – ) glucose are enantiomers.
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SECTION - 3 (Maximum Marks : 16)
(Column Matching)
This section contains 2 column Matching
39. Match the anionic species given in Column I that are present in the ore(s) given in Column II.
(A)
(B)
(C)
(D)
Column I
Carbonate
Sulphide
Hydroxide
Oxide
(P)
(Q)
(R)
(S)
(T)
Column II
Siderite
Malachite
Bauxite
Calamine
Argentite

C
39.
(A)  (PQS),
Siderite
Malachite
Bauxite
Calamine
Argentite
(B)  (T), (C)  (QR),





(D)  (R),
FeCO3
CuCO3 .Cu(OH)2
Al2O3 .2H2O
ZnCO3
Ag2S
40. Match the thermodynamic processes given under Column Iwith the expressions given unber Column
II.
Column I
(A) Freezing of water at 273 K and 1 atm
(B) Expansion of 1 mol of an ideal gas into a
vacuum under isolated conditions
(C) Mixing of equal volumes of two ideal gases at
constant temperature and pressure in an isolated
container
(D) Reversible heating of H2(g) at 1 atm from 300 K
to 600 K, followed by reversible cooling to 300 K
at 1 atm.
40.
(A)  (RT),
JEE Advanced 2015
(B)  (PQS),
(C)  (PQS),
Column II
(P) q = 0
(Q) w = 0
(R)
Ssys < 0
(S)
U = 0
(T)
G = 0
(D)  (PQST),
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