CSIR NET PAPER june2011 june 2019(detailed solutions)
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Detailed Solution
CSIR-UGC-NET
CHEMICAL SCIENCES
June 2011 – June 2019
CHEMICAL SCIENCES
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CONTENTS
DETAILED SOLUTIONS: JUNE 2011 TO JUNE 2019
DETAILED SOLUTIONS
PAGES
1. CSIR-UGC-NET-JRF June 2011
1-30
2. CSIR-UGC-NET-JRF Dec. 2011
31-59
3. CSIR.UGC-NET-JRF June 2012
60-91
4. CSIR.UGC-NET.JRF Dec. 2012
92-126
5. CSIR-UGC-NET-JRF June 2013
127-160
6. CSIR-UGC-NET-JRF Dec. 2013
161-190
7. CSIR-UGC-NET-JRF June 2014
191-220
8. CSIR-UGC-NET-JRF Dec. 2014
221-260
9. CSIR.UGC-NET-JRF June 2015
261-296
10.CSIR-UGC-NET-JRF Dec. 2015
297-328
11.CSIR-UGC-NET-JRF June 2016
329-363
12.CSIR-UGC-NET-JRF Dec. 2016
364-394
13.CSIR-UGC-NET-JRF June 2017
395-415
14.CSIR-UGC.NET-JRF Dec 2017
416-435
15.CSIR-UGC-NET-JRF June 2018
436-463
16.CSIR-UGC.NET-JRF Dec 2018
464-486
17.CSIR-UGC-NET-JRF June 2019
487-519
18.CSIR-UGC.NET-JRF Dec 15, 2019
Coming soon
19.CSIR-UGC-NET-JRF Dec 27, 2019
Coming soon
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SOLUTION
June 2011 – June 2019
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SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
280
SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
CHEMICAL SCIENCES BOOKLET-[C]
PART - B
21.
Ni 2 Ar 3d8 4s 0
4s
3d
Geometry
Possible hybridisation
Octahedral
sp3d 2
Square planar
d sp 2
Tetrahedral
sp3
4d
4p
No. of unpaired electrons
2
0
2
Correct answer: (d)
Compound
22.
Ni CN 4
NiCl4
Hybridisation of Ni 2
Number of unpaired electrons
dsp 2
0 diamagnetic
sp3
2 paramagnetic
2
2
Correct answer: (c)
23.
M
S
C
N ,
M
CN bond stretching
at higher frequency
(wave number)
N
C
S
CN bond stretching at lower
frequency(wave number)
Correct answer: (b)
24.
Charge transfer spectra are possible due to transfer of electron from ligand to metal. This is more
probable if the metal is in a high oxidation state or ligand has reducing properties. Intense colour of
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ceric ammonium nitrate, NH 4 2 Ce NO 3 6 is due to charge transfer from bidentate NO3 to Ce 4
which has C.N = 12.
Correct answer: (b)
25.
Structure of above compounds can be represented as
F
F
F
(1 l.p.)
F
F
S
F
–
F
B
F
F
F
Xe
F
(0 l.p.)
Correct answer: (a)
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I
F
(2 l.p.)
Cl
Cl
Cl
Cl
(2 l.p.)
–
SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
26.
281
(a) O2 must be removed as it give its own polarogram, due to following reactions,
H 2 O 2 ; O 2 4H 4e
2H 2O
O 2 2H 2e
27.
28.
(b) Dropping mercury electrode is the working electrode
(c) Id C
(d) Residual current need not to be made zero.
Correct answer: (d)
Vitamin B12 molecule is built-around a corrin ring containing cobalt (III) atom.
Correct answer: (b)
For rotational spectra, 0
Molecule
Dipole moment
Rotational spectra
H2
0
N2
CO
CO2
Correct answer: (c)
0
No
No
Yes
No
0
0
29.
Ziegler-Natta catalyst is Al C 2 H 5 3 TiCl4 .
Correct answer: (a)
30.
Compound
Number of electron in outer shell of central atom
C H Fe
C H Co
C H Ru
C H Co
Oxidation of
5
5
5 2
5
5 2
5
5 2
18
5
19
5
18
5
5
18
5 2
5
C 5 H 5 Co to 5 C 5 H 5 Co gives an 18 electron compound, so easier to
2
2
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oxidise.
Correct answer: (b)
CH3
31.
H3C
CH3
Cl
W
H3C
CH3
CH3
Fe
K+ Cl
Pt
Cl
W(CH3)6
(Only 6-bond with metal)
Correct answer: (a)
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Ru
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282
H
32.
O
N
H
H
H
H
Bond angle:
104.5º
Shape:
Bent/angular
Hybridisation
sp3
Bond length order:
O H < N–H < C–H
C
H
107º
H
H
H
109.5º
Pyramidal
Tetrahedral
sp3
sp3
Correct answer: (c)
33.
34.
C
F
C
F
C
F
F
F
F
Stabilisation via back-bonding. Due to innert pair effect stability of +2 oxidation state increases
down the group.
Correct answer: (b)
Isomers of [Ru(bpy)2Cl2] can be represented as
II and III are enantiomers.
Correct answer: (b)
e0 Positron
1
;
19
Ne
9 F
14
6
e0 particle
14
1
C
7 N
35.
19
10
36.
Correct answer: (b)
Kurnakov test is used to distinguish between cisplatin and transplatin, by using thiourea as a reagent.
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Cisplatin + Aqueous thiourea
deep yellow solution
Crystallisation
yellow needle shape crystal
of tetrakis(thiourea) platinum (II)
chloride
Transplatin + Aqueous thiourea
Colourless solution
Crystallisation
white needle shape crystal of
trans-bis(thiourea)diaammine
platinum (II) chloride
Correct answer: (c)
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SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
2
37.
283
2
2
2
g
u
g
u
Closed shell
• Only valence shell is considered while writing term symbol.
Term symbol is,
2s 1
or –
g or u
here 1 1 1 0
1 1
1
2 2
Or,
2s 1 3
Reflection under plane containing nuclie = (–) × (+) = (–).
and u×u = g
s
term symbol,
3
g
Correct answer: (b)
38.
From Clausius inequality, dS
dq
T
dq
0
... (1)
T
At constant volume, dw = – PdV = 0
Therefore, from first law of thermodynamics,
dS
Or,
dq du dw ; dq = du
dw 0
At constant entropy ds = 0,
Therefore, from equation (1), we get
dU s,v
T
0
dU s,v 0
or
Here equality holds only for reversible process.
For spontaneous process,
dU s,v 0
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Us,v 0
or
Correct answer: (c)
39.
1
1
1 1
1 1
2 2
Given electronic configuration of the molecule is
k
For zero order reaction, A
P
d A
0
A0 k
k A k
dt
On integration and putting proper limit we get,
A 0 A kt
at
t t 12 and A
... (1)
A 0
2
From equation (1), we get
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u
u
Valence shell
SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
284
40.
A 0
k t 1 2 t1 2
A 0
2
2k
Correct answer: (a)
For an aquous solution at 25ºC, the Debye-Hückel limiting law is
log A Z Z
µ
log 0.509 Z Z
µ
A 0.509
Correct answer (c)
41.
For an asymmetric top, I x I y Iz
Therefore, Total energy (E) = E x E y E z
42.
43.
J 2y
J 2y
J 2x
2I x 2I y 2I z
J
E 2 I
So, above equation will give three rotational constants, Bx , By & Bz .
Correct answer: (c)
Q-band (or PQR contour) is obtained for the bending vibration of C2H2 when its linearity is lost.
Correct answer: (c)
•
The degeneracy associated with the quantum number M J (the orientation of the rotation in
•
•
space) is partly removed when an electric field is applied on a polar molecule (eg. HCl, NH3,etc)
This splitting of states by an electric field is called Stark Effect.
For any linear molecule (eg. AX or BX) in electric field (E), the energy of the state with quantum
number J and MJ is given by,
E J, M J hc B J J 1 a J, M J 2 E 2
here energy depends on 2 square of dipole moment , so large dipole moment will cause larger
splitting.
AX BX
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Correct answer: (b)
For Langmuir adsoption (Monolayer, only),
kp
1 kp
Coverage()
44.
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
=1 (zero order)
high pressure
= kp/(1+kp)
At lower pressure , kp << 1
So, that
kp p
= kp (first order)
At high pressure, kp >>1
Low pressure
1 (Independent of p)
pressure
As Langmuir assumed that binding at a site has no influence on the properties of neighbouring sites;
this means that enthalpy of adsorption is independent of coverage.
Correct answer: (a)
45.
C g
A s B g
G 0 G 0 product G 0 reactant
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G 0 G 0f C G 0f A G 0f B
G 0 0
47.
In standard state G 0f 0
0 0 0 0
46.
285
Or,
RTnK p 0
Or,
nK p 0
Or,
Kp 1
Kp = 1 is possible when, partial pressure of B (PB) is equal to partial pressure of C (PC),
i.e. PB PC 3 atm and Ptotal = PB + PC = 6 atm.
Correct answer: (c)
As per B.O. approximation we can consider the two different motions independent if their energy
difference is large. In other words if energy difference of two different motions is large, the coupling
between them will be minimum.
In case of AX the energy difference of vibrational and rotational energy is large compared to
BX. Hence coupling will be stronger in BX.
Correct answer: (a)
•
•
The entropy of a compound is related to the number of ways in which the molecules can be
distributed among different energy states; the greater the number; the larger is the degree of
disorder, i.e. larger is the entropy.
According to Boltzmann distribution, larger, the energy difference w.r.t. ground state lower will
be population in exited state.
n exited
e / kT , here = ex gs
n ground
ex Bhc , so higher the value of B, lower will be population in different exited state (i.e.
lower disorder or entropy)
•
48.
1
B ,
I
h2
B
82 I
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BO 2 B N 2 B D 2 B H 2
Srot for O2 is highest.
Correct answer: (b)
True normalised wave function for 1s orbital of H-atom is given by
3
2
1 1 r a0
1s
and E1s 0.5 atomic unit
e
a0
2
12
1 1.24
1
1
1.5625 1.6 0.8
Since, E1s
2
2n
2 1
2
2
So, 0.8 0.5
Correct answer: (c)
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286
49.
A is constant of motion (i.e., does not change with time) if
ˆ
A
d
1 ˆ ˆ
If, dt A i A, H t 0
ˆ
A
1 ˆ ˆ
A,H 0
If Aˆ is time independent i.e., t 0 then, for Aˆ to be constant of motion,
i
ˆ ˆ 0
A,H
Or,
ˆ H
ˆ 0 , then also we can say A is a constant of motion.
Note: If A,
Correct answer: (b)
50.
ˆ † , and is defined as
Hermitian conjugate or adjoint of an operator Aˆ is written as A
†
*
d
d d
dx
dx dx
Correct answer: (a)
51.
T
For ideal gas, Joule-Thomson coefficient JT P 0 . So we can not produce cooling or
H
heating by expansion under isoenthalpic process.
Under the given process,
... (1)
PV a constant
... (2)
For ideal gas,
PV nRT
Replacing P from equation (2) in (1), we get
TV a 1 constant
or,
a 1
TV
T2V2a 1
1 1
T2 V1
T1 V2
a 1
T
V
log 2 a 1 log 1
T1
V2
Now, in equation (3)
... (3)
V V
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2
1
V
log 1 ve
V2
So far T2 T1 heating
log
T2
ve
T1
a must be less than 1. Also expansion causes decrease in pressure of gas, this is possible for a > 0, as
PV a constant .
Correct answer: (c)
52.
y 2 4x
After 1% increase in value of x,
... (1)
y 4x
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287
y 2 x½
y
1 x
100
100
y
2 x
% error in y = ẵ ì% error in x.
= ẵ ì 0.1 = 0.05.
Correct answer: (d)
(3)
(4)
(3)
53.
3
Me
(4)
1
(2)
(1) O
3
(1)
2
(2)
4
H
1
1R-Configuration
2
4R-Configuration
4
Correct answer: (a)
180º
O
Me
Me
H
H
O
54.
Correct answer: (b)
55.
Chemical reaction involved in the above transformation can be illustrated as
SH
O
+
S
BF3.Et2O
SH
S
HS
OHBF3
–H
–H
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OBF
HS
O .......BF3
3
HS
S
SH
HS
S
SH
S
Correct answer: (a)
56.
Chemical reaction involved in the above transformation can be illustrated as
O
[3, 3]
sigmatropic
rearrangement
aryl allyl ether
Correct answer: (a)
O
H
OH
Ortho-allyl phenol
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S
SOLVED PAPER : CSIR-UGC-NET/JRF June 2011
288
57.
Chemical reaction involved in the above transformation can be illustrated as
H
O
O
Na2CO3
OH
O
I2/KI
O
O
O
O
H
I
I
Correct answer: (a)
58.
Chemical reaction involved in the above transformation can be illustrated as
OLi
O
e–(NH3)x
Me
NH3
Me
Me
Me
O
Me
Me
Me
Me
Me
Correct answer: (d)
59.
Chemical reaction involved in the above transformation can be illustrated as
N2
H
Me
H
H
H
Me
hv
Me
Me
Me
Me
Carbene
(Intermediate)
Correct answer: (d)
60.
1, 3-trans-disubstitution = a, e substitution.
At C–1 - equatorial -t-Bu; At C–3 - axial - methyl group.
Larger substituent always prefer equatorial position. Hence, correct conformation can be represented as
CH3
H3C
2
H
CH
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Most Stable
1
H3C
3
3
Correct answer: (c)
H
H
+
+
61.
H
H
H
H
(A) Homoaromatic character
(A romatised by bypassing the sp3-carbon atom)
(+)
(B)
+
(C)
• Delocalisation
• sp2 - carbons (planar)
+
• Delocalisation
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• 4n = 4 ( - electron)
n=1
• Antiaromatic character.
Correct answer: (a)
62.
63.
64.
289
• (4n+2) = 2 ( -electron)
n=0
• Aromatic character.
Because C–O of anhydride = 1800 –1900 cm–1.
C–O of ketone = 1720 cm–1. C–O of amide = 1600–1700 cm–1.
Correct answer: (d)
Because the ease of elimination increasing with increased size and branching in the radical.
Correct answer: (b)
Chemical reaction involved in the above transformation can be illustrated as
1
H
5
5
Me
4
1
[1, 5]H shift
(Sigmatropic
rearrangement)
2
3
1
4
H
Me
Me
3
2
Correct answer (c)
R
R
–
LUMO
65.
HOMO
alkene
hv
R
R
R
R
hv
+
R
R
R
HOMO
–
HOMO
R
R
R
R
R
R
R
Cycloaddition
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LUMO
Correct answer: (b)
66.
Chemical reaction involved in the above transformation can be illustrated as
R
C
CH2
BH3
H
+
R
C
H
–
CH2
R
H
C
H
BH
H
Syn-addition
(Markovnikov)
More Stable T.S.
Correct answer: (c)
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H
CH2
BH2
R
C
H
CH2
BH2
(H & BH2–Syn
addition)
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290
67.
max 279 is n–* , which can be differentiated by low 15
Correct answer: (b)
68.
Chemical reaction involved in the above transformation can be illustrated as
MeO
NO2
NO2
NO2
MeO
H
H
NaOEt
E1CB
Carbanion
(Stable)
Correct answer: (d)
69.
Chemical reaction involved in the above transformation can be illustrated as
O
O
H2O2
HOO
H 2O 2 NaOH
O
NaOH
Me
Soft Nucleophile
H 2O
Me
Mechanism:
O
O
OOH
+
O
O
Me
O
OH
Me
Me
Correct answer: (d)
I
(+)
OR
Solvent
(ROH)
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70.
2º Carbocation
(sp2)
A
I
slow
OR
(ROH)
sp2
B
I
NGP
O
O
C
Stable
ROH
Fast
OR
O
(NGP-Anchimeric
assistance in ionisation)
Correct answer: (a)
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291
PART -C
71.
Alkali metals in liq. NH3 act as source of electrons and are supposed to be a good one-electron
reducing agents.
M x y NH 3
M NH 3 x e NH 3 y
These solvated electrons can reduce O2 molecule to superoxide ion.
–
e NH 3 y O 2
O 2 yNH 3
These superoxide ions can combine with solvated metal ion to give Alkali metal superoxides.
M NH 3 x O 2
MO 2 xNH 3
Correct answer: (a)
4
72.
3
II
III
Fe 2 to Fe 3 , Fe CN 6 to Fe CN 6 , NH 2 OH to HNO3 ,SO32 to SO24
KMnO 4 , KIO 4 ,Ce SO 4 2 [All are stronger O.A. than H2O2)
H 2 O 2 oxidise,
H 2O 2 reduce
Note: In the question, only I7 and Ce 4 are in their highest oxidation states, so the only possibility
of reduction by H2O2 exists.
Correct answer: (b)
73.
74.
Compound
Uses
(A)
Na 3 PO 4
Water softener, paint stripper
(B)
Ar3 PO4
Plasticizers
(C)
Et3 PO4
Insecticides
(D)
CaHPO 4 .2H 2 O
Correct answer: (a).
Toothpaste
(A) Correct, As 4f-electrons are too well shielded to interact, so they are like inert gas-type ion,
like those of alkaline earth metals, that attracts ligands only by overall electrostatic forces.
(B) Correct, Lanthanides, owing to the unavailability of orbitals for bonding, high basicity and
rather large size of Ln+3, form few complexes and mostly with oxygen or oxygen plus
nitrogen chelating ligands such as dietones, hydroxy acids EDTA etc. So, they
rarely show isomerism and also lability of ligands make the isolation of isomers is
difficult.
(C) Incorrect, C.N. more than 8 are common, due to their larger size and smaller chelating ligand
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like, NO 3 , SO 42 etc.
2
eg. Ce NO3 6 , C • N 12
(D) Correct, As 4f electrons are well shilded from external fields by the overlying 5s and 5p electrons. Thus the magnetic effect of orbital motion is can not be neglected.
Correct answer: (a)
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292
O –1/2
75.
F
•
•
S
F
O –1/2
O
–1/2
–1/2
(SF4)
Sea-Saw
Cl
S
O
S
Cl
–1/2
F
O
S
S
F
O
Cl
Cl
(S2Cl2)
(SO42–)
Tetrahedral
bent shape around
both s-atom
distorted tetrahedral
(Because all bonds are
not equivalent, so repulsion
will be different between
different bonds.)
Correct answer: (b)
76.
77.
2
For a d5 octahedral complex Mn H 2O 6 , all transitions are Laporte forbidden (orbital forbidden) as well as spin forbidden. Absorptions associated with doubly forbidden transitions are extremely weak, so it is very lightly coloured.
Correct answer: (d)
X-axis as internuclear axis.
2
2
2
2
*2
2
*2
2
2
2
*1
*1
N 2 1s2 1s*2 22s *2
2s 2p x 2p y 2p z ; O 2 1s 1s 2s 2s 2p x 2p y 2p z 2p y 2p z
Correct answer: (a)
78.
In the given case ligand in same, so higher the formal oxidation state of the metal, lower will be
LMCT, So, order is MnO 4 CrO 42 VO34
Correct answer: (a)
79.
80.
Carboxypeptidase enzyme contains Zn+2 ion and it cleaves (hydrolyses) the carboxy terminal amino
acid from a peptide chain.
Correct answer: (b)
g || g g e , gives the evidence that Cu+2 ion is present in tetragonally distorted octahedral field.
ground state
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3d
octahedral field
Tetragonally disorted
octahedral field
Correct answer: (b).
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293
81.
Oxidation of metal is easier for electron rich systems, so electron rich metals undergo oxidative
addition and as reduction is easier for electron-deficient metal, so it undergoes reductive elimination.
Correct answer: (d)
82.
Ti i Pr 4 TiEt 4 TiMe 4 Ti CH 2 Ph 4 . This is correct stability order because Ti CH 2 Ph 4
has no - elimination but Ti i Pr 4 has more - elimination. More - elimination less stable.
Correct answer: (c)
83.
1 atm
Ni 4CO
Ni CO 4
25ºC
200 atm
Fe 5CO
Fe CO 5
200ºC
Correct answer: (c)
84.
Shape
O
C
I —
O
I —
–
I
N
O
O
O
N
O
Number of lone pair
Linear
4
Linear
9
Angular
6
Linear
4
Correct answer: (b)
85.
For, Cu 2 2e
Cu,
E cell E
0
cell
E 0 0.34V
1
0.059
1
0.059
635 103
log
0.34
log
2
2
Cu
2
63.5 g/mol 0.1 Lit
0.054
log10 = 0.31 volt.
2
So, when Ecell = 0.31 volt will reach, then deposition of Cu will start.
0.34
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For Ag e
Ag , E0 = 0.50V
On completion of this reaction,
E cell E 0cell
0.059
1
log
1
Ag
0.1 103
Ag
[Given :
108 0.1 ]
1
3
0.059
E cell 0.50
log 0.1 10 = 0.203 volt
1
108 0.1
So, we will have to fix cathode potential above 0.31 volt so that no. Cu is deposited at the cathode.
We can not reduce the potential upto 0.203 V. So, that all Ag+ has been deposited.
Correct answer: (c)
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294
OC
OC
CO
H
Ru
CO
OC
OC
OC
CO
H
Ru
OC
OC
OC
86.
CO
Ru
CO
CO
Ru
OC
87.
CO
Ru
Ru
CO CO
86-electron cluster H 2 Ru 6 CO 18 displays a distorted octahedral metal geometry..
Here each hydride ligand is connected to 3-Ru atoms.
Correct answer: (d)
During hydroformylation the intermediate CH3–CH2–CH2–Co(CO)4 gets transformed to acyl intermediate CH 3CH 2 CH 2 – COCo CO 3 .
Correct answer: (a)
88.
As we move from 2nd row transition element (Zr) to 3rd row transition element (Hf), size should
increase, but due to lanthanide contraction offset the increase, so Zr and Hf have almost similar size.
Correct answer: (a)
89.
SnF4 is ionic, here Sn+4 = 5sº, so no quadrupole splitting.
SnCl4 is covalent, and Sn is sp3 hybridised, So there is S-electron density in the molecule.
quadrupole splitting is possible.
R3SnCl is covalent, and Sn is sp3, so there is S-electron density in the molecule.
quadrupole splitting is possible.
Correct answer: (c)
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90.
Cr+2
Co+3
Large reorganisation for Co+3
outer sphere, so inner sphere
much faster
+2
HOMO = (of Cr )
Cr+2
Ru+3
Reorganisation of Ru+3 less
than CO+2, so, slower tendency to go
inner sphere
+2
HOMO = (of Cr )
+3
+3
LUMO = (of Ru )
LUMO = (of Co )
Acceleration IS/OS=1010
Acceleration IS/OS=102
Inner sphere mechanism is favorable for totransition only
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91.
92.
295
As Cr 2 / Ru 3 has * to * transition which is responsible for its slower acceleration in going from
outer sphere mechanism to inner sphere mechanis.
Correct answer: (b)
Both Fe(II)–TPP and Fe(III)–TPP are square planar (because of tetradentate-planar, porphyrine type
ligand), and both Fe(II) and Fe(III) are dsp2–hybridised, so there is s-electron density, which is responsible for their quadrupole moment and existence of mossbauer spectra.
So both Fe(II)-TPP and Fe(III)-TPP show increase in s-electron density, but Fe(II) has higher
s-electron density. So Fe(II) has isomer higher shift (0.52 mms–1) than Fe(III) (0.45 mms–1).
Correct answer: (b)
1 k
;
2
where k = force constant of bond
µ= reduced mass.
Ratio of frequency/wave number of Co–D and Co–H bond is,
As frequency of vibration
Co– D
Co– D
Co2H
Co–H
60 1
61 1
60 2
2
62
Co –D Co– H / 2
1840
1300 cm1
2
93.
Correct answer: (a)
Due to Jahn-Teller distortion.
Correct answer: (d)
94.
B2 O 3 3H 2O
2H 3BO3
(orthoboric acid)
N 2 O 5 H 2 O
2HNO3
(Nitric acid)
SO 3 2H 2 O
H 2SO 4
(Sulfuric acid)
P4 O10 6H 2 O
4H 3 PO 4
(orthphosphoric acid)
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Correct answer: (b)
95.
235
92
1
0
U n
142
56
1
Ba 91
36 Kr 30 n
Energy released per atom (E) = m c 2 .
mu mn mBa mKr 3m n c 2
2
E 235.0435 1.00866 141 9164 90.9234 3 1.00866 1.67 1027 3 108 J
Therefore, E per atom 2.8 10 11 J
1 amu 1.67 10 27 kg , c 3 108 m/s
Energy released from 1 mole disintegration of
235
92
U E Per mole
E per atom N A 2.8 1011 6.023 1023
1.68 1013 J 1.68 1010 kJ
Correct answer: (d)
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296
96.
k
Acetaldehyde (CH3CHO)
product, rate = k[CH3CHO]2.
2
In terms of partial pressure , rate = kPCH
3 CHO
1
1
0
kt
Integral rate law expression is, P
PCH3CHO
CH3 CHO
Or,
t 12
1
0
k PCH
3 CHO
0
250
Since given t = 400 sec and PCH
3CHO
400 sec
1
k 250 Torr
and
0
PCH
200 Torr
3CHO
Then
t 12
1
0
CH3 CHO
kP
k 105 Torr 1s1
1
500 sec
10 200
5
Correct answer: (d)
97.
From enzyme kinetics, R k 2 E 0 s 0
k m s 0
1 km 1
1
R k 2 E 0 S0 k 2 E 0
Intercept
Slope
1
R
slope =
km
k2[E0]
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1
intercept = k [E ]
2
0
1
[S]0
We know that,
1
k2
1
1
107
Catalytic efficiency = k
9
Slope
E
E
k
40
2.5
10
m
0
m
0
k 2 E 0
98.
Correct answer: (c)
Given for any H-orbital
Radial function has form r exp r and -part has form exp 3i .
From the solution of Schrodinger equation for H-atom,
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We get part
1 im
e ; where m is integer = 0, 1, 2, 3, ...............
2
1 im
e .
2
(Here m can be +ve or –ve integer including zero).
From given part , m can be either +3 or –3.
Option (a) incorrect,
for n 4 let n 5
3 let 5 this is not possible
m3
Option (b) correct,for n =4, 3 , m = –3 (Can be possible solution)
for n = 4, l > 3 (let 4), m = 3 (Can not be possible)
Option (c) incorrect,
Option (d) incorrect,
for n > 4, l = 3, m = –3
for n > 4, l can be upto 4. (So not possible)
Correct answer (b)
99.
V2 V
Standard deviation in speed
2
2
2
rrms
Vavg
8 RT
8 RT
3RT 8RT
3
3
M
M
M
M
0.67
RT
RT
0.7
M
M
Correct answer: (a)
100.
Second order perturbation for nth state is given by,
E n
2
0m H1 0n
0
0
E n Em
m n
2
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Therefore, perturbation for ground state, (i.e., n = 0)
2
E 0
2
0
10 H ' 00
0
2 H ' 0
E 00 E10
E 00 E 02
V10
2
E 00 E10
V20
2
E 00 E 02
2
22
42
2 4 6
02 04
Correct answer: (a)
101.
n A1
1
6 1 2 1 3 3 1 0 2
6
n A2
1
6 1 2 1 3 3 1 0 2
6
nF
1
6 2 2 1 3 3 0 0 1
6
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297
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298
T 2A1 2A 2 E
Correct answer: (a)
102.
The direct product of two energy state should transform according to x, y or z axis for electronic
transition.
A1 A1 A1 transform with z axis = allowed
A1 E E Transform with x,y axis = allowed
A 2 E E Transform with x,y axis = allowed
A1 A 2 A 2 Not transform as per x,y or z = forbidden
Correct answer: (d)
103.
64
Gd Xe 4f 7 5d1 6s 2 ,
Gd 2 Xe 4f 7 5d1 6s 0
Removal of electrons takes place from the outermost orbital only.
Correct answer: (d)
104.
For 3D
2S + 1 = 3, 2S = 3–1
For term, S, P, D, F
2S = 2, S = 1
Value of L= 0, 1, 2, 3
S = 1, L = 2
J can have values from |L+S| ........... |L–S|
|2+1| ............ |2–1|
3
2
1
Correct answer: (b)
105.
Energy
Delocalisation energy = 2 2 2 4 = 0
Correct answer: (a)
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106.
H
RT
... (1)
2.0 104
T
... (2)
We have nk constant–
Given, nk 3.0
From (1) and (2),
H 0 2 10 4 R 166 kJ / mol
In equation (2) multiplying both sides by (–RT),
... (3)
RTnk 3RT 2 104 R
... (4)
G 0 TS0 H 0
From equation (3) and (4)
S0 3R 24.9 JK 1mol 1
Correct answer: (c)
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107.
1 10 RTnP1
299
Standard state for component 1 is defined at,
(Standard pressure)
P1 1 bar
1 10 R 298 n1
1 10
n1 0
So, 10 is independent pressure, but 0 is dependent on temperature.
Correct answer: (c)
108.
e2 N A
Ci Zi2
0r k BT i
We have,
2e 2 N A
Ci Z
0 k B T
2 r
i
2
i
1
2 0.03 12 0.03 22
C Z
i
i
2 r
molkg 1
2 100
0.18
30 nm 1
10 2
1
Therefore, Debye-Hückel screning length
2
i
30 nm molkg 1
30 nm
1
9
nm 1
10
10
nm
9
Correct answer (a)
109.
For oxidation reduction half-cell
Reduced state (R)
Oxdised state O ne
From modified nernst equation (in terms of formal potential E 0 )
E E 0'
RT R
ln
nF O
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or,
EE
RT 1
ln
nF e 2
or,
E E0
RT
ln e2
nF
or,
E E 0 2
0'
RT
ln e
nF
ln e 1
2RT
nF
Correct answer: (a)
or ,
110.
E E 0'
CO 2 g H 2 g ;
(i) CO g H 2 O g
k eq k1
CO g 3H 2 g ;
(ii) CH 4 g H 2 O g
k eq k 2
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... (1)
... (2)
1
molkg 1
1
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300
Adding equation (1) and (2)
CO 2 g 4H 2 g , k eq k 1k 2
CH 4 g 2H 2O g
Correct answer: (c)
111.
The virial expression for a real gas can be written as
PV
1 BP P CP P 2 ......
... (1)
RT
B
C
PV
1 V V ....
... (2)
RT
V
V
From equation (2),
RT BV CV
P
1
.........
V
V V2
Substitute this ‘P’ in R.H.S. of equation (1),
C
PV
RT BV CV
B
1 BP
1
2 ......... + 1 V V ........
RT
V
V V
V
V
B RT 1
PV
2
1 P
2 BP BV RT CP RT .........
RT
V
V
Compare Ist term on R.H.S. of equation (1) and (3)
Or,
BV BP RT ;
BV BP
here RT
Correct answer: (d)
112.
kT 2 df
Average energy per particle E
f dT
(Where f = single particle, partition function,) here, f
... (1)
AT m
V
df
d AT m
dT dT V
df A
mT m 1
dT V
... (2)
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Put this value in equation (1), we get
E
kT 2
AT m
V
A
m 1
mT mkT
V
Correct answer: (a)
113.
1
Optical density (O.D) = Absorbance = log c
T
1
Solution A: T = 0.5; O.D log
log 2 0.301
0.5
From equation (1), 0.1 1 =0.301 3.01 mol1dm3cm 1
Solution B: O.D c 3.01 0.5 0.1 0.1505
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... (1)
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301
1
1
Solution C: O.D log T log 0.1 1
Therefore, optical density order is C > A > B.
Correct answer: (d)
114.
In Raman Spectra, Stokes lines are seperated by 4 B cm–1 and have wave number value less than the
incident radiation, but the first Stoke’s line is obsorbed at 6B
6B 6 2 cm 1 12 cm 1
Therefore, first Stoke line wave number = wave number of incident – 12cm–1 radiation.
20487 –12 cm 1 20475 cm 1
Correct answer: (b)
115.
Given, Tunelling probability , T e 2kL e10
1
2
2mE
where k
and L barrier length
m
k
If mass if halved m ' 2 , then k '
and L' 2L
2
10
1
2
2
T' e
e 10
Correct answer: (b)
116.
If
2
A † A , then Hermitian operator..
A† A , then antihermitian operator..
Given, A
d
x
dx
†
d
d
†
d
A x A† x
x
dx
dx
dx
d †
d
and x † x
dx
dx
So, A is neither Hermitian nor antihermitian.
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†
†
d
d
d
A A† dxd x
x x x
dx
dx
dx
†
†
AB † B† A †
d
d
x x A A†
dx
dx
AA † is Hermitian. Clearly it can be shown than A† A is also hermitian.
Correct answer: (c)
117.
In Langmuir isotherm, Fractional coverage
kp
1 kp
Case I: At low pressure, kp < < 1
kp directly proportional to pressure
Case II: At high pressure, kp > > 1
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