PAPER-2
PART-A
Select the correct alternative. (More than one is/are correct) [3 × 6 = 18]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.4 Stepwise hydrolysis of P4O10 takes place viaformation of
(A*) Tetrametaphosphoricacid
(B*) Tetrapolyphosphoricacid
(C*) Pyrophosphoric acid
(D*) Orthophosphoricacid
Q.5 Monochlorinationof(S)-2-chlorobutanecangive
(A) (S)-1,2-dichlorobutane (B*) (R)-1,2- dichlorobutane
(C*) (2R, 3S)-2,3-dichlorobutane (D) (2R,3R)-2,3-dichlorobutane
[Sol. 

 


h
/
Cl2
+
+ + ]
Q.6 Select the correct statement about the compound NO[BF4]
(A*) It has 5  and 2  bond
(B)Nitrogen-oxygen bondlengthis higherthannitricoxide(NO)
(C*) It is a diamagnetic specie
(D) B–Fbond length in this compound is lower than in BF3.

PART-B
MATCH THE COLUMN [3 × 8 = 24]
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.4 Matchthecolumn:
ColumnI ColumnII
(A) (P) Conformationofmaximumtorsionalstrain
(B) (Q) Conformationwithstrongintramolecularhydrogenbond
(C) (R) Highestboilingpoint
(D) (S) ConformationofminimumVanderWaalstrain
[Ans. (A) R, (B) Q, (C) P, (D) S ]
Q.5 ColumnI ColumnII
(Co-ordinationcompound) (Type of isomerism shown)
(A) Na2[Pt(SCN)2(Ox)2] (P) Ionizationisomerism
(B) [CrCl2(NH3)4]NO3 (Q) Linkageisomerism
(C) [Pt(NO2)(Gly)(NH3)] (R) Geometricalisomerism
(D) K3[Fe(OH)2(C2O4)2] (S) Opticalisomerism
[Ans. (A) Q,R,S (B) P,R (C) Q,R (D) R,S]
Q.6 Matchthe names ofcarboxylicacids in columnIwith pka valuein column II.
ColumnI ColumnII
(A) Benzoicacid (P) 4.17
(B) Ethanoicacid (Q) 4.14
(C) o-methylbenzoicacid (R) 4.74
(D) p-flourobenzoicacid (S) 3.91
[Ans. (A) P, (B) R, (C) S, (D) Q]
[Sol. With increase in electronwithdrawing groups acidity(Ka) ofcarboxylic acid increases decreasingthe
value of pKa ]

PART-C
SUBJECTIVE: [5 × 10 = 50]
Q.6 Calculate the Hvaporization [CH3COOH( l )] in kJ/mol
Given data
Hsolution [KF. CH3COOH(s)] in glacial acetic acid = –3 kJ/mole
Hsolution [KF(s)] in glacial acetic acid = +35 kJ/mole
The strength of H-bond between F–(g) and CH3COOH (g) = +46 kJ/mole
Lattice enthalpyof KF.CH3COOH(s) = +734 kJ/mole
Lattice Enthalpyof KF(s) = +797 kJ/mole
[Ans. 21 kJ/mole]
[Sol.
–3 = +734 + 46 –797 + 35 – x
x = 21
Hvaporization [CH3COOH( l )]= +21 kJ/mole ]
Q.7 Thegaseousdecomposition reaction
A(g)  2B(g) + C(g)
is observedtofirst orderovertheexcess ofliquidwaterat 25°C, it is found that attheend of10min, the
total pressureof system is 188torr and after verylongtime it is 388 torr.
Calculate rate constant of the reaction (in hr–1)
Given : vapour pressure of H2O at 25° is 28 torr (ln2 = 0.7, ln3 = 1.1, ln10 = 2.3)
[Ans. 1.2 hr–1]
[Sol. A(g)  2B(g) + C(g)
Let initialpressureis P0 P0 0 0
After10min (P0–x) 2x x
A.V.L.T. 0 2P0 P0
as per given (P0 – x) + 2x + x + vapour pressure of H2O = 188
P0 + 2x = 160 ...(1)
& 3P0 + 28 = 388
3P0 = 360
P0 = 120 ...(2)
 2x = 40  x = 20 torr
So, K = 







 x
P
P
n
t
1
0
0
l = 





100
120
n
10
1
l =
10
1
× (ln 4 + ln 3 – ln 10) = 0.02 min–1 = 1.2 hr–1 ]

Q.8 When1moleofA(g)is introducedinaclosed rigid1litrevesselmaintained atconstanttemperaturethe
followingequilibriaareestablished.
A(g) l B(g) + C(g) :
1
C
K
C (g) l D(g) + B(g) :
2
C
K
The pressure at equilibrium is twicetheinitial pressure.Calculatethevalue of
1
2
C
C
K
K
if
eq
eq
]
B
[
]
C
[
=
5
1
[Ans. 4]
[Sol. A(g) l B(g) + C(g) : 1
C
K
1–x (x+y) (x–y)
C (g) l D(g) + B(g) : 2
C
K
(x–y) y (y+x)
ntotal = (1 – x) + (x + y)(x – y) + y
2 = 1 + x + y
 x + y = 1 ...(1)
Also,
y
x
y
x


=
5
1
 5x – 5y = x + y
 4x = 6y  x = 1.5 y ...(2)
From eq. (1) & (2)
1.5y + y = 1  y = 0.4
So, x = 0.6
Now,
1
C
K =
)
x
1
(
)
y
x
)(
y
x
(



=
4
.
0
2
.
0
1
=
2
1
2
C
K =
)
y
x
(
)
y
x
(
y


=
2
.
0
1
4
.
0 
= 2
Thus,
1
2
C
C
K
K
= 4 ]

Q.9 Total entropychange (system + surrounding) for the process H2O (l ,–8°C) H2O (s, –8°C) at 1 atm
and – 8°C is 1 J/K mol.
Find Cp,m (H2O, l) in J/K mol.
Given : Hmelting(at –8°C) = + 5784.95 J/mol, Hmelting(at 0°C) =6006 J/mol
Cp,m(H2O, s) = 36 J/K mol ln 





265
273
= 0.03
[Ans. 75 ]
[Sol. H2O(l) at –8°C  H2O(s) at –8°C
Ssystem = Cp,m(l) 





265
273
n
l –
273
6006
+ Cp,m(s) 





273
265
n
l
Ssystem = (Cp,m(l) – Cp,m(s)) 





265
273
n
l – 22 = (Cp,m(l) – 36) × 0.03 – 22
Ssurr =
265
95
.
5784
= 21.83
Ssystem + Ssurr = Stotal
1 = (Cp,m(l) – 36) × 0.3 – 22 + 21.83
1.17 = (Cp,m(l) – 36) × 0.03
Cp,m(H2O, l) = 75 J/K-mol ]
Q.10 Theequilibriumconstant Kp forthe reaction
N2O4 (g) l 2NO2(g) is 4.5 at temperature T.
What wouldbetheaveragemolarmass (in gm/mol)of anequilibrium mixtureofN2O4 &NO2 formed
bythe dissociation of pure N2O4 at a total pressure of 2 atm at temperature T.
[Ans. 57.5]
[Sol. Let initial mole of N2O4 (g)is one
N2O4 (g) l 2NO2 (g)
Initialmoles 1 0
Atequilibrium 1 –  2
Total no. ofmoles at equilibrium =1 + 
Now, 4
2O
N
P = P
1
1





; 2
NO
P = P
)
1
(
2




Hence Kp =
4
2
2
O
N
2
NO
P
P
= P
)
–
1
(
4
2
2



 4.5 = 2
)
–
1
(
4
2
2



4.5 – 4.5 2 = 82 ; 12.5 2 = 4.5
2 = 9/25 ;  = 3/5 = 0.6
Mole fractionof N2O4: 4
2O
N
x =




1
1
=
6
.
1
4
.
0
= 0.25
 2
NO
x = 0.75
Average molar mass of mixture = 0.25 × 92 + 0.25 × 46 = 57.5 ]