Physical

[STRAIGHT OBJECTIVE TYPE]
Q.1 & Q.2 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 What is the pK
b
of a weak base whose 0.1 M solution has pH =9.5 ?
(A) 7.5 (B) 8 (C) 9 (D) 10
Q.2 The enthalpy change for the reaction,
CH
3
CHO (g) → CH
4
(g) +CO(g) at 300 K is – 17.0 kJ /mol
Calculate the temperature at which ∆
r
H for the reaction will be zero.
[Given : C
p,m
(CH
4
,g)=38 J/K mol ; C
p,m
(CO ,g)=31 J/K mol & C
p,m
(CH
3
CHO, g)=52 J/K mol]
(A) 1300°C (B) 1027°C (C) 700°C (D) None of these
[MULTIPLE OBJECTIVE TYPE]
Q.3 & Q.4 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.3 The equilibrium between, gaseous isomers A, B and C can be represented as
Reaction Equilibrium constant
A (g) l B (g) : K
1
=?
B (g) l C (g) : K
2
=0.4
C (g) l A (g) : K
3
=0.6
If one mole of A is taken in a closed vessel of volume 1 litre, then
(A) [A] +[B] +[C] =1 M at any time of the reactions
(B) Concentration of C is 4.1 M at the attainment equilibrium in all the reactions
(C) The value of K
1
is
24 . 0
1
(D) Isomer [A] is least stable as per thermodynamics.
Q.4 Which of the following arrangement will produce oxygen at anode during electrolysis ?
(A) Dilute H
2
SO
4
solution with Cu electrodes.
(B) Dilute H
2
SO
4
solution with inert electrodes.
(C) Fused NaOH with inert electrodes.
(D) Dilute NaCl solution with inert electrodes.
[SUBJECTIVE TYPE]
Q.1 & Q.2 are "Subjective" questions. Marks will be awarded only if correct bubbles are filled in OMR sheet.
Q.1 Calculate the EMF (in milliVolt) of the cell
Pt | H
2
(g) | RNH
3
Cl (aq) | | HA (aq) | H
2
(g) | Pt
1 bar 0.1 M 0.01 M 0.5 bar
[Given : K
a(HA)
=4 ×10
–6
,
) RNH ( b
2
K =10
–5
, 2.303
F
RT
=0.06 ; log 2 =0.3]
Q.2 A solution of Na
2
S
2
O
3
is to be standardised by titration against iodine liberated from standard KIO
3
solution. The later is prepared by dissolving 4.28 gm of KIO
3
in water and making upto 500 ml. 20 ml
of this solution are then mixed with excess KI solution and the following reaction occurs.
÷
3
IO +5I ¯ +6H
+
→ 3I
2
+3H
2
O
The resulting iodine is titrated with the Na
2
S
2
O
3
solution according to
I
2
+
÷ 2
3 2
O S 2 → 2 I ¯ +
÷ 2
6 4
O S
and it is found that 48 ml are needed.
Calculate the concentration (in millimoles / L) of Na
2
S
2
O
3
solution.
[STRAIGHT OBJECTIVE TYPE]
Q.1 & Q.2 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 What volume of 0.2 M RNH
3
Cl solution should be added to 100 ml of 0.1 M RNH
2
solution to
produce a buffer solution of pH =8.7?
Given : pK
b
of RNH
2
=5 ; log 2 =0.3
(A) 50 ml (B) 100 ml (C) 200 ml (D) none of these
Q.2 For the reaction A(g) +2B(g) l C(g) +D(g) ; K
c
=10
12
.
If the initial moles of A,B,C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What
is the equilibrium concentration of B?
(A) 10
–4
(B) 2 ×10
–4
(C) 4 ×10
–4
(D) 8 ×10
–4
[MULTIPLE OBJECTIVE TYPE]
Q.3 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.3 Select correct statement(s).
(A) An electron's energy increases with increasing distance from nucleus.
(B) The shortest wavelength in the Brackett series of He
+
ion is
4
R
(C) The second orbit in He
+
has radius as the first orbit in hydrogen atom.
(D) The number of d-electrons retained in Fe
2+
ion is 6 (Atomic number of Fe =26)
[MATCH THE COLUMN]
Q.1 is "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I are to be
matched with only one entry of column-II
Q.1 Column I Column II
(A) ln 2 (P) If initial mole is two then mole of radioactive element
left after time t.
(B) 1– e
–λt
(Q) Fraction of nuclei that decay in time t.
(C) e
–λt
(R) Ratio of activity of nucleus at time t to initial activity.
(D)
|
|
.
|


\
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÷
2 / 1
t
t
1
2 (S) Ratio of half life to mean (average) life for radioactive
e l e m e n t .
Q.1 & Q.2 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 What is the pK
b
of a weak base whose 0.1 M solution has pH =9.5 ?
(A) 7.5 (B*) 8 (C) 9 (D) 10
[Sol. pOH =
2
1
[pK
b
– log C]
9 =pK
b
– log (0.1)
pK
b
=8 Ans. ]
Q.2 The enthalpy change for the reaction,
CH
3
CHO (g) → CH
4
(g) +CO(g) at 300 K is – 17.0 kJ /mol
Calculate the temperature at which ∆
r
H for the reaction will be zero.
[Given : C
p,m
(CH
4
,g)=38 J/K mol ; C
p,m
(CO ,g)=31 J/K mol & C
p,m
(CH
3
CHO, g)=52 J/K mol]
(A) 1300°C (B*) 1027°C (C) 700°C (D) None of these
[Sol. As per Kirchoff's law
) T T ( C H H
1 2 m , p r T r T r
1 2
÷ A = A ÷ A
∆
r
C
p,m
=38 +31–52 =17
∴ 0 – (–17 ×1000) =17 (T
2
–300)
T
2
=1300 K or
T
2
=1027°C Ans. ]
[MULTIPLE OBJECTIVE TYPE]
Q.3 & Q.4 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.3 The equilibrium between, gaseous isomers A, B and C can be represented as
Reaction Equilibrium constant
A (g) l B (g) : K
1
=?
B (g) l C (g) : K
2
=0.4
C (g) l A (g) : K
3
=0.6
If one mole of A is taken in a closed vessel of volume 1 litre, then
(A*) [A] +[B] +[C] =1 M at any time of the reactions
(B) Concentration of C is 4.1 M at the attainment equilibrium in all the reactions
(C*) The value of K
1
is
24 . 0
1
(D*) Isomer [A] is least stable as per thermodynamics.
[Sol. (a) [A] +[B] +[C] =1 mole/L since for every reaction ∆n
g
=0
(b)
] B [
] C [
=0.4 ⇒ [C] =0.4 [B]
] C [
] A [
=0.6 [A] =0.6 [C]
[A] +[B] +[C] =1
0.6[C] +
] 4 . 0 [
] C [
+[C] =1
[C]
)
`
¹
¹
´
¦
+ + 1
4 . 0
1
6 . 0
=1
[C] =
(
¸
(

¸

=
1 . 4
1
4 . 16
4
(
¸
(

¸

=
24 . 0
1
] A [
] B [
=K
1
]
Q.4 Which of the following arrangement will produce oxygen at anode during electrolysis ?
(A) Dilute H
2
SO
4
solution with Cu electrodes.
(B*) Dilute H
2
SO
4
solution with inert electrodes.
(C*) Fused NaOH with inert electrodes.
(D*) Dilute NaCl solution with inert electrodes.
[SUBJECTIVE TYPE]
Q.1 & Q.2 are "Subjective" questions. Marks will be awarded only if correct bubbles are filled in OMR sheet.
Q.1 Calculate the EMF (in milliVolt) of the cell
Pt | H
2
(g) | RNH
3
Cl (aq) | | HA (aq) | H
2
(g) | Pt
1 bar 0.1 M 0.01 M 0.5 bar
[Given : K
a(HA)
=4 ×10
–6
,
) RNH ( b
2
K =10
–5
, 2.303
F
RT
=0.06 ; log 2 =0.3]
[Ans. 0087.00 ]
[Sol. E
cell
=
2
06 . 0
log
RHS H
2
LHS
LHS H
2
RHS
) P ( ] H [
) P ( ] H [
2
2
+
+
For R NH
3
Cl K
h
=
5
14
b
w
10
10
K
K
÷
÷
=
=10
–9
α =
1 . 0
10
C
K
9
h
÷
= =10
–4
RNH
3
+
(aq) +H
2
O (l) l RNH
2
(aq) +H
3
O
+
(aq)
At eq
m
C (1–α) – Cα Cα
[H
3
O
+
]
LHS
=Cα
=0.1 ×10
–4
=10
–5
For HA α =
01 . 0
10 4
C
K
6
a
÷
×
= =2 ×10
–2
[H
3
O
+
]
RHS
=Cα =0.01 ×2 ×10
–2
⇒ 2 ×10
–4
E
cell
=
2
06 . 0
log
5 . 0
1
) 10 (
) 10 2 (
2 5
4
×
×
÷
÷
=
) 10 8 ( log
2
06 . 0
2
×
=
2
2
06 . 0
8 log
2
06 . 0
× +
=0.03 ×0.9 +0.06 =0.087 V
or E
cell
=87 mV Ans. ]
Q.2 A solution of Na
2
S
2
O
3
is to be standardised by titration against iodine liberated from standard KIO
3
solution. The later is prepared by dissolving 4.28 gm of KIO
3
in water and making upto 500 ml. 20 ml
of this solution are then mixed with excess KI solution and the following reaction occurs.
÷
3
IO +5I ¯ +6H
+
→ 3I
2
+3H
2
O
The resulting iodine is titrated with the Na
2
S
2
O
3
solution according to
I
2
+
÷ 2
3 2
O S 2 → 2 I ¯ +
÷ 2
6 4
O S
and it is found that 48 ml are needed.
Calculate the concentration (in millimoles / L) of Na
2
S
2
O
3
solution.
[Ans. 0100.00 Ans.
]
[Sol. Molarity of KIO
3
solution =
5 . 0
214 / 28 . 4
=0.04 moles / L
Moles of I
2
produced =
3
1000
20 04 . 0
×
×
=2.4 ×10
–3
Moles of
÷ 2
3 2
O S
req. =2 ×
2
I
n
=2 ×2.4 ×10
–3
=4.8 ×10
–3
Concentration of
÷ 2
3 2
O S =
3
3
10 48
10 8 . 4
×
×
÷
=100 millimoles / L ]
[STRAIGHT OBJECTIVE TYPE]
Q.1 & Q.2 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 What volume of 0.2 M RNH
3
Cl solution should be added to 100 ml of 0.1 M RNH
2
solution to
produce a buffer solution of pH =8.7?
Given : pK
b
of RNH
2
=5 ; log 2 =0.3
(A) 50 ml (B*) 100 ml (C) 200 ml (D) None of these
[Sol. Let V ml of RNH
3
Cl added into RNH
2
solution
[RNH
3
Cl] in resultant solution =
V 100
V 2 . 0
+
×
; [RNH
2
] =
V 100
1 . 0 100
+
×
pOH =pK
b
+log
] RNH [
] RNH [
2
3
+
; 5.3 =5 +log
(
(
(
(
¸
(




¸

|
.
|

\
|
+
×
|
.
|

\
|
+
×
V 100
1 . 0 100
V 100
V 2 . 0
∴ 2 =
10
V 2 . 0 ×
⇒ V =100 ml ]
Q.2 For the reaction A(g) +2B(g) l C(g) +D(g) ; K
c
=10
12
.
If the initial moles of A,B,C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What
is the equilibrium concentration of B?
(A) 10
–4
(B*) 2 ×10
–4
(C) 4 ×10
–4
(D) 8 ×10
–4
[Sol. A(g) +2B(g) l C(g) +D(g)
At eq
n
0.5–x 1–2x 0.5+x 3.5 +x
÷
~
y
÷
~
2y
÷
~
1
÷
~
4
(due to very high value of equilibrium constant)
10
12
=
2
) y 2 ( y
4 1
×
×
; y =10
–4
[B(g)] =2y ⇒ 2 ×10
–4
Ans. ]
[MULTIPLE OBJECTIVE TYPE]
Q.3 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.3 Select correct statement(s).
(A*) An electron's energy increases with increasing distance from nucleus.
(B) The shortest wavelength in the Brackett series of He
+
ion is
4
R
(C) The second orbit in He
+
has radius as the first orbit in hydrogen atom.
(D*) The number of d-electrons retained in Fe
2+
ion is 6 (Atomic number of Fe =26)
[MATCH THE COLUMN]
Q.1 is "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I are to be
matched with only one entry of column-II
Q.1 Column I Column II
(A) ln 2 (P) If initial mole is two then mole of radioactive element
left after time t.
(B) 1– e
–λt
(Q) Fraction of nuclei that decay in time t.
(C) e
–λt
(R) Ratio of activity of nucleus at time t to initial activity.
(D)
|
|
.
|


\
|
÷
2 / 1
t
t
1
2 (S) Ratio of half life to mean (average) life for radioactive
element.
[Ans. (A) S, (B) Q, (C) R , (D)
P]Q.1 & Q.2 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 At 300 K, the vapour pressure of an ideal solution containing 3 mole of A and 2 mole of B is 600 torr.
At the same temperature, if 1.5 mole of A & 0.5 mole of C (non-volatile) are added to this solution the
vapour pressure of solution increases by 30 torr. What is the value of
o
B
P
?
(A) 940 (B) 405 (C*) 90 (D) none of these
[Sol. P =
B
o
B A
o
A
x P x P +
600 =
|
.
|

\
|
+
+
|
.
|

\
|
+ 3 2
2
P
2 3
3
P
o
B
o
A ;
3000 P 2 P 3
o
B
o
A
= +
& 630 =
|
.
|

\
|
+ +
+
|
.
|

\
|
+ + 5 . 0 2 5 . 4
2
P
5 . 0 2 5 . 4
5 . 4
P
o
B
o
A ;
4410 P 2 P 5 . 4
o
B
o
A
= +
1410 P 5 . 1
o
A
=
;
90 P & 940 P
o
B
o
A
= =
Ans. ]
Q.2 The ∆
f
H° (N
2
O
5
, g) in kJ/mol on the basic of the following data is :
2NO(g) +O
2
(g) → 2NO
2
(g) ∆
r
H° =–114 kJ/mol
4NO
2
(g) +O
2
(g) → 2N
2
O
5
(g) ∆
r
H° =–102.6 kJ /mol
∆
f
H° (NO,g) =90.2 kJ /mol
(A*) 15.1 (B) 30.2 (C) – 36.2 (D) none of these
[Sol.
2
1
N
2
(g) +
2
1
O
2
(g) → NO (g) ∆
f
H° =90.2
N
2
(g) +O
2
(g) → 2NO(g) ∆
r
H° =90.2 ×2 ... (1)
2NO(g) +O
2
(g) → 2NO
2
(g) ∆
r
H° =– 114 ... (2)
2NO
2
(g) +
2
1
O
2
(g) → N
2
O
5
(g) ∆
r
H° =
2
6 . 102 ÷
=– 51.3 ... (3)
(1) +(2) +(3)
N
2
(g) +
2
5
O
2
(g) → N
2
O
5
(g) ∆
f
H° (N
2
O
5
, g) =15.1 kJ /mol ]
[MULTIPLE OBJECTIVE TYPE]
Q.3 & Q.4 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.3 For the reaction A → B, the rate law expression is
t d
] A [ d
÷
=k [A]
1/2
. If initial concentration of [A] is
[A]
0
, then
(A*) The integerated rate expression is k =
) A A (
t
2
2 / 1 2 / 1
0
÷
(B*) The graph of A Vs t will be
(C) The half life period
2 / 1
t =
2 / 1
0
] A [ 2
K
(D*) The time taken for 75% completion of reaction
4 / 3
t =
k
] A [
0
[Sol. A → B
t =0 a –
t a–x x
dt
dx
=k (a–x)
1/2 ;
} }
=
÷
x
0
t
0
2 / 1
dt k
) x a (
dx
kt
1
2
1
) x a (
) 1 (
x
0
1
2
1
=
(
(
(
(
¸
(




¸

|
.
|

\
|
+ ÷
÷
÷
+ ÷
;
| | kt ) x a ( 2
0
x
2 / 1
= ÷
| | kt ) x a ( a 2
2 / 1 2 / 1
= ÷ ÷
;
(
(
¸
(


¸

|
.
|

\
|
÷ =
2 / 1
2 / 1
2 / 1
2
a
a 2 kt
=
|
.
|

\
|
÷
2
1
1 a 2
2 / 1
; =
|
|
.
|


\
|
÷
2
1 2
a 2
=
( ) 1 2 a 2 ÷
;
( )
k
1 2 ] A [ 2
t
0
2 / 1
÷
=
t
75%
=
(
(
¸
(


¸

|
.
|

\
|
÷
2 / 1
2 / 1
4
a
a
k
2
; =
(
¸
(

¸

÷
2
1
1 a
k
2
2 / 1
⇒
k
] A [
k
a
0
2 / 1
= ]
Q.4 A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. =45) is dissolved in 100 ml
water and titrated with 0.5 M HCl when
th
5
1
|
.
|

\
|
of the base was neutralised the pH was found to be 9
and at equivalent point pH of solution is 4.5 . Given : All data at 25°C & log 2 =0.3.
Select correct statement(s) .
(A) K
b
of base is less than 10
–6
(B*) Concentration of salt (C) at equivalent point is 0.25 M
(C*) Volume of HCl is used at equivalent point is 100 ml
(D) Weight percentage of base in given sample is 80%.
[Sol. When base is
th
5
1
|
.
|

\
|
neutralised pOH =pK
b
+log
|
|
.
|


\
|
5 a 4
) 5 / a (
RNH
2
+HCl → RNH
3
Cl 5 =pK
b
+log
|
.
|

\
|
4
1
a x pK
b
=5.6
4
5
a
–
5
a
pH =
2
1
[pK
w
– pK
b
– log C] ⇒ 4.5 ×2 =(14 – 5.6 – log C)
log C =–0.6
C =0.25
Let V ml of HCl is used
100 V
V 5 . 0
+
×
=0.25
V =100 ml
M.Moles of acid =M.Moles of base
=0.5 ×100 ⇒ 50
Wt. % of base in sample =
5 . 2
45 10 50
3
× ×
÷
×100 ⇒ 90 ]
[COMPREHENSION TYPE]
Q.5 to Q.7 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 5 to 7
W.H. Nernst derived a mathematical relationship between the emf of a cell (E
cell
) and the concentration
of reactants and products in a redox reaction under non-standard conditions. The nernst equation for
a redox reaction of the type
aA +bB → cC +dD
is E
cell
=
o
cell
E –
nF
RT
ln Q
at 298 K E
cell
=
o
cell
E –
n
0591 . 0
log Q
where Q is the reaction quotient. At equilibrium, there is a no net transfer of electron, so E =0 and Q
=K, where K is the equilibrium constant. We can also apply nernst equation for a half cell to calculate
oxidation or reducting potential.
Q.5 What is the EMF of represented cell at 298
Ag | Ag
+
(aq, 0.1 M) | | H
+
(aq, 0.1 M) | H
2
(g, 0.1 bar) | Pt
Given :
V 8 . 0 E
o
Ag / Ag
÷ =
+
(A) 0.77 V (B) –0.829 V (C*) – 0.77 V (D) None of these
[Sol. Cell reaction is 2Ag (s) +2H
+
(aq) → 2Ag
+
(aq) +H
2
(g)
E
cell
= – 0.80 –
2
2
) 1 . 0 (
1 . 0 ) 1 . 0 (
log
2
0591 . 0 ×
E
cell
=– 0.80 +0.02955
=– 0.77 V Ans. ]
Q.6 Find the solubility product of a saturated aqueous solution of Ag
3
PO
4
at 298 K if the emf of the cell
Ag | Ag
+
(saturated Ag
3
PO
4
) | | Ag
+
(aq, 0.01 M) | Ag is 0.1182 V at 298 K.
(A) 10
–16
(B) 2.7 ×10
–15
(C*) 3.33 ×10
–17
(D) None of these
[Sol. E
cell
=
LHS
RHS
] Ag [
] Ag [
log
1
0591 . 0
+
+
0.1182 =
|
.
|

\
|
x
01 . 0
log
1
0591 . 0
;
2
10
x
01 . 0
=
[Ag
+
]
LHS
=x =10
–4
M
Ag
3
PO
4
(s) l 3Ag
+
(aq)

+
÷ 3
4
PO
(aq)
10
–4
4
10
3
1
÷
×
K
sp
=[Ag
+
]
3
[PO
4
3–
] =(10
–4
)
3

|
.
|

\
|
×
÷4
10
3
1
K
sp
=3.33 ×10
–17


]
Q.7 What would be the reduction potential of an electrode at 298 K, which originally contained 1 M
K
2
Cr
2
O
7
solution in acidic buffer solution of pH =1.0 and which was treated with 50% of the Sn
necessary to reduce all
÷ 2
7 2
O Cr to Cr
3+
. Assume pH of solution remains constant.
Given : V 33 . 1 E
o
H , Cr / O Cr
3 2
7 2
=
+ + ÷
; log 2 =0.3 ;
06 . 0
F
RT 303 . 2
=
(A) 1.285 V (B) 1.193 V (C*) 1.187 V (D) None of these
[Sol.
÷ 2
7 2
O Cr + 14 H
+
+6 e¯ → 2Cr
3+
+7H
2
O
Initial con
n
1 0.1 0
After reaction 0.5 0.1 M 1M
E
RP
=
o
RP
E
–
6
06 . 0
log
14 2
7 2
2 3
] H ][ O Cr [
] Cr [
+ ÷
+
E
RP
=1.33 –
6
06 . 0
log 14
) 1 . 0 ( ) 5 . 0 (
1
=1.33 –
6
06 . 0
log (2 ×10
14
)
=1.187 V ]
[STRAIGHT OBJECTIVE TYPE]
Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 At certain temperature (T) if conductivity of pure water is 5.5 ×10
–7
S cm
–1
then calculate pOH of
pure water at temperature T.
Given :
·
+
ì
H
=350 S cm
2
eq
–1
;
·
ì
¯ OH
=200 S cm
2
eq
–1
(A) 5 (B*) 6 (C) 7 (D) 8
[Sol.
N
1000 k
O H
2
×
= .
·
;
550
1000 10 5 . 5
N
7
× ×
=
÷
=10
–6
K
w
=N
2
or M
2
; pH =pOH =6]
Q.4 Calculate the millimoles of
÷ 2
3
SeO in solution on the basis of following data:
70 ml of
60
M
solution of KBrO
3
was added to
÷ 2
3
SeO solution. The bromine evolved was removed by
boiling and excess of KBrO
3
was back titrated with 12.5 ml of
25
M
solution of NaAsO
2
.
The reactions are given below.
(a)
÷ 2
3
SeO +
÷
3
BrO +H
+
→
÷ 2
4
SeO
+Br
2
+H
2
O (b)
÷
3
BrO +
÷
2
AsO
+H
2
O → Br¯ +
÷ 3
4
AsO
+H
+
(A) 1.6 ×10
–3
(B) 1.25 (C*) 2.5 (D) None of these
[Sol. +4 +5 +6 0
(a)
÷ 2
3
SeO +
÷
3
BrO +H
+
→
÷ 2
4
SeO
+Br
2
+H
2
O
+5 +3 –1 +5
(b)
÷
3
BrO +
÷
2
AsO
+H
2
O → Br¯ +
÷ 3
4
AsO
+H
+
In reaction (b)
gm eq. of
÷
3
BrO =gm. eq. of
÷
2
AsO
2 n 6 n
2 3
AsO BrO
× = ×
÷ ÷
=
2
25
1
1000
5 . 12
× ×
=10
–3
6
10
n
3
BrO
3
÷
=
÷
In reaction (a)
moles of
÷
3
BrO consumed =
6
10
60
1
1000
70
3 ÷
÷ × =10
–3
gm eq. of
÷ 2
3
SeO =gm. eq. of
÷
3
BrO
5 10 2 n
3
SeO
2
3
× = ×
÷
÷
3
SeO
10 5 . 2 n
2
3
÷
× =
÷ Ans. ]
Q.6 How many times solubility of CaF
2
is decreased in 4 ×10
–3
M KF (aq.) solution as compare to pure
water at 25°C. Given K
sp
(CaF
2
) =3.2 ×10
–11
(A) 50 (B*) 100 (C) 500 (D) 1000
[Sol. 4 S
3
=3.2×10
–11
S =2 ×10
–4
Let solubility of CaF
2
is x in KF solution
K
sp
=x ( 2x +4 ×10
–3
)
2
; 3.2 ×10
–11
=x ×10
–6
×16 ; x =2 ×10
–6


]
[REASONING TYPE]
Q.10 & Q.12 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.10 Statement-1 : pH of 10
–7
M NaOH solution is exist between 7 to 7.3 at 25°C.
Statement-2 : Due to common ion effect ionization of water is reduced.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.12 Statement-1 : In general phenolphthalein is used as an indicator for the titration of weak acid
(HA) against strong base (NaOH).
Statement-2 : At equivalent point solution is basic.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[COMPREHENSION TYPE]
Q.14 to Q.16 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 14 to 16
A factory, producing methanol, is based on the reaction :
CO +2H
2
l CH
3
OH
Hydrogen & carbon monoxide are obtained by the reaction
CH
4
+H
2
O l CO +3H
2
Three units of factory namely, the "reformer" for the H
2
and CO production, the "methanol reactor" for
production of methanol and a "separator" to separate CH
3
OH from CO and H
2
are schematically
shown in figure.
CH
4
CO CO
H
2
O H
2
H
2
CH
3
OH
CH
3
OH
+ + +
+
( ) o ( ) | ( ) ¸ ( ) o
Reformer Methanol
Reactor
Separator
Four positions are indicated as α, β, γ and δ. The flow of methanol at position γ is 10
3
mol/sec. The
factory is so designed that
3
2
of the CO is converted to CH
3
OH. Excess of CO and H
2
at position δ
are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the
position (β) mole ratio of CO to H
2
is
3
1
.
CO +2H
2
l CH
3
OH ∆H
r
=– 100 R
Q.14 What is the flow of CO and H
2
at position (β)?
(A) CO : 1500 mol / sec.; H
2
: 2000 mol/sec.
(B) CO : 1500 mol / sec.; H
2
: 3000 mol/sec.
(C) CO : 1000 mol / sec.; H
2
: 2000 mol/sec.
(D*) CO : 1500 mol / sec.; H
2
: 4500 mol/sec.
[Sol. x 3x 0
CO + 2H
2
l CH
3
OH
x – y 3x – 2y y
3
2
x
y
=
y =
3
2
x =10
3
mol/ sec.
x =
2
3
×10
3
mol / sec.
x =1500 mol / sec.
(CO, β) =1500 mol / sec.
(H
2
, β) =4500 mol / sec. ]
Q.15 What is the flow of CO and H
2
at position (γ) ?
(A) CO : 500 mol / sec.; H
2
: 1000 mol/sec. (B*) CO : 500 mol / sec.; H
2
: 2500 mol/sec.
(C) CO : 500 mol / sec.; H
2
: 2000 mol/sec. (D) CO : 500 mol / sec.; H
2
: 1500 mol/sec.
[Sol. (CO, γ) =1500 – 1000 ⇒ 500 mol / sec.
(H
2
, γ) =4500 – 2000 ⇒ 2500 mol / sec. ]
Q.16 Amount of energy released in methanol reactor in 1 minute?
(A)1200 kcal (B*) 12000 kcal (C) 6000 kcal (D) None of these
[Sol. CO +2H
2
l CH
3
OH
q =∆H ×1000 ×60
=6 ×10
4
×100 ×2
=12 ×10
6
cal
q =12000 kcal ]
Q.1 & Q.10 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 The freezing point depression of a 0.10 M solution of formic acid is – 0.2046°C. What is the equilibrium
constant for the reaction at 298 K?
HCOO¯(aq) + H
2
O(l) HCOOH (aq) +OH¯ (aq)
(Given : K
f
(H
2
O) =1.86K kg mol
–1
, Molarity =molality)
(A) 1.1 ×10
–3
(B*) 9 ×10
–12
(C) 9 ×10
–13
(D) 1.1 ×10
–11
[Sol. ∆T
f
=K
f
. m . i
0.2046 =1.86 ×0.1 (1 +α)
α =0.1
HCOOH(aq) H
+
(aq) +OH¯ (aq)
K
a
=
) 1 (
C
2
o ÷
o
=
9 . 0
) 1 . 0 ( 1 . 0
2
×
=
9
1
×10
–2
K
h
=
a
W
K
K
=
2
14
10
9
1
10
÷
÷
×
⇒ 9 ×10
–12
Ans. ]
Q.2 Determine the amount of heat (in kcal) given off at constant volume when 0.5 mol of N
2
& 1.5 mol of
H
2
reacted according to equation at 300 K.
N
2
(g) + 3H
2
(g) → 2NH
3
(g)
300 r
H A =– 380 kcal/mol
(Given : R =2 cal / mol/K)
(A) 378.8 kcal (B) 190 kcal (C) 1888.8 kcal (D*) 189.4 kcal
[Sol. – 190 =∆E +
1000
300 2 ) 1 ( × × ÷
∆E =– 189.4 kcal/mol ]
Q.3 What mass of H
2
(g) is needed to reduce 192 gm of MoO
3
to metal? [At. wt. of Mo =96]
(A*) 8 gm (B) 16 gm (C) 32 gm (D) None of these
[Sol. Equivalents of MoO
3
=
6
M
⇒
6
144
⇒ 24
=equivalents of H
2
24
192
=
1
W
2
H
;
2
H
W =8 gm ]
Q.4 The yield of product in the reaction
2A(g) + B(g) 2C(g) + Q kJ
would be lower at :
(A) low temperature and low pressure (B) high temperature & high pressure
(C) low temperature and to high pressure (D*) high temperature & low pressure
Q.5 Give the correct order of initials T (ture) or F(false) for following statements
I : Lyophobic sols are irreversible sols.
II : Micelles formation takes place only above kraft temperature
III : PO
4
3–
ions have more coagulation value than SO
4
2–
ions for congulation of positive sols.
IV : The values of the colligative properties observed experimentally are very small of colloidal sols
(A) F T T F (B) T F T F (C) T T T T (D*) T T F T
Q.6 The solubility of common salt is 36.0 gm in 100 gm of water at 20°C. If systems I, II and III contain
20.0, 18.0 and 15.0 g of the salt added to 50.0 gm of water in each case, the vapour pressures would
be in the order.
(A) I <II <III (B) I >II >III (C) I =II >III (D*) I =II <III
Q.7 Read the following statements.
Statement-1 : Higher the critical temp of a gas, more readily it will be adsorbed on the solid adsorbent,
Statement-2 : The excluded volume of a gas will be smaller, if
C
C
P
T
is small
(A) Statement-1 is true & statement-2 is false
(B) Statement-1 is false & statement-2 is true
(C*) Both are correct
(D) Both are incorrect.
[Sol.
C
C
P
T
= 2
b 27 / a
Rb 27 / a 8
=
R
b 8
]
Q.8 The conductivity of a saturated solution of Ag
3
PO
4
is 9 ×10
–6
S m
–1
and its equivalent conductivity is
1.50 ×10
–4
S m
2
equivalent
–1
. The K
sp
of Ag
3
PO
4
is
(A*) 4.32 ×10
–18
(B) 1.8 ×10
–9
(C) 8.64 ×10
–13
(D) None of these
[Sol λ =K ×
N
1000
1.50 ×10
–4
×10
4
=9 ×10
–5
×10
–2


×
N
1000
N =6 ×10
–5
S =M =
f
n
N
=
3
10 6
5 ÷
×
=2 ×10
–5
mol / L
Ag
3
PO
4
=3 Ag
+
+PO
4

3 S S
K
sp
=(3S)
3
. S

=27. S
4
=27 ×(2 ×10
–5
)
4
= 4.32 ×10
–18
]
Q.9 96.5 gm of A oxidized as A → A
3+
+3e
–
, when 2F charge is passed through solution with current
efficiency of 90%. What is electrochemical equivalent of A ?
(A) 53.61 (B) 48.25 (C*`) 5.55 ×10
–4
(D) 5.55 ×10
–3
[Sol. w =Z ×n ×q
Z =
96500 2 9 . 0
5 . 96
× ×
=5.55 ×10
–4
]
Q.10 Which graph represents zero order reaction [A(g) → B(g)] :
(A)
t
[B]
(B)
t
dt
] B [ d
(C)
[A]
0
t
1/2
(D*)
t
3/4
[A]
0
[Sol. k =
t
x
t =x .
k
1
4 / 3
t =
k 4
3
. a ]
[REASONING TYPE]
Q.1 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason). Each
question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.1 Statement-1 :
p
n
ratio increases due to emission of apositron
( ) e
0
1 +
Statement-2 : Positron forms in the nucleus by the conversion of neutron into a proton.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C*) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[MULTIPLE OBJECTIVE TYPE]
Q.2 to Q.4 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.2 80.0 gm salt of weak base & strong acid XY is dissolved in water and formed 2 litre of aqueous
solution. The pH of the resultant solution was found to be 5 at 298 K. If XY forms CsCl type crystal
having
+
X
r (radius of X
+
) =1.6 Å & –
Y
r
(radius of Y¯) =1.864 Å then select write statement(s).
(Given : K
b
(XOH) =4 ×10
–5
; N
A
=6 ×10
23
)
(A*) Molar mass of salt is 100 g/mol
(B) % Degree of dissociation of salt is 0.25
(C*) Edge length of AB is 4Å
(D) Density of solid salt XY is 2 in gm/cc
[Sol. XY (aq) → X
+
(aq) +Y¯(aq)
X
+
(aq) +H
2
O(l) XOH (aq) +H
+
(aq)
At equation
C(1 – α) – Cα Cα
[H
+
] =Cα =
C
K
K
b
w
×
; (10
–5
)
2
=
5
14
10 4
10
÷
÷
×
×
M
80
×
2
1
M =100 gm/mol ; % α =
C
K
h
×100 ⇒ 2.5 ×10
–3
a for XY : 3a =2 (r
+
+r¯)
1.732 ×a =2 (1.6 +1.864)
a =4Å = 4 ×10
–8
cm
Effective formula unit of XY =1
d =
A
3
N a
M . Z
×
= 23 3 8
10 6 ) 10 4 (
100 1
× × ×
×
÷ =
384
1000 1×
=2.6 gm/cc ]
Q.3 Select write statement(s)
(A) 8 Cs
+
ions occupy the second nearest neighbour locations of a Cs
+
ion
(B*) Each sphere is surrounded by six voids in two dimensional hexagonal close packed layer
(C) If the radius of cations and amions are 0.3 Å and 0.4 Å then coordinate number of cation in the
crystal is 6.
(D*) In AgCl, the silver ion is displaced from its lattice position to an interstitial position such a defect
is called a frenkel defect
Q.4 The enthalpy of neutralization of 1M solution of HCl with 1M NaOH is – 57.3 kJ/mol. The enthalpy of
solution of a weak base XOH (1M) with same HCl solution is – 54.6 kJ/mole. If molar conductivity of
1M solution of XOH is 200 ×10
–4
Sm
2
mol
–1
& molar conductivity of XOH at infinite dilution is
0.20 Sm
2
mol
–1
then select correct statement(s) [Assume all statements at 298 K]
(A*) ∆H
ionization
of XOH is 3.0 kJ /mol
(B) pH of 1M XOH solution is 12
(C) pK
a
of X
+
is 1
(D*) At equivalent point resultant solution of HCl & XOH is acidic
[Sol. α =
·
.
.
m
m
=
2 . 0
10 200
4 ÷
×
=0.1 ;
acid is 10 % ionised so enthalpy of ionisation of XOH =
9 . 0
7 . 2
=3 kJ/mol
XOH (aq) X
+
(aq) + OH¯ (aq)
C –X X X
K
b
=
C 9 . 0
) C 1 . 0 (
2
= 0.011 ; K
a
=
011 . 0
10
14 ÷
=9.1 ×10
–13
Due to cationic hydrolysis resultant solution is acidic ]
[MATCH THE COLUMN]
Q.1 is "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I are to
be matched with one or more than one entries of column-II and vice versa.
Q.1 Match the Column.
Column I Column II
(A) 2 s (P) Number of radial node =1
(B)
¢
2
r
r
(Q) Number of maxima in radial probability
distribution curve =2
(C)
¢
r
r
(R) Number of angular node =0
(D) 4 d (S) ψ
angular
depends upon θ
[Ans. (A) P,Q,R (B) P,Q,S (C) R (D) P,Q,R ]
[STRAIGHT OBJECTIVE TYPE]
Q.1 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 What is the pK
b
of a weak base whose 0.1 M solution has pH =9.5 ?
(A) 7.5 (B*) 8 (C) 9 (D) 10
Q.2
C
H O
Cl

÷ ÷ ÷ ÷ ÷ ÷
KOH . conc
Products of above reaction is
(A*)
Cl
OH
Cl
CO
2

(B)
Cl
OH

Cl
OH
(C)
Cl
OH
Cl
COH
2
(D)
Cl
CO
2


Cl
CO
2

Q.3 What is oxidation state, magnetic moment and type of hybridisation of central metal cation in following
complex.
[Os(ONO) (O)
2
(O
2
) (SCN) (H
2
O)]

OH
(A) +7, 3 B.M., d
2
sp
3
hybridisation (B) +8, 0 B.M., sp
3
d
2
hybridisation
(C*) +8, 0 B.M., d
2
sp
3
hybridisation (D) +9, 0 B.M., sp
3
d
2
hybridisation
Q.5
O H Zn
O
2
3
+
÷÷ ÷ (A) ÷ ÷ ÷ ÷ ÷ ÷
NaOH . conc
(B)
Product (B) is
(A)
OH CH
|
OH CH
2
2
÷
÷
(B)
CONa
2
 ©
|
CONa
2

©
(C)
COOH
|
CH – ONa
2

©
(D*)
CO Na
2
|
CH – OH
2

©
Q.6 The dipole moment of H
2
O
2
is more than that of H
2
O but H
2
O
2
is not a good solvent because
(A) it has a very high dielectric constant so that ionic compounds cannot be dissolved in it
(B) it does not act as an oxidising agent
(C) it acts as a reducing agent
(D*) it decomposes readily and acts as an oxidising agent in chemical reactions
[COMPREHENSION TYPE]
Q.7 to Q.9 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 7 to 9
Ni
II
(NH
3
)
4
(NO
3
)
2
. 2H
2
O molecule may have two unpaired electrons or zero unpaired electron and
measurement of magnetic moment helps us to predict the geometry.
Q.7 If magnetic moment value is zero then the formula of the complex will be
(A*) [Ni(NH
3
)
4
] (NO
3
)
2
. 2H
2
O (B) [Ni(NH
3
)
2
(H
2
O)
2
](NO
3
)
2
. 2NH
3
(C) [Ni (NH
3
)
4
(H
2
O)
2
] (NO
3
)
2
(D) [Ni (NO
3
)
2
(H
2
O)
2
]
Q.8 If the magnetic moment value is
2 2
and conducts electricity then the formula of the complex is
(A) [Ni(NH
3
)
4
] (NO
3
)
2
. 2H
2
O (B) [Ni(NH
3
)
2
(H
2
O)
2
](NO
3
)
2
. 2NH
3
(C*) [Ni (NH
3
)
4
(H
2
O)
2
] (NO
3
)
2
(D) [Ni (NH
3
)
4
(NO
3
)
2
]. 2H
2
O
Q.9 The higher and lower value of magnetic moment of the given complex corresponds to the following
geometries respectively.
(A) Octahedron and Tetrahedron (B*) Octahedron and square planar
(C) Square planar and Octahedron (D) Octahedron and Octahedron
[REASONING TYPE]
Q.10 & Q.11 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.10 Statement-1 : CH
3
– CH
2
– Cl +NaI
÷ ÷ ÷ ÷ ÷
Acetone
CH
3
– CH
2
– I +NaCl ↓
Above reaction is Finkelstein reaction.
Statement-2 : Acetone is polar protic solvent and solubility order of sodium halides decreases
dramatically in order NaI >NaBr >NaCl. The last being virtually insoluble in
this solvent and a 1° and 2° chloro alkane in acetone is completely driven to
the side of Iodoalkane by the precipitation reaction.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C*) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.11 Statement-1 :
Br
K O – C–Me
(DMSO)
Me
Me
+
Above reaction is Elimination reaction.
Statement-2 : Type of reaction is E
2
and Hoffmann's product is major product in the above
reaction because sterically hindered base is used for elimination reaction.
(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[MULTIPLE OBJECTIVE TYPE]
Q.12 to Q.14 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are
correct.
Q.13
OH
O
(X)
compound (X) is obtained by :
(A*)
OH
O

÷ ÷ ÷ ÷ ÷
NaOH . dil
(B*)
3 4 2 3
CH C — ) CH ( — C CH
| | | |
O O
÷ ÷
÷ ÷ ÷ ÷ ÷
NaOH . dil
(C) H C — ) CH ( — C CH
| | | |
O O
5 2 3
÷ ÷ ÷ ÷ ÷ ÷ ÷
NaOH . dil
(D) H C — ) CH ( — C — H
| | | |
O O
6 2
÷ ÷ ÷ ÷ ÷ ÷
NaOH . dil
[MATCH THE COLUMN]
Q.1 is "Match the Column" type. Column-I contains three entries and column-II contains four entries. Entry of
column-I are to be matched with only one entry of column-II.
Q.1 Column I Column II
(Metal) (Process used in commercial extraction of pure metal)
(A) Ag (P) Vapour phase refining
(B) Cu (Q) Cupellation
(C) Ni (R) Electrolytic reduction
(S) Bessemerisation
[Ans. (A) Q, (B) S, (C) P]
Q.1 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.2 During mixed acid nitration of benzene concentrated nitric acid behaves like:
(A) Acid (B) Base (C) Catalyst (D) Electrophile
Q.3 Which of following species will have highest bond length.
(A)
÷
2
O
(B*)
÷ 2
2
O
(C) F
2
(D)
+
2
O
Q.5 Identify B and C (respectively) in following conversion.

A
÷ ÷ ÷ ÷ ÷ ÷
4 2
SO H . conc
A
©
÷ ÷ ÷ ÷
O H
OsO
3
4
B
A
÷ ÷÷ ÷
©
H
C
(A)
OH
OH OH
&
(B)
OH
OH
&
(C)
OH
OH OH
&
O
(D) None of these
Q.7
Et O C CH CH C O Et
| | ||
O O
2 2
÷ ÷ ÷ ÷ ÷ ÷ ÷
÷ ÷ ÷ ÷

OEt
?
Major product of this reaction is:
(A) (B*)
(C)
OEt C CH CH C EtO
| | | |
O O
÷ ÷ = ÷ ÷
(D)
[Sol. + EtO
r
→ →

←
÷ ÷ ÷ ÷

EtO
]
Q.8 How many pairs of enantiomers are possible for following complex compound.
[M(AB) (CD) ef ]
n ±
(where AB, CD – Unsymmetrical bidentate ligand ; e,f – Monodentate ligands)
(A) 20 (B) 5 (C*) 10 (D) 8
Q.9 Ease of oxidation of following ring systems is in order:
CN
(I)
OH
(II) (III)
(A) I >II >III (B) III >II >I (C) I >III >II (D*) II >III >I
[Sol. Ease of oxidation α electrons density ]
[REASONING TYPE]
Q.10 to Q.13 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.11 Statement-1 : The Friedel-Crafts alkylation of nitro-benzene gives meta alkylated
nitrobenzene
Statement-2 : Nitro group is meta directing during aromatic electrophilic substitution as it
decreases electron density at o & p position
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D*) Statement-1 is false, statement-2 is true.
Q.12 Statement-1 : In general phenolphthalein is used as an indicator for the titration of weak acid
(HA) against strong base (NaOH).
Statement-2 : At equivalent point solution is basic.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.13 Statement-1 : 'S' on reaction with NaOH, produces H
2
S first as one of the product.
Statement-2 : H
2
S is acidic in nature to produce Na
2
S as final product.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[COMPREHENSION TYPE]
Q.14 to Q.19 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 17 to 19
Using dil. H
2
SO
4
gas evolution takes place from several acid radicals. If we consider their smell,
colour and other observation if any, several acid radicals can be detected in this step.
Q.17 Using dil. H
2
SO
4
, which of the acid radical can not be confirmed.
(A)
÷
2
NO
(B)
÷
2 3
CO CH (C*)
÷
3
HSO (D) S
2–
Q.18 When the aqueous suspension of Ag
2
SO
3
and Ag
2
CO
3
are heated, the respective ppts. are obtained
as
(A) Ag
2
O & Ag
2
O (B) Ag
2
O & Ag (C*) Ag & Ag
2
O (D) Ag & Ag
Q.19 Which of the following radical(s) is/are producing same gas on treatment with (Zn +dil. H
2
SO
4
).
I. SO
3
2–
II. HSO
3
¯ III. S
2–
IV. Cl¯
(A) I and II only (B*) I, II and III only (C) I, II, III and IV (D) II, III and IV only
[MATCH THE COLUMN]
Q.1 to Q.3 are "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I
are to be matched with one or more than one entries of column-II and vice versa.
Q.1 Match reactions given in column I with names in II.
Column I Column II
(A)
COOEt
COOEt

÷ ÷ ÷ ÷
EtONa
(P) Knovenagel reaction
(B) CH
2
(COOEt)
2
+
O

A
©
÷ ÷ ÷ ÷ ÷
, O H ) ii (
EtOK ) i (
3
(Q) Perkin reaction
(C)
O
O
O
Br

©
÷ ÷ ÷ ÷
O H ) ii (
Zn ) i (
3
(R) Reformatsky recation
(D)
O
O
OEt
MeO
÷ ÷ ÷ ÷
MeOK
(S) Dieckmann's condensation
Q.2 Column I Column II
(Commercial methods) (Metals)
(A) MS +2O
2

÷÷ ÷
A
MSO
4
(P) Ag
MSO
4
+MS ÷÷ ÷
A
2M +2SO
2
(B) The metal(s) is/are not reduced during smelting step (Q) Au
(C) The ore of the metal(s) is / are concentrated (R) Cu
by froth floatation method
(D) Impure metal +CN¯ solution → soluble complex (S) Pb
Soluble complex +Zn → pure metal
[Ans. (A) R,S; (B) R; (C) P,R,S; (D) P,Q]
Q.3 Match the column :
Column I Column II
(A) CH
2
=CH
2
(P) Br
2
– H
2
O → decolourise
(B) Me – C ≡ C – H (Q) Cold dil alkaline KMnO
4
→ decolourise
(C) Me – C ≡ C – Me (R) NaNH
2
÷÷ ÷
A
ppt.
(D) OH (S) NaHCO
3
→ CO
2
↑
Q.1 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.2
N
©
OH


÷÷ ÷
A
Number of α–H in alkene which is major product in this reaction is
(A) 2 (B) 4 (C) 5 (D) 7
Q.3 Which of the following option is having maximum number of unpaired electrons.
(A) A tetrahedral d
6
ion.
(B) [Co(H
2
O)
6
]
3+
(C) A square planar d
7
ion
(D*) A co-ordination compound with magnetic moment of 5.92 B.M.
Q.5 What are the most likely products of the reaction shown below?

÷ ÷ ÷ ÷
+
O H
3
(A) & (B*) &
(C) & (D) &
Q.6 A compound with the empirical formula Fe(H
2
O)
4
(CN)
2
has a magnetic moment corresponding to
3
2
2 unpaired electrons per iron. What is the correct possible formula for the above complex.
(A) [Fe(CN)
2
(H
2
O)
4
] (B*) [Fe(H
2
O)
6
]
2
[Fe(CN)
6
]
(C) [Fe(H
2
O)
6
] [Fe(CN)
4
(H
2
O)
2
] (D) [Fe(H
2
O)
6
] [Fe(CN)
6
]
[REASONING TYPE]
Q.7 & Q.10 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.7 Statement-1 :
O
O
A
major
+
Statement-2 : 2-butene is more stable than 1-butene as it is having more α – H.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.8 Statement-1 : Froth floatation method is used for benefication of galena though ZnS is present
as impurity.
Statement-2 : NaCN or KCN solution is very good depressant to suppress the floating
property of PbS.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C*) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.9 Statement-1 : H C C Cl
| |
O
3
÷ ÷ ÷ ÷ ÷ ÷
NaOH
Cl
3
C – CH
2
OH +Cl
3
C – COONa
Statement-2 : There are no α – H in this compound, so it can't give aldol.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.10 Statement-1 : ∆
0
increases in the order of [CrCl
6
]
3–
<[Cr(CN)
6
]
3–
<[Cr(C
2
O
4
)
3
]
3–
Statement-2 : Stronger the ligand field higher will be ∆
0
value.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D*) Statement-1 is false, statement-2 is true.
[COMPREHENSION TYPE]
Q.11 to Q.19 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 14 to 16
Dr. Robertson Nehwal worked out following reaction sequence.
÷ ÷ ÷ ÷
NBS
(A) ÷ ÷ ÷ ÷ ÷
KOH . alc
(B)
4
2
CCl
Br
÷ ÷÷ ÷
(C)
2
NaNH
KOH . alc
÷ ÷ ÷ ÷ ÷
(D)
4 2
4
SO H . dil
HgSO
÷ ÷ ÷ ÷ ÷
(E)

(G)
excess
KOH
÷ ÷ ÷ ÷
(F)
Q.14 Select true statement(s):
(A*) when D is treated with D
2
Pd BaSO
4
it gives cis alkene.
(B) when D is treated with Na liq. ND
3
it gives trans alkene.
(C) both
(D) none
Q.15 True about E is:
(A) It gives one oxime with NH
2
OH
(B) It gives foul smelling substance with CHCl
3
& KOH
(C*) It gives yellow ppt. with NaOI.
(D) It gives effervescence with NaHCO
3
.
Q.16 G is:
(A)
O
| |
CH C Ph ÷ ÷
(B)
O
| |
CHO C Ph ÷ ÷
(C)
OH
|
CHO CH Ph ÷ ÷
(D*)
OH
|
COOK CH Ph ÷ ÷
Paragraph for question nos. 17 to 19
When redox reaction occurs within the reactant, in which one component acts as oxidising agent and
other component acts as reducing agent, then it is named as intra-molecular redox reaction, which
usually occur in thermal decomposition of ionic compounds.
Q.17 Which of the following compounds does not give nitrogen gas on heating?
(A) NH
4
NO
2
(B*) (NH
4
)
2
SO
4
(C) NH
4
ClO
4
(D) (NH
4
)
2
Cr
2
O
7
Q.18 Which of the following salt does not give NO
2
gas on heating?
(A) Pb(NO
3
)
2
(B) Hg(NO
3
)
2
(C*) KNO
2
(D) AgNO
3
Q.19 NH
4
ClO
4
+HNO
3
(dilute) → X +HClO
4
X
÷ ÷÷ ÷
heat
Y (gas)
Gas (Y) is :
(A) O
2
(B) N
2
(C) NO
2
(D*) N
2
O
[MULTIPLE OBJECTIVE TYPE]
Q.20 to Q.23 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are
correct.
Q.21 Suitable reagent(s) to carry out following conversion is / are
N
CH
3
C–CH
3
O
→
N
Et
CH
3
(A)
SH CH
|
SH CH
2
2
÷
÷
followed by H
2
Ni (B) N
2
H
4
/ H
2
O
2
(C) N
2
H
4
/ EtONa (D) Zn / Hg / HCl
Q.23 Which of the following compound(s) show(s) optical isomerism.
(A*) [Pt(bn)
2
]
2+
(B*) [CrCl
2
(en)
2
]
+
(C*) [Co(en)
3
] [CoF
6
] ( D * )
[Zn(gly)
2
]
2+
[COMPREHENSION TYPE]
Q.1 to Q.3 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which
ONLY ONE is correct.
Paragraph for question nos. 1 to 3

(A) (B) + (C) + (D)
Orange
solid
Green
Solution
Yellow
solution
Blue
Coloured
solid
White
solid
Pink so
A Anhydrous
CoCl
2
H
2
O / OH¯
2
H
2
O / H
2
+
Et O
2
H
2
O NiCl
2
so
(E)
conc.
HCl
(F) (G) (H)
Red hot Mg
(I) (J)
Q.1 Which of following gas is turning the filter paper dipped into (MnSO
4
+H
2
O
2
) solution brownwish
black.
(A) gas 'C' (B) gas 'D' (C*) gas 'J ' (D) None of these
Q.2 Type of hybridisation in compound (K)
(A) d
2
sp
3
(B*) sp
3
d
2
(C) dsp
2
(D) sp
3
Q.3 Which of the following set of reagent is suitable to get (F) from (G) .
(A) Neutral H
2
O
2
solution (B) H
2
O
2
/ H
+
+amyl alcohol
(C) H
2
O
2
/ OH¯ (D*) H
2
O
2
/ H
+
[REASONING TYPE]
Q.4 to Q.7 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.5 Statement-1 : NH
2
– NH
2
gives Prussian blue colour during Lassaigne test (element detection
test).
Statement-2 : NaCN gives Prussian blue colour of Fe
4
[Fe(CN
6
)]
3
with FeSO
4
, FeCl
3
.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D*) Statement-1 is false, statement-2 is true.
Q.6 Statement-1 : Conc. HNO
3
can be stored in Al-container
Statement-2 : Al reacts with conc. HNO
3
forming Al
2
O
3
which makes impervious coating on it and
does not dissolve any more.
(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.7 Statement-1 : Result of Victor mayer test
1° ROH → Red colour
2° ROH → Blue colour
3° ROH → White or No colour
Statement-2 : Victor mayer test is a method for separations of 1°, 2° and 3° alcohol.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C*) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[MULTIPLE OBJECTIVE TYPE]
Q.8 to Q.12 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct.
Q.9 Which of the following compounds on direct heating can not produce anhydrous form of it.
(A*) FeCl
3
. 6H
2
O (B) CuSO
4
. 5H
2
O (C*) MgCl
2
. 6H
2
O (D*) ZnCl
2
. 2H
2
O
Q.11 H C CH
| |
O
3
÷ ÷ and
3 3
CH C CH
| |
O
÷ ÷ is differentiated by
(A*) Tollen's reagent (B*) Fehling (C) Iodoform (D) NaHSO
3
[MATCH THE COLUMN]
Q.1 & Q.2 are "Match the Column" type. Column-I contains three entries and column-II contains four entries.
Entry of column-I are to be matched with ONLY ONE ENTRY of column-II
Q.1 Match the observations of Column-I with inferences in Column-II :
Column-I Column-II
(A) Canary yellow precipitate with ammonium molybdate (P) NO
3
¯
(B) Brown ring test with dil. H
2
SO
4
(Q)
÷ 2
4
SO
(C) Yellow ppt. with HgCl
2
solution (R) PO
4
3–
(S) NO
2
¯
[Ans. (A) R, (B) S, (C) Q ]
Q.2
Column - I Column-II
(A)
NO
2
(P) Group attached with phenyl ring is +M.
(B)
N
(Q) Rate of electrophilic aromatic substitution is less then benzene ring
(C)
CH
3
(R) Group attached with phenyl ring will show (+H) (Hyperconjugation)
(S) Group attached with phenyl ring will show (–H) effect
(Hyperconjugation)
[Ans. (A) Q , (B) P , (C) R ]
Q.3 is "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I are to
be matched with ONE OR MORE THAN ONE ENTRIES of column-II and vice versa.
[STRAIGHT OBJECTIVE TYPE]
Q.1 & Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.2 If in the [Ma
2
bcde]
n±
complex, 'c' is replaced by 'b' then number of geometrical isomers will be reduced
by
(A) 6 (B*) 3 (C) 2 (D) 4
Q.3 Polymer obtained by reaction of phenol and formaldehyde is
(A*) Bakellite (B) Terylene (C) Nylon-6,6 (D) Urea formaldehyde resin
Q.5 Which isomeric form of complex compound [Ma
4
bc]
n±
gives three isomers as product by replacing
(a) by (b) [where a,b,c are monodentate simple ligands]
(A) Trans (B*) Cis (C) Cis-Trans both (D) None
Q.6
NO
2
Cl

A
÷ ÷÷ ÷
, Fe
Br
2
(X)
A
÷ ÷ ÷ ÷ ÷
, OH CH
ONa CH
3
3
(Y)
Product (Y) of this reaction is
(A)
NO
2
OCH
3
Cl
(B*)
NO
2
OCH
3
Br
(C)
NO
2
OCH
3
(D)
NO
2
OCH
3
Q.8 By which process pure alumina is produced when bauxite is rich in silica.
(A) Hall's process (B) Bayer's process (C*) Serpec's process (D) (A) and (B) both
Q.9 Which of the following pair will form same osazone when it reacts phenyl hydrazine.
(A)
allose D
OH CH
|
OH C H
|
OH C H
|
OH C H
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
and
glucose D
OH CH
|
OH C H
|
OH C H
|
H C HO
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
(B*)
glucose D
OH CH
|
OH C H
|
OH C H
|
H C HO
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
and
mannose D
OH CH
|
OH C H
|
OH C H
|
H C HO
|
H C HO
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
(C)
glucose D
OH CH
|
OH C H
|
OH C H
|
H C HO
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
and
gulose D
OH CH
|
OH C H
|
H C HO
|
OH C H
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
(D)
allose D
OH CH
|
OH C H
|
OH C H
|
OH C H
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
and
alactose g D
OH CH
|
OH C H
|
H C HO
|
H C HO
|
OH C H
|
CHO
2
÷
÷ ÷
÷ ÷
÷ ÷
÷ ÷
Q.11 Unknown sample (A) on heating swells up first and aqueous solution of (A) gives white ppt with
NaOH which becomes soluble with excess of NaOH addition.
(A) Borax (B*) Alum (C) Microcosmic salt (D) None
Q.12
CH
3
OH
OH

÷ ÷ ÷ ÷
PCC
(X) Major
(A)
OH
O
(B)
O
O
(C*)
O
OH
(D)
OH
Q.14 Aqueous solution of (M) +(NH
4
)
2
S → yellow ppt (B)
÷ ÷ ÷ ÷ ÷
2 2 4
S ) NH (
insoluble.
The cation present in (M) is
(A) CdS (B) SnS
2
(C*) Cd
2+
(D) Sn
2+
Q.15 H O C Ph
| |
O
÷ ÷ ÷ +H–O
18

A
÷ ÷÷ ÷
HCl
(X)
Major product (X) is.
(A*)
O
Ph–C–O
18
(B)
O
Ph–C–O
18
(C)
O
Ph–C–O
(D) Ph–O
Q.17 Choose the correct statement
(A) Both Au and Ag dissolve in Aqua regia
(B) Using Na
2
CrO
4
, group-V cations (Ca
2+
, Sr
2+
, Ba
2+
) are well distinguished
(C*) Using conc. H
2
SO
4
Cl¯, Br¯, I¯ are well distinguished
(D) NO
2
¯ and NO
3
¯ are not distinguished by dil. H
2
SO
4
Q.18 HC ≡ CH
4 2
4
SO H
HgSO
÷ ÷ ÷ ÷ ÷ (A)
C 5
OH . dil
°
÷ ÷ ÷ ÷ ÷

(B)
Give the IUPAC name of "B" is
(A) 2-butenal (B*) 3-hydroxy butanal
(C) 3-formyl-2-propanol (D) 4-oxo-2-propanol
Q.20 Which of the following reagent will separate Bi
3+
in one step from mixture having Fe
3+
, Bi
3+
, Cd
2+
(A) dil. HCl +H
2
S (B) Excess NH
4
OH (C) Excess NaOH (D*) KI
Q.21 Compare rate of nitration in the following compounds.
CF
3
(a)
(b)
CH
3
(c)
OH
(d)
(A) b >c >a >d (B) d >c >a >b (C) b >a >c >d (D*) d >c >b >a
Q.23 Which of the following carbide does not release any hydrocarbon on reaction with water.
(A*) SiC (B) Be
2
C (C) CaC
2
(D) Mg
2
C
3
Q.24 Phenol and benzoic acid is separated by
(A*) NaHCO
3
(B) NaOH (C) Na (D) NaNH
2
Q.26 Which of the constitutent is / are responsible for continuous burning in Holme's signal
(A) C
2
H
2
(B) PH
3
(C*) P
2
H
4
(D) All
Q.27 Incorrect statement about given carbohydrate is
OH OH
H H
H
OH
H
CH – OH
2
O
O
OH
H OH
H
OH
H
O
H
CH – OH
2
(A) Above compound is a reducing sugar (B) Above compound undergo mutarotation
(C*) Above compound is a non-reducing sugar (D) Above compound has a glycosidic linkage
Q.29 H
2
S gas is not evolved when
÷ 2
3
SO ion reacts with following reagents.
(A) Zn +dil. H
2
SO
4
(B*) Al +conc. NaOH
(C) Al +dil. HCl (D) None of these
Q.30
) mole 1 (
Zn
O
3
÷
÷÷ ÷ (A)
÷ ÷ ÷ ÷ ÷ ÷
KOH . conc
(B)
End product (B) of above reaction is
(A)
CHO
CHOH
2
(B*)
CO K
2
 ©
CH OH
2
(C)
CH OH
2
CH OH
2
(D)
CO K
2
 ©
CO K
2
 ©
PART-A
[STRAIGHT OBJECTIVE TYPE]
Q.1 to Q.8 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.2
Br
|
CH CH CH CH
3 2 3
÷ ÷ ÷

A
÷ ÷ ÷ ÷ ÷
KOH . alc
X (major)
4
2
CCl
Br
÷ ÷÷ ÷
Products.
Last products of the reaction.
(A) Racemic mixture (B*) Meso compound (C) Diastereomers (D) Optically active
Q.3 To obtain silver from silver amalgam it is heated in vessel which is made of
(A) Cu (B*) Fe (C) Ni (D) Zn
Q.5
÷ ÷ ÷ ÷ ÷
v h / Br
2
Major (X)
A
÷ ÷ ÷ ÷ ÷
/ KOH
Alcoholic
Major (Y)
Peroxide
Br H
÷ ÷ ÷ ÷
÷
Major (Z)
Major final product (Z) is?
(A)
Br
(B)
Br
(C*)
Br
(D)
Br
Q.6 In Fe-extraction, the roasting is adopted though the ore is not having any sulphide because
(A) Haematite is to be decomposed
(B*) All FeO is to be converted into Fe
2
O
3
(C) All Fe
2
O
3
is to be converted into FeO
(D) Slag formation is encouraged.
Q.7
H C C CH
3
÷ ÷ ÷
÷ ÷ ÷ ÷
NaH
(X)
CH –I
2
(Y)
2 2
3
O H
O
÷÷ ÷ (Z)
Final products (Z) of reaction are?
(A) Et–COOH &
COOH
(B*) CH
3
– COOH &
CH –COOH
2
(C) CH
3
– COOH &
COOH
(D) CH
3
– COOH &
COOH
Q.8 In the given molecule choose thecorrect order of their angle.
C
O—H
O
H
y
x
z
(A*) x >y >z (B) x <y <z (C) x =y >z (D) Can't be predicted.
[REASONING TYPE]
Q.9 & Q.10 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Q.9 Statement-1 : Nitric oxide is diamagnetic in liquid state.
Statement-2 : It has fractional bond order.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.10 Statement-1 : In Mn
2
(CO)
10
molecule, there are total 70 electrons in both Mn atoms.
Statement-2 : Mn
2
(CO)
10
molecule acts as oxidising agent.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C*) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
[MULTIPLE OBJECTIVE TYPE]
Q.11 to Q.13 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are
correct.
Q.12 Which of following compound will not undergo S
N
1 reaction ?
(A*)
Br
(B*)
Cl CH C H
2
÷ =
(C*)
Cl
(D)
3
3
3
CH
|
Cl C CH
|
CH
÷ ÷
Q.13
2
M
g

+
O

2
M
g
O
2
÷
2C
+O

2C
O
2 ÷
T
1
T
2
T
3
–1200
–1100
–1000
–900
–800
–700
–600
–500
–400
–300
–200
–100
0
A
G
(
k
J
/
m
o
l
)
Temperature
Correct statements about the plot is / are:
(A*) T
1
and T
2
are melting point & boiling point of Mg respectively.
(B) T
1
and T
2
are melting point & boiling point of MgO respectively.
(C*) Reduction of MgO by coke is possible above T
3
(D*) Mg can be extracted from gaseous products by rapid cooling.
PART-B
[MATCH THE COLUMN]
Q.2 is "Match the Column" type. Column-I and column-II contains four entries each. Entry of column-I are to
be matched with one or more than one entries of column-II and vice versa.
Q.2 Column I Column II
(A)
O
÷ ÷ ÷ ÷
©
O H
3 (P) No-reaction
(B)
3 3
3 3
3
CH CH
| |
CH CH O C CH
|
CH
÷ ÷ ÷ ÷
÷ ÷ ÷ ÷
©
O H
3 (Q)
3
3
3
CH
|
OH C CH
|
CH
÷ ÷
is one of the product of the reaction
(C)
O
CH
3
CH
3
CH
3
C
÷ ÷ ÷ ÷
©
O H
3
(R)
OH
is one of the product of the reaction
(D)
O – CH
CH
3
CH
3
÷ ÷ ÷ ÷
©
O H
3
(S)
3
3
CH
|
OH CH CH ÷ ÷
is one of the product of the reaction
[Ans. (A) P; (B) Q,S; (C) Q,R; (D) R,S]