CLASS XII
SET A
CHEMISTRY
Set : |A -31 |B - |
31.
Phenol is less acidic than
(A) ethanol
(B) methanol
(C) o-nitrophenol
(D) p-cresol.
Solution : Nitro is electron withdrawing group
 (C)
Set : |A - 32|B - |
32.
Schiff’s reagent is
(A) magenta solution decolourised with sulphurous acid
(B) magenta solution decolourised with chlorine
(C) ammonical manganese sulphate solution
(D) ammonical manganese chloride solution
Solution : Schiff’s reagent is the colourless solution obtained by passing SO2 gas in pink
solution of magenta dye (rosanaline hydrochloride).
Set : |A - 33|B - |
33.
is the final product obtained when one of the following is reacted
with base
O
O
||
||
(A)
CH3 – C – CH2 – CH2 – CH2 – C – CH3
O
O
||
||
(B)
CH3 – CH2 – CH2 – C – CH2 – C – CH3
O
O
||
||
(C) CH3 – C – CH2 – CH2 – CH2 – C – CH2 – CH3
O
O
||
||
(D)
CH3 – C – CH2 – CH = CH – C – CH3
Solution :
O
||
C
O
||
C
CH3
CH2
O
C – CH3
CH2
CH2
CH2
CH
CH2
C – CH3

Alkali

CH2
1
CLASS XII
SET A
Set : |A - 34|B - |
34.
In the reaction
oxidi sin g
CH3 – CH = CH – CHO 
 CH3 – CH = CH – COOH, the oxidising agent can be
agent
(A) alkaline KMnO4
(C) Benedict’s solution
(B) acidified K2Cr2O7
(D) all of the above
Benedict (Cu  ) solution
Solution : CH3CH = CH – CHO  CH3 – CH = CH – COOH
Benedict solution (solution of CuSO4, sodium carbonate and sodium citrate) is
specific for oxidation of aldehydes.
Set : |A - 35|B - |
35.
(A)
(B)
(C)
(D)
Solution : Intermediate is
O
O
COOH which decarboxylates to
CH3
CH3
(when -keto acid is heated, CO2 is lost)
 (B)
Set : |A - 36|B - |
36.
For the reaction R – Br  R – O – N = O the suitable reagent is
(A) NaNO2 + HCl
Solution:
(B) HNO2
(C) AgNO2
(D) KNO2
R – Br  R – O – N = O
Alkyl halide
Alkyl nitrites
Alkyl nitrites are prepared by the action of alkyl halide
and potassium nitrite only but in case of silver nitrite the main product is nitro alkane
although a small amount of alkyl nitrite is also formed.
R – Br + KNO2  R – O – N = O + KBr
2
CLASS XII
SET A
(D)
Set : |A - 37|B - |
37.
The following method cannot be considered suitable for the preparation of alkyl
halide:
(A) Halogenation of alkane
(C) ROH and HX
Solution:
(B) ROH and PX3
(D) Alkene and HX
High temp. 400°
(1) R – H + X2 
R – X + HX
UV light
This is most drastic method as it required High temperature or catalyst CuCl2, FeCl3,
FeBr3 etc.
(2)
ROH + PX3  3RX + H3PO3
(3)
ROH + HX  R – X + H2O
H
|
(4) R – C = C – H + HX  R – C = C – H
|
|
|
|
X H
H H
(2), (3) & (4) are very feasible process
 (A)
Set : |A - 38|B - |
38.
In two separate experiments equal quantities of an alkyl halide, C4H9Cl were treated
at the same temperature with equal volume of 0.1 molar and 0.2 molar solutions of
NaOH respectively. In the two experiments, the times taken for the reaction of
exactly 50% of the alkyl halide were the same. The most likely structure of halide is:
(A) CH3CH2CH2CH2Cl
(C) (CH3)2 CHCH2Cl
Solution:
(B) CH3CH(Cl) CH2 CH3
(D) (CH3)3 CCl
CH3
|
(After
CH3 – CH2 – CH2 – CH2Cl  H3C – C rearrangement)
|
CH3
CH3
CH3
|
|
OH
CH3 – C
CH3 – C – Cl
|
|
CH3
CH3
As we already mentioned that Hydrolysis of 3° alkyl
halide is independent of nucleophilic concentration.
 (D)
Set : |A - 39|B - |
39.
Which of the following is an example of basic dye?
(A) Alizarine
(B) Indigo
(C) Malachite
3
(D) Orange – I
CLASS XII
SET A
Solution :
+
N (CH 3)2
C
N(CH 3)2
 (C)
Set : |A - 40|B - |
40.
The pKa of acetylsalicylic acid (aspirin) is 3.5 . The pH of gastric juice in human
stomach is about 2-3 and pH in the small intestine is about 8. Aspirin will be.
(A) Unionized in the small intestine and in the stomach
(B) Completely ionized in the stomach and almost unionized in the small intestine.
(C) Ionized in the stomach and almost unionized in the small intestine
(D) Ionised in the small intestine and almost unionised in the stomach
Solution : More ionized in basic medium and less ionized in acidic medium, common ion
effect
 (D)
Set : |A - 41|B - |
41.
On subjecting mesityl oxide to the iodoform reaction, one of the products is the
sodium salt of an organic acid. Which acid is obtained?
(A) (CH3)2C=CH—CH2COOH
(C) (CH3)2C=CH—COOH
Solution:
(B) (CH3)2CH—COOH
(D) (CH3)2C=CH—CO—COOH
O
O
CH3
CH3
NaOH / I2
C=CH—C—CH3 
 
CH3
C=CH—C—OH + CHI3
CH3
mesityl oxide
 (C)
Set : |A -42|B - |
42.
The quantity of electricity needed to electrolyse completely 1 M solution of CuSO4,
Bi2(SO4)3, AlCl3 and AgNO3 each will be
(A) 2F, 6F, 3F, and 1F respectively
(C)2F, 6F, 1F and 3F respectively
Solution:
(B) 6F, 2F, 3F and 1F respectively
(D) 6F, 2F, 1F and 3F respectively
CuSO4 + 2e–  Cu + SO 4 
Bi2(SO4)3 + 6e–  2Bi + 3SO 4 
AlCl3 + 3e–  Al + 3Cl–
4
CLASS XII
SET A
AgNO3 + e–  Ag + NO 3
Since, in 1M CuSO4 solution, 1M Bi2(SO4)3 solution, 1M AlCl3
solution and 1M AgNO3 solution, 2 mole electron, 6 mole electron, 3 mole electron are
needed to deposit Cu, Bi, Al and Ag at the cathode respectively.
But one mole electron = 1F electricity
That’s why number of Faraday required to deposit 1M of each
CuSO4, Bi2(SO4)3, AlCl3 and AgNO3 solution are 2F, 6F, 3F and F respectively
 (A)
Set : |A - 43|B - |
43.
Emf of the cell Ni | Ni2+ (0.1 M) || Au3+ (1.0 M) | Au will be
1
(A) 1.75 V
(B) + 1.7795 V
(C) + 0.7795 V
(D) – 1.7795 V
Cell reaction : 3Ni + 2Au+3  3Ni+2 + 2 Au
Solution:
E cell  E 0cell 
0.0591
[Ni 2 ] 3
log
6
[ Au3 ] 2
= (0.25 + 1.5) –
= 1.75 –
0.0591
(0.1) 3
log
6
(1) 2
0.0591
log(. 1)
2
= 1.75 + 0.295
 (B)
= + 1.7795 V
Set : |A - 44|B - |
44.
A first order reaction is 87.5% complete in an hour. The rate constant of the reaction
is
(A) 0.0346 min–1
Solution:
(B) 0.0693 h–1
(C) 0.0693 min–1
(D) 0.0346 h–1
For 87.5% change there will be three half-changes. Hence half-time of the
reaction will be
1
60
h =
= 20 min, since half-time of a first order reaction is a constant
3
3
independent of initial concentration.
20 min
20 min
20 min
100% 

 50% 

 25% 

 12.5%
(un reacted)
0.693 0.693
= 0.0346 min–1
k

t1 / 2
20
 (A)
Set : |A - 45|B - |
45.
The vapour pressure of benzene at 80°C is lowered by 10 mm by dissolving 2g of a
non-volatile substance in 78g of benzene. The vapour pressure of pure benzene at
80°C is 750 mm. The molecular weight of the substance will be:
(A) 15
(B) 150
(C) 1500
5
(D) 148
CLASS XII
Solution:
SET A
P0  Ps
w M

Ps
m W
10
2  78

750  10  m  78
 m = 148;
Hint:
[m comes 150 if formula
P0  Ps
w M
is used. But this is only for

Po
m W
dilute solutions.]
(B)
Set : |A - 46|B - |
46.
Y g of non - volatile organic substance of molecular mass M is dissolved in 250 g
benzene. Molal elevation constant of benzene is Kb. Elevation in its boiling point is
given by
(A)
Solution:
M
KbY
4K b Y
M
1000  K b  Y
4K b Y
T =
=
250  M
M
(B)
(C)
KbY
4M
(D)
KbY
M
 (B)
Paragraph for Questions Nos. 47, 48 & 49
The noble gases have closed-shell electronic configuration and are monoatomic gases
under normal conditions. The low boiling points of the lighter noble gases are due to weak
dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to series of compounds with oxidation
numbers +2, +4 and +6. eF reacts violently with water to given e . The compounds of
xenon exhibit rich stereochemistry and their geometries can be deduced considering the
total number of electron pairs in the valance shell.
Set : |A -47 |B - |
47.
Argon is used in are welding because of its
(A) low reactivity with metal
(C) flammability
(B) ability to lower the melting point of metal
(D) high calorific value
Solution: Argon, being a noble gas, will not react with the metals, thus, can be used in are
welding.
Set : |A - 48|B - |
48.
The structure of e is
(A) linear
Solution:
(B) planar
(C) pyramidal
6
(D) T-shaped
CLASS XII
SET A
Set : |A - 49|B - |
49.
eF and eF are expected to be
(A) oxidizing
(B) reducing
(C) unreactive
(D) strongly basic
Solution:
Set : |A - 50|B - |
50.
In the following reaction sequence in aqueous solution, the species X, Y and Z,
respectively, are
S
→
(A) [ g(S
(C) [ g(S
clear solution
) ]
) ]
g S
g S
→
white precipitate
(B) [ g(S
(D) [ g(S
g S
g
→
blac precipitate
) ]
) ]
g S
g S
g S
g
Solution:
Paragraph for Questions Nos. 51,52 & 53
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is
described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons
and three atoms are sandwiched in between them. A space-filling model of this structure,
called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface
surrounded in the same plane by six identical spheres as closely as possible. There spheres
are then placed over the first layer so that they touch each other and represent the second
layer. Each one of these three spheres touches three spheres of the bottom layer. Finally,
the second layer is covered with third layer that is identical to the bottom layer in relative
position ssume radius of every sphere to be ‘r’
Answer the following questions –
Set : |A - 51|B - |
51.
The number of atoms in the HCP unit cell is
(A) 4
(B) 6
(C) 12
Solution:
7
(D) 17
CLASS XII
SET A
Set : |A - 52|B - |
52.
The volume of this HCP unit cell is –
(A)
√ r
(B) 1 √ r
(C) 1 √ r
(D)
(C) 32%
(D) 26%
Solution:
√
r
Set : |A - 53|B - |
53.
The empty space in this HCP unit cell is
(A) 74%
(B) 47.6 %
Solution:
Set : |A - 54|B - |
54.
For a dilute solution containing 2.5 g of a non-volatile non electrolyte solute in 100 g
of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming
concentration of solute is much lower than the concentration of solvent, the vapour
pressure (mm of Hg) of the solution is (take
g mol )
(A) 724
Solution:
(B) 740
(C) 736
(D) 718
Paragraph for Questions Nos. 55, 56 & 57
In the following reaction sequence, product I, J and L and formed. K represents a reagent.
1 Mg ether
1 aB
C
ex
ynal
→
Br
8
CLASS XII
SET A
Set : |A - 55|B - |
55.
The structure of the product I is –
(A)
(C)
(B)
(D)
Solution:
Set : |A - 56|B - |
56.
The structures of compound J and K, respectively, are
(A)
(B)
(C)
(D)
Solution:
9
CLASS XII
SET A
Set : |A - 57|B - |
57.
The structure of product L is
(A)
(B)
(C)
(D)
Solution:
Paragraph for Questions Nos. 58, 59 & 60
The coordination number of i is 4.
(cyano complex)
iCl
C (excess)
iCl
Conc Cl(excess)
B (chloro complex)
Set : |A - 58|B - |
58.
The IUPAC name of A and B are
(A) Potassium tetracyanonickelate (II), potassium tetrchloronickelate (II)
(B) Teracyanopotassiumnickelate (II), tetrachloropotassiumnickelate (II)
(C) Teracyanonickel (II), tetrachloronickel (II)
(D) Potassium tetracyanonickel (II), potassium tetrachloronickel (II)
Solution:
Set : |A - 59|B - |
59.
Predict the magnetic nature of A and B
(A) Both are diamagnetic
(B) A is diamagnetic and B is paramagnetic with one unpaired electrons
(C) A is diamagnetic and B is paramagnetic with two unpaired electrons
(D) Both are paramagnetic
10
CLASS XII
SET A
Solution:
Set : |A - 60|B - |
60.
The hybridization of A and B are
(A) dsp sp
(B) sp sp
(C) dsp dsp
Solution: (a) Discussed above.
11
(D) sp d d sp