Target IIT-JEE 2017
Class: 10 + 2
PAPER – 1 (8.1.2016)
Section – A (Single Correct) Negative Marking [-1]
This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out
of which ONLY ONE is correct.
1.
2.
The compounds of sulphur obtained by reaction of sulphur and conc. hot NaOH are
a. Na2S & Na2S2O3
b. Na2SO4, Na2S
c. Na2S4O6, Na2S
d. Na2S2O3, Na2SO3
A
Compound M on heating leaves no residue and N2 is also not obtained. Aqueous solution of M reacts
with alkali and an alkaline gas is evolved. The resulting solution is treated with Al in alkaline medium to
liberate same gas that produces a deep blue solution with Cu(NO3)2 solution. M is
a. NH4NO3
A
b. NH4NO2
c. NH4Cl
d. NH4Br

Sol. NH 4NO 3  N2O  2H2O ; NO 3 ion on Reduction give NH3; NH 4 ion with alkali give NH3
3. Both chlorine and bromine can be produced in the laboratory by reacting the halides with manganese
(IV) oxide. If the following equation is balanced in an acidic solution, then the MnO2 is acting as ______
and the coefficient in front of H+ is ________
MnO2(s) + ___Cl-(aq) + __ H+(aq) 
__ Mn2+(aq) + __ Cl2(g) + ___H2O(l)
4.
a. oxidizing agent, 2
b. oxidizing agent, 3
c. oxidizing agent, 4
d. reducing agent, 3
C
Compounds A and B are treated with dil. HCl separately. The gases liberated are Y and Z respectively.
Y turns acidified dichromate paper green while Z turns lead acetate paper black. The compounds A
and B are respectively:
a. Na2SO3 and Na2S
c. Na2S and Na2SO3
A
dil. Acid
b. NaCl and Na2CO3
d. Na2SO3 and Na2SO4
dil. Acid
Sol. Na 2SO3  SO 2; Na 2S  H2S
5. Hemiacetals are usually unstable whereas acetals are stable. Which has the most stable hemiacetal?
b. C6H5 – CHO
a.
O
H
c. CH 3  C CH 2  CH 3
d. HO
||
O
O
D
H
O
OH
Sol. HO
O
If a hemiacetal is five or six membered cyclic. They are stable.
Coordinated by: Dr. Sangeeta Khanna, Ph.D (Chemistry), A.P. Singh (Maths), Shiv. R. Goel (Physics)
D:\Important Data\2016\+2\Intelliquest\Test-3 8.1.2017\+2 Paper - 1 8.1.2017.docx
1
Target IIT-JEE 2017
Class: 10 + 2
SECTION – B (More than One Answer Type) No Negative Marking
This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)
out of which One or More than one answer may be correct.
5 × 5 = 25
1.
Which of the following is(are) correctly matched?
a. Copper-Bessemer converter
b. Iron – Blast furnance
c. Chromium – Aluminothermic process
d. Tin – Electroytic reduction
A,B,C
Sol. (A) Self reduction takes place in Bessemer converter.
(B) Slag formation & reduction of haematite to iron take place in blast furnance at different temperatures.
(C) Cr2O3 + Al  Al2O3 + 2Cr.
This is called aluminothermic process.
(D) Tin is obtained by carbon reduction of SnO2 (cassiterite ore)
2. Which of the following is/are true?
3.
a. Among halide ions, iodide ion is the most powerful reducing agent
b. Fluorine is the only halogen which does not show a variable oxidation state
c. HOCl is a stronger acid than HOBr
d. H4XeF6 can oxidise HF to F2
A,B,C,D
In the reaction given below
O
+ H2N – OH
Conc. H2SO4
X
Expected product(s) is/are:
O
O
H
a.
N
H
H
b.
H
N
N
c.
N
d.
O
A,C
O
O
Sol.
4.
NOH
Beckmann’s
Rearrangement
N
NH
H
+
O
The complex [Fe(H2O)5NO]+2 is formed in the brown ring test for nitrates when freshly prepared FeSO 4
solution is added to aq. Solution of NO3– followed by addition of Conc. H2SO4. Select correct statement
about this complex.
a. Colour change is due to charge transfer
b. Its iron is in +1 oxidation state and nitrosyl as NO+
c. Its magnetic moment 3.87 BM confirm 3 unpaired e– in Fe
d. Fe is present in +2 state
A,B,C
Sol. Fe is present in +1 oxidation state
Dr. Sangeeta Khanna Ph.D
2
CHEMISTRY COACHING CIRCLE
D:\Important Data\2016\+2\Intelliquest\Test-3 8.1.2017\+2 Paper - 1 8.1.2017.docx
Target IIT-JEE 2017
Class: 10 + 2
5.
How many of the following pair are stronger than benzoic acid
(a) HCOOH, CH3COOH, CH2  COOH
|
Cl
COOH
COOH
;
;
SH
(c)
OH
Cl
CH3
(b)
COOH
COOH
;
HCOOH ;
NO2
COOH
HO
(d) CH2 – COOH ;
COOH
Cl
OH
;
COOH
B,D
SECTION – C (Paragraph Type) Negative Marking [-1]
This Section contains 6 questions. Each of these questions has four choices A), B), C) and D) out of
which ONLY ONE is correct.
Comprehension 1
Go through the following graph and answer the following questions.
0
– 100
– 200
4Cu + O2 2Cu2O
– 300
2FeO
O2 
2Fe +
– 400
– 500
of O2
–1
G /kJ mol
2C +
 2ZnO
2Zn + O2
– 600
C + O2  CO2
O2 
2CO
– 700
– 800
A
– 900
l O3
 2/3A 2
– 1000
O2
4/3Al +
– 1100

2Mg + O2
2MgO
– 1200
0°C
400°C
800°C
1200°C
1600°C
2000°C
Temperature
Dr. Sangeeta Khanna Ph.D
3
CHEMISTRY COACHING CIRCLE
D:\Important Data\2016\+2\Intelliquest\Test-3 8.1.2017\+2 Paper - 1 8.1.2017.docx
Target IIT-JEE 2017
Class: 10 + 2
1.
At what approximate temperature, zinc and carbon have equal affinity for oxygen.
a. 900°C
b. 1500°C
c. 500°C
d. 1200°C
A
Sol. At about 1000°C, lines cross and thus at this temperature zinc and carbon have equal affinity for
oxygen
2. To make the following reduction process spontaneous, temperature should be:
3MgO + 2Al  Al2O3 + 3Mg
a. < 1000°C
b. > 1200°C
c. < 500°C
d. > 500° but < 1200°C
B
Sol. This will be above a temperature when two curves intersect that is > 1200°C.
3. Which of the following statement is incorrect?
a.
b.
c.
Reduction of calcined / roasted haematite ore to pig iron in blast furnace takes place in the lower
temperature range and in the higher temperature range by CO and C respectively.
The reduction of zinc oxide using coke takes place at higher temperature than that in case of
copper
It is quite easy to reduce oxide ores of copper directly to the metal by heating with coke after 500600 K
CO can act as a Reducing agent at high temperature.
d.
D
Sol. All other statements are true
Comprehension -2
A neutral organic compound A(C10H20O2) neither reduces Tollen’s reagent nor forms precipitate with 2,
4-dinitrophenyl hydrazine, but can be resolved into enantiomers. A on acid hydrolysis forms two
compounds B and C, both are chiral. C neither reduces Fehling’s solution nor forms iodoform with
alkaline iodine solution. C on oxidation with CrO3/HCl/pyridine forms D which is still resolvable into
enantiomers. D on further treatment with aqueous (RO)3Al solution gives back A and on oxidation D
give B.
4.
How many different stereoisomers exist for A?
a. 2
B
Sol.
b. 4
CH3 O
c. 5
CH3
CH3 – CH2 – CH – C – O – CH2 – CH – CH2 – CH3
 (A)

d. 6
CH3
HOH
CH3
CH3 – CH2 – CH – COOH + CH3 – CH2 – CH – CH2OH
(B)
(C)
PCC
CH3
Oxidation
CH2 – CH2 – CH – CHO
(D)
Tishenko
Reaction
CH3
CH3 – CH2 – CH – COOCH2 – CH – CH2CH3
CH3
ester
The above ester has two chiral carbons and no symmetry hence, four stereoisomers.
Dr. Sangeeta Khanna Ph.D
4
CHEMISTRY COACHING CIRCLE
D:\Important Data\2016\+2\Intelliquest\Test-3 8.1.2017\+2 Paper - 1 8.1.2017.docx
Target IIT-JEE 2017
Class: 10 + 2
5.
What is true regarding D?
a. It forms a yellow precipitate with KOH/I2
b. It forms an orange precipitate with 2, 4 – dinitrophenyl hydrazine
c. D is less reactive in aldol condensation & give cannizzaro Reaction inspite of having -hydrogen
d. On treatment with NaHSO3, it forms salt which cannot be separated by fractional crystallization
C
Sol. Due to formation of less stable carbanion Aldol Reaction is slow
6. The correct statement regarding the following reaction is:
H SO
4
D  Cl 2 2 
   chlorinati on
a. Product is achiral
c. Product is a pair of diastereomer
B
b. Product is a racemic mixture
d. No  - halogenation occur with D
CH3
Sol. D + Cl2
+
H
CH3CH2 – C – CHO (Racemic)
Cl
SECTION – D (Integer Type) No Negative Marking
This Section contains 5 questions. The answer to each question is a single digit integer ranging
from 0 to 9.
1. What is the co-ordination number of metals in alums.
Sol. Six; In alums every metal is surrounded by six H2O molecules. 24 water for 4 metal ion
2. How many isomeric ketones can be formulated by C6H12O?
Sol. 7
O
O
||
||
CH 3  C CH 2  CH 2  CH 2  CH 3 ;
O CH3
CH 3  CH 2  C CH 2  CH 2  CH 3
CH3
O
CH 3  C  CH  CH 2  CH 3
O CH 3
CH 3  C CH 2  CH  CH 3
O CH3
||
|
||
|
||
(d  ) ;
|
||
CH 3  C  C  CH 3 ;
|
CH 3  CH 2  C  CH  CH 3
|
CH 3
OH
CO2CH3
C(CH3)2
x CH3MgI
+
H
3.
CO2CH3
C(CH3)2
OH
Number of moles (x) of Grignard reagent consumed in the above reaction is:
Sol. 4
4. In the Lewis structure for the BrF4- ion, how many lone pairs of electrons are placed around the central
atom?
Sol. 2
it is sp3d2;
5.
F
F
F
Br
F
(IO 3 )
Iodate ions
can be reduced to iodine by iodide ions in acidic medium
How many moles of iodine are produced for every mole of iodate ions consumed in the reaction?
Sol. 3; 5I– + IO3- + 6H+  3I2 + 3H2O
Dr. Sangeeta Khanna Ph.D
5
CHEMISTRY COACHING CIRCLE
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