Target IIT-JEE 2017
Class: 10 + 2
PAPER – 2 (8.1.2016)
Section – A (Single Answer Correct Type) 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.
S1: [Co(EDTA)]– has two optical isomers.
S2: [Co(NH3)4(NO2)2]+ show linkage isomers.
S3: [For [PtPy(NH3)(NO2) ClBr], theoretically fifteen different geometrical isomers are possible.
S4: [Cr(H2O)4Cl2]Cl2.2H2O can show hydrate as well as ionization isomerism.
a. F T T F
b. F F T F
c. T F T F
d. T T T F
D
Sol. S1, S2 and S3 are correct statements.
S4: It is an example of only hydrate isomerism not ionization isomerism because such isomerism
occurs owing to exchange of ions between coordination and ionization spheres.
2. A hydride (X) of group 15 element is distinctly basic and has unexpectably high boiling point. It reacts
with NaOCl to gives another hydride (Y), which is a strong reducing agent and is used in organic
analysis. X and Y are:
a. PH3, P2H4
b. NH3, N2H4
c. AsH3, As2H4
d. NH3, NH4Cl
B
Sol. 2NH3 + NaOCl  NaCl + H2O + N2H4
3. Fluorine is above chlorine in Group VII. Which one of the following statements about fluorine and
fluorine compounds is correct?
4.
a. The electronegativity of fluorine is less than that of the other Group VII elements.
b. Fluorine is the most powerful oxidizing agent of the Group VII elements.
c. Hydrogen fluoride is the strongest acid of all the Group VII hydrides.
d. Hydrogen fluoride is easily decomposed into its elements.
B
A hydrocarbon X has molecular formula C5H10X on treatment with B2H6 in H2O2/NaOH gives an
optically active C5H12O which on treatment with CrO3/HCl/pyridine gives C5H10O which is still chiral.
Which of the following can be a product of reductive ozonolysis of X?
O
O
a. CH3 – C – CH3
B
CH3
b. CH3 – CH2 – C – CH3
O
c. CH3 – CH – CHO
d.
H
H
Sol. H3C – CH2 – C= CH2
B2H6
–
H2O2/HO
CH3CH2 – C – CH2OH
CH3
5.
PCC
CH3CH2 – CH – CHO
CH3
CH3
Chiral
Arrange the following in the increasing order of stability of their most stable enol.
O
I.
O
O
II.
C2H5O
O
OC2H5
III.
IV.
O
O
a. I < II < III < IV
b. IV < III < II < I
c. II < I < IV < III
d. III < IV < II < I
B
Sol. Diketo (I) forms highest enol content due to stablisation of enol by intermolecular H – bonding. Electron
donating resonance effect by ester group slightly decreases enol content.
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 - 2 8.1.2017.docx
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Target IIT-JEE 2017
Class: 10 + 2
SECTION – B (Paragraph Type)
This Section contains 2 paragraph. Each of these questions has four choices A), B), C) and D) more
than one may be correct
6 × 4 = 24 Marks
Passage – 1
When aldehydes lacking -hydrogen is treated with concentrated solution of a strong base, cannizaro
reaction take place. In this reaction, one molecule of aldehyde is oxidized while other is reduced. The
widely accepted mechanism is:
O
O
Fast
Step I: A – C – H + HO–
A–C–H
OH
(I)
O
O
Step II: A – C – H + A – C – H
 A – C – OH + A – C – OH
Fast
hy dride transf er
(I) OH
1.
O
O
O
Slow
H
–
A – C – O + A – C – OH
H
H
Which of the following observed fact does not establishes the correctness of the above mechanism?
a. If A – COD is used, reaction occur at same rate provided all other conditions are same
b. In mixed Cannizaro reaction, if one aldehyde is H2CO, it is always oxidised
c. If reaction is carried out in D2O in the presence of NaOD, some C – D bond will be formed
d. If reaction is carried out in D2O in the presence of NaOD, no C – D bond formation occur
A,B,C
Sol. When reaction is carried out in D2O, no C–D bond formation occur. It establishes the correctness of
above mechanism that reaction proceed by H – (hydride) transfer in the slow rate determining step.
2. What is the rate law derived from the above mechanism?
a. R = k [A – CHO] [HO–]2
c. R = k [A – CHO]2 [HO–]
C

O 


|
Sol. Rate = k1 [ArCHO]  Ar  C  H


|

OH 


b. R = k [ A – CHO]2 [HO–]2
d. R = k [A – CHO]2

Intermediate ( I )
[I]
The intermediate “I” can be determined as : K 

[HO ][ ACHO ]
3.
Substituting [I] in Eq (i) gives
R = k1 K[ACHO]2 [HO–] = k [ACHO]2 [OH–]
In the Cannizaro reaction mentioned below:
O
H – C – H + NaOD
D O
H O
2 3 

The possible product(s) is/are:
a. CH3OH
A,B,C
b. CH3OD
c. HCOOD
d. H2CDOH
Coordinated by: Dr. Sangeeta Khanna, Ph.D (Chemistry), A.P. Singh (Maths), Shiv. R. Goel (Physics)
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2
Target IIT-JEE 2017
Class: 10 + 2
–
O
Sol. H – C – H + OD
O
O
H–C–H
CH2O
O
H – C – OD + CH3O–
–
H – C – O + CH3OD
OD
or CH3OH + H2O
–
+
CH3O + H3O
However, C – D bond is not formed in this reaction.
Paragraph – 2
The zero (or 18) group of periodic table consists of six gaseous elements namely He, Ne, Ar, Kr, Xe,
and Rn. On account of their highly stable ns2p6 configuration in the valence shell, these gases have
little tendency to undergo any reaction, hence they were called „inert gases‟. However, due to finding of
number of reaction of these elements, these are correctly called „noble gases‟.
4.
Xe forms more number of compounds than the other noble gases
b. due to its higher electron affinity
d. due to zero electronegativity
5.
a. due to its lower ionization potential
c. due to its electronic structure
A
Which of the following is true for Xenone fluoride
a. These are good oxidising agent
c. They are fluorinating agent
A,B,C
XeF4 + O2F2  [X] + O2
Select the correct statement(s) for [X] & O2F2
b. They are readily hydrolysed
d. There are good reducing agent
6.
a. Partial hydrolysis of [X] gives XeOF4 as one of the product.
b. [X] is sp3d3 hybridised
c. [X] reacts with SiO2 to form SiF4
d. O – O bond length of O2F2 is less than H2O2
A,B,C,D
Sol. XeF4 + O2F2  XeF6  O2
(x)
(A) XeF6 + H2O  XeOF4 + 2HF
XeOF4 + H2O  XeO2F2 + 2HF
XeO2F2 + H2O  XeO3 + 2HF
(C) 2XeF6 + SiO2  2XeOF4 + SiF4
SECTION – C (Matrix Type) No Negative Marking
This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and
five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching
with one or more statement(s) given in Column II.
8 × 2 = 16 Marks
1. Matrix Match
List – I
List - II
A. Detection of CO
P. aq. KI
B. Estimation of O3
Q. I2O5
C. Detection of halogen (X2)
R. Starch paper
D. Estimation of I2
S. Hypo solution
Sol. A  q; B  p, s; C  r, s, may be p; D  s
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 - 2 8.1.2017.docx
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Target IIT-JEE 2017
Class: 10 + 2
2.
Match the Ka value:
A.
List – I
Benzoic acid
B.
O2N
C.
Cl
D.
H3CO
P.
COOH
List - II
3.3 × 10–5
Q. 10.2 × 10– 5
COOH
COOH
R.
30.6 × 10–5
S.
6.4 × 10–5
Sol. A(s); B(r); C(q); D(p);
The acids (A and C) are stronger than benzoic acid (A) because electron withdrawing groups increases
the acidic nature. The acid (D) is weaker acids than benzoic acid (A) because electron releasing
groups decreases the acidic nature. B is stronger acid (higher Ka value) than C since electron
withdrawing effect of NO2 is more than Cl. D is electron donating power of CH3O is greater than CH3
group. Thus the acidic strength order is:
D (Ka = 3.3 × 10–5) < A(6.4 × 10–5) < C(10.2 × 10–5) < B(30.6 × 10–5)
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.
5 × 4 = 20 Marks
1.
Which of the following complexes are diamagnetic?
[Pt (NH 3 ) 4 ] 2  [Co(SCN) 4 ] 2  [Cu(en) 2 ] 2  [HgI 4 ] 2 
square planar
(i)
[NiCl 4 ]2
( v ii)
tetrahedral
(ii)
square planar
(iii)
[Ni(CO )4 ]
[PtCl 2 (NH 3 )2 ]
( v iii)
(ix )
K 3 [Cu(CN )4 ]
[Ni(gly )2 ]
(v )
( v i)
tetrahedral
(iv )
Sol. 5
(i), (iv), (v), (viii), (ix)
(i)  dsp2 ; (ii) sp3 ; (iii) dsp2 (1unpaired e- ); (iv) sp3 ; (v) d10 (sp3 ); (vi) d8 (sp3 ); (vii) d8 (sp3 ), (viii) d10 (sp3 )
( d8 )
(d7 )
(d10 )
2.
Which amongst the following metal carbonyls are inner orbital complexes with paramagnetic property?
(I) Ni(CO)4; (II) Fe(CO)5;
(III) [V (CO)6]–
(IV) Cr (CO)6
(V) [Mn(CO)5]– (VI) [V(CO)6]
Sol. 1
Ni (CO)4 sp3 diamagnetic
Fe(CO)5 dsp3 diamagnetic
[V(CO)6]– d2sp3 diamagnetic [V(CO)6] = d5; d2sp3 have 1 unpaired e–.
Cr(CO)6 d2sp3 diamagnetic
[Mn(CO)5]– = d2sp3 & diamagnetic
3. Titanium shows magnetic moment of 1.73 BM in its compound. What is the oxidation state of titanium
in the compound?
Sol. 3
 = 1.73 i.e. one unpaired electron, Ti (E.C.) = [Ar]18 3d2, 4s2, so it will lose three electrons to have one
unpaired electron and thus the oxidation state of titanium is +3.
4. How many of the following cannot be used for the preparation of H2?
a. Zn + HCl (dil.) 
b. NaH + H2O 
c. Zn + HNO3(dil.) 
d. Fe  H2O 
e. Zn  H2SO 4 
f. Zn  H2SO 4
g. CaH2 + H2O 
h. Cu  HNO 3 
steam
dil.
Conc.
dil.
Coordinated by: Dr. Sangeeta Khanna, Ph.D (Chemistry), A.P. Singh (Maths), Shiv. R. Goel (Physics)
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Class: 10 + 2
Target IIT-JEE 2017
Sol. 3
c, f, h
5. The value of x in the complex HxCo(CO)4, on the basis of E.A.N. rule is (At. No. Co = 27):
Sol. At. No. of Co = 27
At. No. of next noble gas = 36 (Kr)
 Number of electrons to be provided by H – atom
x =36 – (27 + 4 ×2)
= 36 – 35 = 1
Formula of complex compound is HCo(CO)4. Therefore value of x is 1.
Coordinated by: Dr. Sangeeta Khanna, Ph.D (Chemistry), A.P. Singh (Maths), Shiv. R. Goel (Physics)
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