JEE ADVANCED - 2016Paper - I

CHEMISTRYCODE - 6

SOLUTIONS

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PART II : CHEMISTRYSECTION 1 (Maximum Marks : 15)

• This section contains FIVE questions.• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is

correct.• For each question, darken the bubble corresponding to the correct option in the ORS.• For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 If none of the bubble is darkened.Negative Marks : -1 In all other cases.

19. P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal

thickness, dr, at a distance r from the nucleus. The volume of this shell is 24 .r dr The qualitative

sketch of the dependence of P on r is

A) B) C) D)

Sol: C

20. One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from

1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy if surroundings

( ) 1 in J KsurrS −∆ is (1L atm = 101.3J)

A) 5.763 B) 1.013 C) -1.013 D) -5.763

Sol: C

Resystem Re

vv

QS Q W

T∆ = = = − ( )P V P V= − − ∆ = ∆

( )system

3 2 1 3

300 300

P VS

T

+ −∆ +∆ = = = lit-atm

3 101.3251.013

300systemS J+ ×∆ = =

S 1.013 .surr J∆ =−

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21. Among ( ) [ ] ( ) [ ]2

4 3 2 3 6 2 2 24 4, , , , and ,Ni CO NiCl Co NH Cl Cl Na CoF Na O CsO

− the total

number of paramagnetic compounds is

A) 2 B) 3 C) 4 D) 5

Sol: B

[ ]2

4NiCl−

Oxidation state of Ni = + 2

[ ] 8 228 3 4Ni Ar d s

[ ]2 83Ne Ar d+

As ct is a weak ligand hence electrons do not pair

3d 4s 4p

2Ni Ar+

[ ]2

4NiCl−∴ is paramagnetic

[ ]3

6CoF−

O.S of Co = + 3

[ ] 7 227 3 4Co Ar d s

[ ]3 63 andCo Ar d F+ − is a weak ligand, electrons do not pair

3d 4s 4p

[ ]3

6CoF−∴ is paramagnetic

( )4Ni CO O.S of Ni = 0

[ ] 2 84 3 andNi Ar S d CO is a strong logand, hence electrons in ( )4Ni CO are paired

[ ] 2 84 3Ni Ar S d

3d 4s

In the presence of CO pairing takes place

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3d 4s 4p

Hence ( )4Ni CO is diamagnetic

( )3 24Co NH Cl Cl O.S of Co = + 3

E C of 3Co+

[ ] 63Ar d

3d 4s 4p

As four 3NH & 2Cl are there, but 3NH is a stronf logand, pairing of e− takes place

3d 4s 4p

Hence ( )3 24Co NH Cl Cl is diamagnetic

22 2 2Na O O −→

as per M.O.E.D. diagram

32o pσ

2 2x yp pπ π

2 2x yp pπ π

32 pσ

52 pO−

52 pO−

all electrons are paired, hence 2 2Na O is diamagnetic

2 2CsO O −→

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32o pσ

2 2x yp pπ π

2 2x yp pπ π

32 pσ

52 pO−42 p

O−

As one unpaired e− is present is 2 2,O CsO− is paramagnetic

22. The increasing order of atomic radii of the following Group 13 elements is

A) Al Ga In Tl< < < B) Ga Al In Tl< < <

C) Al In Ga Tl< < < D) Al Ga Tl In< < <Sol: B

Atomic radius of Gallium is less than aluminium due to inproper screaning of outer s- electron byinner d-electrons

23. On complete hydrogenation, natural rubber produces

A) ethylene-propylene copolymer B) vulcanised rubber

C) polypropylene D) polybutylene

Sol: A

SECTION 2 (Maximum Marks: 32)

This section contains EIGHT questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of thesefour option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened.

Partial Marks : + 1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.

zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : -2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three willresult in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) willresult in -2 marks, as a wrong option is also darkened.

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24. The product(s) of the following reaction sequence is(are)

A) B) C) D)

Sol: B

3

|| ||O O

COCH C O C CH− − −2NH 3

||O

NH C CH− −

||O

NH C− −||O

NH C− −

R

Br

Br

Br

Br

Br

Br

Br

2NH2/H H O+

2NH

2NH Cl⊕ −

1 /NaNO HCl

273 278k−

/Cu HBr

Br Br

3 /KBrO HBr

3CH 3CH

2NH Cl⊕ −

N

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25. The correct statement(s) about the following reaction sequence is (are)

( ) ( ) ( )32

3

/i)9 12 ii)

Cumene major minorCHCl NaOHO

H OC H +→ → +P Q R

2

NaOHPhCH Br

→Q S

A) R is steam volatile

B) Q gives dark violet coloration with 1% aqueous 3FeCl solution

C) S gives yellow precipitate with 2, 4-dinitrophenylhydrazine

D) S gives dark violet coloration with 1% aqueous 3FeCl solution

Sol: B, C

2O

/H HO+

OH

P

3 /CHCl NaOlOH OH

CHO

( )Q ( )R

CHOCHO

2O CH− −NaOH

2CH Br−

CHO

OH

QS

3 3CH CH CH− −

P - gives violet colour with 2FeCl

Q, R - gives violet colour with 3FeCl

S - Gives 2, 4 DNP test

Q, R - Gives 2, 4 DNP test

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26. The crystalline form of borax has

A) tetranuclear ( ) 2

4 5 4B O OH

− unit

B) all boron atoms in the same plane

C) equal number of 2 3 andsp sp hybridized boron atoms

D) one termuninal hydroxide per boron atom

Sol: A, C, D

27. The reagent(s) that can selectively precipitate 2S − from a mixture of 2 24 andS SO− − aqueous solution

is(are)

A) 2CuCl B) 2BaCl

C) ( )3 2Pb OOCCH D) ( )2 5

Na Fe CN NO Sol: A

A) 2CuCl givesCuS

ppt+ 4

Soluble

CuSO

B) 2BaCl givesBaS

ppt +4B a S O

p p t

C) ( )3 2Pb OOCH gives

PbS

ppt +4P b S O

p p t

D) ( )2 5Na Fe CN NO gives

Violet

Colour +

2 4

soluble

Na SO

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28. A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibitsupward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratioless than 1, the possible mode(s) of decay is(are)

A) ( )decay emission − − B) orbital or K-electron capture

C) neutron emission D) decay + − (positron emission)

Sol: B, D

29. Positive Tollen’s test is observed for

A) B) C) D)

Sol: A, B, C

2|

H

CH CH C O= − =2

TR CH CH COOH→ = −

C CH HO O

||OH

CH||O

C

O

TR

TR

TR

|| ||OO

C CH− − −

30. The compound(s) with TWO lone pairs of electrons on the central atom is(are)

A) 5BrF B) 3ClF C) 4XeF D) 4SF

Sol: B, C

BrF

F

FF

F

5BrF ( )1 loan pair Cl

F

F

F

F

3ClF ( )2 loan pair

XeF

F F

F

4XeF ( )2 loan pair S

F

F

F

4SF

F

( )1 loan pair

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31. According to the Arrhenius equation,

A) a high activation energy usually implies fast reaction.

B) rate constant increases with increase in temperature. This is due to a greater number ofcollisions whose energy exceeds the activation energy.

C) higher the magnitude of activation energy, stronger is the temperature dependence of the rateconstant.

D) the pre-exponential factor is a measure of the rate at with collisions occur, irrespective of theirenergy.

Sol: B, C, D

SECTION 3 (Maximum Marks: 15)

• This section contains FIVE questions.

• The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, bothinclusive.

• For each question, darken the bubble corresponding to the correct integer in the ORS.

• For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.

Zero marks : 0 In all other cases.

32. In the following monobromination reaction, the number of possible chiral products is

Sol: 7, 2

Brominatian can occur on 1, 3, 4 or 6th carbon.

2 5C H2 5C H

3CH

H

H H

HH

H

H

HBr

Br

Br

Br

Br

Br

( )3

1

CH

( )3

2

CH

(3

3

CH

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2 2 3CH CH CH 2 2 2CH CH CH Br

H HBr Br

( )2

4

CH Br

( )3

5

CH

Chiral Chiral

33. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is same as its

molarity. Density of this solution at 298 K is 32.0 g cm .− The ratio of the molecular weight of the

solute and solvent, ,solute

solvent

MW

MW

is

Sol: 9

1000

1000 . of solute

Mm

d M M wt=

− ×

as m =M

1000

1000 2 . of solute

mm

m M wt

×=× − ×

2000 . . of solute 1000m M wt− × =

2000 1000 . . of solutemx M wt− =

0.1 1000 . . of solute1000

0.9 . . of solvent

M w

M wt

×=×

1 . of solute1

9 . of solvent

M wt

M wt= ×

. . of solute9

. . of solvent

M wt

M wt∴ =

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34. The number of geometric isomers possible for the complex [ ] ( )2 2 2 2 2CoL Cl L H NCH CH O− −= is

Sol: 4

M2CH O−

2CH

2NHCl

Cl

2NH2CH

2CHO

M

HC

2CH

Cl

Cl

2NH2CH

2CHO

O

212NH

( )1 ( )2

M M

2 2C CH H

2 2C CH H

O

O

O O

C Cl l

2

2

N

N

H

H2 2

C CH H

C Cl l

( )4

2 2C CH H

2 2N NH H

M

2CH

2CH

O

Cl

2NH

2CH

Cl

O

2NH

( )3

2CH

(5)

35. In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphateanions to produce X moles of a sulphur containing product. The magnitude of X is

Sol: 6

2 22 3 4 4 2 23 8 6 15 8 2S O MnO SO H O MnO OH− − − −+ → + + +

36. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. Theabsolute temperature of an ideal gas is increased 4 times and its pressure is incrased 2 times. As aresult, the diffusion coefficient of this gas increases x times. The value of x is

Sol: 4