jms-3 paper -2 sol

8/2/2019 Jms-3 Paper -2 Sol

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JMS 3 PAPER - 2

http://www.chemistrycrest.com/ Page 1

PAPER-2

Maximum Marks: 80

Question paper format and Marking scheme:

1. In Section I (Total Marks: 24), for each question you will be awarded 3 marks if you darken ONLY thebubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,minus one (1) mark will be awarded.

2. In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darken ALL thebubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. There are no negativemarks in this section.

3. In Section III (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLY thebubble corresponding to the correct answer and zero marks otherwise There are no negative marks in thissection.

4. In Section IV (Total Marks: 16), for each question you will be awarded 2 marks for each row in whichyou have darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marksotherwise. Thus each question in this section carries a maximum of 8 Marks. There are no negative marksin this section.


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JMS 3 PAPER - 2


SECTION I (Total Marks : 24)(Single Correct Answer Type)

This Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

1. The plot of compressibility factor (Z) versus pressure for the same gas given at different temperatures.

3T

2T

1T

4T

Ideal gas

P(atm)

Z

1.0

The correct order of temperatures is

A) T4 > T3 > T2 > T1 B) T2 > T3 > T1 > T4

C) T1 > T2 > T3 > T4 D) T3 > T2 > T1 > T4

Sol. (C)As the temperature increased, minima of the curve shifts upwards.

2. Dipole moment of the following molecules are in the order

C C=

Cl

H

CH3

H

(1)

C C=

Cl

NO2

H C3

(2)

Cl

C C=

ClH C3

(3)

Cl Cl

C C=

H

(4)

Cl

Cl H

(A) 1 > 4 > 2 > 3 (B) 2 > 3 > 1 > 4

(C) 3 > 2 > 4 > 1 (D) 2 > 4 > 1 > 3


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JMS 3 PAPER - 2


Sol. (B)

C C=

Cl

NO2

H C3

(2)

Cl

Cl

C C=

ClH C

3

Cl

(3)

(Net vector in thedirection of NO

2 )

(Net vector in thedirection of Cl )

- NO2 is a stronger - I group than Cl .For 4, dipole moment is zero due to cancellation of bond moments two C Cl bonds as they are

oriented in opposite direction.

3.

( ) ( ) ( )2 2 22Cl g Ba OH X aq BaCl H O+ + +

2 4 4 X H SO Y BaSO+ +

2 2365KY Z H O O

> + +

Y, Z respectively and magnetic behaviour exhibited by Z are

A)4 2, HClO ClO , diamagnetic B) 3 2, HClO ClO , paramagnetic

C)3 2, HClO Cl O , diamagnetic D) 4 2 7, HClO Cl O , paramagnetic

Sol. (B)

( ) ( )( )

2 3 2 22 22 5 6

X

Cl Ba OH Ba ClO BaCl H O+ + +

( )

( ) ( )

3 2 4 3 42

3 2 2 2365

2

2 2K

ZY

Ba ClO H SO HClO BaSO

HClO ClO H O O>

+ +

+ +

Z has odd e-.So, exhibits paramagnetism


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JMS 3 PAPER - 2


3CH

32CHCH

H

H

HO

Cl

3CH

32CHCH

H

H

HO

Cl

4. NH4Cl crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. Ifthe radius of the Cl ion is 181 pm, radius for NH4

+ ion is

(A) 154.1 pm (B) 92.6 pm(C) 366.3 pm (D) none of these

Sol. (A) In body centred cubic lattice, oppositely charged ions touch each other along the body diagonal.

2 2 3

3

2

3

2

3 387 335.152

181

335.15 181.0 154.15

c a

c a

a

c

r r a

r r a

a

pm

given r pm

r pm

+ =

+ =

=

= =

=

= =

5. The isomers given are

and

(A) Identical (B) Diastereomers

(C) Enantiomers (D) none of the above

Sol. (A)Configuration at both the chiral carbons is same. So, they are identical or homomers

If Configuration at both the chiral carbons is opposite, they are enatiomers. ( R, R & S,S )

If Configuration at one of the chiral carbons is opposite and same at the other carbon , they are

diastereomers. ( R, S & S,S or R, R & R ,S )

HO H

ClH

CH3

CH CH2 3

1

4

3

1

4

3

HO H

ClH

CH CH2 3

CH3

1

4

3

1

4

3

R

R

R

R

22


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JMS 3 PAPER - 2


6.

560 K 470 K

inert atmosphere under pressure Z X Y The three allotropes ' ', ' 'X Y and ' 'Z of phosphorus in the above changes are respectively,

A) white, black and red B) black, white and red

C) red, black and white D) red, violet and black

Sol. (A)X = White phosphorus

Y = Black phosphorus

Z = Red phosphorus

7. An optically active Aldose A upon treatment with 4NaBH is converted into an optically inactivealditol (B). A on Ruff degradation gives C whose alditol (D) is optically inactive. Ruff degradation ofC gives optically active D-glyceraldehyde A,B,C & D areA) -D- Glucopyranose, D-glucitol, D-ribose, ribitolB) D- ribose, ribitol, D-erythrose, erythritol

C) D glucopyranose, L-Dlucitol, L-ribose, ribitol

D) D fructofuranose, D-Manitol, D-ribose, ribitol

Sol. (B)


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8. Acidic hydrolysis is minimum in

(A)

O COR

(B)

O COR

NO2 (C)

O C R

NO2

|

O

(D)

O C R

OC H2 5

|

O

Sol. (D)In acid catalysed hydrolysis. the mechanism operated is

If G is Electron withdrawing group then it functions as a good leaving group.

SECTION II (Total Marks : 16)(Multiple Correct Answer(s) Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE OR MORE may be correct.

9. The major products of following reaction sequence are :

( )

H SO2 4

1 (I)

( )

LiAlH4

2 (II)

( )3

H(II)

(A) I = (B) I = (C) III = (D) III =


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Sol. (B, D)

H SO2 4

(1) Pinacol (PinacoloneRearrangment)

( )H O

2

(3) H

10. Mg3B2( ) HCl aq [X] + MgCl

[X] + H2O( ) HCl aq [Y] + H2

Regarding [X] and [Y], the correct option/s is/are

(A) [X] is BCl3 and [Y] is H3BO3

(B) [X] is B2H6 and [Y] is H3BO3

(C) [X] on reaction with air and [Y] on strong heating (red heat) give same compound.

(D) In [Y], B attains octet by accepting OH from water

Sol (B, C, D)

(A) Mg3B2( ) HCl aqMgCl2 + B2H6 [X], X is not BCl3, it B2H6

(B) B2H6 + 6H2O

2H3BO3 [Y] + H2 ,(C) Both [X] and [Y] gives B2O3 ; B2H6 + 3O2 B2O3 + 3H2O ;

H3BO3Red hot B2O3

(D) B(OH)3(aq) + 2H2O B(OH)4 + H3O

+ (aq)


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11. Which one of the following is/are optically active?

(A) (B)

H C3

H C3 Cl

Cl

NO2

H C

3

CH3

(C) (D)

C C C= =

H C3

H H

CH3

CH3

H

H

CH3

Br

Br

Sol. (C, D)

A) has no asymmetric carbons and also plane of symmetry.In tetra ortho-substituted compounds, the

planes of the rings are perpendicular to each other.

B) does not have any asymmetric carbon.

C) does not have any plane of symmetry.

D) has asymmetric carbon and has no plane symmetry.It can be represented as

CH3

CH3

Br

Br H

H


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12. The true statements among the following areA) Magnesium with dil. HNO3 produces hydrogen

B) Sodium nitrate on heating at 500C gives oxygen

C) Sodium carbonate liberates CO2 on heating at 100oC

D) Sodium when heated in excess of oxygen gives sodium monoxide.

Sol. (A, B)

3 3 2 22 ( ) Mg HNO Mg NO H + +

3 2 22 2 NaNO NaNO O +

Na2CO3 does not decompose on heating

2 2 22 ( ) Na O excess Na O+

SECTION-III (Total Marks : 24)(Integer Answer Type)

This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to9. The bubble corresponding to the correct answer is to be darkened in the ORS.

13. Amongst the following , no of molecules that are coloured due to charge transfer phenomenon are2

4CrO , 22 7Cr O

, 4MnO , Cu2O, ZnO, CuSO4, MnS

Sol. (4)

2

4CrO , 2

2 7Cr O ,

4MnO , Cu2O are coloured due to charge transfer phenomenon

14. The standard reduction potential of H2O2 is 1.776V and standard oxidation potential of H2O2 is 0.682V. The overall cell potential for disproportionation of H2O2 is almost equal to

Sol. (1)

For H2O2 disproportionation,

02 2 2 ; 1.7762 2 2

H O H e H O E V red+ + + =

( )02 2 ; 0.6822 2 2

H O O H e E oxi

+ + + =

0 0 02 2 2 ; 1.0942 2 2 2

H O H O O E E E V red ox

+ = + =


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15. Amongst the following , no of species having linear shape is

3 2 2I , ICl , BeCl , 52 6XeF , XeF , BrF , HC CH .Sol : (5)

16. In 2HI(g) H2 (g) + I2 (g), 2 mole HI, 4 mole H2 and 4 mole of I2 are found at equilibrium. If

some hydrogen is now introduced into the container such that at equilibrium concentration of HIand I2 become equal. Unit place of number of moles of H2 added is

Sol. (0)

( )2

4 4 42

eqk = =

2Hi H2 + I2

Av.eq. 2 4 4

After addition 2 4+x 4

Av. New eq. 2+2y 4+x-y 4-y

So, 2+2y=4-y y = 2/3

And eq

10 10

3 3

k 4 410 10

3 3

x +

= = =

So ,10 10 30

4 103 3 3

x = = =

moles

17. ( )3 7C H Br A gives ( )3 6C H B with alcoholic potash. (B) on oxidation gives ( )2 4 2C H O C + carbon

dioxide and water, (B) with hydrobromic acid gives (D), an isomer of (A). The difference in thepositions of Br in (A) and (D) is

Sol. (1)alc.KOH

CH CH CH Br CH CH=CH3 2 2 3 2

CH CHBrCH3 3

(B)

CH COOH+CO +H O3 2 2

(C)

(D)

2-bromopropane

1-bromopropane(A)

Difference in position of Br in (A) and (D) is = 2 1 = 1


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18. An electron is accelerated from rest and it has wavelength of 1.41 . Unit place of potentialthat should be dropped so that wavelength associated with electron becomes 1.73 .

Sol. (5)For an e-,

0150= AV

( )1 2

150V = 75

1.41V=

and( )

2 2

150V = 50

1.73V=

Hence potential should be dropped by 25 V.

SECTION-IV (Total Marks : 16)(Matrix-Match Type)

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and fivestatements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching withONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches withthe statements given q and r, then for the particular question, against statement B, darken the bubblescorresponding to q and r in the ORS.

19.

Column-I Column-II

(A) Acetone + CHCl3 (P) P total < PA + PB

(B) CS2 + Acetone (Q) Vmix > 0

(C) Phenol and aniline (R)

Hmix is not equal to 0

(D) Toluene + Benzene (S) Maximum boiling azeotropes

(T) A - B Interaction= A - A & B - B

Sol. A-P, R, S ; B-Q, R ; C-P, R ,S; D-T


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(A & C) Acetone + CHCl3 shows ve deviation

Phenol + aniline shows ve deviation

For a non ideal solution with ve deviation.

A - B Interaction > A - A & B B

P total < PA + PB

Vmix < 0

H

mix

< 0

It forms a max boiling azeotrope

(B) CS2 + Acetone shows + ve deviation

For a non ideal solution with + ve deviation.

A - B Interaction < A - A & B B

P total > PA + PB

Vmix > 0

Hmix

> 0

It forms a min boiling azeotrope

(D) Toluene + Benzene is an ideal solution

For an ideal solution,

A - B Interaction = A - A & B B

Ptotal

= PA

+ PB

Vmix = 0

Hmix

=0

Do not form azeotropes


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JMS 3 PAPER - 2


20. Match the reactions in column I with appropriate types of steps/reactive intermediate involved inthese reactions as given in column II.

Column-I Column-II

(A)CPBAm

(P) product contains one ring

(B)2Br

OH2

aq

NaOH

(Q) product contains two rings

(C)3 2

H C CH CH =

22ICH

etherCu/Zn

(R) product is an epoxide

(D) ( ) h2 2 23H C = CH CH CHN (S) product is a hydrocarbon

(T) A carbene (or) carbene likespecies participates

Sol. A - Q, R ; B - Q,R ; C - P, S, T ; D - Q, S,T

(A)

m- CPBA is m-Chloro per benzoic acid.

m CPBA O

(B)

Anti addition

Br2

H O2

Br

OH

aq.NaOH O


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JMS 3 PAPER - 2


(C)

Simmons-Smith reaction in which a carbene participates

CH I2 2H C CH CH H C CH CH

3 2 3 2 =

CH2

(D)

jms-3 paper -2 sol

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