jms-4 paper -1 solutions

8/2/2019 JMS-4 PAPER -1 SOLUTIONS

1/15

JMS 4 PAPER 1

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

PAPER-1Maximum Marks: 80

Question paper format and Marking scheme:

1. In Section I ( Total Marks: 21), for each question you will be awarded 3 marks if you darken ONLY the

bubble 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 the

bubble(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: 15), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

4. In Section IV (Total Marks: 28), for each question you will be awarded 4 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in thissection.


2/15

JMS 4 PAPER 1


SECTION I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

1. Correct acidic strength order of the following is

COOH

1

COOH

Cl

2

COOH

3

CH3

COOH

4

OMe

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

Sol. (C)

.

Except NH2, almost all osubtituted benzoic acids are strong than benzoic acid due to ORTHO

EFFECT.

Reason : Benzoic acid is resonance hybrid. Carbonyl gp is coplanar with the ring.

Ortho subtituent tends to prevent this coplanarity. Therefore, resonance is prevented.

So, the O atom of OH group has greater +ve charge resulting in increased acidic strength.

2. Match the following

ListI List II

(Sol) (Nature)

1) Starch sol P) Lyophilic

2) Gold sol Q) Lyophobic

3) Sulphur R) Multimolecular

4) Natural Rubber S) Macromolecular

T) Negatively charged

A) 1 - P,T; 2 - R,T; 3 - Q; 4 - S,T

B) 1 - P,S,T; 2 - Q,R,T; 3 - R; 4 - P,T

C) 1 - S,T; 2 - Q,T; 3 - R; 4 - P,T

D) 1 - P,Q,T; 2 - S,R,T; 3 - P,S; 4 - P,T


3/15

JMS 4 PAPER 1


Sol. (B)

Lyophilic :Lyophilic means liquid-lovingEg : gum, gelatine, starch, Rubber in a suitable liquid

Lyophobic : Lyophobic means liquid-hating.Eg : metals ( Gold sol) , their sulphides, etc.,

Multimolecular :On dissolution of a substance, large number of atoms or smaller molecules aggregatetogether to form species having size in the colloidal range , forming multimolecular colloids.

Eg : Gold sol , Sulphur sol

Macromolecular : Macromolecules in suitable solvents form solutions in which the size of the

macromolecules is in the colloidal range, forming macromolecular colloids.

Ex : starch, cellulose, proteins, enzymes ,polythene,nylon, polystyrene, synthetic rubber, etc.

Negatively charged : Metals, e.g., copper, silver, gold sols.

Metallic sulphides, e.g., As2S

3,Sb

2S

3, CdS sols.

Acid dye stuffs, e.g., eosin,congo red sols.Sols of starch, gum, gelatin,clay, charcoal, etc.

3. Compound (A),5 10 4C H O is oxidized by 2 2Br H O to the acid, 5 10 5C H O which readily forms a

lactone. (A) forms a triacetate with2Ac O and a hydrazone with 2PhNHNH . A is oxidized by 4HIO

when only one molecule of it is consumed. The structure of A is

A)CHO

(CHOH)3

CH2OH

B)

CHO

( )2

CHOH

2CH OH

CO

C) CHO

C

C

C

2CH OH

H

H OH

OH

O

D)

( )2HOH C CH CH OH CHO

2CH OH

Sol. (D)

A forms a hydrazone with2PhNHNH . So, it may be an aldehyde or ketone

A is oxidized by2 2Br H O to acid which forms a lctone , So it must be an aldehyde

A forms a triacetate. So, it must have three OH groups

A is oxidized by 4HIO when only one molecule of it is consumed. So it must contain either two OH

groups on adjacent carbons or OH or NH2 and carbonyl group

So, It must be D because, it has one OH and CHO on adjacent carbons.Remaining have more than

One such arrangements


4/15

JMS 4 PAPER 1


4. To 500 ml of 2 M impure H2SO4 sample, NaOH solution 1 M was slowly added & the following

plot was obtained. The percentage purity of H2SO4 sample and slope of the curve respectively are

vol. of NaOH added(L)

1.5

1.0

0.5

)Hof(mol +

A) 50%,1

3 B) 75%,

1

3

C) 75%, 1 D) none of these

Sol. (C)

Mole of H2SO4 =500

2 11000

x = , then Mole of H+ = 1 x 2 = 2

But, given mole of H+

= 1.5

Therefore, % purity =1.5

100 75%2

x =

Required mole of OH-= mole of H

+= 1.5 mole

Therefore, vol of NaOH =1.5

1.51

n

litM = =

Slope = tan 1350 = - 1

5. IP/EA ratio for bromine is10

3. Its electronegativety is 2.8. Its electron affinity on Mullikens scale

n kJ/mol is,

A) 81 B) 93 C) 162 D) 351

Sol. (D)

10

3

544

10

3

544

13

3 2.8544

351.5

IP EA

IP EAEN

EA EA

EN

EA

EA

=

+=

+=

=

=


5/15

JMS 4 PAPER 1


6. +

NaOD / D O

2 ?

(A) (B)

(C) (D)

Sol. (D)It is cannizzaro reaction

Ph CH OD + Ph C O

2

O

7. Match the following

ColumnI Column II

A) Wurtzite structure P) Anions occupy F.C.C lattice

B)Zinc Blende Structure Q) Closest packed structure

C)Antifluorite structure R) Anions occupy HCP lattice

D)Rocksalt structure S) Cations occupy half/alternate tetrahedral voids

A) A-R,S; B-P, R,S; C-S; D-P,

B) A-Q,R,S; B-P,Q,S; C-P,Q; D-P,Q

C) A-Q, B-P; C-S; D-R

D) A-P,Q,S; B-R,; C-S; D-P,

Sol. (B)


6/15

JMS 4 PAPER 1


Wurtzite type structure:Eg: ZnS

( )

-2occupy latticepointsof

+2Zn occupyhalf or alternateTV's

S HCP

Zinc blende (or) Sphalerite type

Eg: ZnS2

S occupy lattice points of FCC

2Zn

+ occupy half (or) alternative TVS

Anti fluorite type structure:

Eg: Na2O

+Na allTVs

-2O lattice points of FCC

Rock salt type structure

Eg: NaCl

Cl-

occupy lattice points of FCC+Na occupy all sOV

SECTION II (Total Marks : 16)

(Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of

which ONE OR MORE may be correct.

8. Which of the following statements are true.

A) The ratio of the velocity of electron in first three Bohr orbits of hydrogen atom is 2:3:6 respectivelyB) The ratio of magnitude of Total Energy:Kinetic Energy:Potential Energy for any orbit is 1:1:2

C) The wavelength of a green light is 500 nm, then its frequency is 6 x 1014 HzD) The ratio of de-broglie wavelength of a H atom, He atom and CH4 molecule moving with equal

kinetic energy is 4:2:1

Sol. (B, C, D)

A)False

1 2 3

1 1: ; 1: : 6 : 3 : 2

2 3

zv

n

v v v

= =


7/15

JMS 4 PAPER 1


B)2

2

2

Potential Energy

Kinetic Energy2

Total Energy2

1 1: : : :1 1:1: 2

2 2

ze

rze

r

ze

r

TE KE PE

=

=

=

= =

C)8

6 9

14

3 10

0.5 10 500 10 5006 10

C x

x x m nmx

= = = = =

D)

4

2. .

1 1 1 1 1: : : : 1: : 4 : 2 :1

2 41 4 16H He CH

h

m KE =

= = =

9. Among the following orbital interactions, those which represents hyper conjuction interaction.

A) B)

C) D)

Sol. (A, B)

A has ( C), positive charge conjugation

B has odd e-conjugation


8/15

JMS 4 PAPER 1


10. Which of the following is true about the complex [PtCl2(NH3)(OH2)] ; [Atomic no. of Pt = 78]

(i) It will have two geometrical isomeric forms, cis and trans

(ii) The hybridisation state of Pt(II) is sp3

(iii) It is a square planar complex

(iv) It is a diamagnetic complex

(v) It can show hydrate isomerism

(vi) It is a tetrahedral complex(A) (i), (iii) (B) (ii),(v) (C) (ii),(vi) (D)(iv)

Sol. (A, D)[PtCl2(NH3)(OH2)] shows geometrical isomerism ( Ma2bc type)

Pt (II) is 5d8. It forms square planar complex which is diamagnetic

11. The polarimeter readings in an experiment to measure the rate of inversion of canesugar (1st order reaction) were as follows

time (min) : 0 30 angle (degree) : 30 20 15

Identify the true statement (s) log 2 = 0.3, log 3 = 0.48, log 7 = 0.84 (A) The half life of the reaction is 82.7 min

(B) The solution is optically inactive at 131 min.(C) The equimolar mixture of the products is dextrorotatory

(D) The angle would be 7.5 at half time

Sol. (A,B, D)

0 30 30 20 15

0r ra

a x r r t

=

Half life of the reaction :

( )

( )

30 152.303log

30 20 15k

=

12.0303 45log 0.008377 min30 35

= =

1/2

0.69382.7min

0.008377t = =

The solution will be optically inactive ( r = 0 ) :

( )

2.303 450.008377 log

0 15t=

131.1 mint =


9/15

JMS 4 PAPER 1


The angle at half time :

( )30 15 215 / 2

a

x a a = =+

at half time

45 452 15

15 2x

x = + =

+

022.5 15 7.5x = =

SECTION-III (Total Marls : 15)(Paragraph Type)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions and

based on the other paragraph 3 multiple choice questions have to be answered. Each of these questions

has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

Paragraph for Question Nos. 12 to 13

A chemist opened a cupboard to find four bottles containing water solutions, each of which had

lost its label. Bottles 1, 2, 3 contained colourless solutions, while bottle 4 contained a blue solution.

The labels from the bottles were lying scattered in the floor of the cupboard. They were :

Copper (II) sulphate, Hydrochloric acid

Lead nitrate, Sodium carbonate

By mixing samples of the contents of the bottles, in pairs, the chemist made the following

observations.

Bottle 1 + Bottle 2 White precipitate

Bottle 1 + Bottle 3 White precipitateBottle 1 + Bottle 4 White precipitate

Bottle 2 + Bottle 3 Colourless gas evolved

Bottle 2 + Bottle 4 No visible reaction

Bottle 3 + Bottle 4 Blue precipitate

12. Bottle 3 contains(A) copper (II) sulphate (B) hydrochloric acid

(C) lead nitrate (D) sodium carbonate

Sol. (D)

13. When bottle 1 is mixed with bottle 4, white precipitate is observed, which is

(A) PbSO4 (B) PbCO3(C) PbCl2 (D) Pb(NO3)2

Sol. (A)

Solutions for Q No 12 to 13 :

12. Bottle 2 + bottle 3 colourless gas (carbon dioxide) sodium carbonate + acidIt suggests that bottle 2 and 3 contain sodium carbonate and HCl.


10/15

JMS 4 PAPER 1


Bottle 3 + 4 blue precipitate. It confirms Cu2+

in either of the bottles , considering available sulphate,

chloride, nitrate and carbonate.

So, the precipitate must be carbonate,

Hence, 3 carbonate (sodium carbonate)

4 Cu2+

2 HCl

Left out , 1 - Pb(NO3)2

13. Pb(NO3)2 + Cu SO4 Cu(NO3)2 + PbSO4

Soluble white ppt.Bottle 1 Bottle4

Paragraph for Question Nos. 14 to 16

A is a hydrocarbon with molecular formula C6H10. It reacts with Br2 to give B. With Zn/ethanol B

gives A and with alcoholic KOH, B gives C. C, A differ from each other with respect to sites of

unsaturation.

14. Compound A maybe

(A) (B)CH3

CH3

(C) (D)

CH = CH CH CH CH = CH2 2 2 2

CH2

Sol. (B)

15. Which of the following gives a dialdehyde on ozonolysis?

(A) only A (B) A and B

(C) Only C (D) A and C

Sol. (D)


11/15

JMS 4 PAPER 1


16. Which of the following reactions cannot generate either A or B or C?

(A) (B)

OH

+H

CH2

+H

OH

(C)

P

Br2

OH

OH (D)

+H

OH

Sol. (D)

Solutions for Q No 14 to 16 :

14.

Br2

Br

Br

KOH

alcohol

(A) (B) (C)

Zn

ethanol

15. Being cyclo alkenes, both A and C give dialdehydes on ozonolysis, which involves replacement of c = cby two c = o groups.

16. Option (A) :

OH

+H

(C)

Option (B) :

CH2

+H

OH

(A)


12/15

JMS 4 PAPER 1


Option (C) :

P

Br2

OH

OHBr

Br

(B)

Option (D) :

Not possible because a carbocation is never formed at the vinylic carbon , unless otherwise the leaving

group is a super leaving group, - OTf.

SECTION-IV (Total Marks : 28)

(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a single digit integer, ranging

from 0 to 9. The bubble corresponding to the correct is to be darkened in the ORS.

17. (No of POP bonds in P4O6) + (No of BOB bonds in borax) (No of SOS bonds in form ofSO3) is

Sol. (8)

No of POP bonds in P4O6 = 6

No of BOB bonds in borax = 5


13/15

JMS 4 PAPER 1


No of SOS bonds in form of SO3 ( S3O9) = 3

Therefore, it is 6 + 5 3 = 8

18. The total number of optically active alkynes having molecular formula C3FClBrI is

Sol. (8)

19. Among the following diazonium salts, those less reactive than (I) towards coupling reaction are

Sol. (3)


14/15

JMS 4 PAPER 1


Diazo coupling : Arene diazonium ions are weak electrophiles. They react with highly reactive aromatic

compounds i.e., with phenols, & 30 amines to yield azo compounds.This electrophilic aromatic

substitution reaction is called Diazo coupling reaction.

+I groups decrease the reactivity of a diazonium cation towards coupling as they decreaseelectrophilicity

-I groups increase the reactivity of a diazonium cation towards coupling as they increase electrophilicity

III has no group whereas V and VI have +I groups.Remaining have I groups.

20. No of metals that can be extracted by carbon reduction method is

Na, Al, Mg, Sn, Fe, Pb, Zn, Ca

Sol. (4)

The metals which are less electro positive and do not form carbides with carbon are reduced by carbon

reduction methodEg : oxides of Sn, Fe, Pb, Zn

Ores of Na, Al, Mg, Ca are reduced by electrolytic reduction as their reduction by carbon needs veryhigh temperature and at such high temperature, formation of carbides may occur.

21. 3000 ml sample of H2O2 has 5.6 volume strength. If this sample liberates 5.6 litre of O2 at 1 atm

& 273 K, then molarity of resultant H2O2 solutio is 1/x . x is (Assuming no change in volume of

H2O2 solution)

Sol. (3)

Volume strength of H2O2 = 5.6So, volume of O2 liberated by 3 lit of this is 3 x 5.6 lit

Strength of it remaining after liberation of 5.6 lit of O2 = 3 x 5.6 5.6 = 2 x 5.6

Vol of O2 liberated at STP = Vol strength of H2O2 x Vol of H2O2 solution

New volume strength =2 5.6

3

x

Volume strength = Molarity x 11.2

So the molarity of new H2O2 =

2 5.613

11.2 3

x

M=


15/15

JMS 4 PAPER 1


22 . Bleaching powder reacts with iodide ion according to the following unbalnced equation

OCI

+ I

+ H

+

I2 + Cl

+ H2OA 0.6 g sample of bleaching powder requires 35.24 mL of 0.1084 M Na2S2O3 to titrate the

liberated iodine. The percentage of Cl in the sample is nearly z. Sum of the digits of z is

ol : (5)

Using simplified formula,

% of available Cl =3.55 0.1084 35.24

230.6

x x=

23. The number of tripeptides formed by three different amino acids are

Sol. (6)

Since each amino acid has one N-terminal end and one C-terminal end, therefore, three different amino

acids form 2 3 6 = different tripeptides.

Let the three amino acids be 1 ,2 and 3

1 2 3, 2 3 1, 3 1 2, 3 2 1, 1 3 2, 2 1 3.

jms-4 paper -1 solutions

Documents