jms-4 paper -2 solutions

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

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JMS 4 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 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 negative

marks in this section.

3. In Section III (Total Marks: 24), 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 this

section.

4. In Section IV (Total Marks: 16), for each question you will be awarded 2 marks for each row in which

you have darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks

otherwise. Thus each question in this section carries a maximum of 8 Marks. There are no negative marks

in this section.


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JMS 4 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 of

which ONLY ONE is correct.

1. H for the reaction ( ) ( )2

H H| |

H C N g +2H g H C N H|

H

(g) is 150 KJ

Calculate the bond energy of C N bond. [Given bond energies of C H = 416 KJ mol1;

H H = 436 KJ mol1; C N = 293 KJ mol1; N H = 369 KJ mol1]

A) 762 KJ mole1 B) 481 KJ mole1

C) 841 KJ mole1 D) 741 KJ mole1

Sol. (C)

( ) ( )2

H H| |

H C N g +2H g H C N H|

H

[ ] [ ][ ] [ ]

H = ( ) ( ) 2 ( ) 3 ( ) ( ) 2 ( )

150 416 ( ) 2(436) 3(416) 293 2(369)

150 416 ( ) 872 2279

150 ( ) 991

( ) 150 991 841 /

BE C H BE C N BE H H BE C H BE C N BE N H

BE C N

BE C N

BE C N

BE C N kj mol

+ + + +

= + + + +

= + +

=

= + =

2. Given are two reactions:

3CH OH

1NuCH I Rate constant k

3

2NuCH I Rate constant k 3

DMF

The difference in the value of k1 and k2 is highest when Nu is

(A)Cl (B) Br (C) NCS (D) 3CH S

Sol. (A)

In DMF, none of these nuclephiles get solvated. But in CH3OH, the negative nuclephiles get solvated.

Cl being the smallest, is solvated to the maximum extent. Therefore, in 3CH OH , it is the poorest

nucleophile. So, the difference between k1 and k2 is highest.


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JMS 4 PAPER - 2


3. ColumnI Column IIA) Borax

A

P) BNB) B2H6 + H2O Q) B2H6

C) B2H6 + NH3 R) H3BO3

D) BCl3 + LiAlH4 S) NaBO2 + B2O3

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

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

Sol. (A)060

2 4 7 2 2 4 7 2 2 4 7 2 2 310 5 2

c

Na B O H O Na B O H O Na B O NaBO B O

+, Glassy mass (Borax bead

test)

' '

2 6 3' ' ( )high T

xB H excess NH BN+ (Boron nitride)

2 6 2 3 3 26 2 6B H H O H BO H+ +

3 4 2 6 33 2 3 3BCl LiAlH B H AlCl LiCl+ + +

4. Statement-1 : HIO4 decomposes . 1, 2 glycols but not 1,3 or higher glycolsStatement 2 : Only 1, 2 glycol forms cyclic esters which subsequently undergo clevage to form

carbonyl compound.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Sol. (A)

Periodic acid cleavage of a 1,2 - glycol involves a cyclic periodate intermediate which involves the

following mechanism. It is not possible with 1,3 or higher glycols


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JMS 4 PAPER - 2


6. S1 : The graphite is diamagnetic and diamond is paramagnetic in nature.

S2 : Graphite acts as a metallic conductor along the layers of carbon atoms and as semi-conductor

perpendicular to the layers of the carbon atoms.S3 : Graphite is less denser than diamond

S4 : C60 is called as Buckminster fullerene(A) F T T T (B) T F T T (C) F F T T (D) F T T T

Sol. (A)

Both graphite and diamond are diamagnetic.

7.

Indicate the correct stereochemical relationship amongst the

following monosacharides

I) II) III) IV)

A) I and II are anomers; III and IV are epimers

B) I and III are epimers; II and IV are anomers

C) I and II are epimers; III and IV are anomers

D) I and III are anomers; I and II are epimers

Sol : (D)

Anomers differ in stereochemistry at 1st carbon. (anomeric carbon)

I and III have difference stereochemistry at 1st

carbon.So, they are anomers

Epimers differ in stereochemistry at carbon other than 1stand II have difference in stereochemistry at 4th carbon. So they are C - 4 epimers.

8. Match the following

I Bauxite (a) Lead

II Carnallite (b) Copper

III Malachite (c) Magnesium

IV Galena (d) Halls Process(A) I a, II b, III c, IV d (B) I d, II c, III b, IV a

(C) I b, II a, III d, IV c (D) I d, II b, III c, IV a

Sol. (B)


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JMS 4 PAPER - 2


Bauxite is Al2O3 .2H20.It is of two types. Red bauxite which contains Iron oxide impurities,purifiedby Baeyers and Halls process and the other one is white bauxite which contains SiO2 impurities, purified

by serpeck processCarnallite contains Mg.It is KCl MgCl26H2O

Malachite is CuCO3. Cu(OH)2

Galena containslead.It is PbS

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 of

which ONE OR MORE may be correct.

9. Figure displays the plot of the compressibility factor Z verses p for a few gases

IV

III

I

II

Which of the following statements is/are correct for a vander waals gas ?

A) The plot I is applicable provided the vander waals constant a is negligible.B) The plot II is applicable provided the vander waals constant b is negligible.

C) The plot III is applicable provided the vander waals constants a and b are negligible.

D) The plot IV is applicable provided the temperature of the gas is much higher than its critical

temperature.

Sol. (A, B, C)

I shows positive deviation from ideal behavior. So, a is negligible

II shows negative deviation from ideal behavior, So, b is negligible

III shows ideal behavior. So a and b are negligible

10. Which of the following molecule/s gives only acids on hydrolysisA) PCl3 B) SF4 C)NCl3 D) P4O6

Sol. (A, B, C, D)

3 2 3 33 0 3PCl H H PO HCl+ +

4 2 22 0 4SF H SO HF + +

3 2 36 0 3NCl H NH HOCl+ +

4 6 2 3 36 0 4P O H H PO+


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JMS 4 PAPER - 2


11. Given log3 = 0.4771, log2 = 0.3010. To measure the activity of a given radioactive sample, a count

rate meter is used. Two recordings separated in time by 1 minute show a count rate ratio of 2.25.

The decay constant (min1

) of the sample is :(A) 0.83 min

1(B) equal to its t1/2 numerically in min.

(C) 1 min1 (D) varying with time

Sol. (A, B)

k =1

t2.303 log

final

initial rate

rate

k =1

12.303 log 2.25 = 2.303 x log(1.5)2

= 2 x 2.303 x log1.5 = 0.83 min1

t1/2 = 0.6930.83

= 0.83 min

12. Identify the set in which product formed is correctly matched with given reagent. The

reactant is 1-tert-butyl cyclohexene.

Sol. (A, B, C)

A) Syn addition. But, equitorial is more stable with bulky - C(CH3)3.

B) Anti addition

C) Anti addition

D) Syn addition. So, T should be equatorial (axial given)

Cis : Both either upper or lower

Trans :One upper and one lower


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JMS 4 PAPER - 2


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 to

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

13. pH of 6.66 x 104 M solution of Al(OH)3 is xy .

x + y is...

Given the first dissociation of Al(OH)3 is complete, second dissociation is 50% and third is

insignificant

Sol. (2)

( ) ( )3 2Al OH Al OH OH+ + (Complete dissociation)

46.66 10 M 46.66 10 M

( ) ( )2

2Al OH Al OH OH

+ + + (50% dissociation)

46.66 10 M 43.33 10 M

Total 4 4 36.66 10 3.33 10 10OH = + =

pOH = 3;

pH = 11

14. 0.5g of fuming H2SO4(oleum) is diluted with water. This solution is completely neutralized by 26.7

ml of 0.4 N NaOH. The percentage of free SO3 in the sample is nearly Z X 4. z is

Sol. (5)

No of eq of H2SO4 = no of eq of NaOH

=26.7

0.4 0.01068

1000

x =

Wt of H2SO4 = 0.01068 x 49 = 0.52332

Wt of H2O added = 0.52332 0.5 = 0.02332

H2O + SO3 H2SO4

18 80

0.02332 ?

= 0.1036

Therefore % of free SO3 =0.1036

100 20.720.5

x =


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JMS 4 PAPER - 2


15. For an ionic compound AX3(s) formed between a metal A and a non-metal X (outermost shell

configuration of X = ns2np5). Unit place of enthalpy of formation (magnitude) of AX3(s)

in k cal mol1 using following data.

(Non-metal X is found to exist in nature as a diatomic gas)

Hsublimation

A(s) = 100 Kcal/mol

1

HIE

A(g) = 60 Kcal/mol

2

HIE

A(g) = 150 Kcal/mol

3

HIE

A(g) = 280 Kcal/mol

Hdiss X2(g) = 80 Kcal/mol

He.g

X(g) = 110 Kcal/mol

HLattice energy A X3(s) = 470 Kcal/mol

Sol. (0)

For the ionic compound AX3 (s)

( ) 2( ) 3( )

3

2s g s

A X AX+

sub 1 2 3 .3H ( H H H ) H 3 H H2

3100 (60 150 280) (80) 3( 110) 470

2

90

f IE IE IE e g LEH = + + + + + +

= + + + + +

=

diss


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JMS 4 PAPER - 2


16. Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of

MnO2. Passing a current of 27A for 24 hours gives 1 kg of MnO2. The current efficiency is n. The

unit place value of n is

Sol. (5)

87 24 60 601000

2 96500

25.67

cx x xx

c amp

=

=

Therefore current efficiency =

25.67

100 95%27 x =

17. Decomposition of A follows first order kinetics by the following equation.

4A(g) B(g) + 2C(g)

If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg.

What is half-life of A in min ? (Assume only A is present initially)

Sol. (5)

4A(g)

B(g) + 2C(g)t = 0 800

t = 10 min 800 4p p 2p

800 4p + p + 2p = 800 - p = 650

p = 150

Pressure of A = 800 -4 x 150 =200,

So, 2 x t1/2 = 10 minutes

t1/2 = 5 minutes

18. Among K2CO3, MgCO3, CaCO3, BeCO3, Na2CO3, BaCO3, SrCO3, the no of carbonates,thosemore thermally stable than MgCO3 are

Sol. (5)

Na2CO3, K2CO3 ,CaCO3, SrCO3 ,BaCO3,

Thermal stability of alkaline earth metal cabonates increases down the group. So CaCO3, SrCO3 and

BaCO3 are more stable than MgCO3

Thermal stability of alkali metal cabonates increases down the group.


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JMS 4 PAPER - 2


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 andfivestatements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with

ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with

the statements given q and r, then for the particular question, against statement B, darken the bubblescorresponding to q and r in the ORS.

19. Match the reactions mentioned in Column-I with the nature of the reaction/characteristic(s) of the

products mentioned in Column-II

Column I Column II

(A) strongly heated3 2H PO

(p) One of the products is a reductant

(B) 2i)H O3 ii)

PCl

(q) One of the products is tribasic

(C)2 2

NO H O+ (r) Dehydration

(D)3 4 10

HNO P O + (s) One of the products has central atom in+5 oxidation state

(t) Disproportionation

Sol : (A p,q,s, t;B p,q,s, t;C p,s, t;D r,s)

3 2 3 3 3

3 3 3 4 3

3 2 3 3

3 3 3 4 3

2 2 2 3

3 4 10 3 2 5

A) 3H PO 2H PO PH

4H PO 3H PO PH

B) PCl 3H O H PO 3HCl

4H PO 3H PO PHC) 2NO H O HNO HNO

D) 4HNO P O 4HPO 2N O

+

+

+ +

+

+ +

+ +

Note :

Substance in highest oxidation state acts only as oxidant

Substance in lowest oxidation state acts only as reductant

Substance in intermediate oxidation state can act as both oxidant and reductant


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JMS 4 PAPER - 2


20. Match the reactions given in Column-I with the facts given in Column-II.

Column-I Column-II

(A) +PyH CrO Cl3CH CH CH OH

3 2 2 CH Cl2 2

(p) Net result is

reduction

(B) K Cr O2 2 7

H SO ,H O2 4 2

CH CH CH OH3 2 2

(q) A ketone is formed

(C)

CH3

LiAlH4

etherCH CH CH Cl

3 2

(r) Net result is

oxidation

(D) ( )3 3PCl

3ether

CH CH OH3

CH COK +

CH2

CH3

(s) A hydrocarbon is

formed

(t) An acid is formed

Sol. A-r; B-r,t; C-p,s; D-s

A)This oxdising agent is PCC ( pyridinium chloro chromate), an agent that does not over oxidise theresulting aldehyde.

B)

K Cr O2 2 7CH CH CH OH CH CH COOH

3 2 2 3 2H SO ,H O2 4 2

C)

This is reduction and a hydrocarbon results.


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JMS 4 PAPER - 2


D)

( )( )

PCl33CH CH OH CH CH Cl CH CH CH CH

3 3 2 2 33 3

CH CO K

CH C OH

+

=

CH CH2 3

CH CH2 3

(an alkene)

This reaction is not at all a redox reaction. Because at one site, carbon is losing a strong electronegative

atom chlorine (getting reduced) and the next adjacent site is losing one hydrogen (getting oxidized). The

net result is not a redox reaction.

The bulky base gives Hofmann elimination to give but-1-ene.

jms-4 paper -2 solutions

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