jms i paper 1 solutions

8/2/2019 Jms i Paper 1 Solutions

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JMS – I PAPER - 1

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PAPER-1

Maximum 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 thebubble(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: 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 thebubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this

section.



JMS – I 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 of

which ONLY ONE is correct.

1. In 3 2 1ψ , the sum of angular momentum, spherical nodes and angular nodes is

(A)6 4

2

h π

π

+(B)

63

2

h

π + (C)

6 2

2

h π

π

+(D)

6

2

h

π

Sol. (A)

3 2 1ψ represents, n= 3, l = 2, m = 1

Angular momentum of e− in an orbital =

( )12

62

hl l

h

π

π

+

=

Spherical nodes = 1 0n l− − =

Angular nodes = 2l =

6 0 2

2

hsum

π

∴ = + +

=6 4

2

h π

π

+

2. Which of the following is incorrect about solubility trend in group 2.

Least soluble Most soluble

A) Carbonates BeCO3 BaCO3

B) Fluorides BaF2 BeF2

C) Hydroxides Be(OH)2 Ba(OH)2

D) Sulphates BaSO4 BeSO4

Sol. (A)

Solubility of carbonates in water decreases on moving down the group

BeF2 is very soluble in water owing to the high solvation energy of Be in forming

[Be ( H2O ) 4 ]2-

.Other fluorides are almost insoluble.

Solubility of hydroxidess in water increases on moving down the group

Solubility of sulphates in water decreases on moving down the group



JMS – I PAPER - 1


3.

, Reactant ‘A’ is

(A)

( )5

|| ||CH - C - CH - C - CH

3 2 3

O O

(B)

( )

|| ||CH - C - CH - C -H

3 2 4

O O

(C) ( )|| ||

H- C - CH - C -H2 5

O O

(D) ( )||

CH - C - CH -CH3 2 24

O

OH −

Sol. (B)

Intra molecular aldol condensation



JMS – I PAPER - 1


4. At the point of intersection of two curves as shown, the concentration of B is given for the

reaction A nB→ as

(A) 0

2

nA(B) 0

1

A

n −(C) 0

1

nA

n +(D)

0

1.

1

n A

n

−

+

Sol. (C)

At the point of intersection

[ ] [ ]left formed

A B=

A nB→

a-x nxa-x = nx

( )1a n x= +

0

1 1

Aa x

n n

= =+ +

∴B formed = 0

1

nAnx

n=

+

5.

O

( ) ( )CH CO HCH N

3 32 2CH C CH CH A major B major3 2 3

− − − → →.

Product‘B’is

(A)

O||

CH C CH CH CH3 2 2 3

O− − − − − (B)

O||

CH C CH CH CH3 2 2 3

O− − − − −

(C)

O||

CH CH C CH CH3 2 2 3

O− − − − − (D)

O||

CH C CH CH3 2 3

O− − − −

Sol. (B)



JMS – I PAPER - 1


6. Assertion : The radii of noble gases are larger than that of the radii of precedent halogens.Reason : Radii of Noble gases are measured as van der waal’s radii as they do not form

covalent bonds like halogens and rcovalent < rvan der waal’s.

(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(B) Both Assertion and Reason are true but Reason is not correct explanation of Assertion.

(C) Assertion is true but Reason is false

(D) Assertion is false but Reason is true

Sol. (A)

Being monoatomic, noble gases have very large values of non-bonded radii.So, radii of noblegases should be compared not with the covalent radii but with the van der Waals radii of other

elements.

7. Arrange the following in the decreasing order of their acidic nature

1) 4–Nitro benzoic acid

2) 4–Nitro phenol

3) 4–Methoxy benzoic acid

4) p–CresolA) 1 > 2 > 4 > 3 B) 3 > 1 > 2 > 4

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

Sol. (C)

Greater the acid dissociation constant( Ka) or lesser the PKa , greater is the acidic nature

Pka values of the given compounds ( For comparison sake)

3.4 ( - NO2 is strong –I & strong – M effecting group)



JMS – I PAPER - 1


4.5 ( - OCH3 is with + M > - I)

7.1 (Acids > phenols)

10.2 ( - CH3 is + I & Hyperconjugation)

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. Identify the correct statements A) In solid state,N2O5 exists as NO2

+ and NO3- ions

B) In solid state PCl5 contains PCl3-and PCl2

+ions

C) Solid PBr5 has PBr2+ and PBr3

-

D) Solid Beryllium chloride has chain structure

Sol. (A,D)

Solid N2O5 is a salt, consisting of separated anions and cations. Cation is linear nitronium ion

NO2+

and anion is planar nitrate NO3−

ion. So, this solid is also called nitronium nitrate.

In gaseous and liquid phases, PCl5 has a trigonal bipyramidal structure. The three equatorial

P–Cl bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is

due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond

pairs.

In solid state, PCl5 exists as an ionic solid,[PCl4]+[PCl6]

–in which the cation, [PCl4]

+is

tetrahedral and the anion, [PCl6]–is octahedral.

Solid PBr5 contains PBr4+, Br - .

Beryllium chloride has a chain structure in the solid state



JMS – I PAPER - 1


9.

+H O

3A+B →O .

A and B cannot be differentiated by

(A) Iodoform (B) Fehling (C) NaHSO3 (D) 2,4 – DNP

Sol. (B,C,D)

Both A and B give Fehling’s test being aldehydes ), react with NaHSO3 & give 2,4 – DNP test(

being carbonyl compounds)

A fails to give Iodofom test as it has no methyl keto group or methyl carbinol group

10. Which of the following is/are correct regarding Borax.A) Borax is correctly represented as Na2[B4O5(OH)4].8H2O

B) In Borax, Out of 10 water molecules of crystallization, 8 are attached to B atoms and 2 are attached to

Na+

.C) Borax has two B atoms in triangular geometry and two are in tetrahedral geometry.

D) Borax on heating with CoO on bunsen flame forms a blue coloured Co(BO3)2bead

Sol. (A,C)



JMS – I PAPER - 1


Borax is Sodium tetraborate decahydrate ( Na2B4O710H2O). It contains [B4O5(OH)4]2-

tetranuclear units and therefore its correct formula is Na2[B4O5 (OH)4].8H2O

. Borax on heating with CoO on bunsen flame forms a blue coloured cobalt metaborate Co(BO2)2bead.

11. Which of the following statement/s is/are incorrect A) Maltose has C1–C4 glycosidic linkage between glucose units and is reducing due to

hemi acetal groups at C1 of both units

B) Amylopectin has α–D–glucose units with chains made of C1–C4 glycosidic linkages

with branches at C1–C6 of different chains

C) Lactose has α–glycosidic linkage between C1 of galactose and C4 of glucose and is

reducing due to glucose unitD) Sucrose is a non reducing sugar with glycosidic linkage between C1 of α–glucose and C2 of

α–fructose

Sol. (A,C,D)

Maltose :

4-O-α-D-Glucopyranosyl-D-glucose

:



JMS – I PAPER - 1


Amylopectin :

Sucrose :

α-D-glucopyranosyl-(1→2)-β-D-fructofuranoside



JMS – I PAPER - 1


Lactose :

β-D-galactopyranosyl-(1→4)-D-glucose

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

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions andbased 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 of which ONLY ONE is correct.

Paragraph for Question Nos. 12 and 13

A metal complex having ( )Cr NH Cl Br3 24

has been isolated in two forms (A) and (B). The

form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous

ammonia , whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

12. Complexes A and B are respectively

(A) ( )Cr NH Cl Br3 24

and ( )Cr NH Br Cl3 24

(B) ( )Cr NH Br Cl3 24

and ( ) 2Cr NH

3 4Cl Br

(C) ( )Cr NH ClBr Cl3 4

and ( )Cr NH Cl Br3 24

(D) ( )Cr NH Cl Br3 24

and ( )Cr NH ClBr Cl3 4



JMS – I PAPER - 1


Sol. (C)

( ) ( )Cr NH Cl Br +AgNO AgCl Cr NH ClBr NO3 3 3 34 4

→ ↓ +

White ppt

( ) ( )Cr NH Cl Br+AgNO AgBr Cr NH Cl NO3 2 3 3 2 34 4

→ ↓

Pale yellow

13. EAN and magnetic moment of Cr is A and B are

(A) 33, 15 BM (B) 36, 15 BM (C) 36, 24 BM (D) 36, 24 BM

Sol. (A)Oxidation state of ‘Cr’ in both A & B = +3

And coordination no. of Cr is 6

EAN = 24 - 3 + 6 × 2 = 33∴

No. of unpaired electrons in 3Cr + in the complex = 3

∴Magnetic moment = ( )3 3+2 BM

= 15 BM

Paragraph for Question Nos. 14 to 16

Passage:

O

O

( )1 aq NaOH →

( ) +2 HA

CHCl3 →

NaOHB

PhNH2 C+D →

(14) Which statements about compound A are correct?(1) Compound A is used in the formation of Phenolpthalein

(2) Compound A is used in the formation of Aspirin

(3) Compound A gives color with FeCl3

(A) 1,2,3 (B) 1,2 (C) 2,3 (D) 1,3 .

Sol. (D)

OH

A =



JMS – I PAPER - 1


Phenolphthalein is synthesized by condensation of pthalic anhydride with two equivalents of

phenol under acidic conditions

Phenol gives violet coloured complex with neutral FeCl3. Phenol form characteristic colourediron complexes when treated with neutral ferric chloride solution. Eg. Phenol & resorcinol -

violet colour, catechol-green etc.

15. Compound B will be

(A)

OH

C-H

O

(B)

C-H

O

O O

(C)

OH

O

C H

(D)

Sol. (A)



JMS – I PAPER - 1


OH

C-H

O

B =

Reimer –Tiemann reaction.

( salicylaldehyde )

16. Compounds C and D are(A) Identical (B) optical isomers

(C) Geometrical isomers (D) functional isomers

Sol. (C)



JMS – I PAPER - 1


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.

Ratio of X to Y is

(A)1 (B) 2 (C) 3 (4) 6

Sol. (2)



JMS – I PAPER - 1


X = 6

Y = 3

2 X

Y =

With alcohols : ( 1 eq )

R – MgX + H – OR → R – H + Mg(OR)X

With alkyl halides : ( 1 eq )

R – MgX + X’ – R’ → R – R’ + MgXX’

With thiols : ( 1 eq )

R – MgX + H – SR → R – H + Mg(SR)X

With Esters : ( 2 eq )

→



JMS – I PAPER - 1


With acid halides : ( 2 eq )

→ → HOH

→ −HCl O

||CR|R

−

′

18. For an isomerisation reaction, A B⇌ , the temperature dependence of equilibrium

constant is given by log K =2000

4.0T

− . The value of 0

2.303

S∆at 300 K in cal is equal to

Sol. (8)0 2.303 logG RT K ∆ = −

0 0 0 0 0

log2.303 2.303 2.303 2.303

G H T S S H K

RT RT R RT

∆ ∆ − ∆ ∆ ∆= − = − = + −

2000log 4.0K

T = −

0

42.303

S

R

∆∴ =

0

4 2 82.303

S∆= × =

19. The number of stable acyclic isomers having molecular formula CH3NO is

( Unstable isomers having –OH at double bonded carbon should not be considered)

Sol. (3 )

CH3NO

Degree of unsaturation or Index no = 12

+∑n(V-2)

Where n = no of atoms, V = valency of atom



JMS – I PAPER - 1


1(4 2) 3(1 2) 1(3 2) 1(2 2)1 1

2

− + − + − + −+ =

Therefore possible functional groups are Amides, Nitroso compounds and OximesAmide :

Nitroso compound :

Oxime : CH2 = N- OH ( No geometriucal isomerism possible)HO – CH = NH is unstable, so not considered

20. An organic compound A reacts with water to form two acids B and C. Compound A also

reacts with NaOH to form D and E(two salt solutions). B and D give yellow precipitate with

BaCl2 solution whereas C and E give white precipitate with AgNO3 solution, insoluble in

dil.HNO3. Unit place of molecular weight of A is

Sol. (5 )

Mol. Wt of CrO2Cl2 is 155



JMS – I PAPER - 1


21. A gaseous equilibrium, ( ) ( ) ( )2 2 A g B g C g+⇌ is set at temperature T, with only A as the

starting material. The total equilibrium pressure is P atm and C

PP n= . If

1

81

pK

P = .

Calculate the value of n.

Sol. (9)2

2

. B C p

A

P PK

P=

2 B C P P=

( ) A B C P P P P= − +

( )2 C C P P P= − +

3 C P P= −

ButC

PP

n= =

3 31 A PP P Pn n ∴ = − = −

22

B C

PP P

n= =

( )

2

22

2 23

2

4.

4.

33P

P P

P nn nK n nn

Pn

∴ = =−−

( ) ( )2 2

4 1 4

813 3

PK

P n n n n= ⇒ =

− −

( )2 23 4 81 9 3 3 2 2 9 6n n − = × = × × × × = ×

9n∴ =



JMS – I PAPER - 1


22. Number of six membered rings – number of 5 membered rings in Buckminsterfullerene is

Sol : (8)

C60 molecule is called Buckminsterfullerene .It has soccer ball like shape.It contains twenty

six- membered rings and twelve five membered rings. A six membered ring is fused with six or

five membered rings but a five membered ring can only fuse with six membered rings.

23. An organic compound containing methoxy groups was heated with conc.HI . The methyl

iodide evolved was treated with ethanolic AgNO3 to get precipitate of AgI.The precipitate

was purified,dried and weighed. If the mass of AgI obtained was 470 g, the no of methoxy

groups in the compound is

Sol : (2)

Application of Zeisel’s determination

3 3( ) R OCH n nCH I nAgI = =

235 gm of AgI is equivalent to 1 methoxy group

Therefore 470 gm = 2 methoxy groups

jms i paper 1 solutions

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