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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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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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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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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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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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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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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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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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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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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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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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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.
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