JEE Main - 2018Paper - I

CHEMISTRYCODE- C

CHEMISTRY Analysis

Sl.No. UNIT NAME Q. Nos. Correct Wrong

1. ATOMIC STRUCTURE

2. PERIODIC TABLE

3. CHEMICAL BONDING 10,12,15

4. STATES OF MATTER

5. STOICHIOMETRY 23

6. THERMODYNAMICS 7

7. CHEMICALAND IONIC EQUILIBRIUM 1,5,6,28,26

8. HYDROGENAND ITS COMPOUNDS 17

9. ALKALI AND ALKALINE EARTH METALS

10. P-BLOCK ELEMENTS ( BORON, CARBON FAMILY) 11,3

11. PRACTICAL CHEMISTRY

12. ENVIRNMENTAL CHEMISTRY 20

13. ORGANIC CHEMISTRY 4

14. HYDRO CARBONS 22

15. SOLID STATE 14

16. SOLUTIONS

17. ELECTROCHEMISTRY & CHEMCIAL KINETICS 9,29

18. SURFACE CHEMISTRY

19. METALLURGY

20. OXYGEN FAMILY

21. NITROGEN FAMILY 8

22. HALOGEN FAMILY

23. D & F-BLOCK ELEMENTS, COORDINATION COMPOUNDS 30,21,16

24. POLYMERS

25. BIOMLECULES 18,19

26. CHEMISTRY IN EVERY DAY LIFE

27. ORGANIC COMPOUNDS CONTAINING C, H AND O

28. ORGANIC COMPOUNDS CONTAINING NITROGEN 4,27,24

29. INERT GASES 13

30. HALO ALKANES - ALKYL & ARYL HALIDES31. PURIFICATION AND ANALYSIS ORG COMPOUNDS 25

32. CHEMICAL KINETICS 2

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[CODE -C ]

1. Which of the following slats is the most basic in aqueous solution?

a) 3CH COOK b) 3FeCl c) ( )3 2Pb CH COO d) ( )3

Al CN

Sol : A

Salts of weak acid and strong base have basic aqueous solution

2. Which of the following compoundswill be suitable for Kjeldahl’s method for nitrogen estimation?

a) b)

c) d)

Sol: A

Nitrogen in organic compound should be linked to C & H. The method is not used in the case ofnitro, azo and azoxy compounds and to the compounds containing nitrogen in the ring (e.g. pyridinequinoline, isoquinoline, etc.). The method is extensively used for estimating nitrogen in food, fertilizersand agricultural products.

3. Which of the following are Lewis acids?

a) 3 3AlCl and SiCl b) 3 4PH and SiCl

c) 3 3BCl and AlCl d) 3 3PH and BCl

Sol: C

Group 13 element form electron deficient compounds and hence act as Lewis acids Both B and Al

have only 6 electronsin their valence shell in 3 3&BCl AlCl

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4. Phenol on treatment with 2CO in the presence of NaOH followed by acidification produces

compounds X as major product. X on treatment with ( )3 2CH CO O in the presence of catalytic

amount of 2 4H SO produces:

a) b)

c) d)

Sol: D

O

( ) 21 /CO NaOH

( ) 32 H O+

( )3 2CH Co O

2 4H SO

OH OHCOOH COOH

O C 3CH

Kolbes reaction

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5. An alkali is treated against an acid with methyl orange as indicator, which of the following is acorrect combination

Base Acid End Point

a) Strong Strong Pinkish red to yellow

b) Weak Strong Yellow to pinkish red

c) Strong Strong Pink to colourless

d) Weak Strong Colourless to pink

sol: B

Less than Hp 3.1it is red, greater than 4.4 yellow

6. An aqueous solution contains 0.01 2M H S and 0.02 M HCl . If the equilibrium constants forthe

formation of HS − from 2H S is 71.0 10−× and that of 2S − from HS − ions is 131.2 10−× then

the concentration of 2S − ions in aqueous solution is

a) 203 10−× b) 216 10−× c) 195 10−× d) 85 10−×Sol: A

72

2 13

2 202

10

1.2 10

2 1.2 10

H S H SH K

SH H S K

H S H S K

+ −

+ − −

+ − −

+ =

+ = ×

− − − − − − − − − − − − − − − − − − − − −

+ = ×

[ ]

2 2

2

H SK

H S

+ − =

[ ]2 2

200.2

1.2 100.1

S −−

× = [ ]H HCl+ = as HCl is a strong acid and dissociation of 2H S is

suppressed ( H + ion concentration will be mainly due to HCl a strong acid)

2 203 10S M− − = ×

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7. The combustion of benzene (I) gives 2CO (g) and ( )2H O I Given that heat of combustion of

benzene at constant volume is -3263.9 1kJ mol− at 025 C heat of combination of benzene at con

stant pressure will be ( )1 18.314R JK mol− −=

a) -452.46 b) 3260 c) -3267.6 d) 4152.6

Sol: C

( )6 6 2 2215/ 2 6 3 ; 3263.9gC H O CO H O U kJ+ → + ∆ = −

15 36

2 2n

−∆ = − =

3

3

33263.9 8.314 298 10

2

3263.9 3716.358 10

3263.9 3.716

3267.6

H U nRT

kJ

−

−

∆ = ∆ + ∆

= − + − × × × = − − ×= − −= −

8. The compound that does not produce nitrogen gas by the thermal decomposition is

a) ( )4 2 72NH Cr O b) 4 2NH NO c) ( )4 42

NH SO d) ( )3 2Ba N

Sol: C

( )4 4 4 4 32NH SO NH HSO NH∆→ +

9. How long (approximate) should water be electrolysed by passing through 100amperes current sothat the oxygen released can completely burn 27.66g of diborane? (Atomic weight of B=10.8u)

a) 0.8hours b) 3.2 hours c) 1.6 hours d) 6.4 hours

Sol: B

2 6 2 2 3 23 3B H O B O H O+ → +

27.6gm of 2 6B H requries 23 32gm O×

Ectw

F=

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8 1003 32

96500

t× ×× =

3 32 96500

8 100t

× ×=×

=11580sec

Time in hours =11580

3.213600

hours=

10. Total numbers of lone pair of electrons in 3I − ion is

a) 6 b) 9 c) 12 d) 3

Sol: B

: : : : : : :I I I I I I−

− + → − →

9 lone pairs

11. When metal ‘M’ is treated with NaOH a white gelatinous precipitate ‘X’ is obtained which is solublein excess of Na OH Compound ‘X’ when heated strongly gives an oxide which is used in chromatgraphyas an adsorbent. Then meatl ‘M’

a) Ca b) Al c) Fe d) Zn

Sol : B

( )2 242 2 2 2Al NaOH H O Na Al OH H + + → +

12. According to molecular orbital theory, which of the folloiwng will not be a viable molecule?

a) 2He+ b) 2H − c) 22H − d) 2

2He +

Sol: C

22H − will have configaration

*2 21 1S S

Bond order =2 2

02

− =

Hence 22H − is not formed

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13. The increasing order of basicity of the following compounds is

a. b.

c. d.

a) b < a < c < d b) b < a < d < c c) d < b < a < c d) a < b < c < d

Sol : A

14. Which type of ‘defect’ has the presence of cation in the interstitial sites?

a) Vacancy defect b) Frenkel defecct

c) Metal deficiency defect d) Schottky defect

Sol : B

15. Which of the following compounds contain(s) no covalent bond(s)?

3 2 2 6 2 4, ,KCI PH O B H H SO

a) 2 4,KCI H SO b) KCI c) 2 6,KCI B H d) 2 6 3, ,KCI B H PH

Sol: B

Since it is an ionic compound.

16. The oxidation state of Cr in ( ) ( )2 3 6 66 2[ ] ,Cr H O Cl Cr C H and

( ) ( ) ( ) ( )2 22 2 3K Cr CN O O NH respectively are

a) +3 ,+2, and +4 b) +3,0 and +6 c) +3,0 and +4 d) +3,+4 and +6

Sol: C

( ) ( )2 6 3 6 0 1 3 0

3

Cr H O Cl x

x

+ × + − × = = +

( )6 6 2Cr C H

Benzene is a neutral molecule hence oxidiation State of Cr =0

( ) ( ) ( ) ( )2 2 32 2k Cr CN O O NH

4X = +

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17. Hydrogen peroxide oxidises ( ) 4

6Fe CN

− to ( ) 3

6Fe CN

− in acidic medium but reduces

( ) 3

6Fe CN

− to ( ) 4

6Fe CN

− in alkaline medium. The other products formed are respectively

a) ( ) ( )2 2 2H O O and H O OH −+ + b) ( )2 2 2H O and H O O+

c) ( )2 2H O and H O OH −+ d) ( )2 2 2H O O and H O+

Sol : B

( ) ( )4 3

2 2 26 6

HFe CN H O Fe CN H O+− −

+ → +

( ) ( )3 4

2 2 2 26 6

OHFe CN H O Fe CN H O O+− −

+ → + +

18. Glucose on prolonged heating with HI gives:

a) 1-Hexene b) Hexanoic acid c) 6-iodhexanal d) n-Hexane

Sol; D

cos HIGlu e n hexane∆→ − , complete reduction takes place

19. The predominant from of histamine present in human blood is ( ), 6.0apK Histidine =

a) b)

c) d)

Sol : C

Since aPk is 6.0, it is weakly acidic in nature exist as cation

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20. The recommended concentration of fluoride ion in drinking water is upto 1ppm as fluoride ion is

requried to make teeth enamel harder by converting ( ) ( )4 2 23 .Ca PO Ca OH to

a) ( ) ( )2 23 .CaF Ca OH b) ( )4 22

3 .Ca PO CaF

c) ( ){ }223 .Ca OH CaF d) [ ]2CaF

Sol: B

It converts to fluorapatite ( )( )4 223 Ca PO CaF

21. Consider the following reaction and statments:

( ) ( )3 2 3 3 34 3Co NH Br Br Co NH Br NH

+ − + → + I) Two isomers are produced if the rectant complex ion is a cis iosmer

II) Two isomers are produced if the rectant complex ion is a trans- isomer

III) Only one iosmer is produced if the rectant complex ion is a trans-isomer

IV) Only one isomer is produced if the rectant complex ion is a cis-isomer

The correct statments are :

a) I and III b) III and IV c) II and IV d) I and II

Sol: A

3NH 3NH 3NHBr BrBr Br3H N 3H N

Br Br Br

CO

CO CO

CO CO

3NH 3NH3NH

3NH 3NH3NH

3NH 3NH 3NH

3NH 3NHBr−

Br−

Meridonial

Br

Facial

Br Br

Br BrBr

3H N

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22. The trans alkenes are formed by the reduction of alkynes with

a) 4NaBH b) 3/ .Na liq NH

c) Sn HCl− d) 2 4/ ,H Pb C BaSO

Sol : B

Birch Reduction

23. The ratio of mass percent of C and H of an organic compound ( )X Y ZC H O is 6 : 1 If one mol

ecule of the above compound ( )X Y ZC H O contains half as much oxygen as requried to burn one

molecule of compound X YC H completely to 2CO and 2H O The empirical formula of compounds

X Y ZC H O is :

a) 2 4C H O b) 3 4 2C H O c) 2 4 3C H O d) 3 6 3C H O

Sol : C

12100 %

4100 %

xof c

M

of HM

× =

× =

% 6

% 112

100 6

1100

C

Hx

My

M

=

×=

×

2Y x=

2Y x= confirm with options 1,3, and 4 but based on no.of oxygen atom present as

2 4 2 2 23 2 2C H O CO H O+ → +

2 4 3 2 2 2

32 2

2C H O O CO H O+ → +

No of oxygen atom is X Y ZC H O is half the oxygen atoms consumed in combination of X YC H

corresponds to 2 4 3C H O

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24. Phenol reacts with methyl chlorformate in the presence of NaOH to form product A. A reacts with

2Br to form product B. A and B are respectively

a)

b)

c)

d)

Sol: B

O HΘ

→ 3- -||

Cl C OCH

O

→

OH OΘ

Br

3

||O

O C CH− −3

||O

O C O CH− − −

2Br

ortho,para directing grow

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25. The major product of the following reaction is

a) b)

c) d)

Sol: C

Here 2E mechanism for elimination of HX takes place

26. Which of the following lines correctly show the temperature dependence of equilibrium constant, K,for an exothermic reaction

a) B and C b) C and D c) A and D d) A and B

Sol: D

0

0 0 0

ln

G RTmk

GK

RT

G H T S

∆ = −∆= −

∆ = ∆ − ∆

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0 0

lnH T S

KRT

∆ − ∆= −

As reaction is exothermic H Ve∆ = −

0 0H I SlmK

R T R

∆ ∆ = +

y mx c= +

The graph is a straight line with +ve slope

Hence graph A

0

0

H ve

S ve

∆ =−∆ =+

Hence graph B

0

0

H ve

S ve

∆ = −∆ = −

27. The major product formed in the following reaction is:

a) b)

c) d)

0

0

H ve

S ve

∆ = −∆ = +

0

0

H ve

S ve

∆ = −∆ = −

1/T

A

Bln k

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Sol : C

O

O

HI

Heat

3|H

O CH⊕

−

O

I

OH

3 3 2CH OH CH CH I+ +

I −

I −

28. An aqueous solution contains an unknown concentration of 2Ba + .When 50mL of a 1M solution of

2 4Na SO is added, 4BaSO just beigns to precipitate The final volume is 500mL The solubility

product of 4BaSO is 101 10−× What is the original concentration of 2Ba +

a) 92 10 M−× b) 91.1 10 M−× c) 101.0 10 M−× d) 95 10 M−×Sol: B

24 1 1No moles of SO M V− = in 50ml

24 500mn of SO in ml−

22 14

5 101000 10

500M of SO

−− −×= × =

2810 500

BaM in ml solution+

−=

2 2450 500m mn Ba in ml n Ba in ml+ +=

1 21 2

11

1000 1000450 500

101000 1000

V VM M

M −

× = ×

× = ×

91 1.1 10M M−= ×

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29. At 0518 C the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of

363 Torr, was 1.00Torr 1s− when5% had reacted and 0.5Torr 1s− when 33% had reacted The order

of the reaction is

a) 3 b) 1 c) 0 d) 2

Sol: D

1 1xR KP=

11

51.00

100

xp

K P× = × −

1

95

100

x

k p = ×

2 2xR KP=

67 3630.5

100

x

k× = ×

1 1

2 2

95 3631 100

0.5 67 363100

x

x

x

R P

R P

× = = =

×

( )952 1.41

67

xx = =

1

22 2 2x

x

= = =

30. For 1 molal aqueous solution of the following compounds, whihc one will shoe the highest freezingpoint?

a) ( )2 2 25Co H O Cl Cl H O b) ( )2 2 24

.Co H O Cl Cl H O

c) ( )2 3 233Co H O Cl H O d) ( )2 36

Co H O Cl Colligative property

Sol: C

1f f

f

T ik m

T i

∆ =

∆

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For ( )2 2 25[ ]Cl .Co H O Cl H O i=3

( )2 2 24[ ]Cl.Co H O Cl H O i=2

( )2 2 24[ ]3Co H O Cl H O i=1

( )2 36[ ]ClCo H O i=4

Greater the i value

Greater the fT∆ value

Hence lower the fT

Hence fT∆ = d > a > b > c

Hence fT = c > b > a > d