Chemistry question paper of JEE MAIN 2018 with answer key NEET Notes | EduRev

NEET : Chemistry question paper of JEE MAIN 2018 with answer key NEET Notes | EduRev

 Page 1


ALLEN
CODE-C
1
1. Which of the following salts is the most basic
in aqueous solution ?
(1) CH
3
COOK (2) FeCl
3
(3) Pb(CH
3
COO)
2
(4) Al(CN)
3
Ans. (1)
Sol. CH COOK + H O 
32
CH COOK + H O 
32
CH COOH   +   KOH
3
Weak acid Strong base
Hence nature of solution is basic
2. Which of the following compounds will be
suitable for Kjeldahl's method for nitrogen
estimation ?
(1)
NH
2
(2) 
NO
2
(3) 
N Cl
2
–
+
(4) 
N
Ans. (1)
Sol. Kjeldahl's is suitable for Aniline. This method
is used for quantitative analysis of N compound
in organic substance (NH
3 
/ NH
4
+
).
3. Which of the following are Lewis acids ?
(1) AlCl
3 
and SiCl
4
(2) PH
3 
and SiCl
4
(3) BCl
3 
and AlCl
3
(4) PH
3 
and BCl
3
Ans. (3)
Sol. BCl
3
 & AlCl
3
 
both have vacant p-orbital &
incomplete octet. So they act as Lewis acid.
4. Phenol on treatment with CO
2
 in the presence
of NaOH followed by acidification produces
compound X as the major product. X on
treatment with (CH
3
CO)
2
O in the presence of
catalytic amount of H
2
SO
4 
produces :
(1)
O
COH
2
CH
3
O
(2) 
C
CH
3
O
O
O
OH
PART A – CHEMISTRY
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
(3) 
COH
2
COH
2
O
O
CH
3
(4) 
COH
2
O
O
CH
3
Ans. (4)
Sol. 
OH
(i) CO , NaOH
2 
(ii) acidification
OH
COH
2
Salicylic acid
        (X)
OH
COH
2
+ (CH CO)
32
O + conc H SO
24
O–C–CH
3
COH
2
O
               Aspirin
(non-narcotic analgesic)
(X)
This is called Kolbe reactions.
This step is called acylation reaction.
5. An alkali is titrated against an acid with methyl
orange as indicator, which of the following is
a correct combination ?
Base  Acid End point
(1) Strong  Strong Pinkish red
to yellow
(2) Weak  Strong Yellow to
pinkish red
(3) Strong  Strong Pink to
colourless
(4) Weak  Strong Colourless to
pink
Ans. (2)
Sol. Methyl orange shows Red(pinkish) color in
Acidic medium & yellow color in basic
medium since original solution is basic so
initial color Þ yellow
& Titrated with acid so
Final color Þ pinkish (red)
Page 2


ALLEN
CODE-C
1
1. Which of the following salts is the most basic
in aqueous solution ?
(1) CH
3
COOK (2) FeCl
3
(3) Pb(CH
3
COO)
2
(4) Al(CN)
3
Ans. (1)
Sol. CH COOK + H O 
32
CH COOK + H O 
32
CH COOH   +   KOH
3
Weak acid Strong base
Hence nature of solution is basic
2. Which of the following compounds will be
suitable for Kjeldahl's method for nitrogen
estimation ?
(1)
NH
2
(2) 
NO
2
(3) 
N Cl
2
–
+
(4) 
N
Ans. (1)
Sol. Kjeldahl's is suitable for Aniline. This method
is used for quantitative analysis of N compound
in organic substance (NH
3 
/ NH
4
+
).
3. Which of the following are Lewis acids ?
(1) AlCl
3 
and SiCl
4
(2) PH
3 
and SiCl
4
(3) BCl
3 
and AlCl
3
(4) PH
3 
and BCl
3
Ans. (3)
Sol. BCl
3
 & AlCl
3
 
both have vacant p-orbital &
incomplete octet. So they act as Lewis acid.
4. Phenol on treatment with CO
2
 in the presence
of NaOH followed by acidification produces
compound X as the major product. X on
treatment with (CH
3
CO)
2
O in the presence of
catalytic amount of H
2
SO
4 
produces :
(1)
O
COH
2
CH
3
O
(2) 
C
CH
3
O
O
O
OH
PART A – CHEMISTRY
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
(3) 
COH
2
COH
2
O
O
CH
3
(4) 
COH
2
O
O
CH
3
Ans. (4)
Sol. 
OH
(i) CO , NaOH
2 
(ii) acidification
OH
COH
2
Salicylic acid
        (X)
OH
COH
2
+ (CH CO)
32
O + conc H SO
24
O–C–CH
3
COH
2
O
               Aspirin
(non-narcotic analgesic)
(X)
This is called Kolbe reactions.
This step is called acylation reaction.
5. An alkali is titrated against an acid with methyl
orange as indicator, which of the following is
a correct combination ?
Base  Acid End point
(1) Strong  Strong Pinkish red
to yellow
(2) Weak  Strong Yellow to
pinkish red
(3) Strong  Strong Pink to
colourless
(4) Weak  Strong Colourless to
pink
Ans. (2)
Sol. Methyl orange shows Red(pinkish) color in
Acidic medium & yellow color in basic
medium since original solution is basic so
initial color Þ yellow
& Titrated with acid so
Final color Þ pinkish (red)
ALLEN
JEE(MAIN)-2018
2
6. An aqueous solution contains 0.10 M H
2
S and
0.20 M HCl. If the equilibrium constants for the
formation of HS
– 
from H
2
S is 1.0 × 10
–7 
and that
of S
2– 
from HS
– 
ions is 1.2×10
–13 
then the
concentration of S
2– 
ions in aqueous solution
is :
(1) 3×10
–20
(2) 6×10
–21
(3) 5×10
–19
(4) 5×10
–8
Ans. (1)
Sol. [H
2
S] = 0.10 M
[HCl] = 0.20 M  Þ [H
+
]
 
= 0.2 M
(1)H
2
S
 
? HS
–
 + H
+
K
1
 = 1.0 × 10
–7
(2) HS ? S
2
–
 + H
+
K
2
 = 1.2 × 10
–13
So ,
H
2
S ? S
2–
 + 2H
+
= K
1 
× K
2
= 1.2 × 10
–20
So,
[S
2–
] = 
-20
2
+2
1.2 ×10 ×[H S]
[H]
= 
-20 -1
-2
1.2 ×10 ×10
4 ×10
= 3 × 10
–20 
M
* All the [H
+
] will come from strong acid [HCl]
only.
7. The combustion of benzene (l) gives CO
2
(g)
and H
2
O(l). Given that heat of combustion
of benzene at constant volume is
–3263.9 kJ mol
–1 
at 25° C ; heat of combustion
(in kJ mol
–1
) of benzene at constant pressure
will be - (R = 8.314 JK
–1
 mol
–1
)
(1) –452.46 (2) 3260
(3) –3267.6 (4) 4152.6
Ans. (3)
Sol. C
6
H
6
( l) + 
15
2
O
2
(g) ® 6CO
2
(g) + 3H
2
O( l)
Dn
g
 = 6–7.5 = –1.5 (change in gaseous mole)
DU or DE = –3263.9 kJ
DH = DU + Dn
g
RT
Dn
g
 = –1.5
R = 8.314 JK
–1
 mol
–1
T = 298 K
So DH = –3263.9 + (–1.5) 8.314 × 10
–3 
× 298
     = –3267.6 kJ
DH  = Heat at constant pressure
DU / DE = Heat at constant volume
R = gas constant
8. The compound that does not produce nitrogen
gas by the thermal decomposition is
(1) (NH
4
)
2
Cr
2
O
7
(2) NH
4
NO
2
(3) (NH
4
)
2
SO
4
(4) Ba(N
3
)
2
Ans. (3)
Sol. 1. (NH
4
)
2 
Cr
2
O
7 232
CrON
D
¾¾®+ + 4H
2
O
2. NH
4
NO
2 
D
¾¾® N
2 
+ 2H
2
O
3. (NH
4
)
2
SO
4 
D
¾¾® 2NH
3 
+ H
2
SO
4
4. Ba(N
3
)
2 
D
¾¾®
 
Ba+3N
2
In reaction (3)  NH
3
 is evolved where as in
reaction 1, 2 and 4  N
2
 is evolved.
9. How long (approximate) should water be
electrolysed by passing through 100 amperes
current so that the oxygen released can
completely burn 27.66 g of diborane ?
(Atomic weight of B = 10.8 u)
(1) 0.8 hours (2) 3.2 hours
(3) 1.6 hours (4) 6.4 hours
Ans. (2)
Sol.   B
2
H
6
 + 3O
2 
® B
2
O
3
 + 3H
2
O
moles of O
2 
required = 3x moles of B
2
H
6
= 3 × 
27.6
27.6
= 3
It
96500
´
= moles of O
2
 × 4
100 t
34
96500
´
=´
t = 
3 4 96500
100
´´
 sec.
= 3.2 hours.
10. Total number of lone pair of electrons in I
3
– 
ion
is
(1)6 (2) 9
(3) 12 (4) 3
Ans. (2)
Sol.
Total number of lone pair in 
3
I
-
is 9.
Page 3


ALLEN
CODE-C
1
1. Which of the following salts is the most basic
in aqueous solution ?
(1) CH
3
COOK (2) FeCl
3
(3) Pb(CH
3
COO)
2
(4) Al(CN)
3
Ans. (1)
Sol. CH COOK + H O 
32
CH COOK + H O 
32
CH COOH   +   KOH
3
Weak acid Strong base
Hence nature of solution is basic
2. Which of the following compounds will be
suitable for Kjeldahl's method for nitrogen
estimation ?
(1)
NH
2
(2) 
NO
2
(3) 
N Cl
2
–
+
(4) 
N
Ans. (1)
Sol. Kjeldahl's is suitable for Aniline. This method
is used for quantitative analysis of N compound
in organic substance (NH
3 
/ NH
4
+
).
3. Which of the following are Lewis acids ?
(1) AlCl
3 
and SiCl
4
(2) PH
3 
and SiCl
4
(3) BCl
3 
and AlCl
3
(4) PH
3 
and BCl
3
Ans. (3)
Sol. BCl
3
 & AlCl
3
 
both have vacant p-orbital &
incomplete octet. So they act as Lewis acid.
4. Phenol on treatment with CO
2
 in the presence
of NaOH followed by acidification produces
compound X as the major product. X on
treatment with (CH
3
CO)
2
O in the presence of
catalytic amount of H
2
SO
4 
produces :
(1)
O
COH
2
CH
3
O
(2) 
C
CH
3
O
O
O
OH
PART A – CHEMISTRY
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
(3) 
COH
2
COH
2
O
O
CH
3
(4) 
COH
2
O
O
CH
3
Ans. (4)
Sol. 
OH
(i) CO , NaOH
2 
(ii) acidification
OH
COH
2
Salicylic acid
        (X)
OH
COH
2
+ (CH CO)
32
O + conc H SO
24
O–C–CH
3
COH
2
O
               Aspirin
(non-narcotic analgesic)
(X)
This is called Kolbe reactions.
This step is called acylation reaction.
5. An alkali is titrated against an acid with methyl
orange as indicator, which of the following is
a correct combination ?
Base  Acid End point
(1) Strong  Strong Pinkish red
to yellow
(2) Weak  Strong Yellow to
pinkish red
(3) Strong  Strong Pink to
colourless
(4) Weak  Strong Colourless to
pink
Ans. (2)
Sol. Methyl orange shows Red(pinkish) color in
Acidic medium & yellow color in basic
medium since original solution is basic so
initial color Þ yellow
& Titrated with acid so
Final color Þ pinkish (red)
ALLEN
JEE(MAIN)-2018
2
6. An aqueous solution contains 0.10 M H
2
S and
0.20 M HCl. If the equilibrium constants for the
formation of HS
– 
from H
2
S is 1.0 × 10
–7 
and that
of S
2– 
from HS
– 
ions is 1.2×10
–13 
then the
concentration of S
2– 
ions in aqueous solution
is :
(1) 3×10
–20
(2) 6×10
–21
(3) 5×10
–19
(4) 5×10
–8
Ans. (1)
Sol. [H
2
S] = 0.10 M
[HCl] = 0.20 M  Þ [H
+
]
 
= 0.2 M
(1)H
2
S
 
? HS
–
 + H
+
K
1
 = 1.0 × 10
–7
(2) HS ? S
2
–
 + H
+
K
2
 = 1.2 × 10
–13
So ,
H
2
S ? S
2–
 + 2H
+
= K
1 
× K
2
= 1.2 × 10
–20
So,
[S
2–
] = 
-20
2
+2
1.2 ×10 ×[H S]
[H]
= 
-20 -1
-2
1.2 ×10 ×10
4 ×10
= 3 × 10
–20 
M
* All the [H
+
] will come from strong acid [HCl]
only.
7. The combustion of benzene (l) gives CO
2
(g)
and H
2
O(l). Given that heat of combustion
of benzene at constant volume is
–3263.9 kJ mol
–1 
at 25° C ; heat of combustion
(in kJ mol
–1
) of benzene at constant pressure
will be - (R = 8.314 JK
–1
 mol
–1
)
(1) –452.46 (2) 3260
(3) –3267.6 (4) 4152.6
Ans. (3)
Sol. C
6
H
6
( l) + 
15
2
O
2
(g) ® 6CO
2
(g) + 3H
2
O( l)
Dn
g
 = 6–7.5 = –1.5 (change in gaseous mole)
DU or DE = –3263.9 kJ
DH = DU + Dn
g
RT
Dn
g
 = –1.5
R = 8.314 JK
–1
 mol
–1
T = 298 K
So DH = –3263.9 + (–1.5) 8.314 × 10
–3 
× 298
     = –3267.6 kJ
DH  = Heat at constant pressure
DU / DE = Heat at constant volume
R = gas constant
8. The compound that does not produce nitrogen
gas by the thermal decomposition is
(1) (NH
4
)
2
Cr
2
O
7
(2) NH
4
NO
2
(3) (NH
4
)
2
SO
4
(4) Ba(N
3
)
2
Ans. (3)
Sol. 1. (NH
4
)
2 
Cr
2
O
7 232
CrON
D
¾¾®+ + 4H
2
O
2. NH
4
NO
2 
D
¾¾® N
2 
+ 2H
2
O
3. (NH
4
)
2
SO
4 
D
¾¾® 2NH
3 
+ H
2
SO
4
4. Ba(N
3
)
2 
D
¾¾®
 
Ba+3N
2
In reaction (3)  NH
3
 is evolved where as in
reaction 1, 2 and 4  N
2
 is evolved.
9. How long (approximate) should water be
electrolysed by passing through 100 amperes
current so that the oxygen released can
completely burn 27.66 g of diborane ?
(Atomic weight of B = 10.8 u)
(1) 0.8 hours (2) 3.2 hours
(3) 1.6 hours (4) 6.4 hours
Ans. (2)
Sol.   B
2
H
6
 + 3O
2 
® B
2
O
3
 + 3H
2
O
moles of O
2 
required = 3x moles of B
2
H
6
= 3 × 
27.6
27.6
= 3
It
96500
´
= moles of O
2
 × 4
100 t
34
96500
´
=´
t = 
3 4 96500
100
´´
 sec.
= 3.2 hours.
10. Total number of lone pair of electrons in I
3
– 
ion
is
(1)6 (2) 9
(3) 12 (4) 3
Ans. (2)
Sol.
Total number of lone pair in 
3
I
-
is 9.
ALLEN
CODE-C
3
11. When metal 'M' is treated with NaOH, a white
gelatinous precipitate 'X' is obtained, which is
soluble in excess of NaOH. Compound 'X'
when heated strongly gives an oxide which is
used in chromatography as an adsorbent. The
metal 'M' is
(1) Ca (2) Al
(3) Fe (4) Zn
Ans. (2)
Sol. Al + 3H
2
O
NaOH
¾¾¾® Al(OH)
3 
¯ +3/2H
2(g)
            (x)
      white gelatinous ppt.
                      ¯
      soluble in excess of NaOH
      and form Na[Al(OH)
4
]
2Al(OH)
3 
D
¾¾® Al
2
O
3
+3H
2
O
used as
adsorbent in chromatography
So metal is Al.
12. According to molecular orbital theory, which of
the following will not be a viable molecule ?
(1)
2
He
+
(2) 
2
H
-
(3) 
2
2
H
-
(4) 
2
2
He
+
Ans. (3)
Sol. The electronic configuration of H
2
2T 
is
Þ ( ) ( )
2
2
1s , *1s ss
Bond order of 
2 ba
2
NN 22
H0
22
- -
= ==
T
.
Hence 
2
2
H
T
does not exist, due to zero bond
order.
13. The increasing order of basicity of the following
compounds is :
(a) (b) 
(c) (d) 
(1) (b) < (a) < (c) < (d)
(2) (b) < (a) < (d) < (c)
(3) (d) < (b) < (a) < (c)
(4) (a) < (b) < (c) < (d)
Ans. (2)
Sol. Order of base nature depends on electron
donation tendency.
In compound 
NH
 nitrogen is sp
2
hybridized so least basic among all given
compound.
compound 
NH
2
NH
 is very strong
nitrogeneous organic base as lone pair of one
nitrogen delocalize in resonance and make
another nitrogen negativly charged and
conjugate acid have two equivalent resonating
structure.
Thus it is most basic in given compouds.
 NHCH
3
 (secondary amine)  more basic
than 
NH
2
 (primary amine)
14. Which type of 'defect' has the presence of
cations in the interstitial sites ?
(1) Vacancy defect
(2) Frenkel defect
(3) Metal deficiency defect
(4) Schottky defect
Ans. (2)
Sol. In Frenkel defect, some of ion (usually cation
due to their small size) missing from their
correct position and occupies position in
interstitial.
Page 4


ALLEN
CODE-C
1
1. Which of the following salts is the most basic
in aqueous solution ?
(1) CH
3
COOK (2) FeCl
3
(3) Pb(CH
3
COO)
2
(4) Al(CN)
3
Ans. (1)
Sol. CH COOK + H O 
32
CH COOK + H O 
32
CH COOH   +   KOH
3
Weak acid Strong base
Hence nature of solution is basic
2. Which of the following compounds will be
suitable for Kjeldahl's method for nitrogen
estimation ?
(1)
NH
2
(2) 
NO
2
(3) 
N Cl
2
–
+
(4) 
N
Ans. (1)
Sol. Kjeldahl's is suitable for Aniline. This method
is used for quantitative analysis of N compound
in organic substance (NH
3 
/ NH
4
+
).
3. Which of the following are Lewis acids ?
(1) AlCl
3 
and SiCl
4
(2) PH
3 
and SiCl
4
(3) BCl
3 
and AlCl
3
(4) PH
3 
and BCl
3
Ans. (3)
Sol. BCl
3
 & AlCl
3
 
both have vacant p-orbital &
incomplete octet. So they act as Lewis acid.
4. Phenol on treatment with CO
2
 in the presence
of NaOH followed by acidification produces
compound X as the major product. X on
treatment with (CH
3
CO)
2
O in the presence of
catalytic amount of H
2
SO
4 
produces :
(1)
O
COH
2
CH
3
O
(2) 
C
CH
3
O
O
O
OH
PART A – CHEMISTRY
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
(3) 
COH
2
COH
2
O
O
CH
3
(4) 
COH
2
O
O
CH
3
Ans. (4)
Sol. 
OH
(i) CO , NaOH
2 
(ii) acidification
OH
COH
2
Salicylic acid
        (X)
OH
COH
2
+ (CH CO)
32
O + conc H SO
24
O–C–CH
3
COH
2
O
               Aspirin
(non-narcotic analgesic)
(X)
This is called Kolbe reactions.
This step is called acylation reaction.
5. An alkali is titrated against an acid with methyl
orange as indicator, which of the following is
a correct combination ?
Base  Acid End point
(1) Strong  Strong Pinkish red
to yellow
(2) Weak  Strong Yellow to
pinkish red
(3) Strong  Strong Pink to
colourless
(4) Weak  Strong Colourless to
pink
Ans. (2)
Sol. Methyl orange shows Red(pinkish) color in
Acidic medium & yellow color in basic
medium since original solution is basic so
initial color Þ yellow
& Titrated with acid so
Final color Þ pinkish (red)
ALLEN
JEE(MAIN)-2018
2
6. An aqueous solution contains 0.10 M H
2
S and
0.20 M HCl. If the equilibrium constants for the
formation of HS
– 
from H
2
S is 1.0 × 10
–7 
and that
of S
2– 
from HS
– 
ions is 1.2×10
–13 
then the
concentration of S
2– 
ions in aqueous solution
is :
(1) 3×10
–20
(2) 6×10
–21
(3) 5×10
–19
(4) 5×10
–8
Ans. (1)
Sol. [H
2
S] = 0.10 M
[HCl] = 0.20 M  Þ [H
+
]
 
= 0.2 M
(1)H
2
S
 
? HS
–
 + H
+
K
1
 = 1.0 × 10
–7
(2) HS ? S
2
–
 + H
+
K
2
 = 1.2 × 10
–13
So ,
H
2
S ? S
2–
 + 2H
+
= K
1 
× K
2
= 1.2 × 10
–20
So,
[S
2–
] = 
-20
2
+2
1.2 ×10 ×[H S]
[H]
= 
-20 -1
-2
1.2 ×10 ×10
4 ×10
= 3 × 10
–20 
M
* All the [H
+
] will come from strong acid [HCl]
only.
7. The combustion of benzene (l) gives CO
2
(g)
and H
2
O(l). Given that heat of combustion
of benzene at constant volume is
–3263.9 kJ mol
–1 
at 25° C ; heat of combustion
(in kJ mol
–1
) of benzene at constant pressure
will be - (R = 8.314 JK
–1
 mol
–1
)
(1) –452.46 (2) 3260
(3) –3267.6 (4) 4152.6
Ans. (3)
Sol. C
6
H
6
( l) + 
15
2
O
2
(g) ® 6CO
2
(g) + 3H
2
O( l)
Dn
g
 = 6–7.5 = –1.5 (change in gaseous mole)
DU or DE = –3263.9 kJ
DH = DU + Dn
g
RT
Dn
g
 = –1.5
R = 8.314 JK
–1
 mol
–1
T = 298 K
So DH = –3263.9 + (–1.5) 8.314 × 10
–3 
× 298
     = –3267.6 kJ
DH  = Heat at constant pressure
DU / DE = Heat at constant volume
R = gas constant
8. The compound that does not produce nitrogen
gas by the thermal decomposition is
(1) (NH
4
)
2
Cr
2
O
7
(2) NH
4
NO
2
(3) (NH
4
)
2
SO
4
(4) Ba(N
3
)
2
Ans. (3)
Sol. 1. (NH
4
)
2 
Cr
2
O
7 232
CrON
D
¾¾®+ + 4H
2
O
2. NH
4
NO
2 
D
¾¾® N
2 
+ 2H
2
O
3. (NH
4
)
2
SO
4 
D
¾¾® 2NH
3 
+ H
2
SO
4
4. Ba(N
3
)
2 
D
¾¾®
 
Ba+3N
2
In reaction (3)  NH
3
 is evolved where as in
reaction 1, 2 and 4  N
2
 is evolved.
9. How long (approximate) should water be
electrolysed by passing through 100 amperes
current so that the oxygen released can
completely burn 27.66 g of diborane ?
(Atomic weight of B = 10.8 u)
(1) 0.8 hours (2) 3.2 hours
(3) 1.6 hours (4) 6.4 hours
Ans. (2)
Sol.   B
2
H
6
 + 3O
2 
® B
2
O
3
 + 3H
2
O
moles of O
2 
required = 3x moles of B
2
H
6
= 3 × 
27.6
27.6
= 3
It
96500
´
= moles of O
2
 × 4
100 t
34
96500
´
=´
t = 
3 4 96500
100
´´
 sec.
= 3.2 hours.
10. Total number of lone pair of electrons in I
3
– 
ion
is
(1)6 (2) 9
(3) 12 (4) 3
Ans. (2)
Sol.
Total number of lone pair in 
3
I
-
is 9.
ALLEN
CODE-C
3
11. When metal 'M' is treated with NaOH, a white
gelatinous precipitate 'X' is obtained, which is
soluble in excess of NaOH. Compound 'X'
when heated strongly gives an oxide which is
used in chromatography as an adsorbent. The
metal 'M' is
(1) Ca (2) Al
(3) Fe (4) Zn
Ans. (2)
Sol. Al + 3H
2
O
NaOH
¾¾¾® Al(OH)
3 
¯ +3/2H
2(g)
            (x)
      white gelatinous ppt.
                      ¯
      soluble in excess of NaOH
      and form Na[Al(OH)
4
]
2Al(OH)
3 
D
¾¾® Al
2
O
3
+3H
2
O
used as
adsorbent in chromatography
So metal is Al.
12. According to molecular orbital theory, which of
the following will not be a viable molecule ?
(1)
2
He
+
(2) 
2
H
-
(3) 
2
2
H
-
(4) 
2
2
He
+
Ans. (3)
Sol. The electronic configuration of H
2
2T 
is
Þ ( ) ( )
2
2
1s , *1s ss
Bond order of 
2 ba
2
NN 22
H0
22
- -
= ==
T
.
Hence 
2
2
H
T
does not exist, due to zero bond
order.
13. The increasing order of basicity of the following
compounds is :
(a) (b) 
(c) (d) 
(1) (b) < (a) < (c) < (d)
(2) (b) < (a) < (d) < (c)
(3) (d) < (b) < (a) < (c)
(4) (a) < (b) < (c) < (d)
Ans. (2)
Sol. Order of base nature depends on electron
donation tendency.
In compound 
NH
 nitrogen is sp
2
hybridized so least basic among all given
compound.
compound 
NH
2
NH
 is very strong
nitrogeneous organic base as lone pair of one
nitrogen delocalize in resonance and make
another nitrogen negativly charged and
conjugate acid have two equivalent resonating
structure.
Thus it is most basic in given compouds.
 NHCH
3
 (secondary amine)  more basic
than 
NH
2
 (primary amine)
14. Which type of 'defect' has the presence of
cations in the interstitial sites ?
(1) Vacancy defect
(2) Frenkel defect
(3) Metal deficiency defect
(4) Schottky defect
Ans. (2)
Sol. In Frenkel defect, some of ion (usually cation
due to their small size) missing from their
correct position and occupies position in
interstitial.
ALLEN
JEE(MAIN)-2018
4
15. Which of the following compounds contain(s)
no covalent bond(s) ?
KCl, PH
3
, O
2
, B
2
H
6
, H
2
SO
4
(1) KCl, H
2
SO
4
(2) KCl
(3) KCl, B
2
H
6
(4) KCl, B
2
H
6
, PH
3
Ans. (2)
Sol.
PH
3
O
2
BH
26
H SO
24
All are covalent compounds 
KCl is ionic compound.
16. The oxidation states of
Cr in [Cr(H
2
O)
6
]Cl
3
,[Cr(C
6
H
6
)
2
], and
K
2
[Cr(CN)
2 
(O)
2
(O
2
)(NH
3
)] respectively are :
(1) +3, +2, and +4
(2) +3, 0, and +6
(3) +3, 0, and +4
(4) +3, +4, and +6
Ans. (2)
Sol.   (i) [Cr(H
2
O)
6
]Cl
3
:Hexaaquachromium(III) chloride
x + 6 × 0 + (–1) × 3 = 0
x3 =+
(ii) [Cr(C
6
H
6
)
2
] : bis(h
6
–benzene)chromium(0)
y + 2 × 0 = 0
y0 =
(iii) K
2
[Cr(CN)
2 
(O)
2 
(O
2
) (NH
3
)] :
Potassium  amminedicyanidodioxidoperoxidochromate(VI)
2 × 1 + z + 2 × (–1) + 2 × (–2) + (–2) + 0 = 0
z6 =+
The oxidation states of Cr in [Cr(H
2
O)
6
]Cl
3
,
[Cr(C
6
H
6
)
2
], and K
2
[Cr(CN)
2 
(O)
2 
(O
2
) (NH
3
)]
respectively are +3, 0 and +6.
17. Hydrogen peroxide oxidises [Fe(CN)
6
]
4– 
to
[Fe(CN)
6
]
3– 
in acidic medium but reduces
[Fe(CN)
6
]
3–
 to [Fe(CN)
6
]
4– 
in alkaline
medium. The other products formed are,
respectively :
(1) (H
2
O + O
2
) and (H
2
O + OH
–
)
(2) H
2
O and (H
2
O + O
2
)
(3)H
2
O and (H
2
O + OH
–
)
(4) (H
2
O + O
2
) and H
2
O
Ans. (2)
Sol. (i) ()
4
21
22
6
Fe CN H O 2H
-
+ -+
éù++
ëû
¯
()
3
32
2
6
Fe CN 2HO
-
+-
éù+
ëû
(ii) ()
3
31
22
6
Fe CN H O 2OH
-
+ --
éù++
ëû
¯
()
4
20
22
6
Fe CN O 2HO
-
+
éù ++
ëû
18. Glucose on prolonged heating with HI gives :
(1) 1–Hexene
(2) Hexanoic acid
(3) 6-iodohexanal
(4) n-Hexane
Ans. (4)
Sol.
HO
H
H
CHO
OH
H
CH OH
2
OH
OH H
HI/
Reduction
D
n-Hexane
Glucose
Page 5


ALLEN
CODE-C
1
1. Which of the following salts is the most basic
in aqueous solution ?
(1) CH
3
COOK (2) FeCl
3
(3) Pb(CH
3
COO)
2
(4) Al(CN)
3
Ans. (1)
Sol. CH COOK + H O 
32
CH COOK + H O 
32
CH COOH   +   KOH
3
Weak acid Strong base
Hence nature of solution is basic
2. Which of the following compounds will be
suitable for Kjeldahl's method for nitrogen
estimation ?
(1)
NH
2
(2) 
NO
2
(3) 
N Cl
2
–
+
(4) 
N
Ans. (1)
Sol. Kjeldahl's is suitable for Aniline. This method
is used for quantitative analysis of N compound
in organic substance (NH
3 
/ NH
4
+
).
3. Which of the following are Lewis acids ?
(1) AlCl
3 
and SiCl
4
(2) PH
3 
and SiCl
4
(3) BCl
3 
and AlCl
3
(4) PH
3 
and BCl
3
Ans. (3)
Sol. BCl
3
 & AlCl
3
 
both have vacant p-orbital &
incomplete octet. So they act as Lewis acid.
4. Phenol on treatment with CO
2
 in the presence
of NaOH followed by acidification produces
compound X as the major product. X on
treatment with (CH
3
CO)
2
O in the presence of
catalytic amount of H
2
SO
4 
produces :
(1)
O
COH
2
CH
3
O
(2) 
C
CH
3
O
O
O
OH
PART A – CHEMISTRY
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
(3) 
COH
2
COH
2
O
O
CH
3
(4) 
COH
2
O
O
CH
3
Ans. (4)
Sol. 
OH
(i) CO , NaOH
2 
(ii) acidification
OH
COH
2
Salicylic acid
        (X)
OH
COH
2
+ (CH CO)
32
O + conc H SO
24
O–C–CH
3
COH
2
O
               Aspirin
(non-narcotic analgesic)
(X)
This is called Kolbe reactions.
This step is called acylation reaction.
5. An alkali is titrated against an acid with methyl
orange as indicator, which of the following is
a correct combination ?
Base  Acid End point
(1) Strong  Strong Pinkish red
to yellow
(2) Weak  Strong Yellow to
pinkish red
(3) Strong  Strong Pink to
colourless
(4) Weak  Strong Colourless to
pink
Ans. (2)
Sol. Methyl orange shows Red(pinkish) color in
Acidic medium & yellow color in basic
medium since original solution is basic so
initial color Þ yellow
& Titrated with acid so
Final color Þ pinkish (red)
ALLEN
JEE(MAIN)-2018
2
6. An aqueous solution contains 0.10 M H
2
S and
0.20 M HCl. If the equilibrium constants for the
formation of HS
– 
from H
2
S is 1.0 × 10
–7 
and that
of S
2– 
from HS
– 
ions is 1.2×10
–13 
then the
concentration of S
2– 
ions in aqueous solution
is :
(1) 3×10
–20
(2) 6×10
–21
(3) 5×10
–19
(4) 5×10
–8
Ans. (1)
Sol. [H
2
S] = 0.10 M
[HCl] = 0.20 M  Þ [H
+
]
 
= 0.2 M
(1)H
2
S
 
? HS
–
 + H
+
K
1
 = 1.0 × 10
–7
(2) HS ? S
2
–
 + H
+
K
2
 = 1.2 × 10
–13
So ,
H
2
S ? S
2–
 + 2H
+
= K
1 
× K
2
= 1.2 × 10
–20
So,
[S
2–
] = 
-20
2
+2
1.2 ×10 ×[H S]
[H]
= 
-20 -1
-2
1.2 ×10 ×10
4 ×10
= 3 × 10
–20 
M
* All the [H
+
] will come from strong acid [HCl]
only.
7. The combustion of benzene (l) gives CO
2
(g)
and H
2
O(l). Given that heat of combustion
of benzene at constant volume is
–3263.9 kJ mol
–1 
at 25° C ; heat of combustion
(in kJ mol
–1
) of benzene at constant pressure
will be - (R = 8.314 JK
–1
 mol
–1
)
(1) –452.46 (2) 3260
(3) –3267.6 (4) 4152.6
Ans. (3)
Sol. C
6
H
6
( l) + 
15
2
O
2
(g) ® 6CO
2
(g) + 3H
2
O( l)
Dn
g
 = 6–7.5 = –1.5 (change in gaseous mole)
DU or DE = –3263.9 kJ
DH = DU + Dn
g
RT
Dn
g
 = –1.5
R = 8.314 JK
–1
 mol
–1
T = 298 K
So DH = –3263.9 + (–1.5) 8.314 × 10
–3 
× 298
     = –3267.6 kJ
DH  = Heat at constant pressure
DU / DE = Heat at constant volume
R = gas constant
8. The compound that does not produce nitrogen
gas by the thermal decomposition is
(1) (NH
4
)
2
Cr
2
O
7
(2) NH
4
NO
2
(3) (NH
4
)
2
SO
4
(4) Ba(N
3
)
2
Ans. (3)
Sol. 1. (NH
4
)
2 
Cr
2
O
7 232
CrON
D
¾¾®+ + 4H
2
O
2. NH
4
NO
2 
D
¾¾® N
2 
+ 2H
2
O
3. (NH
4
)
2
SO
4 
D
¾¾® 2NH
3 
+ H
2
SO
4
4. Ba(N
3
)
2 
D
¾¾®
 
Ba+3N
2
In reaction (3)  NH
3
 is evolved where as in
reaction 1, 2 and 4  N
2
 is evolved.
9. How long (approximate) should water be
electrolysed by passing through 100 amperes
current so that the oxygen released can
completely burn 27.66 g of diborane ?
(Atomic weight of B = 10.8 u)
(1) 0.8 hours (2) 3.2 hours
(3) 1.6 hours (4) 6.4 hours
Ans. (2)
Sol.   B
2
H
6
 + 3O
2 
® B
2
O
3
 + 3H
2
O
moles of O
2 
required = 3x moles of B
2
H
6
= 3 × 
27.6
27.6
= 3
It
96500
´
= moles of O
2
 × 4
100 t
34
96500
´
=´
t = 
3 4 96500
100
´´
 sec.
= 3.2 hours.
10. Total number of lone pair of electrons in I
3
– 
ion
is
(1)6 (2) 9
(3) 12 (4) 3
Ans. (2)
Sol.
Total number of lone pair in 
3
I
-
is 9.
ALLEN
CODE-C
3
11. When metal 'M' is treated with NaOH, a white
gelatinous precipitate 'X' is obtained, which is
soluble in excess of NaOH. Compound 'X'
when heated strongly gives an oxide which is
used in chromatography as an adsorbent. The
metal 'M' is
(1) Ca (2) Al
(3) Fe (4) Zn
Ans. (2)
Sol. Al + 3H
2
O
NaOH
¾¾¾® Al(OH)
3 
¯ +3/2H
2(g)
            (x)
      white gelatinous ppt.
                      ¯
      soluble in excess of NaOH
      and form Na[Al(OH)
4
]
2Al(OH)
3 
D
¾¾® Al
2
O
3
+3H
2
O
used as
adsorbent in chromatography
So metal is Al.
12. According to molecular orbital theory, which of
the following will not be a viable molecule ?
(1)
2
He
+
(2) 
2
H
-
(3) 
2
2
H
-
(4) 
2
2
He
+
Ans. (3)
Sol. The electronic configuration of H
2
2T 
is
Þ ( ) ( )
2
2
1s , *1s ss
Bond order of 
2 ba
2
NN 22
H0
22
- -
= ==
T
.
Hence 
2
2
H
T
does not exist, due to zero bond
order.
13. The increasing order of basicity of the following
compounds is :
(a) (b) 
(c) (d) 
(1) (b) < (a) < (c) < (d)
(2) (b) < (a) < (d) < (c)
(3) (d) < (b) < (a) < (c)
(4) (a) < (b) < (c) < (d)
Ans. (2)
Sol. Order of base nature depends on electron
donation tendency.
In compound 
NH
 nitrogen is sp
2
hybridized so least basic among all given
compound.
compound 
NH
2
NH
 is very strong
nitrogeneous organic base as lone pair of one
nitrogen delocalize in resonance and make
another nitrogen negativly charged and
conjugate acid have two equivalent resonating
structure.
Thus it is most basic in given compouds.
 NHCH
3
 (secondary amine)  more basic
than 
NH
2
 (primary amine)
14. Which type of 'defect' has the presence of
cations in the interstitial sites ?
(1) Vacancy defect
(2) Frenkel defect
(3) Metal deficiency defect
(4) Schottky defect
Ans. (2)
Sol. In Frenkel defect, some of ion (usually cation
due to their small size) missing from their
correct position and occupies position in
interstitial.
ALLEN
JEE(MAIN)-2018
4
15. Which of the following compounds contain(s)
no covalent bond(s) ?
KCl, PH
3
, O
2
, B
2
H
6
, H
2
SO
4
(1) KCl, H
2
SO
4
(2) KCl
(3) KCl, B
2
H
6
(4) KCl, B
2
H
6
, PH
3
Ans. (2)
Sol.
PH
3
O
2
BH
26
H SO
24
All are covalent compounds 
KCl is ionic compound.
16. The oxidation states of
Cr in [Cr(H
2
O)
6
]Cl
3
,[Cr(C
6
H
6
)
2
], and
K
2
[Cr(CN)
2 
(O)
2
(O
2
)(NH
3
)] respectively are :
(1) +3, +2, and +4
(2) +3, 0, and +6
(3) +3, 0, and +4
(4) +3, +4, and +6
Ans. (2)
Sol.   (i) [Cr(H
2
O)
6
]Cl
3
:Hexaaquachromium(III) chloride
x + 6 × 0 + (–1) × 3 = 0
x3 =+
(ii) [Cr(C
6
H
6
)
2
] : bis(h
6
–benzene)chromium(0)
y + 2 × 0 = 0
y0 =
(iii) K
2
[Cr(CN)
2 
(O)
2 
(O
2
) (NH
3
)] :
Potassium  amminedicyanidodioxidoperoxidochromate(VI)
2 × 1 + z + 2 × (–1) + 2 × (–2) + (–2) + 0 = 0
z6 =+
The oxidation states of Cr in [Cr(H
2
O)
6
]Cl
3
,
[Cr(C
6
H
6
)
2
], and K
2
[Cr(CN)
2 
(O)
2 
(O
2
) (NH
3
)]
respectively are +3, 0 and +6.
17. Hydrogen peroxide oxidises [Fe(CN)
6
]
4– 
to
[Fe(CN)
6
]
3– 
in acidic medium but reduces
[Fe(CN)
6
]
3–
 to [Fe(CN)
6
]
4– 
in alkaline
medium. The other products formed are,
respectively :
(1) (H
2
O + O
2
) and (H
2
O + OH
–
)
(2) H
2
O and (H
2
O + O
2
)
(3)H
2
O and (H
2
O + OH
–
)
(4) (H
2
O + O
2
) and H
2
O
Ans. (2)
Sol. (i) ()
4
21
22
6
Fe CN H O 2H
-
+ -+
éù++
ëû
¯
()
3
32
2
6
Fe CN 2HO
-
+-
éù+
ëû
(ii) ()
3
31
22
6
Fe CN H O 2OH
-
+ --
éù++
ëû
¯
()
4
20
22
6
Fe CN O 2HO
-
+
éù ++
ëû
18. Glucose on prolonged heating with HI gives :
(1) 1–Hexene
(2) Hexanoic acid
(3) 6-iodohexanal
(4) n-Hexane
Ans. (4)
Sol.
HO
H
H
CHO
OH
H
CH OH
2
OH
OH H
HI/
Reduction
D
n-Hexane
Glucose
ALLEN
CODE-C
5
19. The predominant form of histamine present in
human blood is (pK
a
, Histidine = 6.0)
(1)
(2) 
N
H
N
H
NH
2
Å
(3) 
(4) 
Ans. (3)
Sol. Structure of Histamine
N
N
NH
2
H
9.75 pka
more basic
Blood is slightly basic in nature (7.35 pH)
approx.
At this pH terminal NH
2
 will get protonated due
to more basic nature.
\ Predominant structure of Histamine is
N
N
NH
3
H
+
20. The recommended concentration of fluoride
ion in drinking water is up to 1 ppm as fluoride
ion is required to make teeth enamel harder by
converting [3Ca
3 
(PO
4
)
2
·Ca(OH)
2
] to :
(1) [3(CaF
2
)·Ca(OH)
2
]
(2) [3(Ca
3
(PO
4
)
2
·CaF
2
]
(3) [3(Ca(OH)
2
]·CaF
2
]
(4) [CaF
2
]
Ans. (2)
Sol. [3Ca (PO ) .Ca(OH) ] + 2F
3 4 2 2 
1
[3Ca (PO ) .CaF ] + 2OH
3 422
1
(drinking 
water 
upto 1ppm)
(Harder teeth 
enamel)
21. Consider the following reaction and
statements :
[Co(NH
3
)
4
Br
2
]
+
 + Br
– 
® [Co(NH
3
)
3
Br
3
] + NH
3
(I) Two isomers are produced if the reactant
complex ion is a cis-isomer.
(II)Two isomers are produced if the reactant
complex ion is a trans-isomer.
(III) Only one isomer is produced if the reactant
complex ion is a trans-isomer.
(IV)Only one isomer is produced if the reactant
complex ion is a cis-isomer.
The correct statements are :
(1) (I) and (III) (2) (III) and (IV)
(3) (II) and (IV) (4) (I) and (II)
Ans. (1)
Sol.
Co
3+
NH
3
NH
3
NH
3
NH
3
Br
–
Br
–
Co
3+
NH
3
NH
3
HN
3
Br
–
Br
–
HN
3
+Br
–
Cis trans
Co
3+
Br
–
NH
3
NH
3
NH
3
Br
–
Br
–
facial
Co
3+
NH
3
NH
3
Br
–
NH
3
Br
–
Br
–
meridionial
Co
3+
NH
3
NH
3
Br
–
NH
3
Br
–
Br
–
+Br
–
Only meridionial
–NH
3
–NH
3
Read More
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