Download Chemistry - medicine.careers360.com

NEET-II (2016)
NEET–II (2016) TEST PAPER WITH ANSWER & SOLUTIONS
(HELD ON SUNDAY 24th JULY, 2016)
136.
Hot concentrated sulphuric acid is a moderately
strong oxidizing agent. Which of the following
reactions does not show oxidizing behaviour ?
140.
(1) Br– > CN – > NH 3 > C 6 H 5 –
(1) C + 2H 2 SO 4 CO 2 + 2SO 2 + 2H 2 O
(2) CN– > Br – > C 6 H 5– >NH 3
(2) CaF 2 + H 2 SO 4  CaSO 4 + 2HF
(3) NH 3 > CN – > Br– > C 6 H 5–
(3) Cu + 2H 2 SO 4  CuSO 4 + SO 2 + 2H 2 O
(4) CN– > C 6 H 5– > Br – > NH 3
Ans. (4)
(4) 3S + 2H 2 SO 4  3SO 2 + 2H 2 O
Ans. (2)
Sol. CaF 2 + H 2 SO 4 CaSO 4 + 2HF
In this reaction, oxidation number of none of the
atom is not changed. Hence H 2 SO 4 is not acting as
oxidising agent.
137. Which of the following pairs of d-orbitals will have
electron density along the axes ?
(2) d xy ,d x 2 y2
(3) d z2 ,d xz
(4) d xz ,d yz
Ans. (1)
Sol. dz2 and dx 2 –y2 has electron density concentrated on
the axis.
138. The correct geometry and hybridization for XeF
are :
(3) octahedral, sp 3 d2
(4) trigonal bipyramidal, sp 3 d
Ans. (3)
XeF 4 , AB 4 L2  sp 3d 2
 geometry  octahedral
 shape  square planar
Among the following which one is a wrong
statement?

139.
(1) SeF 4 and CH 4 have same shape
(2) I3+ has bent geometry
(3) PH 5 and BiCl 5 do not exist
(4) p-d bonds are present in SO 2
Ans. (1)
Sol. (1) SeF 4 –sp 3 d, lp = 1, shape = see-saw
CH 4 –sp 3 , lp = 0, shape =tetrahedral
(2) I3 –sp 3 , lp =2, shape = bent/angular
8
Trans effect order –
CN  C 6H 5 Br  NH 3
141.
Which one of the followng statements related to
lanthanons is incorrect ?
(1) All the lanthanons are much more reactive than
aluminium
(2) Ce(+4) solutions are widely used as oxidizing
agent in volumetric analysis
(3) Europium shows +2 oxidation state.
(4) The basicity decreases as the ionic radius
decreases from Pr to Lu.
Ans. (1)
Sol. (1) Lanthanon's are less reactive than aluminium due
to high IP (Lanthenoid contraction)
(2) Ce +4 is good oxidising agent and easily converted
into Ce +3
(3) Eu(63) = 4f 7 5d 0 6s 2 , Eu +2 = 4f 7
(4) In lanthenoids series 'Ce' to Lu ionic radius regular
decreases and covalent character increase, basic
character of hydroxide decrease
142. Jahn-Teller effect not observed in high spin
complexes of :(1) d4
(2) d9
(3) d7
(4) d8
Ans. (4)
Sol. John Teller effect explain axial distortion in perfect
octahedral geometry. It is present in d 4 high spin,
d7 low spin and d 9 configuations which have odd
number of electrons in eg set.
A weak John Teller effect in also present in d 7 high
spin complex which has odd number of electrons
in the set.
143. Which of the following can be used as the halide
component for Friedel-Crafts reaction ?
(1) Chloroethene
(2) Isopropyl chloride
(3) Chlorobenzene
(4) Bromobenzene
Ans. (2)
4

(2) square planar, sp 3 d2
Sol.
Sol.

(1) d z2 ,d x 2  y 2
(1) Planar triangle, sp 3 d3
The correct increasing order of trans-effect of the
following species is :
(3) PH 5 = d-orbital contraction absent
BiCl 5 = due to inert pair effect
(Bi +5 act as OA, Cl – act as RA)
(4) SO 2 : O=S=O
P -d, P –P both type bonds are present
CH 3
Sol.
+ CH
C
H CH 3
Anhy AlCl 3
CH Cl 

CH 3
cumene
But in chlorobenzene, Bromobenzene, chloroethene
lone pair of halogen are delocalised with  bonds,
so attain double bond character.
CODE-YY
144.
In which of the following molecules, all atoms are
coplanar ?
(1)
CH 3
CH 3
CN
C=C
CN
146.
In pyrrole
4
–—
(2)
5
3
..
N1
2
H
The electron density is maximum on :-
(3)
(4)
(1) 2 and 4
(2) 2 and 5
(3) 2 and 3
(4) 3 and 4
Ans. (2)
Ans. (3)
3
4
H
H H
..
N
5
H
2
1
H
H
H
H H
H
145.
(I)
H
(1)
O
Cl
COOH
H2
H (CH2)6 –NH
C (
C)
N
2C
C
H2
O
H2
H2
C H C H
C
C
NH 2
(4)
CH 3
H2
H2
C H C H
C
C
NH 2
NH 2
(CH2 )4
H
(III)
(IV)
(II)
(V)
Which of the following compounds shall not
produced propene by reaction with HBr followed
by elimination of direct only elimination reaction ?
H2
(4) H 3C–C–CH 2OH
Ans. (1)
Sol.
H 2 C CH 2
C
H2
(CH2 )6
NH 2
Hexamethylene diamine
CH 2 =C=O
O
C NH
(CH 2 )6
NH
H Br
CH 3 CH 2 CH 2
n
Elimination
H 3 C CH CH 2
Br
CH 3 —CH 2 —CH 2 —OH
Polymerisation
(CH2 )4
H
66
Adipic acid
C
H
H 2C—CH 2
C
(3)
H2
66
COOH + H 2 N
O
H
H2
(2) H C–C–CH
3
2Br
n
Ans. (2)
Sol.
HOOC
+
N
(1) H 2 C=C=O
6

(3)
CH 3
147.

NH 2
(2)
H2
H2
C H C H
C
C
6
+
N
Maximum electron density at (2) and (5)
as resonating structures III & IV are more stable than
(II) & (V) so are major contributor.
All carbons are sp 2 hybridised
Which one of the following structures represents
nylon 6,6 polymer ?
H2
H2
C H C
C
C
+
N

Sol.
+
N
H Br
H Br
Elimination
O
Br
H2C C
H 3 C CH CH 2
OH
H 3C C
E lim ination
CH 3 –CH 2 –CH 2 –Br 
 CH 3 –CH=CH
Br
2
9
NEET-II (2016)
148.
Which one of the following nitro-compounds does
not react with nitrous acid ?
CHO
H 3C
(1) H 3C–C–NO 2
H 3C
Sol.
H 3C
O
(4)
H 3C
H 3C
H 3C
H
OH
H
CH 2 OH
HO
HO
CHO
H
H
H
HO
CH
151.
NO 2
Ans. (1)
3°-Nitro compound does not react with HNO
because of absence of –H
149.
The central dogma of molecular genetics states that
the genetic information flows from :(1) DNA  RNA Proteins
H
L-Threose
In the given reaction
+
Sol.
OH
CH 2 OH
L-Erythrose
H2
C
OH
D-Threose
CH 2 OH
NO 2
H
CH 2 OH
CHO
NO 2
H2
C
C
H2
HO
HF


 P
0 C

(3)
C
H
OH
D-Erythrose
CH 3
(2)
CHO
H
2
the product P is :-

(2) DNA  RNA Carbohydrates
(3) Amino acids Proteins DNA
(1)
(2)
(4) DNA  Carbohydrates  Proteins
Ans. (1)
Transcription
Translation
DNA 
 RNA 
 Protein
150.
The correct corresponding order names of four
aldoses with configuration given below
CHO
CHO

Sol.
H
OH
HO
H
OH
H
CH 2OH
CHO
HO
H
H
HO
H
HO
F
(3)
H
(4)
OH
CH 2OH
CHO
Ans. (1)
OH
H
CH 2OH
CH 2OH
respectively, is :(1) L-erythrose, L-threose, D-erythrose, D-threose
(2) D-erythrose, D-threose, L-erythrose, L-threose
(3) L-erythrose, L-threose, L-erythrose, D-threose
(4) D-threose, D-erythrose, L-threose, L-erythrose
Ans. (2)
10
F
H
Sol.
+

Carbocation
(HF)
H
+

ESR
(alkylation)
[Friedel Craft reaction]
CODE-YY
152.
A given nitrogen-containing aromatic compound A
reacts with Sn/HCl, followed by HNO 2 to give an
unstable compound B. B, on treatment with phenol,
forms a beatiful coloured compound C with the
molecular formula C 12 H 10 N2 O. The structure of
compound A is :CN
154.
The correct structure of the product A formed in
the reaction
O
H 2(gas, 1 atmosphere)
CONH 2
(1)
A is :-
Pd/carbon, ethanol
(2)
NH 2
OH
NO 2
(3)
(4)
OH

(2)
(1)
Ans. (4)
OH
NH 2
N 2· Cl
HNO 2
Sn+HCl
Reduction
(A)
Aniline
(B)
Benzene Diazonium
chloride
Sol.
N N
OH
(4)
Ans. (4)
O
Ph—OH
Consider the reaction
155.
CH 3 CH 2 CH 2 Br + NaCN CH 3 CH 2 CH 2 CN + NaBr
O
H 2 gas, (1 atmosphere)


Pd /Carbon,Ethanol
Sol.

p-Hydroxy azo benzene (red colour dye)
153.
(3)

NO 2
O

Which among the given molecules can exhibit
tautomerism ?
This reaction will be the fastest in

O
(1) N,N'-dimethylformamide (DMF)
Ph
Ph
O
(2) water
O
I
II
III
(3) ethanol
(1) Both I and II
(4) methanol
(3) III only
Ans. (1)
Sol.
CH 3 –CH 2 –CH 2 Br + NaCN CH 3 CH 2 CH 2 CN+
N
O
2 path, which is favoured
by polar aprotic solvents like DMF, DMSO, etc.
DMF (Dimethyl formamide)
H C N Me
O Me
(4) Both I and III
Ans. (3)
NaBr
This reaction follows by S
(2) Both II and III
Sol.
H
Keto form
OH
H
Enol form
11
NEET-II (2016)
156.
The correct
order of strengths of the carboxylic
159.
acids
COOH
COOH
I
(1) 0.086 S cm
COOH
II
(3) 2.88 S cm 2 /mol
III
(4) 11.52 S cm
is
2 /mol
Ans. (4)
(1) III > II > I
(2) II > I > III
(3) I > II > III
(4) II > III > I
Sol.
C = 0.5 mol / dm
 = 5.76 × 10
T = 298 K
Ans. (4)
m 
Acidic Strength
COOH
COOH
O
more(—I)
O
>
>
less (—I)
160.
COOH
–3
3
S cm –1
  1000 5.76 10


M
0.5
3
 11.52 Scm 2 /mol
The decomposition of phosphine (PH 3 ) on tungsten
at low pressure is a first-order reaction. It is because
the

Sol.
2 /mol
(2) 28.8 S cm 2 /mol
O
O
The molar conductivity of a 0.5 mol/dm 3
solution of AgNO 3 with electrolytic conductivity of
5.76 × 10 –3 S cm –1 at 298 K is
(+I)
(1) rate is independent of the surface coverage
(2) rate of decomposition is very slow
157.
The compound that will react most readily with
gaseous bromine has the formula
(3) C3 H 6
Ans. (3)
Sol.
(2) C2 H 4

(1) C4 H 10
(4) C2 H 2
Sol.
The decomposition of PH 3 on tungsten at low
pressure is a first order reaction because rate is
proportional to the surface coverage.
161.
The coagulation values in millimoles per litre of the
Gaseous Bromine reacts with alkene to give allylic

CH 3 –CH=CH
2

 H2C C
H
Br
Br2  gas 
CH 2
electrolytes used for the coagulation of As
Which one of the following compounds shows the
III. (MgSO4 ) = 0.22
The correct
(2) Concentrated acetic acid
(3) H2O2
Ans. (1)
In acetic acid, H 2 O 2 and HCN inter molecular
hydrogen bo nd present but in cellulose
intramolecular hydrogen bond present.
12
are
II. (BaCl 2 ) = 0.69,
I. (NaCl) = 52,
(1) Cellulose
(4) HCN
2 S3
given below :
presence of intramolecular hydrogen bond ?
Sol.
(4) rate is inversely proportional to the surface
coverage
Ans. (3)
substituted product by free radical mechanism
158.
(3) rate is proportional to the surface coverage
order of their coagulating power is
(1) III > II > I
(2) III > I > II
(3) I > II > III
(4) II > I > III
Ans. (1)
Sol.
Coagulation power 
1
coagulation value
So, the order is III > II > I
CODE-YY
162.
During the electrolysis of molten sodium chloride,
of the strong electrolyte barium hydroxide is
gas using a current of 3 amperes is
(1) 2
(2) 3
(1) 220 minutes
(2) 330 minutes
(3) 0
(4) 1
(3) 55 minutes
(4) 110 minutes
Ans. (2)
Sol.
Ba OH 2 Ba
166.
 110 min.
n = 3 and l = 1 ?
(2) 14
(3) 2
Ans. (3)
(4) 6
(1) 0.77%
(2) 1.6%
(3) 0.0060%
(4) 0.013%
i
is
Pyridine (C 5 H 5 H 5 N) is a weak base
Kb = C 2
For a sample of perfect gas when its pressure is
changed isothermally from p
–9 )
Ans. (4)
Total 2 electron can fit in the orbital of 3p
164.
that forms
pyridine solution (K b for C 5 H 5 N = 1.7 × 10
Sol.
n=3, l =1 3p
5 H 5 N)
pyridinium ion (C5 H 5 N H) in a 0.10 M aqueous

Sol.
The percentage of pyridine (C
+
How many electrons can fit in the orbital for which
(1) 10
aq  2OH aq
 
solution i =3
35.5
3 t(sec)
96500
t (s) = 6433.33 sec
t(min) = 107.22 min
2
Van't Hoff factor = total number of ions present in

0.1 71 
Ba(OH)2 is strong electrolyte, so its 100% dissociation
occurs in solution
2Cl   Cl 2 g   2e 
E
W
 it
96500
163.
The van't Hoff factor (i) for a dilute aqueous solution
the time required to produce 0.10 mol of chlorine
Ans. (4)
Sol.
165.
to p f, the entropy

1.7 10
0.1
9
change is given by
 pi 
(2) S = RT ln  
 pf 

 pf 
(1) S = nRT ln  
 pi 
 pf 
(3) S = nR ln  
 pi 
 pi 
(4) S = nR ln  
 pf 
4
100
%  0.013%
167.
In calcium fluoride, having the fluorite structure, the
coordination numbers for calcium ion (Ca
2+ )
and
fluoride ion (F –) are
S  nC pm n
Tf
P
nR n i
Ti
Pf
For isothermal T i = T f, ln1 = 0
S  nR n
4
%  1.30 10
Ans. (4)
Sol.
  1.30 10
Pi
Pf
(1) 8 and 4
(2) 4 and 8
(3) 4 and 2
(4) 6 and 6
Ans. (1)
Sol.
In CaF 2 , the coordination numbers for
Ca +2 = 8
F  4
13
NEET-II (2016)
168.
If the E°cell for a given reaction has a negative value,
which of the following gives the
relationships for the values of
(1) G° < 0; K
eq
>1
(2) G° < 0; K
eq
<1
(3) G° > 0; K
eq
<1
(4) G° > 0; K
eq
>1
correct
Sol.
G° and K eq ?
a
a S
 E0cell  ve
171.
 G 0  2.303RT log K
eq
 K eq  1
Which one of the following is incorrect
for ideal
(1) P = Pobs – P
calculated by Raoult's law
=0
Sol.

(2) G mix = 0
(3) H mix = 0
(4) U mix = 0
H mix  0

For an ideal solution
U mix  0
S mix  0
According to G mix  H mix  T S mix
172.
(3) 40, 30
(4) 60, 40
Let atomic weight of x is A
n xy2  0.1 
x
and y is A y
10
A x  2A y
A x  2A y  100 ...(1)
n x3 y2  0.05 
9
3A x  2A y
3A x + 2A y = 180 ...(2)
on solving eq. (1) and (2)
Ax = 40, A y = 30
Incorrect answer, is G mix =0
The solubility of AgCl(s) with solubility product
(1) 3.75 × 10
1.6 × 10
(3) 6 × 10
–10
(1) 1.6 × 10
in 0.1 M NaCl solution would be
M
–11
(2) zero
(3) 1.26 × 10
(4) 1.6 × 10
Ans. (4)
14
(2) 30, 20
The number of electrons delivered at the cathode
during electrolysis by a current of 1 ampere in
60 seconds is (charge on electron = 1.60 × 10 –19 C)
 G mix  0
170.
(1) 20, 30
Ans. (3)
solution ?
Sol.

Suppose the elements X and Y combine to form
two compounds XY 2 and X 3 Y2 . When 0.1 mole of
XY2 weighs 10 g and 0.05 mole of X 3 Y2 weighs
9 g, the atomic weights of X and Y are

G 0  ve  G  0
Ans. (2)
 Cl  aq
0
0.1 S
Ksp = 1.6 × 10 –10 = [Ag + ] [Cl–] = S (0.1+S)
Ksp is small, S is neglected with respect to 0.1 M
1.6 × 10 –10 = S × 0.1
S = 1.6 × 10 –9 M
 G 0  nF E 0cell
169.
Na (aq
0
0.1M





 Ag aq   Cl aq 
0
0
S
S 0.1
AgCl  s 
Ans. (3)
Sol.
NaCl aq  
0.1M
0
–9
M
M
23
(2) 7.48 × 10
(4) 6 × 10
20
Ans. (1)
Sol.
–5
20
Q = ne
i.t = n.e
n
1  60
 3.75 10
1.6 10 19
20
electrons
23
CODE-YY
173.
Boric acid is an acid because its molecule
(1) accepts OH
–
177.
NO 2, NO 3 and NH 4 respectively are
from water releasing proton
(2) combines with proton from water molecule
+
(3) contains replaceable H
ion
(4) gives up a proton
3
(2) sp 2 , sp and sp
3
(3) sp, sp 3 and sp
2
Ans. (1)
Sol.
B(OH)3 + H 2 O  [B(OH)4 ]– + H +
174.
AlF3 is soluble in HF only in presence of KF. It is due
Sol.
to the formation of
(1) AlH3
(2) K[AlF3 H]
(3) K3 [AlF3 H 3 ]
(4) K3 [AlF6 ]
178.
Sol.
AlF3 + 3KF  K3 [AlF6 ]
175.
Zinc can be coated on iron to produce galvanized
iron but the reverse is not possible. It is because
Linear
NO 3  sp 2
Trigonal planar
NH 4  sp 3
Tetrahedral
Which of the following fluoro-compounds is most
likely to behave as a Lewis base ?
(2) SiF4
(3) BF3
(4) PF 3
Sol.
PF 3 act as Lewis base due to present of lone pair
on P atom.
179.
Which of the following pairs of ions is isoelectronic
and isostructural ?

iron
(1) CF4
Ans. (4)
(1) zinc has lower negative electrode potential than
(2) zinc has higher negative electrode potential than
iron
NO 2  sp

Ans. (4)
(3) zinc is lighter than iron
(4) zinc has lower melting point than iron
(1) SO 23 , NO 3
(2) ClO 3, SO 32 
(3) CO 23 , NO 3
(4) ClO 3, CO 32 
Ans. (2 & 3)

Sol.
(1) sp, sp2 and sp
(4) sp 2 , sp 3 and sp
Ans. (1)
Ans. (2)
The hybridizations of atomic orbitals of nitrogen in
Zinc has higher negative electrode potential than
Sol.
(2) In SO 23 ,ClO 3 , No. of electrons = 42,
Shape : Pyramidal
iron, so iron cannot be coated on zinc.
176.
(3) In CO 3 2,NO 3 , No. of electrons = 32
The suspension of slaked lime in water is known as
(1) milk of lime
Shape : trigonal planar
(2) aqueous solution of slaked lime
180.
In context with beryllium, which one of the following
statements is incorrect ?
(3) limewater
(1) Its salts rarely hydrolyze.
(4) quicklime
(2) Its hydride is electron-deficient and polymeric.
(3) It is rendered passive by nitric acid.
Ans. (1)
Sol.
Aqueous solution of slaked lime
Suspension solution of slaked lime
 lime water
 milk of lime
(4) it forms Be 2 C.
Ans. (1)
Sol.
Be salts are covalent nature, so easily hyrolysed.
15