CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022

SINCE 1984

PHYSICS & CHEMISTRY COMBINED TEST

SCHEDULE-2021-2022

Test ID : 000 Paper ID : CPT-17

li ity-cumte
b NEW tu
Li
st I
In
GH

t
gh
T

i
INSTITUTE

L
w
a
e
l

es t
Na

TEST DATE : 15-03-2022

SOLUTION

NLI / 1
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022

PHYSICS
SECTION-A SECTION-A
1. (3) [NCERT-] 1. (3) [NCERT-]

B' = n2B B' = n2B

= 22B = 22B

= 4B = 4B

2. (1) [NCERT-] 2. (1) [NCERT-]

Vector form of Biot-savart's law

   
 µ0 id l  r  µ0 id l  r
dB = dB =
4π r 3 4π r 3

3. (1) [NCERT-] 3. (1) [NCERT-]

q q
r r

Magnetic field at centre

u te
B
µ0 2i
stit µ 2i
4 r
t In B  4 r 0

igh

µ0 qn L 
µ qn
2 r
ew 0

2 r
N
4  10 7 qn 4  10 7 qn
 
2 r 2 r

2nq 2nq
  10 7   10 7
r r

4. (3) [NCERT-] 4. (3) [NCERT-]

Magnetic induction at O
O

µ0 3 i
B µ0 3 i
4 2 R B
4 2 R

3µ0 i 3µ0 i
 
8 R 8 R

5. (4) [NCERT-] 5. (4) [NCERT-]

µ0 2 q µ0 2 q
B B
4 R t 4 R t

µ0  2  2  1.6  10 19 µ0  2  2  1.6  10 19

4  0.8  2 4  0.8  2

NLI / 2
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
6. (3) [NCERT-] 6. (3) [NCERT-]

µ0N µ0N
B i B i
l l

10 10
 4  107  5  4  107  5
102 102

 2  103 Tesla  2  103 Tesla

7. (4) [NCERT-] 7. (4) [NCERT-]

B = µ0 ni B = µ0 ni

hence B  i Bi

8. (4) [NCERT-] 8. (4) [NCERT-]

The magnetic induction at O due to the current in 'O' AB O

portion AB will be zero because O lies on AB

9. (1) [NCERT-] 9. (1) [NCERT-]

u te
stit
t In i
gh
i

Li
Magnetic lines
Magnetic lines
of ew of
force N force

10. (4) [NCERT-] 10. (4) [NCERT-]

Magnetic field due to long straight wire

µ0 2i µ0 2i
B B
4 d 4 d

B1 d2 40 B1 d2 40
   
B2 d1 10 B2 d1 10

B1 0.04 B1 0.04
B2   B2  
4 4 4 4
= 0.01 T = 0.01 T

11. (1) [NCERT-] 11. (1) [NCERT-]

NLI / 3
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
12. (1) [NCERT-] 12. (1) [NCERT-]
Magnetic field at centre due to circular coil

µ0 2Ni
B µ0 2Ni
4 r B
4 r
 10 7  2  3.14  0.1  1000  10 7  2  3.14  0.1  1000
0.1 0.1
13. (1) [NCERT-] 13. (1) [NCERT-]
µ0 2i
B µ0 2i
4 R B
4 R

B A iA RB I 2R B A iA RB I 2R
       
BB iB R A 2I R BB iB R A 2I R

14. (4) [NCERT-] 14. (4) [NCERT-]

4
1 Tesla = 10 gauss 4
1 Tesla = 10 gauss
15. (4) [NCERT-] 15. (4) [NCERT-]

2l 2l

u te
stit
i
0
2l
t In i
0
2l

igh
L
ew
Magnetic field due to one side of the square at O
centre O N
µ0 i
µ0 i B1  2 sin 45
B1  2 sin 45 4 l
4 l
Hence, magnetic field at centre due to all side

µ0 2in µ0 2in
B  4B1  B  4B1 
l l
16. (3) [NCERT-] 16. (3) [NCERT-]

B A iA rB B A iA rB
   
BB iB rA BB iB rA

1 i1 2 1 i1 2
   
3 i2 1 3 i2 1

i1 1 i1 1
 
i2 6 i2 6

17. (4) [NCERT-] 17. (4) [NCERT-]

At mid point, magnetic field due to both the wires

are equal & opposite So Bnet = 0
Bnet = 0
NLI / 4
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
18. (1) [NCERT-] 18. (1) [NCERT-]

a a

45° 45°
i a i a
45° 0 45° 0

Magnetic field due to one side of square

µ0 i µ0 i
B (sin 45  sin 45) B (sin 45  sin 45)
4 a / 2 4 a / 2
Magnetic field at centre due to all sides
Bnet = 4B Bnet = 4B

µ0 2 2i µ0 2 2i
=4 =4
4 a 4 a

19. (1) [NCERT-] 19. (1) [NCERT-]

mv 2mqV mv 2mqV
r  r 
qB qB qB qB

u te m
titR  m
R1 mx R
 (q,V,B  Same)
1 x
(q,V,B  Same)
R2 my s
In
2 y

t
mx  R1 
2
igh m R 
 
2


my  R2 
L m
x

R 
1

ew y 2

20. (2)
N[NCERT-] 20. (2) [NCERT-]
, 0o, 180°, 90°
If  is neither 0°, 180°, 90° then path is helical.
21. (1) [NCERT-]
21. (1) [NCERT-]
F = q V B sin 
F = q V B sin 
 = 0°
 = 0°
F = 0. F = 0.
22. (4) [NCERT-] 22. (4) [NCERT-]
     
F  q(V  B) F  q(V  B)

23. (1) [NCERT-134] 23. (1) [NCERT-134]

 0  l l l   1 1  0  l l l   1 1
B      0    B      0   
4  r r r   2r 4r  4  r r r   2r 4r 

24. (3) [NCERT-] 24. (3) [NCERT-]

mv mv
r r
eB eB

e v e v
 
m Br m Br

NLI / 5
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
25. (1) [NCERT-] 25. (1) [NCERT-]
Force acts perpendicular to the velocity in a
magentic field so speed of electron will remain
same
26. (2) [NCERT-] 26. (2) [NCERT-]

qBr0 qBr0
v v
m m

2
1  qBr0 
2 1  qBr0 
K max  m
K max  m
2  m  2  m 

27. (4) [NCERT-] 27. (4) [NCERT-]

Magnetic force on charge will be zero hence path

will be straight line
28. (3) [NCERT-] 28. (3) [NCERT-]

w = Fd cos 90° w = Fd cos 90°

=0 =0

29. (2) [NCERT-] 29. (2) [NCERT-]

By the fleming's left hand rule

u te
30. (2) [NCERT-] 30. (2)it
s t [NCERT-]
M=iA
t In M = i A
= ef  r2
igh = ef  r 2

L e  v  r
=
e  v  r 2
ew =
2

2 r N 2 r

31. (4) [NCERT-] 31. (4) [NCERT-]

   
In equilibrium angle between M & B is zero. It is M&B
happened when plane of the coil is perpendicular 
 B
to B
32. (1) 32. (1)

33. (1) [NCERT-]

For no force on wire C 33. (1) [NCERT-]

FCD = FCB 'C'

µ0 2  15  5 µ 2  5  10 FCD = FCB
l  0 l
4 x 4 (15  x) µ0 2  15  5 µ 2  5  10
l  0 l
4 x 4 (15  x)

NLI / 6
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
34. (3) [NCERT-] 34. (3) [NCERT-]
 = MB sin 90°  = MB sin 90°
= ir2 B = ir2 B
35. (1) [NCERT-] 35. (1) [NCERT-]
 
L0 L0
     
F  i(L  B) F  i(L  B)
 
F =0 F =0
SECTION-B SECTION-B
36. (3) [NCERT-] 36. (3) [NCERT-]

µ0 2M µ0 2M
Baxis  Baxis 
4 d3 4 d3

µ0 M µ0 M
Bbroad  Bbroad 
4 d3 4 d3
37. (1) [NCERT-142] 37. (1) [NCERT-142]
Inside the conductor, B  r and outside B  1/r . B r B  1/r
There fore the B – r curve inside the conductor is a
B–r
straight line inclined to r axis, while outside it, it is te
a rectangular hyperbola.
t itu
38. (2) [NCERT-] 38.ns(2) [NCERT-]
t I
W =nW
1 2
g h W =nW 1 2

MB( 1 – cos 90°) = n MB(1 – cos 60°) Li MB( 1 – cos 90°) = n MB(1 – cos 60°)
w
Ne
n=2 n=2

39. (2) [NCERT-] 39. (2) [NCERT-]

     
  MB   MB
= 50 i × (0.5i + 3j) = 50 i × (0.5i + 3j)

= 150 K̂ N.m = 150 K̂ N.m

40. (3) [NCERT-144] 40. (3) [NCERT-144]

 0  dl sin   0  dl sin 
| dB | | dB |
4 r2 4 r2
dl = x = 10–2 m, I = 10 A, r = 0.5 m = y, dl = x = 10–2 m, I = 10 A, r = 0.5 m = y,
 = 90o; sin  = 1  = 90o; sin  = 1

10 7  10  102 10 7  10  102
| dB |  4  10 8 Tesla | dB |  4  10 8 Tesla
25  10 2 25  10 2

41. (4) [NCERT-] 41. (4) [NCERT-]

Bv Bv Bv Bv
tan    tan   
BH 3B v BH 3B v

 = 30°  = 30°

42. (1) [NCERT-] 42. (1) [NCERT-]

NLI / 7
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
43. (2) [NCERT-] 43. (2) [NCERT-]
In sum position

1 1
T T
M1  M2 M1  M2

in difference position

1 1
T T
M1  M2 M1  M2

3 2 2M  M 3 2 2M  M
 
T 2 2M  M T 2 2M  M

T = 3 3 second T= 3 3

44. (3) [NCERT-139] 44. (3) [NCERT-139]

0 2 i(2r)2 0 2 i(2r)2
By = . By = .
4  [(2r)2  d2 ]3/2 4  [(2r)2  d2 ]3/2

0 2 ir2 0 2 ir2
Bx = . Bx = .
4  [r2  d2 / 4]3/2 t e 4 [r2  d2 / 4]3/2
tu
By 4[4r2  d2 ]3/ 2 sti B 4[4r  d ] 2 2 3/ 2

Bx

[4r 2  d2 ]3/2 [8] t In B  [4r  d ] [8]
y
2 2 3/2

gh
x

Li
By 4 1 B 4 1
  ew y
 
Bx 8 2 N Bx 8 2

45. (4) [NCERT-139] 45. (4) [NCERT-139]

As r = –p/qB As r = –p/qB
so y = 2r = –2p/qB so y = 2r = –2p/qB
46. (3) [NCERT-134] 46. (3) [NCERT-134]
Field is zero, due to symmetrical current
distruibution .
47. (4) [NCERT-136] 47. (4) [NCERT-136]

Bcentre Bcentre
Bx = Bx =
8 8

0 2 lR 2 1 0 2 l 0 2 lR2 1 0 2 l
 
4 (R  x )
2 2 3/ 2
8 4 R 4 (R  x )
2 2 3/ 2
8 4 R
8 R3 = (R2 + x2)3/2 8 R3 = (R2 + x2)3/2
(2R)3 =(R2 + x2)1/2 (2R)3 =(R2 + x2)1/2
2R =(R2 + x2)1/2 2R =(R2 + x2)1/2
4R2 = (R2 + x2) or x2 = 3R2 4R2 = (R2 + x2) or x2 = 3R2

x= R 3 x= R 3

NLI / 8
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
48. (1) [NCERT-135 ] 48. (1) [NCERT-135 ]

mg = I lB mg = I lB

mg mg
B B
l l

0.2  9.8 0.2  9.8

  0.65 Tesla   0.65
2  1.5 2  1.5

49. (4) [NCERT-147] 49. (4) [NCERT-147]

   
  = 0 × Total current enclosed
B.d   = 0 ×
B.d

= 0 [I1 + I2 + I3 + I4 + I5] = 0 [I1 + I2 + I3 + I4 + I5]

= 0 [I + 2 + 3 + (– 1) + (–4)] = 0 [I + 2 + 3 + (– 1) + (–4)]
= 0 Wb m–1 = 0 Wb m–1

50. (1) [NCERT-171] 50. (1) [NCERT-171]

In the line of current carrying wire the magnetic field

is zero.

u te
stit
t In
igh
L
ew
N

NLI / 9
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022

CHEMISTRY
SECTION-A SECTION-A

51. (3) [NCERT-XII-247] 51. (3) [NCERT-XII-247]

Ethylenediaminetetraacetate ion [EDTA–4] is an [EDTA–4]

important hexadentate ligand. It can bind through
two nitrogen and four oxygen atom to a central
metal ion.

52. (1) [NCERT-XII-256] 52. (1) [NCERT-XII-256]

–2
[Ni(Cl)4] – 2.82 B.M. [Ni(Cl)4]–2 – 2.82 B.M.
–3
[Co(C2O4)3] – 0 B.M. [Co(C2O4)3]–3 – 0 B.M.
[FeF6]–3 – 5.92 B.M. [FeF6]–3 – 5.92 B.M.
[Mn(CN)6]–4 – 1.73 B.M. [Mn(CN)6]–4 – 1.73 B.M.
53. (2) [NCERT-XII-258] 53. (2) [NCERT-XII-258]

[FeF6]–4 complex, Fe is in +2 oxidation state. [FeF6]–4 , Fe

+2
3d 4s

te
3d 4s
u
–
As F is a weak field ligand, it cause no pairing of stit
electrons there fore, electronic configuration of Fe
t In F +2 –

gh
–4
in [FeF ] k is t e 4 2 [FeF6]–4 Fe+2
6 2g g .

Li 4
is t2g eg 2
.
54. (4) [NCERT-XII-255]
w
Ne
54. (4) [NCERT-XII-255]
8 2
[Ni(CO) ] = 3d 4s
4
[Ni(CO)4] = 3d8 4s2
CO is strong field ligand, so unpaired electrons
get paired in [Ni(CO)4] CO [Ni(CO)4]

3d 4s 4p CO
xx x x x x x x 3d 4s 4p CO
CO COCOCO Ni xx x x x x x x
sp3 (Tetrahedral) CO CO CO COCOCO Ni
Dia-magnetic in nature CO CO CO
sp3 (Tetrahedral)
Dia-magnetic in nature CO
55. (1) [NCERT-XII-258]
55. (1) [NCERT-XII-258]
According to spectro chemical series, order of
increasing field strength is :

SCN– < F – < C2O4–2 < CN– < CO

SCN– < F – < C2O4–2 < CN– < CO
56. (3) [NCERT-XII-252]
56. (3) [NCERT-XII-252]
+3 en +3
en +3 +3
en en
en Co Co en
en Co Co en
en en
en en
(dextro) (Laevo)
Mirror (dextro) (Laevo)
Mirror

NLI / 10
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
57. (2) [NCERT-XII-259] 57. (2) [NCERT-XII-259]

4 4 4 4
t   0   27000 = 12000 cm–1 t   0   27000 = 12000 cm–1
3 9 3 9

58. (2) [NCERT-XII-259] 58. (2) [NCERT-XII-259]

Increasing order of crystal field splitting energy is

H2O < NH3 < CN–
H2O < NH3 < CN–
Thus increasing order of crystal field splitting
energy for the given complexes is

[Co(H2O)6]+3 < [Co(NH3)6]3+ < [Co(CN)6]–3

[Co(H2O)6]+3 < [Co(NH3)6]3+ < [Co(CN)6]–3
Thus, increasing order of wavelength of absorption
is

[Co(CN)6]–3 < [Co(NH3)6]+3 < [Co(H2O)6]+3 [Co(CN)6]–3 < [Co(NH3)6]+3 < [Co(H2O)6]+3

59. (2) [NCERT-XII-245] 59. (2) [NCERT-XII-245]

Cl– is a weak field ligand but in [PtCl4]–2 compound Cl– [PtCl4]–2

Cl– is a strong field ligand with Pt, so hybridisation Cl–
is dsp2 and square planer. dsp2

60. (1) [NCERT-XII-262] 60.

u te
(1) [NCERT-XII-262]

The presence of +ve charge on the metal carbonyl stit +ve

would resist the flow of the metal electron charge
t In * CO CO
to * orbitals of CO. This could increase the CO h
g
Li
bond order and hence absorb at a higher frequency
compared to its absorption in a neutral w metal
carbonyl. N e
61. (1) [NCERT-XII-249]
61. (1) [NCERT-XII-249] Fe+2 = 3d64s2 = 3d64s0
Fe+2 = 3d64s2 = 3d64s0 CN– = (d2sp3)
CN– = stronge field ligand so hybridisationis (d2sp3)
[Fe(CN)6]–4 = (II)
[Fe(CN)6]–4 = Hexacyanferrate(II) ion.

62. (4) [NCERT-XII-251] 62. (4) [NCERT-XII-251]

[Co(NH3)4Cl2] has 2 geometrical isomers. [Co(NH3)4Cl2]

Cl Cl Cl Cl
H3N H3N Cl H3N NH3 H3N Cl
NH3 Co Co
Co Co H3N NH3 H3N NH3
H3N NH3 H3N NH3
Cl NH3 Cl NH3
Trans Cis Trans Cis
Cis forms has Cl–Co–Cl angle of 900. Cl–Co–Cl 900

63. (2) [NCERT-XII-266] 63. (2) [NCERT-XII-266]

More the number of unpaired electron, higher in

the magnetic moment.

Fe+2 = 3d64s0 Configuration Fe+2 = 3d64s0

No. of unpaired electrons = 4 =4

NLI / 11
 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
64. (4) [NCERT-XII-252] 64. (4) [NCERT-XII-252]

Octahedral complexes having bidentate ligands of M(aa)2b2 M(aa)2bc

the type M(aa)2b2 or M(aa)2bc can exist in cis and
trans-isomeric form. Where, aa is a symmetrical aa
bidentate ligand.
65. (1) [NCERT-XII-263]
65. (1) [NCERT-XII-263]
[(Ph3P)3RhCl]
Wilkinsion catalyst is [(Ph3P)3RhCl].
66. (2) [NCERT-XII-248]
66. (2) [NCERT-XII-248]
(+3, 0, +2)
(+3, 0, +2)
[Cr(H2O)6]Cl3
[Cr(H2O)6]Cl3
x+0×6+3×(–1)=0 x=+3
x+0×6+3×(–1)=0 x=+3
[Cr(C6H6)2]
[Cr(C6H6)2]
y+2×0=0 y=0
y+2×0=0 y=0
[NiCl2 (PPh3)2]
[NiCl2 (PPh3)2]
z+2×(–1)+0×2 = 0
z+2×(–1)+0×2 = 0

z = +2
u te
z = +2

67. (2) [NCERT-XII-263] stit

67. (2) [NCERT-XII-263]

Cis - [PtCl (NH ) ] is used to inhibit the growth of t

In - [PtCl (NH ) ]
2 3 2
g h 2 3 2

Li 68. (1)
tumours.

68. (1) w
[NCERT-XII-245]
e
[NCERT-XII-245]

[Ni(NH3)6]+2 Octahedral
N [Ni(NH3)6]+2

[Pt(NH3)4]+2 Square planar [Pt(NH3)4]+2

[Zn(NH3]4]+2 Tetrahedral [Zn(NH3]4]+2

69. (3) [NCERT-XII-258] 69. (3) [NCERT-XII-258]

Degenerate orbitals of [Cr(H2O)6]+3 [Cr(H2O)6]+3

Cr
+3 dx2-y2 dz2 dx2-y2 dz2
Cr+3

dxy dyz dzx dxy dyz dzx

Hence according to the options given degenerate dyz

orbitals are dxy, dyz and dzx. dzx
70. (3) [NCERT-XII-250] 70. (3) [NCERT-XII-250]
K3[Cr(C2O4)3] K3[Cr(C2O4)3]
Potassium trioxalato chromate(III)
(III)

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 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
71. (3) [NCERT-XII-230] 71. (3) [NCERT-XII-230]

Factual

72. (3) [NCERT-XII-231] 72. (3) [NCERT-XII-231]

Mn2O7 = Mn+7 = Acidic Mn2O7 = Mn+7 =

V2O5 = V+5 = Amphoteric V2O5 = V+5 =
CrO = Cr+2 = Basic
CrO = Cr+2 =
73. (2) [NCERT-XII-229]
73. (2) [NCERT-XII-229]
Ti3+  3d1  coloured ions
Ti3+  3d1 
Ti22  3d2 4s2
ti  3d2 4s2
74. (4) [NCERT-XII-224]
74. (4) [NCERT-XII-224]
Cr3+  t2g3 eg0  symmetrical filling
Cr3+  t2g3 eg0 
75. (3) [NCERT-XII-204]
75. (3) [NCERT-XII-204]
XeF6 + 2H2O  XeO2F2 + 4HF XeF6 + 2H2O  XeO2F2 + 4HF
.. ..
:F:
.. .. .. : :F: ..
:F
..
F
..
te .. ..F: ..
:F
..
itu Xe ..F.. :
.. Xe .. ..
19 lone pairs
:F
.. .. :
F
t :F 19
:..
F: s ..

t In F:
:..

76. (3)
h
[NCERT-XII-197]ig 76. (3) [NCERT-XII-197]
L
2NaOH + Cl2  NaCl + NaClO + H O
ew2
2NaOH + Cl  NaCl + NaClO + H O 2 2

Cl–
ClO – N Cl– ClO–

77. (1) [NCERT-XII-234] 77. (1) [NCERT-XII-234]

2KMnO4+5SO2+2H2OK2SO4+2MnSO4+2H2SO4 2KMnO4+5SO2+2H2OK2SO4+2MnSO4+2H2SO4

Dark purple Colourless

78. (1) [NCERT-XII-0] 78. (1) [NCERT-XII-0]

P4+8SOCl2  4PCl3 + 4SO2 + 2S2Cl2
P4+8SOCl2  4PCl3 + 4SO2 + 2S2Cl2
79. (2) [NCERT-XII-179]
79. (2) [NCERT-XII-179]

Factual
80. (2) [NCERT-XII-189]
80. (2) [NCERT-XII-189]
H2S2O8
H2S2O8 Peroxodisulphuric acid
O O–O O
O O–O O
S S
S S
O OH OH O
O OH OH O

81. (2) [NCERT-XII-168]

81. (2) [NCERT-XII-168]

Factual

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 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
82. (3) [NCERT-XII-219] 82. (3) [NCERT-XII-219]
Mn  Least melting point Mn 
83. (2) [NCERT-XII-229] 83. (2) [NCERT-XII-229]
TiF62–  Ti+4  3d0  colourless TiF62–  Ti+4  3d0 
Cu2Cl2  Cu+1  3d10  colourless
Cu2Cl2  Cu+1  3d10 
84. (4) [NCERT-XII-228]
84. (4) [NCERT-XII-228]
 = 2.84; n = 2 m = 2.84; n = 2
Ni2+  3d8 Ni2+  3d8

2 unpaired electrons 2 unpaired electrons

85. (1) [NCERT-XII-205] 85. (1) [NCERT-XII-205]

XeF4 + O2F2  XeF6 + O2 XeF4 + O2F2  XeF6 + O2

Xe+4  Xe+6 O+1  O20 Xe+4  Xe+6 O+1  O20

oxidation Reduction

SECTION-B SECTION-B

86. (4) ut
e [NCERT-XII-173]
tit P O
86. (4) [NCERT-XII-173]
P O is more acidic than P O because with n s P4O6
increase in oxidation state, acidity of the oxide t I
4 10 4 6 4 10

increases. Likewise N O is more acidic than N O g. h

2 5
L
Oxide of a more electronegative element is more i 2 3 N O ,N O 2 5 2 3

acidic.
e w NO, PO 2 3 4 6

N2O3 is more acidic than P4O6. N 87. (4) [NCERT-XII-174]

87. (4) [NCERT-XII-174]

The correct reaction is NH4Cl + naNO2  N2+2H2O+NaCl
88. (2) [NCERT-XII-178]
NH4Cl + NaNO2  N2+2H2O+NaCl
N 4 +5
88. (2) [NCERT-XII-178]
Covalency of N is 4 in N2O5 and its oxidation state N2 O 5 N 4
is +5. +5

As clear from the structure of N2O5, covalency of N2 O 5 N 4

nitrogen is 4 but the oxidation state is +5. +5
x–2 x–2
N2O5 is a neutral compound. Let the oxidation state N2 O 5 
of N2O5 be x.
N +x
(2 × x) + (–2 × 5) = 0
(2 × x) + (–2 × 5) = 0
2x – 10 = 10 2x – 10 = 10
x=5 x=5

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 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
89. (4) [NCERT-XII-180] 89. (4) [NCERT-XII-180]

Both chromium and aluminium become passive Cr Al HNO3

on reaction with conc. HNO3 because of the
formation of oxide on the surface of the metal.

90. (3) [NCERT-XII-186] 90. (3) [NCERT-XII-186]

H2O has maximum boiling point because it exhibits H2 O

hydrogen bonding. On moving down the group size
of atom increases and hence magnitude of van
der waals forces increases. Therefore, the correct –
order of boiling point is H2O >H2Te > H2Se > H2S.
H2O > H2Te > H2Se > H2S.
91. (1) [NCERT-XII-194]
91. (1) [NCERT-XII-194]

5SO2  2MnO4  2H2O  5SO24  4H  2Mn2

 2H2O  5SO24  4H  2Mn2
Pink coloured Colourless
5SO2  2MnO4
Pink coloured Colourless
7+ 2+
The reduction of Mn to Mn result in
decolorization of KMnO4.
Mn7+ Mn2+ KMnO4
92. (3) [NCERT-XII-198]
92. (3) [NCERT-XII-198]
The correct order of bond dissociation enthalpy
among halogen is Cl2 > F 2 > Br2 > I 2. Bond
dissociation enthalpy of Cl2 is greater than F2 due
u te
to repulsion between lone pairs in F2 which is
stit Cr > F 2 2
> Br2 > I2
smaller than Cl2.
t In (3)
93. [NCERT-XII-228]
93. (3) [NCERT-XII-228] h
g
Li
Magnetic moment  n (no. of unpaired electron) 
e w
Co 2+
3d 7
n=3 Mn 3d N n=5 2+ 5
Co2+ 3d7 n=3 Mn2+ 3d5 n=5
Mn3+ 3d4 n=4 Cr3+ 3d n=3 Mn3+ 3d4 n=4 Cr3+ 3d n=3
94. (1) [NCERT-XII-185]
94. (1) [NCERT-XII-185]

O O
|| || O O
HO  P  P  OH || ||
| | HO  P  P  OH
OH OH | |
OH OH

(H4P2O6) Hypophosphoric acid contains one P–P

bond. (H4P2O6) P–P

95. (4) [NCERT-XII-200] 95. (4) [NCERT-XII-200]

96. (3) [NCERT-XII-185] 96. (3) [NCERT-XII-185]

Strong reducing behaviour of H3PO2 is due to

presence of one –OH group and two P–H bonds. H3PO2 –OH
P–H
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 CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022
97. (3) [NCERT-XII-228] 97. (3) [NCERT-XII-228]

Xn+ : Fe2+ Xn+ : Fe2+

n=2 n=2

No. of unpaired electron = 4 =4

= 24 B.M. = 24 B.M.

98. (1) [NCERT-XII-232] 98. (1) [NCERT-XII-232]

Factual

99. (4) [NCERT-XII-222] 99. (4) [NCERT-XII-222]

Factual

100. (2) [NCERT-XII-238] 100. (2) [NCERT-XII-238]

+7 is the maximum oxidation state. +7

u te
stit
t In
igh
L
ew
N

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CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022

u te
stit
t In
igh
L
ew
N

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CPT-17 [Solution] CRACK NEET 2022 / 15-March-2022

u te
stit
t In
igh
L
ew
N

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