Final Round Test Series NEET - 20201

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Final Round Test Series NEET - 2020Class : XII + Repeater Date : 28.03.2020 Test No. : 12

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HINTS AND SOLUTIONS

01. (1) we know A2.

22

21

2

1

A

A

1

4

2

1

A

A A

1 : A

2= 2 : 1

02. (3)max

= 221 = 24 = 9 .

min = 221 – = 2–4 = .

03. (3)

04. (3)

05. (2)

06. (2)

07. (1)

08. (2)

09. (1)

10. (3)

11. (2)

2x

6

10

2 1 107.692

105460 10

12. (1)

13. (3)

14. (4)

15. (3)

16. (2) Wave nature

17. (2) Principle of Superposition.

18. (2)

Given I1 : I2 = 100 : 1

1

2

I

I = 10 : 1

Imax = 2

1 2I I = (10 + 1)2 = 121

Imin = 2

1 2I I = (10 – 1)2 = 81

max

min

I

I = 121 : 81

19. (4) In coherent sources initial phase remains constant.

20. (2) Phase difference changes with time.

21. (3) Given I1 = I & I

2 = 4I

221.max

III = 9I

2

min. 1 2I I I = I

22. (3)

23. (3)

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24. (3)D

d

25. (2)

26. (1) = 2

D

y.d

y = 41

dD

= 2

I’ = 4I cos2 2

= 2I

27. (1)

t 1.33

1.5

2 × 600t3

4

t = 225 nm.

28. (2) In the ray diagram shown

AQC ~ DMC

x

/ 2 =

3d

d x =

3

2

O

D

E

CIddQ

A

B

path

x

Total length = 3

2

3

23

(option = B)

29. (1)

30. (2)

31. (3)

32. (3)

33. (2)u

1

v

1

f

1

34. (4) f = R/2

35. (2)

36. (3)

37. (2)

38. (1)21 f

1

f

1

f

1

20

1

20

1 so f = 10 cm

39. (4)

40. (3) Red colour have maximum wavelength and least

refractive index.

41. (2)

42. (1)1 1 1

f u v

1

10

1

30

1

v

1 1

30

1

10v

1 3

30

2

30

1

15cm

V = – 15cm (Real image)

43. (4)u

1

v

1

f

1

f=10cm

10cm

4cm

ve

F

10

1

v

1

10

1

10

1

10

1

v

1

V = + 5cm

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cm2h10

5

4

hi

44. (1) Applying Snell's law 1 sin 60° = 3 sinr

r = 30°

From symmetry r' = r = 30°

Again applying snell's law at second surface

1 sin e = 3 sinr e = 60°

Deviation at first surface = i – r = 60° – 30° = 30°

Deviation at second surface

= e – r = 60° – 30° = 30°

Therefore total deviation = 60°

45. (4) As the prism is an equilateral one, A = 60°. As

the ray of light passes symmetrically, the prism is

in the position of minimum deviation.

So, r = A

2 =

60

2

= 30°

also, i = 49°

= sin

sin

i

r =

sin

sin

49

30

=

0 7547

0 5

.

. = 1.5

46. (1)

Lesser the electronegativity difference between the two

bended atom, higher is the covalent character. Since

the elctronegativity difference between N & Cl is mini-

mum, NCl3 has maximum covalent character, N is more

electronegative then P, As and Sb.

47. (3)

N2O

3 is an acidic oxide and present as a blue coloured

liquid at –30°C. N2O and NO are neutral whereas NO

2

is brown coloured.

48. (1)

NH3 has maximum melting point among the hydrides of

group 15 because NH3 molecules are associated

through H-bonding. Thus H-bonding is absent in other

hydrides of group 15, as a result their melting point

is lower than that of NH3.

49. (3)

NO and NO2 are both odd electron species. Due to the

presence of odd electron, they are paramagnetic in

monomeric state.

50. (2)

Hydrides of group 15 are basic due to their ability to

donate lone pair of electrons. On moving down the group,

the size of atom increases and consequently charge

density decreases. Therefore the lone pair is less avail-

able for donation. Hence basicity decreases down the

group.

51. (4)

Both chromium and aluminium become passive on re-

action with conc. HCl because of the formation of ox-

ide on the surface of the metal.

52. (1)

Because of small size of oxygen atom, addition of addi-

tional electron is not much favourable. On moving down

the group size of p-orbitals increases and it can easily

accommodate the additional electron. Therefore among

group 16 elements, O has lowest electron affinity. Higher

the stability of anion formed, higher is the negative value

of electron affinity.

53. (3)

H2O has maximum boiling point because it exhibits hy-

drogen bonding. On moving down the group size of

atom increases and hence magnitude of van der Waals

forces increases. Therefore, the correct order of boil-

ing point is H2O < H

2Te < H

2Se < H

2S.

54. (4)

CO, and NO neutral oxide whereas Al2O

3 is an ampho-

teric oxide. N2O3, SO

2 and N

2O

5 are acidic oxide. Ox-

ide of non-metal in higher oxidation state are acidic in

nature.

55. (3)

S2 exist at elevated temperature and it is paramagnetic

like O2 due to the presence of unpaired electrons.

56. (3)

The bleaching action due to reduction is temporary. Since

the bleaching action of SO2 is due to reduction,

its bleaching action is temporary.

57. (2)

All halogens are coloured due to the absorption of dif-

ferent quanta of radiation in visible region. This results

in excitation of outer electrons to higher energy level

which gives colour of the halogen. Cl2 is a greenish yel-

low gas.

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58. (3)

The correct order of bond dissociation enthalpy among

halogen is Cl2 > F

2 > Br

2 > I

2. Bond dissociation en-

thalpy of Cl2 is greater than F

2 due to repulsion between

lone pairs in F2 which is smaller than Cl

2

59. (4)

Generally, oxoacids in which the central atom is in higher

oxidation state are stronger acid but in case of

oxoacids of phosphorous the order is just reversed be-

cause in H3PO

2 only one ionizable –OH group is present

and two and three –OH group in H3PO3 and H3PO

4

respectively. Since in H3PO

4 and 1–OH group is there,

it has higher tendency to release H+ than H3PO

3 and

H3PO

4.

60. (1)

The activity of oxyacids of chlorine as oxidizing agents

is related inversely to the number of O atoms in the

molecule. Since HOCl hypochlorous acid contains only

1–OH group, it is strongest oxidizing oxyacids of chlo-

rine.

61. (1)

Silver Iodide (AgI) is insoluble in aqueous solution of

ammonia.

62. (4)

NCl5 does not exist neither do [BCl

6]3– because of the

non availability d orbital in N and B. PH5 does not

exist because H is more electropositive than P and it

cannot bring +5 oxidation state of phosphorus. Since

Cl is more electronegative than P, PCl5 exist.

63. (4)

When axial and equatorial bonds are interchanged in a

trigonal bipyramidal compound, this is known as pseudo

reaction or pseudo rotation. In solid state PF5 exists as

covalent compound unlike PCl5 which exists as

[PCl5+][PCl6]–. PH5 does not exist because H is not

much electronegative to make the d orbital contact which

is required for the formation of PH5.

64. (3)

Cl2O

7 is more acidic than SO

2, P

2O

5 and SiO

2

because Cl is present in a very high oxidation state of

+7. Higher the oxidation state of central atom in an ox-

ide, highers the acidity.

65. (2)

A conjugate base of a strong acid is weak whereas con-

jugate base of a weak acid is strong. Since the acidic

strength of group 17 hydrides is in the order HF < HCl <

HBr < HI, HF is the weakest acid and hence its conju-

gate base F– is strongest base.

66. (2)

SO2 is readily decolourises acidified KMnO

4.

67. (4)

Fact.

68. (2)

In oxides, oxidation state of oxygen is –2, but in OF2,

oxidation state of oxygen is +2 because F is more

electronegative than oxygen.

69. (4)

Hg does not have abnormal electronic configuration.

70. (1)

Fact.

71. (3)

Ni has the lowest IE1 value because of the sheilding of

3d electrons.

72. (4)

All the given options are correct.

73. (4)

Zn does not show variable oxidation state due to 3d104s2

configuration (i.e. very stable).

74. (2)

The highest oxidation state of Mn with F is +4. Because

F is very highly electronegative eg. MnF4.

75. (2)

Oxygen stabilisers higher oxidation states because of

its tendency to form double bonds.

76. (4)

Reducing nature of metal depends on

Enthalpy of atomisation

Ionization enthalpies

Hydration energy (fact).

77. (4)

Sc is non typical transition element because in +3 state,

there is no d-electron.

78. (1)

CrO is basic because it is in the least oxidation state of

Cr.

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79. Answer (1)

Pm is radioactive (fact).

80. (3)

Zr and Hf have about the same radius due to lanthanide

contraction.

81. (1)

Size of lanthanide decrease because of poor screening

of 4f electrons.

82. (3)

Misch metal composition – (50% Ce, 25% La, small

amount of Nd and Pr blended with iron oxide).

83. (2)

Most stable oxidation state is +3 (fact).

84. (4)

Both (1) & (2) are correct (fact).

85. (2)

Generally, we consider Cu, Ag, Au as coinage metals

and they come under transition metals.

86. (3)

Among the given, Mn exhibit highest oxidation state,

i.e. +7.

87. (2)

Mn2O

7 is acidic in nature (fact).

88. (3)

Zn2+ does not show any colour because no any transi-

tion is possible there due to 3d10 configuration.

89. (1)

Fact.

90. (3)

Fe, Ru, Os belong to group 8 due to their electronic

configuration.