Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which of the following species is not electrophilic in nature? 

  • Cl+

  • BH3

  • H3O+

  • H3O+


C.

H3O+

Electrophiles are electron deficient species. Among the given H3O+ has lone pair of electron for donation, thus, it is not electron deficient and hence, does not behave like an electrophile.

744 Views

2.

Three moles of an ideal gas expanded spontaneously into the vacuum. The work done will be

  • infinte

  • 3 J

  • 9 J

  • 9 J


D.

9 J

W = pext. ΔV
For vaccum,

ptext = 0
therefore,
W = 0  ΔV = 0

572 Views

3.

A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be
(h = 6.6 x 10-34 Js)

  • 6.6 x 10-32 m

  • 6.6 x 10-34 m

  • 1 x 10-35 m

  • 1 x 10-35 m


C.

1 x 10-35 m

straight lambda space equals space straight h over mv space equals space fraction numerator 6.6 space straight x space 10 to the power of negative 34 end exponent over denominator 0.66 space straight x space 100 end fraction space equals space 10 to the power of negative 35 end exponent
747 Views

4.

Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is 

  • Mg< Ca < Cl < P

  • Cl < P < Mg < Ca

  • P < Cl < Ca < Mg

  • P < Cl < Ca < Mg


B.

Cl < P < Mg < Ca

With the increase in the number of electron in the same shell, the atomic radii decrease due to increase in effective nuclear charge. However, atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases.

The electronic configuration of the given elements is

Mg12 = [Ne] 3s2
Ca20 = [Ar] 4s2
P15 = [Ne]3s2 3p3
Cl17 = [Ne] 3s2 3p5

In Mg, P and Cl, the number of electrons is increasing in the same shell thus, the order of their atomic radii is

Cl <P < Mg

In Ca, the electron is entering in the higher shell, thus, it has the highest atomic radii among the given. Thus, the order of radii is

Cl < P < Mg < Ca

1211 Views

5.

Some statement about heavy water are given below

A) Heavy water is used as a moderator in nuclear reactors.

B) Heavy water is more associated than ordinary water.

C) Heavy water is more effective solvent than ordinary

  • A and B

  • A, B and C

  • B and C

  • B and C


A.

A and B

Heavy water is used as a moderator in nuclear reactors. Its boiling point is higher as compared to the ordinary water. Thus, it is more associated as compared to ordinary water. The dielectric constant is however higher for H2O, thus, H2O is a more effective solvent than heavy water (D2O).

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6.

Among the following which one has the highest cation to anion size ratio?

  • CsI

  • CsF

  • LiF

  • LiF


B.

CsF

The order of size of given cations is 

Li+ < Na+ < Cs+

and the order of the size of given anions is 


I- > F-

Thus, when the cation is largest and anion is smallest the cation to anion size ratio is maximum. Hence cation to anion size ratio is maximum
531 Views

7.

In which of the following molecules the central atom does not have sp3 hybridisation?

  • CH4

  • SF4

  • BF4-

  • BF4-


B.

SF4

When the number of hybrid orbitals, H  is 4, the hybridization is sp3 hybridization
straight H space equals space 1 half left square bracket space straight V space plus space straight M space minus space straight C space plus space straight A right square bracket
where comma space straight V space equals space number space of space monovalent space atoms
straight C space equals space total space positive space charge
straight A space equals space negative space charge

straight a right parenthesis space For space CH subscript 4
straight H space equals space 1 half left square bracket space 4 space plus space 4 space minus 0 space plus space 0 right square bracket space equals space 4 space
sp cubed

straight b right parenthesis space space For space SF subscript 4 comma end subscript
straight H space equals space 1 half left square bracket space 6 plus 4 minus 0 plus 0 right square bracket space equals space 5 space
sp cubed straight d

straight c right parenthesis space For space BF subscript 4 superscript minus comma
straight H space equals space 1 half left square bracket space 3 plus space 4 minus 0 plus 1 right square bracket space equals space 4 comma space thus space sp cubed

straight d right parenthesis space For space NH subscript 4 superscript plus
straight H space equals space 1 half left square bracket 5 plus 4 minus 1 plus 0 right square bracket space equals space 4 space thus space sp cubed

Thus, only in SF4 the central atom does not have sp3 hybridisation.

1202 Views

8.

The reaction,

2A (g) + B (g)  ⇌ 3C (g) + D (g) 

is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of  D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

  • [(0.75)3 (0.25)] / [(1.00)2 (1.00)]

  • [(0.75)3 (0.25) ] / [(0.50)2(0.75)]

  • [(0.75)3(0.25)] / [(0.50)2(0.75)]

  • [(0.75)3(0.25)] / [(0.50)2(0.75)]


B.

[(0.75)3 (0.25) ] / [(0.50)2(0.75)]

space space space space space space space space space space space space space space space space 2 space straight A space left parenthesis straight g right parenthesis space space plus space straight B space left parenthesis straight g right parenthesis space space space space rightwards harpoon over leftwards harpoon space space space space space space space 3 straight C space left parenthesis straight g right parenthesis space plus space straight D space left parenthesis straight g right parenthesis
initial space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space 0
At space equil space space space space 1 minus 0.50 space space space space space space space space 1 minus 0.25 space space space space space space space space space space 75 space space space space space space space space space space space space 0.25

straight K space equals space fraction numerator left parenthesis 0.75 right parenthesis cubed left parenthesis 0.25 right parenthesis over denominator left parenthesis 0.50 right parenthesis squared left parenthesis 0.75 right parenthesis end fraction
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9.

Match List I (equation) with List II (types of the process) and select the correct option.

List I (Equations)
List II (Types of process)

A

KP > Q

1

Non-spontaneous

B

ΔGo < RT In Q

2

Equilibrium

C

KP = Q

3

Spontaneous and endothermic

D

T> ΔH/ ΔS

4

Spontaneous

  • A
    B
    C
    D
    1
    2
    3
    4
  • A B C D
    3 4 2 1
  • A B C D
    4 1 2 3
  • A B C D
    4 1 2 3

C.

A B C D
4 1 2 3

A) When KP > Q, the reaction goes in the forward direction, ie, the reaction is spontaneous.

B) Given ΔGo < RT lnQ, thus, ΔGo = + ve and hence, the reaction is non-spontaneous.

C) At equilibrium, Kp = Q


D) T > ΔH/ΔS or TΔS >ΔH

Thus condition is true for spontaneous endothermic reactions (as ΔG greater than equal to ΔH- TΔS)
414 Views

10.

The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129o C is (Atomic masses C = 12.01, H = 1.01 and R = 8.314 JK-1 mol-1)

  • 215216 Pa

  • 13409 Pa

  • 41648 Pa

  • 41648 Pa


C.

41648 Pa

Given,

Volume , V = 0.03 m3
Temperature, T = 129 + 273 = 402 K
mass of methane, w = 6.0 g
mol. Mass of methane, M = 12.01 + 4 x 1.01 = 16.05
From, ideal gas equation,

pV = nRT

straight p space equals fraction numerator 6 over denominator 16.05 end fraction space straight x space fraction numerator 8.314 space straight x space 402 over denominator 0.03 end fraction space space equals space 41648 space Pa

992 Views