Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The values of ΔH and ΔS for the reaction, C(graphite) + CO2(g) → 2 CO (g) are 170 kJ and 170 JK-1 respectively. This reaction will be spontaneous at

  • 710 K 

  • 910 K

  • 1110 K

  • 1110 K


C.

1110 K

For spontaneous process, ΔG < 0

increment straight G space equals space increment straight H minus straight T increment straight S
Given comma space increment straight H space equals space 170 space kJ
increment straight S space equals space 170 space JK to the power of negative 1 end exponent
straight T space equals ?
increment straight G space equals space increment straight H minus space straight T increment straight S

0 less than 170 space straight x space 10 cubed space minus space straight T space straight x space 170
straight T greater than 1000
straight T minus 1110 space straight K

614 Views

2.

Amongst the elements with following electronic configuration, which one of them may have the highest ionisation energy?

  • [Ne] 3s2 3p3

  • [Ne] 3s2 3p2

  • [Ar] 3d10, 4s2 4p3

  • [Ar] 3d10, 4s2 4p3


A.

[Ne] 3s2 3p3

Ionisation energy usually increases from left to right in a period with a decrease in atomic size and decrease from up to down in a group with an increase in atomic size.

[Ne] 3s2 3p3     = Group V

[Ne] 3s2 3p2     = Group IV


[Ar] 3d10, 4s2 4p3 = Group V

[Ne] 3s2 3p1 = Group III

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3.

Which of the following molecules acts as a Lewis acid?

  • (CH)3B

  • (CH3)O

  • (CH3)P

  • (CH3)P


A.

(CH)3B

According to Lewis concept " Acids are electron acceptor and bases are electron donor" Since, electron deficient compounds also have an ability to accept electrons, these are regarded as acids.

Trimethylborane (CH)3B have incomplete octet thus, act as Lewis acid.



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4.

According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?

  • straight N subscript 2 superscript minus space less than space straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript
  • straight N subscript 2 superscript 2 minus end superscript less than space straight N subscript 2 superscript minus space less than space straight N subscript 2
  • straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript space less than straight N subscript 2 superscript minus
  • straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript space less than straight N subscript 2 superscript minus

A.

straight N subscript 2 superscript minus space less than space straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript

Bond Order = fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
Where
Nb = number of electrons in bonding MO

Na = number of electrons in antibonding MO


straight N subscript 2 space left parenthesis space 7 plus 7 space equals space 14 right parenthesis space equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight sigma 2 straight p subscript straight z squared

BO space equals space fraction numerator 10 minus 4 over denominator 2 end fraction space equals space 3

straight N subscript 2 superscript minus space left parenthesis 7 plus 7 plus 1 space equals space 15 right parenthesis
equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight pi asterisk times 2 straight p subscript straight x to the power of 1

BO space equals space fraction numerator 10 minus 5 over denominator 2 end fraction space equals space 2.5
straight N subscript 2 superscript 2 minus end superscript space left parenthesis space 7 plus 7 plus 2 space equals space 16 right parenthesis
equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared space σp subscript straight z squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight sigma 2 straight p subscript straight x to the power of 1 space almost equal to straight pi asterisk times 2 straight p subscript straight y to the power of 1
BO space equals space fraction numerator 10 minus 6 over denominator 2 end fraction space equals space 2
Hence space comma space the space increasing space order space of space straight B. straight O space is comma
straight N subscript 2 superscript 2 minus end superscript space less than space straight N subscript 2 superscript minus space less than straight N subscript 2

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5.

Oxidation number of P in PO43-, of S in SO42- and that of Cr in Cr2O72- are respectively,

  • +5,+6 and +6

  • +3, + 6 and +5

  • +5,+3 and +6

  • +5,+3 and +6


A.

+5,+6 and +6

(i) Sum of oxidation states of all atoms = charge of ion.

(ii) oxidation number of oxygen = -2

Let the oxidation state of P in PO43- is x.
 PO43- 

x + 4 (-2) = - 3
x-8 = - 3
x = +5

Let the oxidation state of S in SO42- is y
y + 4(-2) = -2
y-8 = - 2
y = +6

Let the oxidation state of Cr in Cr2O72- is z.
2 x z+7(-2) = -2
2z-14 = - 2
z=+6
Hence, oxidation state of P, S and Cr are +5, +6 and +6

855 Views

6.

The ionisation constant of ammonium hydroxide is 1.77 x 10-5 at 298 K. Hydrolysis constant of ammonium chloride is 

  • 5.65 x 10-10

  • 6.50 x 10-12

  • 5.65 x 10-13

  • 5.65 x 10-13


A.

5.65 x 10-10

Given comma space straight K subscript straight a space left parenthesis NH subscript 4 OH right parenthesis space equals space 1.77 space straight x space 10 to the power of negative 5 end exponent

NH subscript 4 OH space leftwards harpoon over rightwards harpoon space NH subscript 4 to the power of plus space plus space OH to the power of minus

straight K subscript straight a space equals space fraction numerator left square bracket NH to the power of plus subscript 4 right square bracket left square bracket OH to the power of minus right square bracket over denominator left square bracket NH subscript 4 OH right square bracket end fraction space equals space 1.77 space straight x space 10 to the power of negative 5 end exponent space... space left parenthesis straight i right parenthesis

Hydrolysis of NH4Cl takes place as,

NH4Cl + H2O → NH4OH + HCl
or 
NH4+ +H2O → NH4OH + H+

Hydrolysis constant, Kh

straight K subscript straight h space equals space fraction numerator left square bracket NH subscript 4 OH right square bracket left square bracket straight H to the power of plus right square bracket over denominator left square bracket NH subscript 4 to the power of plus right square bracket end fraction space... space left parenthesis ii right parenthesis

or space straight K subscript straight h space equals space fraction numerator left square bracket NH subscript 4 OH to the power of plus right square bracket left square bracket OH to the power of minus right square bracket over denominator left square bracket NH subscript 4 to the power of plus right square bracket left square bracket OH to the power of minus right square bracket end fraction space... space left parenthesis iii right parenthesis

from space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis iii right parenthesis

straight K subscript straight h space equals space straight K subscript straight w over straight K subscript straight a space left square bracket space because space left square bracket OH to the power of minus right square bracket left square bracket straight H to the power of plus right square bracket space equals space straight K subscript straight w right square bracket
equals space fraction numerator 10 to the power of negative 14 end exponent over denominator 1.77 space straight x space 10 to the power of negative 5 end exponent end fraction
space equals space 5.65 space straight x space 10 to the power of minus to the power of 10
2024 Views

7.

Which one of the element with the following outer orbital configuration may exhibit the largest number of oxidation states?

  • 3d5,4s2

  • 3d5,4s1

  • 3d5,4s2

  • 3d5,4s2


C.

3d5,4s2

A number of oxidation states exhibited by d-block elements is the sum of the number of electrons (unpaired) in d-orbitals and number of electrons in s - orbital.


a) 3d3, 4s2 ⇒ Oxidation state = 3+2 = 5

b) 3d5, 4s1 ⇒ Oxidation state = 5+1 = 6

c) 3d5, 4s2 ⇒ Oxidation state = 5+2 = 7

d) 3d4, 4s2 ⇒ Oxidation state = 2+ 2 = 4 

Hence, an element with 3d5, 4s2 configuration exhibits the largest number of oxidation states. 

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8.

From the following bond energies:

H-H bond energy: 431.37 kJ mol-

C=C  bond energy: 606.10 kJ mol-

C- C bond energy: 336.49 kJ mol-

C-H bond energy: 410.50 kJ mol-

Enthalpy for the reaction,


will be


  • 1523.6 kJ  mol-

  • -243.6 kJ mol-

  • -1200. kJ mol-

  • -1200. kJ mol-


C.

-1200. kJ mol-

For a reaction



increment straight H subscript straight r space equals space left square bracket space 4 space straight x space BE subscript left parenthesis straight C minus straight H right parenthesis end subscript space plus space 1 space straight x space BE subscript straight C equals straight C space end subscript space plus space 1 space straight x space BE subscript left parenthesis straight H minus straight H right parenthesis end subscript right square bracket minus 6 straight x space BE subscript left parenthesis straight C minus straight H right parenthesis end subscript space plus space 1 space straight X space BE subscript left parenthesis straight C minus straight C right parenthesis end subscript
equals space left parenthesis space 4 space straight x space 410.50 space plus space 1 space straight x space 606.10 space plus space 1 space straight x space 431.37 right parenthesis minus left square bracket 6 space straight x space 410.50 right parenthesis space plus 1 space straight x space 336.49 right parenthesis right square bracket
space equals negative space 120.0 space kJ space mol to the power of minus

934 Views

9.

Maximum number of electrons in a subshell of an atom is determined by the following

  • 4l + 2

  •  2l +1

  • 4l-2

  • 4l-2


A.

4l + 2

Total number of subshells = (2l +1)
therefore, 
Maximum number of electrons in the subshell  = 2 (2l +1) = 4l + 2

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10.

Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and Conc. H2SO4.In the mixture, nitric acid acts as a/an:

  • reducing agent

  • acid

  • base

  • base


C.

base

Proton donor is acids and proton acceptor is bases.
Conc. H2SO4 and conc. HNO3 react in the following manner:

HNO3 + H2SO4 → H2NO3+ +HSO4-
H2NO3+ → NO2+ +H2O
Hence, in this reaction HNO3 acts as a base and H2SO4 as an acid.

757 Views