Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

During the change of O2 to O2- ion, the electron adds on which one of the following orbitals?

  • π* orbital

  • π orbital

  • σ* orbital

  • σ* orbital


A.

π* orbital

During the change of O2 to O2-the electron adds on π* orbital. Molecular orbital configuration of O2 (total e- =16)
 straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 1 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 1

and space configuration space of space straight O subscript 2 superscript minus space left parenthesis 8 plus 8 plus 1 space equals space 17 right parenthesis
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 2 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 1

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2.

In the replacement reaction



The reaction will be most favourable if M happens to be

  • Na

  • K

  • Rb

  • Rb


C.

Rb

For the reaction,


The reaction will be faster with Rb because lattice energy of RbF is less than LiF, NaF, KF.

1028 Views

3.

Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

  • NO< O2-<C22- < He2+

  •  O2- <NO< C22- < He2+

  • C22- < He2+ <O2- <NO

  • C22- < He2+ <O2- <NO


D.

C22- < He2+ <O2- <NO

Bond space order space space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
left parenthesis where space straight N subscript straight b space equals space number space of space electrons space in space bonding space molecular space orbital
straight N subscript straight a space equals space number space of space electrons space in space anti minus bonding space molecular space orbital right parenthesis
In space NO comma space total space electrons space equals space 7 plus 8 space equals space 15
therefore space configuration space of space NO
equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 1
therefore space BO space equals space fraction numerator 8 minus 3 over denominator 2 end fraction space equals space 2.5
space straight O subscript 2 superscript minus space left parenthesis 8 plus 8 plus 1 space equals space 17 right parenthesis
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 2 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 1
therefore space BO space equals space fraction numerator 8 minus 5 over denominator 2 end fraction space equals space 1.5
In space straight C subscript 2 superscript 2 minus end superscript comma space total space electrons space equals space 12 plus 2 space equals space 14
therefore space comma space configuration space of space straight C subscript 2 superscript 2 minus end superscript
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2
therefore space BO space equals space fraction numerator 8 minus 2 over denominator 2 end fraction space equals space 3
In space He subscript 2 superscript plus space total space electrons space equals space 4 minus 1 space equals space 3
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s to the power of 1
therefore space BO space equals space fraction numerator 2 minus 1 over denominator 2 end fraction space equals space 0.5
Hence space comma space correct space order space space of space bond space order space is space
He subscript 2 superscript plus space less than space straight O subscript 2 superscript minus space less than space NO space less than space straight C subscript 2 superscript 2 minus end superscript
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4.

Give that the equilibrium constant for the reaction,

2SO2 (g) + O2 (g)  ⇌ 2SO3 (g)

has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction t the same temperature?

SO3 (g)  ⇌ SO2 (g) +1/2 O2 (g)

  • 1.8 x 10-3

  • 3.6 x 10-3

  • 6.0 x 10-2

  • 6.0 x 10-2


C.

6.0 x 10-2

2SO2 (g) +O2 (g)  ⇌ 2SO3 (g)

Equilibrium constant for this reaction,

straight K equals space fraction numerator left square bracket SO subscript 3 right square bracket squared over denominator left square bracket SO subscript 2 right square bracket squared left square bracket straight O subscript 2 right square bracket end fraction
SO subscript 3 space left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space SO subscript 2 space left parenthesis straight g right parenthesis space plus 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis
Equilibrium space constant space for space this space reaction
straight K apostrophe space equals space fraction numerator left square bracket SO subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of begin display style 1 half end style end exponent over denominator left square bracket SO subscript 3 right square bracket end fraction
On space squaring space eq. space left parenthesis ii right parenthesis thin space both space sides comma space we space have space
left parenthesis straight K to the power of apostrophe right parenthesis squared space equals space fraction numerator left square bracket SO subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket over denominator left square bracket SO subscript 3 right square bracket end fraction
space equals space 1 over straight K
equals 1 over 278
straight K apostrophe space equals space square root of 1 over 278 end root
equals square root of 0.003597 end root
equals space 5.99 space straight x space 10 to the power of negative 2 end exponent
almost equal to space 6 space straight x space 10 to the power of negative 2 end exponent

2073 Views

5.

The orbital angular momentum of p-electron is given as

  • fraction numerator straight h over denominator square root of 2 straight pi end root end fraction
  • square root of 3 fraction numerator h over denominator 2 pi end fraction
  • square root of 3 over 2 end root space h over pi
  • square root of 3 over 2 end root space h over pi

A.

fraction numerator straight h over denominator square root of 2 straight pi end root end fraction

Orbital angular momentum = 
square root of straight l left parenthesis straight l plus 1 right parenthesis end root space x fraction numerator h over denominator 2 pi end fraction
therefore space F o r space p minus e l e c t r o n comma space l italic space italic equals italic 1
italic therefore italic space O r b i t a l italic space a n g u l a r italic space m o m e n t u m italic comma
italic equals italic space italic space square root of l italic left parenthesis l italic plus italic 1 italic right parenthesis end root space x fraction numerator h over denominator italic 2 pi end fraction
italic equals italic space square root of italic 2 italic space x fraction numerator h over denominator italic 2 pi end fraction
italic equals fraction numerator h over denominator square root of italic 2 pi end fraction

615 Views

6.

For real gases van der Waal's equation is written as,

open parentheses straight p space plus an squared over straight V squared close parentheses open parentheses straight V minus nb close parentheses space equals space nRT
where, a and b are van der Waal's constants.
Two sets of gases are

I. O2, CO2, H2 and He
II. CH4, O2, and H2

The gases given in set- I increasing order of 'b' and gases given in set -II in decreasing order of 'a' are arranged below. Select the correct order from the following.

  • (I) He < H2 < CO< O2
    (II)  CH4>  H2> O2

  • (I) O2 < He < H2 < CO2
    (II) H2> O2 > CH4

  • (I) O2 < He < O2 < CO2
    (II) CH4 > O2 > H2  

  • (I) O2 < He < O2 < CO2
    (II) CH4 > O2 > H2  


C.

(I) O2 < He < O2 < CO2
(II) CH4 > O2 > H2  

From experimental values of a and b, sequence of Van der Waals constant (b) is

Gas

H2

He

O2

CO2

Value of b

0.0273

0.0318

0.0273

0.0427

Hence correct order is

H2 < He < O2 < CO2

And sequence of van der Waals constant (a) is

Gas

CH4

O2

H2

Value of a

2.25

1.36

0.244

Hence, correct order is

(II) CH4 >O2> H2

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7.

Given the reaction between two gases represented by A2 and B2 to give the compound AB (g).
A2(g) +B2 (g)  ⇌ 2AB (g)
At equilibrium the concentration
of A2 = 3.0 x 10-3 M
of B2 = 4.2 x 10-3 M
of AB = 2.8 x 10-3 M

If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be

  • 2.0

  • 1.9

  • 0.62

  • 0.62


C.

0.62

A2 (g) +B2 (g)  ⇌ 2AB (g)
The equilibrium constant is given by

straight K subscript straight c space equals space fraction numerator left square bracket AB right square bracket squared over denominator left square bracket straight A subscript 2 right square bracket left square bracket straight B subscript 2 right square bracket end fraction space equals
straight K subscript straight c space equals space fraction numerator left parenthesis 2.8 space straight x space 10 to the power of negative 3 end exponent right parenthesis squared over denominator left parenthesis 3.0 space straight x space space 10 to the power of negative 3 end exponent right parenthesis left parenthesis 4.2 space straight x space 10 to the power of negative 3 end exponent right parenthesis end fraction
straight K subscript straight c space equals space fraction numerator 7.84 over denominator 12.6 end fraction
space equals space 0.62

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8.

Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be

  • 0.83

  • 1.50

  • 3.3

  • 3.3


D.

3.3

For space monoatomic space gases comma
straight C subscript straight p over straight C subscript straight v space equals space straight gamma space equals space fraction numerator begin display style 5 over 2 end style straight R over denominator begin display style 3 over 2 end style straight R end fraction space equals space 1.67
therefore space Equal space volumes space of space two space monoatomic space gases space are space mixed.
therefore space straight gamma space will space remain space the space same
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9.

A structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

  • ABO2

  • A2BO2

  • A2B3O4

  • A2B3O4


D.

A2B3O4

According to ccp,
Number of O2- ions = 4
So, tetrahedral void = 8
and octahedral void = 4
Since A ions occupied 1/4th of the tetrahedral void.
Therefore,
Number of A ions = 1/4 x 8 = 2
Again, B ions occupied all octahedral void.
Therefore, Number of B ions = 4 
A: B:O = 2:4:4
 = 1:2:2
Structure of oxide= AB2O2

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10.

A  certain gas takes three times as long to effuse out as helium. Its molecular mass will be 

  • 27 u

  • 36 u 

  • 64 u

  • 64 u


B.

36 u 

From Graham's diffusion law,

straight r subscript 1 over straight r subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root
O r
V subscript 1 over t subscript 1 space x space t subscript 2 over V subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root
because space V o l u m e space i s space s a m e
therefore space 3 over 1 equals space square root of straight M subscript 2 over 4 end root
space or space 9 equals straight M subscript 2 over 4
Or space straight M subscript 2 space equals space 9 space straight x space 4 space equals space 36 space straight u

553 Views