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Karnataka Board PUCPUC Science Class 11

Which of the following pair is expected to have the same bond order?

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Question

Which of the following pair is expected to have the same bond order?

Options

  • \[\ce{O2, N2}\]

  • \[\ce{O^{+}2 , N^{-}2}\]

  • \[\ce{O^{-}2 , N^{+}2}\]

  • \[\ce{O^{-}2 , N^{-}2}\]

MCQ
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Solution

\[\ce{O^{+}2 , N^{-}2}\]

Explanation:

(i) The electronic configuration of \[\ce{O2}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2 ,π2p_y^2, π^∗2p_x^1, π^∗2p_y^1`

The bond order of \[\ce{O2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 6]` = 2.0

The electronic configuration of \[\ce{N2}\]:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2`

The bond order of \[\ce{N2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 4]` = 3.0

So, the bond order of \[\ce{O2, N2}\] will not be equal.

(ii) The electronic configuration of \[\ce{O^{+}2}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2 ,π2p_y^2, π^∗2p_x^1`

The bond order of \[\ce{O^{+}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 5]` = 2.5

The electronic configuration of \[\ce{N^{-}2}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2, π^∗2p_x^1`

The bond order of \[\ce{N^{-}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 5]` = 2.5

Hence, the bond order of \[\ce{O^{+}2, N^{-}2}\] will be equal.

(iii) The electronic configuration of \[\ce{O^{-}2}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2, π2p_y^2, π^∗2p_x^2, π^∗2p_y^1`

The bond order of `\[\ce{O^{-}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 7]` = 1.5

The electronic configuration of \[\ce{N^{+}2}\] is:

 `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^1`

The bond order of \[\ce{N^{+}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[9 - 4]` = 2.5

Hence, the bond order of \[\ce{O^{-}2, N^{+}2}\] will not be equal.

(iv)The electronic configuration of \[\ce{O^{-}2}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2, π2p_y^2, π^∗2p_x^2, π^∗2p_y^1`

The bond order of \[\ce{O^{-}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 7]` = 1.5

The electronic configuration of  \[\ce{N^{-}2}\] will be:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2, π^∗2p_x^1`

The bond order of \[\ce{N^{-}2}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 5]` = 2.5

Hence, the bond order of \[\ce{O^{-}2, N^{-}2}\] will not be equal.

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Chapter 4: Chemical Bonding and Molecular Structure - Multiple Choice Questions (Type - I) [Page 50]

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NCERT Exemplar Chemistry [English] Class 11
Chapter 4 Chemical Bonding and Molecular Structure
Multiple Choice Questions (Type - I) | Q 67 | Page 50
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