In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

Question

Aicharya · Accepted Answer

According to molecular orbital theory

Bond order

= \frac{(e^{-} i n b o n d i n g m o l e c u l a r o r b i t a l) - (e^{-} i n A n t i b o n d i n g M o l e c u l a r o r b i t a l)}{2}

Molecular orbital configuration.

(A)

O_{2} ⟹ σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < \overset{}{σ} 2 p_{x}^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < \overset{*}{σ} 2 p_{y}^{1} = \overset{*}{π} 2 p_{y}^{1}

B . O . = \frac{6 - 2}{2} = 2

O_{2}^{+} ⟹ σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < \overset{}{σ} 2 p_{x}^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < \overset{*}{π} 2 p_{y}^{1}

B . O . = \frac{6 - 1}{2} = 2.5

Both are paramagnetic because having unpaired

e^{-}

bond order increase

O_{2} \to O_{2}^{+}

(B) N O = σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < \overset{}{σ} 2 p_{x}^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < \overset{*}{π} 2 p_{y}^{1} = \overset{*}{π} 2 p_{z}^{0}

B . O . = \frac{6 - 1}{2} = 2.5

1

Unpaired

e^{-}

so paramagnetic

N O^{+} ⟹ σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < σ 2 p x^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2}

B . O . = \frac{6 - 0}{2} = 3

N O^{+}

doesn't have unpaired

e^{-}

so diamagnetic

N O \to N O^{+}

{Bond order increases and change magnetic character paramagnetic to diamagnetic character change}

(C)

O_{2} \to σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < \overset{}{σ} 2 p_{x}^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < \overset{*}{π} 2 p_{y}^{1} = \overset{*}{π} 2 p_{y}^{1}

B . O . = \frac{6 - 2}{2} = 2

Having

2 -

unpaired

e^{-}

so paramagnetic

O_{2}^{-} \to σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < \overset{}{σ} 2 s^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < \overset{*}{π} 2 p_{y}^{2} = \overset{*}{π} 2 p_{y}^{1}

B . O . = \frac{6 - 3}{2} = 1.5

Having 1- unpaired

e^{-}

so paramagnetic bond order decrease

(D)

N_{2} = σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < π 2 p y^{2} = π 2 p_{y}^{2} < σ 2 p_{x}^{2}

B . O . = \frac{6 - 0}{2} = 3

No unpaired

e^{-}

so diamagnetic

N_{2}^{+} \to σ 1 s^{2} < \overset{*}{π} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < π 2 p_{y}^{2} = π 2 p_{z}^{2} < σ 2 p_{x}^{1}

B . O . = \frac{5 - 0}{2} = 2.5

Having one unpaired

e^{-}

so paramagnetic

Bond order increases

N_{2}

to

N_{2}^{+}

In which of the following processes, the bond order has increased and paramagnetic character has changed to…