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  4. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species: N 2 ; N 2+ ; O 2 ; O 2+?

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Q. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $N _2 ; N _2^{+} ; O _2 ; O _2^{+}$?

JEE MainJEE Main 2023Chemical Bonding and Molecular Structure

Solution:

$N _2$
$\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 \frac{\sigma 2 p _{ z }^2}{ HOMO }$
$N _2^{+}-\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma^* 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 \frac{\sigma 2 p _{ z }^1}{ HOMO }$
$O _2-\sigma 1s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \sigma 2 p _{ z }^2 $
$\pi 2 p _{ x }^2=\pi 2 p _{ y }^2$
$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^1( HOMO )$
$O _2^{+}-\sigma 1 s ^2 \sigma^* 1s ^2 \sigma 2 s ^2 \sigma ^* 2 s ^2 \sigma 2 p _{ z }^2 \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 $
$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^0( HOMO )$
$ N _2 \Rightarrow 0$ unpaired $ e ^{-} \text {in HOMO }$
$ N _2^{+} \Rightarrow 1 $ unpaired $ e ^{-} \text {in HOMO } $
$ O _2 \Rightarrow 2 $ unpaired $ e ^{-} \text {in HOMO } $
$ O _2^{+} \Rightarrow 1 $ unpaired $e ^{-} \text {in HOMO }$