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Bond Order by MOT

Bond order of...

Question

Bond order of $O_{2}, O_{2}^{+}, O_{2}^{-}$ and $O_{2}^{2 -}$ is in order_________.

A

O_{2}^{-} < O_{2}^{2 -} < O_{2} < O_{2}^{+}

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B

O_{2}^{2 -} < O_{2}^{-} < O_{2} < O_{2}^{+}

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C

O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 +}

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D

O_{2} < O_{2}^{+} < O_{2}^{-} < O_{2}^{2 -}

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Solution

The correct option is C $O_{2}^{2 -} < O_{2}^{-} < O_{2} < O_{2}^{+}$
As we know,
bond order $=$ (no of bonding electron - no of anti bonding electron/2
$O_{2} = 2$
$O_{2}^{+} = 2.5$
$O_{2}^{-} = 1.5$
$O_{2}^{- 2} = 1$

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