The order of $O - O$ bond length in $O_{2}, O_{2} (A s F_{6})$ and $K O_{2}$ is :

O_{2} (A s F_{6}) < O_{2} < K O_{2}

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O_{2} < K O_{2} < O_{2} (A s F_{6})

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O_{2} < K O_{2} = O_{2} (A s F_{6})

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K O_{2} < O_{2} (A s F_{6}) < O_{2}

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Solution

The correct option is A $O_{2} (A s F_{6}) < O_{2} < K O_{2}$
$B o n d l e n g t h \propto \frac{1}{B o n d o r d e r}$
In $K (O_{2})$ or $O_{2}^{-},$ bond order is $1.5$ .
In $O_{2}$ , bond order is $2$ .
In $O_{2} (A s F_{6})$ or $O_{2}^{+}$ , bond order is $2.5$ .
Hence, the correct order is $O_{2} (A s F_{6}) < O_{2} < K O_{2}$ .

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