The correct order of $C l - O$ bond lengths in $C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}$ and $C l O_{4}^{-}$ is:

C l O^{-} < C l O_{2}^{-} < C l O_{3}^{-} < C l O_{4}^{-}

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C l O_{4}^{-} < C l O_{3}^{-} < C l O_{2}^{-} < C l O^{-}

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C l O_{3}^{-} < C l O_{4}^{-} < C l O_{2}^{-} < C l O^{-}

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C l O_{4}^{-} = C l O_{3}^{-} = C l O_{2}^{-} = C l O^{-}

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Solution

The correct option is B $C l O_{4}^{-} < C l O_{3}^{-} < C l O_{2}^{-} < C l O^{-}$
Bond length $\propto \frac{1}{B o n d o r d e r}$

Bond order of $C l O^{-} = 1$ ,
Bond order of $C l O_{2}^{-} = 1.5,$
Bond order of $C l O_{3}^{-} = 1.66$ ,
Bond order of $C l O_{4}^{-} = 1.75$ .

Thus the order of bond length is $C l O_{4}^{-} < C l O_{3}^{-} < C l O_{2}^{-} < C l O^{-}$ , thus correct option is B.

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