The correct order of increasing bond angles in the following species is:

C l_{2} O < C l O_{2} < C l O {_{2}}^{-}

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C l O_{2} < C l_{2} O < C l O {_{2}}^{-}

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C l_{2} O < C l O {_{2}}^{-} < C l O_{2}

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C l O {_{2}}^{-} < C l_{2} O < C l O_{2}

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Solution

The correct option is C $C l_{2} O < C l O {_{2}}^{-} < C l O_{2}$
The correct order of increasing bond angles in the following species is
${C l}_{2} O < C l O_{2}^{-} < C l O_{2}$
This is because, In $C l O_{2}^{-}$ there are two lone pairs of electrons which repel each other due to which two oxygen atom come closer leading to decrease in bond angle. Therefore the bond angle in $C l O_{2}^{-}$ is less than $118 °$ which is bond angle in $C l O_{2}$ which has less number of electrons on chlorine.

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