Which one of the following represents the correct increasing order of bond angles in given molecules?

H_{2} O < O F_{2} < O C l_{2} < C l O_{2}

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O C l_{2} < C l O_{2} < H_{2} O < O F_{2}

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O F_{2} < H_{2} O < O C l_{2} < C l O_{2}

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C l O_{2} < O F_{2} < O C l_{2} < H_{2} O

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Solution

The correct option is C $O F_{2} < H_{2} O < O C l_{2} < C l O_{2}$
Since $F$ is more electronegative than $O$ in $O F_{2}$ , the bond pair of electrons lie more towards $F$ whereas in $H_{2} O$ the bonded pair lies more towards $O$ which increases the bond angle in $H_{2} O$ . $O C l_{2}$ has a larger bond angle than $O F_{2}$ and $H_{2} O$ as the size of $C l$ atom is very large which results in higher repulsions and hence more bond angle. $C l O_{2}$ has double bonds which results in a further increase in bond angle due to repulsion.

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