The bond angle in $O F_{2}$ out of $O F_{2}, C l_{2} O, B r_{2} O$ is minimum. It is because in case of $O F_{2}$ :

Electrons are nearer to fluorine due to high electronegativity of

F

compared to

C l

B r

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Lone pair-lone pair repulsion decreases bond angle

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Both (a) and (b) are correct

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None of the above is correct

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Solution

The correct option is A Electrons are nearer to fluorine due to high electronegativity of $F$ compared to $C l$ to $B r$
The bond angle in $O F_{2}$ out of $O F_{2}, C l_{2} O, B r_{2} O$ is minimum. It is because in case of $O F_{2}$ electrons are nearer to fluorine due to high electronegativity of $F$ compared to $C l$ to $B r$ and because of this repulsion between electron decreases; so as bond angle.

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