The correct order of bond angles in $O C l_{2}$ , $H_{2} O$ and $O F_{2}$ would be

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Solution

The O-X (where X = Cl, F, H) bond will have its bond pair closer to X when X is F, since F has higher electronegativity. So the electrons move towards F and the distance between them, as a consequence, decreases. And therefore the bp-bp repulsion decreases, leading to a decrease in bond angle. Whereas this effect is less pronounced in $O C l_{2}$ . `Whereas this effect applies to $H_{2} O$ as well but the value lies in between of $O F_{2} a n d O C l_{2}$ Therefore, the bond angle in $O F_{2} < H_{2} O < O C l_{2}$

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