The `M-O-M` bond angles in `M_(2)O` (where M is halogen) is in the order

Options

A. `Br_(2)O gt Cl_(2)O gt F_(2)O`
B. `F_(2)O gt Br_(2)O gt Cl_(2)O`
C. `F_(2)O gt Cl_(2)O gt Br_(2)O`
D. `Cl_(2)O gt F_(2)O gt Br_(2)O`

Answer

A

Explanation

In `OF_(2)` repulsion between lone pairs is greater than that between bond pair since electrons are away from O and nearer to F. In `Cl_(2)O`, bonding electrons are nearer to O than to Cl, so the bond angle is greater than `109^(@)28'`. In `Br_(2)O`, the bonding electrons are more closer to oxygen than in `Cl_(2)O`, so the bond angle is largest `(116^(@))`.