Explain the observed bond angle order.
`Cl_(2)O(110.8^(@)) gt H_(2)O (104.5^(@)) gt F_(2)O (103.2^(@))`

Explanation

`Cl_(2)O, H_(2)O` and `F_(2)O` all `sp^(3)` hybridised (2 bond pairs `+` 2 lone pairs) having tetrahedral geometry but due to the presence of 2 lone pairs `(l p)` on oxygen atom, all of them have bent shape or V-shaoe.
Therefore, the expected angle should be `109.5^(@)`. But the to repulsion between `l p-l p, l p-bp` and `bp-bp`, the angle varies as shown below:

a. In Case of `Cl_(2)O`, delocalisation of lone pair `( p)` of electrons of oxygen to the vacant d-orbital of chlorine decreases the repulsion by `l p` and increases the repulsion between bp (bond pairs0. The bond angle thus becomes bery large `(110.8^(@))`.
b. In case of `F_(2)O`, because of more electronegative F atom, bond pairs (bp) of electrons in `O-F` bond are shifted towards F, whereas in `H_(2)O, bp` of electrons are drawn towards oxygen atom. SDo in `F_(2)O`, the bond pairs (bp) being displaced away from the central atom has very little tendency to open the angle (i.e. `bp-bp` repulsion decreases and `lp-lp` repulsion closes the angle).
But in `H_(2)O, bp` of electrons are closer to each other, and have more `bp-bp` repulsion than that in `F_(2)O`, which opens the angle.
That is why, bond angle of `F_(2)O` is less than that of `H_(2)O`.
Hence, the bond angle order is `Cl_(2) gt H_(2)O gt F_(2)O`.