How will you account for `104.5^(@)` bond angle in water ?

Explanation

In water, oxygen has `sp^(3)` - hydridisation and the bond angle of HOH should have been `109^(@)28`. In `H_(2)O` the oxygen atom is surrounded by two shared pairs and two lone pairs of electrons. From VSEPR theory , lone pair - lone pair repulsions are stronger than bond angle of HOH in water slightly decreases form the regular tetrahedral angle of `109^(@). 28'` to `145.5^(@)`