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

Explanation

In water, O undergoes `sp^(3)`-hybridization and hence the `angleHOH` should have been `109^(@)-28'`.
In `H_(2)O`, the oxygen atom is surrounding by two shared pairs and two lone pairs of electrons. But acconding to VSEPR theory, lone pair-lone pair repulsions are stronger than bond pair-bond pair repulsions. As a result, the HOH bond angle in water slightly decreases from the regular tetrahedral angle of `109^(@)-28'` to `104.5^(@)` .