Question

How will you account for the ${104.5}^{\circ }$ bond angle in water?

Answer 1

In $H_{2}O$ molecule, the oxygen atom is surrounded by two bonded electron pairs and two lone pairs of electrons. In water, oxygen atom has $s{p}^3-$hybridization and hence the $\angle HOH$ (bond angle) should have been ${109}^{\circ }{28}^{\prime }.$

According to VSEPR theory, lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions.

As a result, the $\angle HOH$ bond angle in water slightly decreases from the regular tetrahedral angle of ${109}^{\circ }{28}^{\prime }$ to${104.5}^{\circ }.$