Question

Which of the following accounts for the bond angle of ${104.5}^o$ in water molecules?

• A. $s{p}^3$ hybrid state of oxygen
• B. high electronegativity of H-atoms
• C. low ionisation energy of H-atoms
• D. lone pair-lone pair repulsion and lone pair-bond pair repulsions

Answer 1

The correct option is D lone pair-lone pair repulsion and lone pair-bond pair repulsions

Bond angles are affected by repulsions which are $lp-lp>lp-bp>bp-bp.$

In $H_{2}O$ molecule, $O$ has 2 bond pairs and 2 lone pairs. So, the steric number is 4 and the hybridisation is $s{p}^3$. Hence, the geometry is tetrahedral and the shape is bent with a bond angle of ${104.5}^{\circ }$ due to lone pair-lone pair repulsion and lone pair-bond pair repulsions.