Question

The $H-O-H$ angle in water molecule is about:

• A. ${90}^o$
• B. ${180}^o$
• C. ${102}^o$
• D. ${105}^o$

Answer 1

The correct option is D ${105}^o$

$H-O-H$ angle in water is slightly less than the typical tetrahedral angle. It is ${104.5}^o$. If all the substituents were same (such as in $C{H}_4$), the bond angle would be ${109.5}^o$. In water, $H-O-H$ bond angle decreases from ${109.5}^o$ to ${104.5}^o$ as lone pair-lone pair repulsion is more than bond pair-bond pair repulsion and lone pair-bond pair repulsion.