Question

For $H_{2}O$ and $H_{2}S$ given -

Compound	Hybridisation of	Hybridisation	bond angle
$H_{2}O$	$O$	$s{p}^3$	${104.5}^0$
$H_{2}S$	$S$	$s{p}^3$	${92.1}^0$

Difference in bond angle is due to-

• A. The larger size of $S$ atom as compared to $O$ atom which decreases the orbital overlap and hence the hybridization tends to be purely p orbitals.
• B. Liquid state of $H_{2}O$ as compared to gaseous state of $H_{2}S$
• C. Both of the above
• D. None of the above

Answer 1

The correct option is A The larger size of $S$ atom as compared to $O$ atom which decreases the orbital overlap and hence the hybridization tends to be purely p orbitals.

There are 3 types of repulsion- (lp-lp), (lp-bp), (bp-bp) thus bond angle in $H_{2}O$ < ${109.5}^{\circ }$ i.e. ${104.5}^{\circ }$ and in $H_{2}S$ ${90}^{\circ }$. It is due to larger size of $S$ atom which minimises the repulsion and allows the bonds in $H_{2}S$ to be purely p-type.