Question

The bond angle in $H_{2}O$ is ${105}^0$ and in $H_{2}S$ it is ${90}^0$. It is due to:

• A. the larger size of $S$ atom as compared to $O$ atom which minimises repulsion and allows the bonds in $H_{2}O$ to be purely $p$-type
• B. liquid state of $H_{2}O$ as compared to gaseous state of $S{O}_2$
• C. both (a) and (b)
• D. none of the above is correct

Answer 1

Answer: (A)

the larger size of $S$ atom as compared to $O$ atom which minimises repulsion and allows the bonds in $H_{2}O$ to be purely $p$-type

The bond pair-bond pair is smaller in magnitude, still the bond pair-bond pair repulsion (steric effect) between the hydrogen in water is bigger than in hydrogen sulfide because oxygen is smaller. Going down the group the bond pair-bond pair repulsion decreases in magnitude much faster than lone pair-bond pair, therefore the smaller angle. Bigger size, in this case wins over smaller electronegativity (which would imply the opposite). If the central atom becomes big enough, the bond angle won't change much downwards.