Question

The bond angle in $H_{2}O$ is ${105}^o$ 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 minimizes repulsion and allows the bonds in $H_{2}S$ 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

Answer 1

Answer: (A)

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

Size of O is very small as compared to S. So electron density in O is very high which causes repulsion which is less in S. So bond angle increases in $H_{2}O$