Question

Which compound has bond angle nearly to ${90}^0$?

• A. $H_{2}O$
• B. $H_{2}S$
• C. ${NH}_3$
• D. ${CH}_4$

Answer 1

Answer: (B)

$H_{2}S$

$H_{2}S$ is $s{p}^3$ hybridized with 2 bond pair and 2 lone pair. But due to lone pair lone pair repulsions, the bond angle decreases from ${109.5}^o$ to ${92}^o$.

approx. Ammonia and water are $s{p}_3$ hybridized. $N{H}_3$ has angle around ${107}^o$ and $H_{2}O$ has angle around ${104.5}^o$. Oxygen being highly electronegative than sulphur, keeps the bond pair electrons of its bond with hydrogen near itself, due to which the bond angle is higher than $H_{2}S$. In $C_4$,the bond angle is ${104.5}^o$.

Although S-atom in $H_{2}S$ molecule shows hybrid state of $s{p}^3$, but the presence of two lone pairs on sulphur atom, decreases the bond angle to ${90}^o$ and the molecule becomes V-shaped.

Option B is correct answer for this quetion.