Question

Which one of the following compounds has bond angle as nearly ${90}^{\circ }$?

• A. $N{H}_3$
• B. $H_{2}S$
• C. $H_{2}O$
• D. $S{F}_6$

Answer 1

Answer: (B), (D)

$H_{2}S$
$S{F}_6$

$H_{2}S$ 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 and $H_{2}O$ has angle around 104.5. 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 $S{F}_6$, we have 6 bond pairs in $s{p}^3{d}^2$ hybridization and the bond angle is ${90}^o$.