Question

When $N{H}_3$ is treated with $HCl,H-N-H$ bond angle:

• A. increases
• B. decreases
• C. remains same
• D. depends upon temperature

Answer 1

The correct option is A increases

$N{H}_3+HCl⟶N{H}_{4}Cl$

In $N{H}_3,H-N-H$ bond was ${107.8}^o$ which is less than the regular tetrahedral due to $lp-bp$ repulsion. When $N{H}_3$ reacts with $HCl.N{H}_4^{+}C{l}^{-}$ is formed where $N{H}_4$ is having tetrahedral geometry without any lone pair, hence the $H-N-H$ bond angle inceases from $107.8\%$ to $109.5\%$