Question

The nitrogen atoms in $N{H}_3,N{H}_2^{-}$ and $N{H}_4^{+}$ are all surrounded by eight electrons. When these species are arranged in increasing order of $H-N-H$ bond angle, correct order is

• A. $N{H}_3,N{H}_2^{-},N{H}_4^{+}$
• B. $N{H}_4^{+},N{H}_2^{-},N{H}_3$
• C. $N{H}_3,N{H}_4^{+},N{H}_2^{-}$
• D. $N{H}_2^{-},N{H}_3,N{H}_4^{+}$

Answer 1

The correct option is D $N{H}_2^{-},N{H}_3,N{H}_4^{+}$

Lone pair-lone pair repulsion is more than lone pair-bond pair or bond pair-bond pair repulsion.

${NH{}^{-}}_2$ has two lone pairs and two bond pairs but due to presence of lone pairs, angle $H-N-H$ is less than ${109}^o$.
${NH{}^{}}_3$ has three bond pairs and one lone pair. SInce one lone pair is less, angle is more than above case.
${NH{}^{+}}_4$ had four bond pairs. Angle is equal to ${109}^o$

Hence, the correct increasing order of $H-N-H$ bond angle is $N{H}_2^{-},N{H}_3,N{H}_4^{+}$.