Question

The nitrogen atoms in $N{H}_3,{NH}_2^{-}$ and $N{H}_4^{+}$ are all surrounded by eight electrons. When these species are arranged according to $HNH$ bond angle, the 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

Answer: (D)

$N{H}_2^{-}<N{H}_3<N{H}_4^{+}$

If an atom has lone pairs and bonded electron pairs, the greater lone- pair lone- pair electron repulsions will cause bond angle to decrease between bonded atoms.

$N{H}_2^{-}$ has $2$ bond pairs and $2$ lone pairs of electrons.

$N{H}_3$ has $3$ bond pairs and $1$ lone pair.

$N{H}_4^{+}$ has $4$ bond pairs and no lone pairs.

Bond- pair bond- pair electron repulsions < bond- pair lone- pair electron repulsions < lone- pair lone- pair electron repulsions.

Greater lone- pair lone- pair repulsions among electrons causes the $H-N-H$ bond angle to decrease.

This effect is greatest for $N{H}_2^{-}$, followed by $N{H}_3$.

Hence, the correct order in terms of $HNH$ bond angle is as follows :

$N{H}_2^{-}<N{H}_3<N{H}_4^{+}$

Therefore, the correct option is D.