Question

The bond angle of $N{H_3}$ , $NH_4^ +$ and $NH_2^ -$ are in the order:A: $NH_2^ - > N{H_3} > NH_4^ +$B: $NH_4^ + > N{H_3} > NH_2^ -$C: $N{H_3} > NH_2^ - > NH_4^ +$D: $N{H_3} > NH_4^ + > NH_2^ -$

As stated above more the number of lone pairs less is the bond angle due to lone pair-lone pair repulsions. Nitrogen usually makes $3$ bond pairs. In $NH_4^ +$ nitrogen has $4$ bond pairs, this fourth bond nitrogen is made with its lone pair. This means there is no lone pair available in $NH_4^ +$. We know less the number of lone pairs more is the bond angle. So the bond angle of $NH_4^ +$ will be the largest. In $N{H_3}$ nitrogen have three bond pairs and one lone pair and in $NH_2^ -$ nitrogen is having only two bond pairs. This means the number of free electrons in $NH_2^ -$ is more than the number of free electrons in $N{H_3}$ (as there are three bonds of nitrogen normally). This means the bond angle of $NH_2^ -$ will be less than that of $N{H_3}$ (as stated above more the number of lone pairs less is the bond angle. So the order of the bond angle will be: $NH_4^ + > N{H_3} > NH_2^ -$
Correct answer is option B that is $NH_4^ + > N{H_3} > NH_2^ -$.