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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: $N{H_3} > NH_4^ + > NH_2^ -$

Last updated date: 25th Jun 2024
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Hint: For the formation of bond angle there must be at least three atoms and two bonds. Bond angle is the angle which is formed between three atoms across at least two bonds. This angle is different for different compounds.

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 $NH_4^ + > N{H_3} > NH_2^ -$.