Question

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

Answer 1

In the given question, the definition of bond angle should be known and the number of pairs of electrons which are non-bonding and remains free need to be determined.

Complete step by step answer:
Non-bonding pairs will have an effect on the shape and angle of the bonds between the bonded pair and non-bonded pairs of electrons. Here, the bond between nitrogen and hydrogen will remain common in all the molecules mentioned, but the change will be seen if the number of nonbonding pairs are present.
The theory of the Electron pair repulsion says that the lone pair-lone pair repulsion > lone-pair bond-pair repulsion > bond-pair bond pair repulsion.
The number of electrons in the valence shell are five in the nitrogen atom and the hydrogen will have one electron so it will share the electron forming covalent bond.
As you can see the structure of $N{H_3}$, there are three hydrogen atoms, three electrons of nitrogen will take part in bond formation with three hydrogen, and two electrons will remain unpaired. Therefore, the bond pair is three and the lone pair is one so the angle will be $107^\circ$ instead of $120^\circ$because of the presence of the one lone pair of electrons. The two non-bonding electrons will be of nitrogen.
In $NH_4^ +$, the bond pair are four and the lone pair is zero the angle will not get distorted but remain same hence the angle will be same as that of the tetrahedral molecule which is $109.5^\circ$. Here, the number of electrons in the outer shell will be four and nitrogen will be converted into ${N^ + }$, hence, it will get bonded with four hydrogen atoms by sharing the electron. No electron will remain non-bonded.
In $NH_2^ -$, the number of bonding pairs are two and the number of nonbonding pairs are two , hence the angle will get smaller than it is due to the repulsion is more in the lone pair than that of the bonding pair and hence the angle will be $104.5^\circ$instead of $180^\circ$. Here the number of electrons in the outer orbital will be six in nitrogen atom and hence nitrogen will be having negative charge ${N^ - }$ hence it will have two nonbonding electron pair after two electrons will be shared with two hydrogen forming two covalent bond.
The structure of the molecule $N{H_3}$, $NH_4^ +$and $NH_2^ -$ are as follows:
Thus, the correct option is B
Note:
The electron repulsion theory will make the difference in the structure. Do not consider the structure as the normal structure as it will change the bond angle of the molecule. $A{B_3},A{B_4}and{\rm{ }}A{B_2}$ type will have different bond angle as there will be no nonbonding pair of electron will repel and change the angle in the molecule.