$The bond angles of {NH}_{3}, {NH}_{4}^{\oplus} and {\overset{⊖}{NH}}_{2} are in the order:$

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${NH}_{3} > \overset{\oplus}{N} H_{4} > \overset{⊖}{{NH}_{2}}$

${NH}_{3} > \overset{⊖}{{NH}_{2}} > \overset{\oplus}{{NH}_{4}}$

${NH}_{2}^{⊖} > {NH}_{3} > {\overset{\oplus}{NH}}_{4}$

$\overset{\oplus}{{NH}_{4}} > {NH}_{3} > \overset{⊖}{{NH}_{2}}$

answer is B.

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Detailed Solution

${\overset{\oplus}{NH}}_{4} \Rightarrow s p^{3} hybridised with no l p of e^{-'} s, so shape is tetrahedral and$ bond angle : 109^o 28'.
${\ddot{NH}}_{3} \Rightarrow s p^{3} hybridised and is pyramidal in shape with one$ lp of e^-s, so due to lp-bp and bp-bp repulsion bond angle decrease to 107°.
$\overset{⊝}{N} H_{2} \Rightarrow Hybridisation = \frac{1}{2} (5 + 2 + 1) = 4 = s p^{3}$ $\overset{}{: {NH}_{2}}$ is sp³ hybridised with two lp of e^-'s,so due to lp-lp, lp-bp and bp-bp repulsion bond angle is further decreases $to {104.5}^{\circ} (like H_{2} O :) and have V -shape.$