The bond angles of $N H_{3}, N H_{4}^{+} a n d {N H_{2}}^{-}$ are in the order:

{N H_{2}}^{-} > N H_{3} > {N H_{4}}^{+}

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{N H_{4}}^{+} > N H_{3} > {N H_{2}}^{-}

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N H_{3} > {N H_{2}}^{-} > N H_{4}^{+}

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N H_{3} > N H_{4}^{+} > {N H_{2}}^{-}

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Solution

The correct option is B ${N H_{4}}^{+} > N H_{3} > {N H_{2}}^{-}$
If an atom has lone pairs and bonded electron pairs, the greater lone pair-lone pair 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.
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 is:
$N H_{4}^{+} > N H_{3} > {N H_{2}}^{-}$

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