The bond angle 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 $N H_{4}^{+} > N H_{3} > N H_{2}^{-}$
${N H_{4}}^{+}$ has a tetrahedral shape with a bond angle of ${109.5}^{o}$

$N H_{3}$ has a trigonal pyramidal shape with a bond angle less than ${109.5}^{o}$ due to the lone pair pushing the bonds closer.

$N H_{2}$ has a bent shape with two lone pairs of electrons which pushes the bond more close thus reducing it further.

Hence, B is the correct answer.

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