The conjugate acid of $N H_{2}^{-}$ ion is:

N H_{3}

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N H_{4}^{+}

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N_{2} H_{4}

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N H_{2} O H

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Solution

The correct option is A $N H_{3}$
Acid + base $\to$ conjugate base + conjugate acid.

Conjugate acid is one which donates the proton of the corresponding base.

So, in $N H_{2}^{-}$ case, $N H_{3}$ loses the proton.

$N H_{2}^{-} + H^{+} \to N H_{3}$

Option A is correct.

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