The conjugate base of $N H_{3}$ is

N H_{4}^{+}

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N H_{2}^{-}

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N O_{2}^{-}

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N_{2}

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Solution

The correct option is B $N H_{2}^{-}$
$N H_{3} ⇄ N H_{2}^{-} + H^{+}$
$The conjugate base of N H_{3} i s N H_{2}^{-} which is known as the azanide anion.$

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