Predict the correct order of reducing nature of group $15$ hydrides.

N H_{3} < P H_{3} < A s H_{3} < S b H_{3} < B i H_{3}

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N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}

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P H_{3} < A s H_{3} < N H_{3} < S b H_{3} < B i H_{3}

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P H_{3} < A s H_{3} < N H_{3} > S b H_{3} > B i H_{3}

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Solution

The correct option is A $N H_{3} < P H_{3} < A s H_{3} < S b H_{3} < B i H_{3}$
Down the group the bond dissociation enthalpy of $M - H$ bond (where, $M =$ group $15$ element) decresaes.
Order of bond dissociation enthalpy:
$N - H > P - H > A s - H > S b - H > B i - H$

Therefore, $B i H_{3}$ can loose hydrogen very easily and hence has the highest reducing power.
Order of reducing power:
$N H_{3} < P H_{3} < A s H_{3} < S b H_{3} < B i H_{3}$

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