The incorrect statement regarding $15^{t h}$ group hydride $E H_{3}$ . $[E = N, P, A s, S b, B i]$

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

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E - H Bond dissociation enthalpy : N - H > P - H > A s - H > S b - H > B i - H

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

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Basicity : N H_{3} > P H_{3} > A s H_{3} > S b H_{3} \geq B i H_{3}

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Solution

The correct option is C $Reducing character : N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}$
(a) The stability of hydrides decreases from $N H_{3}$ to $B i H_{3}$ which can be observed from their bond dissociation enthalpy.
(b) Bond dissociation enthalpy in $kJ/mol$
$N - H 389 > P - H 322 > A s - H 297 > S b - H 255 > B i - H$
(c) With the increase in the size of central atom $E$ , $E - H$ bond becomes weaker going down the group.
$Reducing character \propto \frac{1}{E - H Bond dissociation enthalpy}$
Thus correct order for reducing charcter is:
$N H_{3} < P H_{3} < A s H_{3} < S b H_{3} < B i H_{3}$
(d) Basicity decreases in the order $N H_{3} > P H_{3} > A s H_{3} > S b H_{3} \geq B i H_{3}$

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