Question

The incorrect statement regarding ${15}^{th}$ group hydride $E{H}_3$. $[E=N,P,As,Sb,Bi]$

• A. $\text{Thermal stability}:N{H}_3>P{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$
• B. $\text{Basicity}:N{H}_3>P{H}_3>As{H}_3>Sb{H}_3\ge Bi{H}_3$
• C. $E-H\text{Bond dissociation enthalpy}:N-H>P-H>As-H>Sb-H>Bi-H$
• D. $\text{Reducing character}:N{H}_3>P{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$

Answer 1

The correct option is D $\text{Reducing character}:N{H}_3>P{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$

(a) The stability of hydrides decreases from $N{H}_3$ to $Bi{H}_3$ which can be observed from their bond dissociation enthalpy.
(b) Bond dissociation enthalpy in $\text{kJ/mol}$
$\underset{389}{N-H}>\underset{322}{P-H}>\underset{297}{As-H}>\underset{255}{Sb-H}>Bi-H$
(c) With the increase in the size of central atom $E$, $E-H$ bond becomes weaker going down the group.
$\text{Reducing character}\propto \frac{1}{E-H\text{Bond dissociation enthalpy}}$
Thus correct order for reducing charcter is:
$N{H}_3<P{H}_3<As{H}_3<Sb{H}_3<Bi{H}_3$
(d) Basicity decreases in the order $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3\ge Bi{H}_3$