Question

Reducing nature of the hydrides of group 15 follow the order

• A. $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$
• B. $N{H}_3<P{H}_3<As{H}_3<Sb{H}_3<Bi{H}_3$
• C. $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3=Bi{H}_3$
• D. $N{H}_3<P{H}_3<Bi{H}_3<Sb{H}_3<As{H}_3$

Answer 1

The correct option is B $N{H}_3<P{H}_3<As{H}_3<Sb{H}_3<Bi{H}_3$

As we descend the group, the size of the atom increases while the electronegativity decreases. So, can we say that Nitrogen exerts a greater degree of pull over the N – H bond than phosphorus in P – H bond? That seems logical! Further, with the increase in the size of the central atom, the orbital-orbital overlap between the element and Hydrogen will decrease.
Extending the logic further, can we infer that the strength of the E – H bond weakens as we go down the group? That is a correct assessment. As we go down the group, the bond dissociation energy decreases. In other words, the lower we go, the weaker the E – H bond! The weaker the E – H bond, the easier is the abstraction of Hydrogen. So the reducing nature of the hydrides increases as we go down the group.
$As{H}_3$ is more reducing that $P{H}_3$ which is more reducing than $N{H}_3$. So the increasing order of reducing nature is:
$N{H}_3<P{H}_3<As{H}_3<Sb{H}_3<Bi{H}_3$
The order of the stability is just the reverse