Question

The correct statement for the following scheme of reactions is:
$4NCl3+3LiAlH4\to A+3LiCl+3AlC{l}_3$
$4PC{l}_3+3LiAlH4\to B+3LiCl+3AlC{l}_3$
$4AsC{l}_3+3LiAl{H}_4\to C+3LiCl+3AlC{l}_3$
$4SbC{l}_3+3LiAl{H}_4\to D+3LiCl+3AlC{l}_3$

• A. D is more basic than C
• B. D is more basic than B
• C. B is less basic than A
• D. A is more basic than B, C and D.

Answer 1

The correct option is D A is more basic than B, C and D.

These reactions lead to the formation of hydrides of the respective elements, i.e., $N{H}_3,P{H}_3,As{H}_3,Sb{H}_3$. As we move down the group, the electron density increases because the size increases and the tendency to donate electrons decreases. Therefpre, basicity decreases.
The basicity order is:
$N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$