Question

Among $N{H_3},P{H_3},As{H_3}$ and $Sb{H_3}$ which one is a stronger reducing agent?

Answer 1

Hint:Compound or substance that can reduce other substances or in other words donate electrons to other substances is called a reducing agent. The strength of a reducing agent depends upon how easily it can lose or donate electrons.

Complete step-by-step answer:The compounds given to us are $N{H_3},P{H_3},As{H_3}$ and $Sb{H_3}$
The central atoms of all these four compounds belong to the fifteenth group. We know that as we go down the group, the size of the atom increases. This means that the larger the size of the atom the farther the valence electrons from the nucleus.
Electrons that are farther away from the nucleus experience less attractive force from the nucleus and this means that the electrons can be removed or lost easily.
For the given compounds, the atomic size of the central atom increases as follows.
$N < P < As < Sb$ and hence Sb has the largest atomic size of them all. This means that it is easier to remove an electron from the valence shell of Sb when compared to As, P and N.
Hence the increasing order of strength of reduction of the given compounds can be written as \[N{H_3} < P{H_3} < As{H_3} < Sb{H_3}\]
Therefore, the compound $Sb{H_3}$ or Antimony hydride is a stronger reducing agent when compared to $N{H_3},P{H_3}$ and $As{H_3}$ .

Note: It is to be noted that the strength of a reducing agent increases as the size of the central atom of that particular compound or agent increases. This means that the reducing strength of a compound is directly proportional to its size.