Question

Among $N{{H}_{3}},P{{H}_{3}},As{{H}_{3}}$ and $Sb{{H}_{3}}$ which is the strongest reducing agent?

Answer 1

Nitrogen$N$, Phosphorus $P$ , Astatine $As$ and Antimony $S{{b}_{{}}}$ are Group 5A elements of the Modern Periodic Table and it is worth remembering that as we go down the group the size of the atom increases. From the electrochemistry of elements, we know that an element is a better reducing agent if it is able to supply Hydrogen $H$ atoms (or consume Oxygen $O$).

Complete answer:
The correct answer is $Sb{{H}_{3}}$. It is the strongest reducing agent among the given hydrides.
As already discussed in the hint, the atomic sizes of Group 5A elements are in the order:
$N{{H}_{3}}>P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}$
Hence, the bond lengths of the element and their hydrides $M-H$ is in the order:

\[N-H>P-H>As-H>Sb-H$$ And hence the stability of the hydrides is of the order: $$N{{H}_{3}}>P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}$$ Thus, $$Sb-H$$ bond is weakest and most likely to break and $$Sb{{H}_{3}}$$ is most likely to supply hydrogen atoms and hence it is the best reducing agent among them. **Additional Information:** Reducing agent: The more exact definition of reducing agent is the compound that loses electrons in an electrolytic reaction. Amongst $$N,P,As,Sb$$ , Antimony $$Sb$$ offers the weakest attractive forces of nucleus to the outermost electrons due to its large atomic size. Thus, it is more susceptible to lose electrons than the rest of the other elements and hence its compounds will be better reducing agents. This tendency of losing electrons is known as electropositivity. **Note:** The basicity of Group 5A hydrides decreases as we go down the group which is also due to the same reason, more tendency of losing electrons. Moreover, due to decreasing electronegativity, polarity of the $$M-H$$ decreases and the bond becomes more covalent, hence becoming more soluble as we down the group.