Question

The correct order of reducing abilities of hydrides of ${\text{V}}$ group elements is:
A. ${\text{N}}{{\text{H}}_{\text{3}}}{\text{ > P}}{{\text{H}}_{\text{3}}}{\text{ > As}}{{\text{H}}_{\text{3}}}{\text{ > Sb}}{{\text{H}}_{\text{3}}}{\text{ > Bi}}{{\text{H}}_{\text{3}}}$
B. ${\text{N}}{{\text{H}}_{\text{3}}}{\text{ < P}}{{\text{H}}_{\text{3}}}{\text{ < As}}{{\text{H}}_{\text{3}}}{\text{ < Sb}}{{\text{H}}_{\text{3}}}{\text{ < Bi}}{{\text{H}}_{\text{3}}}$
C. ${\text{N}}{{\text{H}}_{\text{3}}}{\text{ < P}}{{\text{H}}_{\text{3}}}{\text{ > As}}{{\text{H}}_{\text{3}}}{\text{ > Sb}}{{\text{H}}_{\text{3}}}{\text{ > Bi}}{{\text{H}}_{\text{3}}}$
D.${\text{Sb}}{{\text{H}}_{\text{3}}}{\text{ > Bi}}{{\text{H}}_{\text{3}}}{\text{ > As}}{{\text{H}}_{\text{3}}}{\text{ > N}}{{\text{H}}_{\text{3}}}{\text{ < P}}{{\text{H}}_{\text{3}}}$

Answer 1

Reduction is the loss of hydrogen atom. The easier it is to lose the hydrogen bonded to the element, more is the reducing ability.

Complete step by step answer:
The stability of the molecule depends directly on the bond dissociation enthalpy of the compound. It is the energy required to break a bond and form two different entities. The process is endothermic. Thus, we can conclude that more the stability, higher is the bond dissociation enthalpy and more is the difficulty to remove hydrogen atoms from the hydride.
Thus, for the reducing ability to be high, the bond dissociation enthalpy of the hydride must be low.
The bond dissociation energy increases as the electronegativity difference between the bonded atoms decreases.
We know that, down the group, electronegativities of elements decrease.
Thus, we can conclude that the stability of the hydrides decreases down the group due to the decrease in the bond dissociation enthalpy of the bond with hydrogen leading to an increase in the reducing character. As a result, ${\text{N}}{{\text{H}}_{\text{3}}}$is a mild reducing agent and ${\text{Bi}}{{\text{H}}_3}$ has the strongest reducing character among all hydrides of the nitrogen group.
The trend of reducing abilities of the hydrides of group 15 are:
${\text{N}}{{\text{H}}_{\text{3}}}{\text{ < P}}{{\text{H}}_{\text{3}}}{\text{ < As}}{{\text{H}}_{\text{3}}}{\text{ < Sb}}{{\text{H}}_{\text{3}}}{\text{ < Bi}}{{\text{H}}_{\text{3}}}$

The correct option is B.

Note:
The periodic table can be used to derive relationships between various physical and chemical properties and behaviours of elements. The modern periodic table provides an essential framework that helps in the assumption of the characteristics and properties of the elements yet to be discovered.