Question

The reducing power of divalent species decreases in the order

• A. $G{e}^{2+}>S{n}^{2+}>P{b}^{2+}$
• B. $S{n}^{2+}>G{e}^{2+}>P{b}^{2+}$
• C. $P{b}^{2+}>S{n}^{2+}>G{e}^{2+}$
• D. None of these

Answer 1

The correct option is A $G{e}^{2+}>S{n}^{2+}>P{b}^{2+}$

Inert pair effect may be defined as inertness of inner ns sub-shell towards chemical reaction. It is seen in group 3A and 4A group especially (5s and 6s electrons). Therefore, Pb shows mainly +2 oxidation state frequently as compared to +4 oxidation state. Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and only few compounds in +2 state. Tin forms compounds in both oxidation states.
So as we move down the group, due to inert pair effect lower oxidation state become more stable.
So,
The stability of + 2 O.S. follows the order
$P{b}^{2+}>S{n}^{2+}>G{e}^{2+}$
Higher the stability of +2 oxidation states, lesser will be its reducing power.
Therefore, order of reducing power is
$G{e}^{+2}>S{n}^{+2}>P{b}^{+2}$