Question

Ge(II) compounds are powerful reducing agents whereas Pb(IV) compounds are strong oxidants, because

• A. $\text{Pb is more electropositive than Ge}$
• B. $\text{Ionization potential of lead is less than that of Ge}$
• C. $\text{Ionic radii of}P{b}^{2+}\text{and}P{b}^{4+}\text{are larger than those of}G{e}^{2+}\text{and}G{e}^{4+}$
• D. $G{e}^{4+}\text{is more stable than}G{e}^{2+}\text{whereas}P{b}^{2+}\text{is more stable than}P{b}^{4+}$

Answer 1

The correct option is D $G{e}^{4+}\text{is more stable than}G{e}^{2+}\text{whereas}P{b}^{2+}\text{is more stable than}P{b}^{4+}$

Relative stability of +2 oxidation state progressively increases for heavier elements. Thus, Germanium forms stable compounds in +4 state and only few compounds in +2 state.
Ge(II) tends to acquire Ge (IV) state by loss of electrons. Hence it is reducing in nature. Pb shows mainly +2 oxidation state frequently as compared to +4 oxidation state. This is due to the inert pair effect. Pb (IV) tends to acquire Pb (II) O.S. by gain of electrons. Hence it is oxidising in nature.