It is because of inability of `ns^(2)` electrons of the valence shell to participate in bonding that:

Options

A. `Sn^(2+)` is oxidising while `Pb^(4+)` is reducing
B. `Sn^(2+)` and `Pb^(2+)` are both oxidising and reducing
C. `Sn^(4+)` is reducing while `Pb^(4+)` is oxidising
D. `Sn^(2+)` is reducing while `Pb^(4+)` oxidising

Answer

D

Explanation

Since inert pair effect becomes more pronounced down the group, therefore +4 oxidation stable of Sn is more stable than its +2 oxidatation state and hence `Sn^(2+)` loses two electrons and thus acts a reducing agent.
`underset(" Less stable ")(Sn^(2+))rarr underset(" More stable ")(Sn^(4+)) + 2e^(-)`
In contrast, +2 oxidation state of Pb is more stable than its +4 oxidation state, therefore `Pb^(4+)` accepts two electrons and thus acts as an oxidising agent.
`underset(" Less stable ")(Pb^(4+)) + 2e^(-) rarr underset(" More stable ")(Pb^(2+))`
Thus, option (d) is correct.