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

Options

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

Answer

A

Explanation

`Sn^(2+)` is reducing while `Pb^(4+)` is oxidising