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

Options

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

Answer

D

Explanation

`Sn^(+2)toSn^(+4)`
`(R.A.) Sn^(2+) lt Sn^(+4)` Stability order
`Pb^(+4)toPb^(+2)`
`(O.A.) Pb^(+2)gtPb^(+2)` Stability order (inert pair effect)