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It is because of inability of ns^2 elect...

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

A

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

B

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

C

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

D

`Sn^(2+) and Pb^(2+)` are both oxidizing and reducing

Text Solution

Verified by Experts

The correct Answer is:
B
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