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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^(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

Text Solution

Verified by Experts

The correct Answer is:
A

Inability of `ns_(2)` electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect
As a result, Pb(II) is more stable than Pb(IV)
Sn(IV) is more stable than Sn(II)
`:.` Pb(IV) is easily reduced to Pb(II)
` :.` Pb(IV) is oxidising agent
Sn(II) is easily oxidised to Sn(IV)
`:.` Sn(II) is reducing agent
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