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Question

It is because of the inability of ns2 electrons of the valence shell to participate in bonding that


A

Sn2+ is oxidizing while Pb2+ is reducing

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B

Sn2+ and Pb2+ are both oxidizing and reducing

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C

Sn4+ is reducing while Pb4+is oxidizing

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D

Sn2+ is reducing while Pb4+is oxidizing

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Solution

The correct option is C

Sn4+ is reducing while Pb4+is oxidizing


As we go down the group, the outermost s electrons of the ns2 np2 configuration tend not to participate in bond formation. Hence, the stability if +4 oxidation state decreases and the stability of the +2 oxidation state increases (as we descend). This is called inert pair effect.

This implies that Pb4+ would want to grab two electrons and become the more stable Pb2+ and itself get reduced. Hence Pb4+ is oxidizing. Likewise, for tin, Sn4+ is more stable than Sn2+ and hence, the latter ( as in - Sn2+) is reducing.


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