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

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

Verified Solution

Answer & Solution

Correct Answer

(D) \(Sn^{2+}\) is reducing while \(Pb^{4+}\) is oxidising.

Step-by-step Solution

Detailed explanation

Inability of \(n s^{2}\) 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 \(Pb(IV)\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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