Question

It is because of inability of ns² 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

Answer 1

The Correct Option is A

To solve this question, we need to understand the concept of the inert pair effect. The inert pair effect refers to the reluctance of the ns² pair of electrons in the valence shell to participate in bonding, especially observed in heavier elements of Group 14 (also known as the carbon group) as we move down the group.

Here, Sn (Tin) and Pb (Lead) are relevant elements from Group 14. In their respective oxidation states:

• Sn^{2+}: tin has two additional electrons it could potentially lose to achieve a +4 oxidation state.
• Pb^{4+}: lead has already lost all four of its valence electrons, the +2 state is more stable due to the inert pair effect.

The inert pair effect is more significant in lead than in tin, making Pb^{4+} more likely to reduce to a more stable +2 oxidation state by gaining electrons, hence it acts as an oxidizing agent.

Meanwhile, Sn^{2+} can lose two more electrons to achieve a more stable +4 oxidation state, making Sn^{2+} act as a reducing agent.

Therefore, the correct answer is:

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

Justification for ruling out other options:

• \(Sn^{2+}\) is oxidizing while \(Pb^{4+}\) is reducing: Incorrect. From our explanation above, Pb^{4+} is an oxidizing agent, not a reducing agent.
• \(Sn^{2+}\) and \(Pb^{2+}\) are both oxidizing and reducing: This is incorrect as it contradicts the stable states defined by the inert pair effect.
• \(Sn^{4+}\) is reducing while \(Pb^{4+}\) is oxidising: This option is irrelevant as it doesn't involve Sn^{2+} and misses the main concept of the question.