Question

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

Answer 1

The Correct Option is D

The question focuses on the inert pair effect, which explains the reluctance of $ns^{2}$ electrons in the valence shell to participate in bonding, particularly in heavier p-block elements like lead (Pb) and tin (Sn). Let's break down the options to understand the correct answer:

1. The inert pair effect is stronger in lead ($Pb$) compared to tin ($Sn$) due to lead being a heavier element. As a result, lead's $ns^{2}$ electrons are less likely to participate in bonding, leading to a stable $Pb^{2+}$ state.
2. For tin, $Sn^{2+}$ can be further oxidized to the more stable $Sn^{4+}$, thus it acts as a reducing agent, tending to lose electrons and take on a positive oxidation state.
3. For lead, the $Pb^{4+}$ state is less stable owing to the inert pair effect, so it tends to gain electrons to reach a stable $Pb^{2+}$ state, acting as an oxidizing agent.

Given this understanding, let's evaluate the provided options:

• $Sn^{2+}$ is oxidizing while $Pb^{4+}$ is reducing – Incorrect because $Sn^{2+}$ is actually reducing.
• $Sn^{2+}$ and $Pb^{2+}$ are both oxidizing and reducing – Incorrect; they are not both simultaneously oxidizing and reducing, as their stability preferences differ.
• $Sn^{4+}$ is reducing while $Pb^{4+}$ is oxidizing – Incorrect since $Sn^{4+}$ does not act as a reducing agent.
• $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidizing – Correct. $Sn^{2+}$ acts as a reducing agent, while $Pb^{4+}$, due to its instability, acts as an oxidizing agent.

Therefore, the correct answer is: $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidizing.