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 addresses the behavior of $Sn^{2+}$ and $Pb^{4+}$ in relation to their oxidation and reduction properties due to the inert pair effect. The inert pair effect is the tendency of the $ns^{2}$ electrons in heavier p-block elements to be reluctant to participate in bonding. This reluctance leads to the elements under study exhibiting certain oxidation states more readily than others.

1. The inert pair effect is notable in heavier elements like lead (Pb) and influences the stability of oxidation states.
2. In the case of tin (Sn) and lead (Pb), the lower oxidation state (+2) for Pb is more stable due to the inert pair effect, leading to Pb in the +4 oxidation state acting as an oxidizing agent, which means it can accept electrons from another species to be reduced to $Pb^{2+}$.
3. Conversely, tin readily exists in the +2 state as $Sn^{2+}$ and prefers losing additional electrons to reach a more stable +4 state, making it act as a reducing agent, donating electrons to another species.
4. Considering the options, $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidizing correctly describes the behavior of these ions in terms of their redox properties due to the inert pair effect.

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