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NEET CHEMISTRYP Block ElementsMedium

Question

It is because of the inability of ns2ns^2 electrons of the valence shell to participate in bonding that:

A

Sn2+\text{Sn}^{2+} is reducing while Pb4+\text{Pb}^{4+} is oxidising agent

B

Sn2+\text{Sn}^{2+} is oxidising while Pb4+\text{Pb}^{4+} is reducing agent

C

Sn2+\text{Sn}^{2+} and Pb2+\text{Pb}^{2+} are both oxidising and reducing agent

D

Sn4+\text{Sn}^{4+} is reducing while Pb4+\text{Pb}^{4+} is oxidising agent

Step-by-Step Solution

Due to the inert pair effect, the reluctance of valence ns2ns^2 electrons to participate in bonding increases down the group. Therefore, in the carbon family (Group 14), the stability of the +4+4 oxidation state decreases and that of the +2+2 oxidation state increases as we move from Ge\text{Ge} to Pb\text{Pb}. Consequently, for Tin (Sn\text{Sn}), the +4+4 state is more stable than the +2+2 state, making Sn2+\text{Sn}^{2+} a reducing agent as it readily oxidises to Sn4+\text{Sn}^{4+}. Conversely, for Lead (Pb\text{Pb}), the +2+2 state is more stable than the +4+4 state. Thus, Pb4+\text{Pb}^{4+} acts as a strong oxidising agent because it readily gets reduced to Pb2+\text{Pb}^{2+}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from P Block Elements. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYP Block Elementsbecauseinabilityelectronsvalenceparticipate

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