1. Sn²⁺ is oxidising while Pb⁴⁺ is reducing 2. Sn²⁺ and Pb²⁺ are both oxidising and reducing 3. Sn⁴⁺ is reducing while Pb⁴⁺ is oxidising 4. Sn²⁺ is reducing while Pb⁴⁺ is oxidising

Answer: 4. Sn²⁺ is reducing while Pb⁴⁺ is oxidising Key idea: the inert pair effect makes the +2 state more stable than +4 in heavy group 14 elements. Because the ns² electrons do not take part in bonding, mathrmPb²⁺ is more stable than mathrmPb⁴⁺ . So mathrmPb⁴⁺ readily gains electrons to become mathrmPb²⁺ : it is an oxidising agent. mathrmSn²⁺ is easily oxidised to the more stable mathrmSn⁴⁺ : it is a reducing agent. This matches option 4.

Topic: Electronic configuration and general trends, Group 13 and 14· p-Block Elements

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

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

2. \(Sn^{2+}\) and \(Pb^{2+}\) are both oxidising and reducing

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

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

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

1 answer

Best answer

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

Key idea: the inert pair effect makes the +2 state more stable than +4 in heavy group 14 elements.

Because the \(ns^2\) electrons do not take part in bonding, \(\mathrm{Pb^{2+}}\) is more stable than \(\mathrm{Pb^{4+}}\). So \(\mathrm{Pb^{4+}}\) readily gains electrons to become \(\mathrm{Pb^{2+}}\): it is an oxidising agent. \(\mathrm{Sn^{2+}}\) is easily oxidised to the more stable \(\mathrm{Sn^{4+}}\): it is a reducing agent. This matches option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

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

1 answer

Your answer