Due to the inability of ${ns}^{2}$ electrons of the valence shell to participate in bonding, which of the following statement is true.

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${Sn}^{2 +}$ is reducing agent while ${Pb}^{4 +}$ is oxidizing agent

${Sn}^{2 +}$ and ${Pb}^{2 +}$ are both oxidizing agents

${Sn}^{4 +}$ is reducing agent while ${Pb}^{4 +}$ is oxidizing agent

${Sn}^{2 +}$ is oxidizing agent while ${Pb}^{4 +}$ is reducing agent

answer is A.

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Detailed Solution

Inert pair effect is redundancy of s-electrons to participate in bonding. This effect is more pronounced in heavier metals. Therefore, inert pair effect increases on moving down the group.

Both Pb and Sn are group-14 elements and can show both +2 and +4 oxidation states. But due to the inert pair effect, stability of +2 oxidation state increases down the group.

Lead is heavier metal than Sn, this indicates that Pb (II) is more stable than Pb (IV) whereas Sn (IV) is more stable than Sn (II).

This depicts that Pb moves from +4 to+2 oxidation state.

${Pb}^{+ 4} \overset{}{\to} {Pb}^{+ 2}$