It is because of inability of ns2 electrons of the valence shell to participate in bonding that [2017]

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Q.
It is because of inability of ${ns}^{2}$ electrons of the valence shell to participate in bonding that [2017]

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

Ans.
(4)

The inertness of $s$ -subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for Sn, +4 oxidation state is more stable, whereas, for Pb, +2 oxidation state is more stable, i.e., ${Sn}^{2 +}$ is reducing while ${Pb}^{4 +}$ is oxidising.

Solution

Q.
It is because of inability of ${ns}^{2}$ electrons of the valence shell to participate in bonding that [2017]

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

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Solution

Q. It is because of inability of ns2 electrons of the valence shell to participate in bonding that [2017]

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

Ans. (4) The inertness of s-subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for Sn, +4 oxidation state is more stable, whereas, for Pb, +2 oxidation state is more stable, i.e., Sn2+ is reducing while Pb4+ is oxidising.

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Q.
It is because of inability of ${ns}^{2}$ electrons of the valence shell to participate in bonding that [2017]

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