The size of the species $P b, P b^{2 +}, a n d P b^{4 +}$ decreases as:

P b^{4 +} > P b^{2 +} > P b

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P b > P b^{2 +} > P b^{4 +}

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P b > P b^{4 +} > P b^{2 +}

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P b^{4 +} > P b > P b^{2 +}

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Solution

The correct option is D $P b > P b^{2 +} > P b^{4 +}$
For the same element, the atomic number and ionic size keep decreasing as the positive charge increases. Thus $P b$ is the largest and ${P b}^{4 +}$ is smallest among the given set. Hence correct order is $P b > {P b}^{2 +} > {P b}^{4 +}$

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Similar questions

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

P b > {P b}^{2 +} > {P b}^{4 +}

P b^{4 +} + S n^{2 +} \to S n^{4 +} + P b^{2 +}

S n^{4 +} + P b^{2 +} \to P b^{4 +} + S n^{2 +}

E_{P b^{4 +} / P b^{2 +}}^{o} = 1.69 V, E_{S n^{4 +} / S n^{2 +}}^{o} = 0.15 V

P b^{4 +}

P b^{2 +}

During the working of the cell :

P b - H g (1.0 M) | P b^{2 +} (a q) (1 M) | P b^{2 +} (a q) (1 M) | | P b - H g (0.5 M) |

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