Compound that formally contains $P b^{4 +}$ are easily reduced to $P b^{2 +}$ . The stability of the lower oxidation state is due to:

inert pair effect

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ionic radius

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charge

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covalent character

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Solution

The correct option is A inert pair effect
As we go down the group in p-block, there is an increasing tendency for the s $^{2}$ pair not to be used in bonding i.e. the s-electrons remain inert and do not participate in bonding. This is called an inert pair effect. Hence, the oxidation state is decreased by 2 and so Pb $^{2 +}$ is found to be more stable than Pb $^{4 +}$ .

Hence, the correct option is $A$

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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 +}

It is because of the inability of $n s^{2}$ electrons of the valence shell to participate in bonding that

G e (I I)

P b (I V)

Stability of ions of $G e, S n, a n d P b$ will be in the order:

P b^{4 +}

S n^{4 +}

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