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Elastomer

8. Why Pb2+ i...

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8. Why Pb2+ is more stable than Pb4+ ?

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P b^{4 +}

less stable than

P b^{2 +}

P b^{4 +}

is less stable than

P b^{2 +}

. This is due to:

Q. The correct code for stability of oxidation states of Sn and Pb is :

[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 I_{4}

doesn't exist and converts into

P b I_{2}

I_{2}

spontaneously at room temperature but

P b C l_{4}

needs heating to convert into

P b C l_{2}

C l_{2}

P b^{2 +}

is more stable than

P b^{4 +}

due to inert pair effect.

Q. Assertion :Statement

-

P b I_{4}

doesn't exist and converts into

P b I_{2}

I_{2}

spontaneously at room temperature but

P b C l_{4}

needs heating to convert into

P b C l_{2}

C l_{2}

. Reason: Statement

-

P b^{2 +}

is more stable than

P b^{4 +}

due to inert pair effect.

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