You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Carbon Family

Why is Pb4+ l...

Question

Why is $P b^{4 +}$ less stable than $P b^{2 +}$ ?

Due to poor shieding of

6 s^{2}

electron by inner d and f electron

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Due to poor shieding of

6 p^{2}

electron by inner s and f electron

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Due to poor shieding of

6 f^{14}

electron by inner electron

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Due to poor shieding of

5 d^{10}

electron by inner f electron

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A Due to poor shieding of $6 s^{2}$ electron by inner d and f electron
Lead · Electron configuration
$[X e] 4 f^{14} 5 d^{10} 6 s^{2} 6 p^{2}$
Due to poor shieding of $6 s^{2}$ electron by inner d and f electron it is difficult to remove $6 s^{2}$ electron thus they are inert.
Hence, in case of Pb, the +2 oxidation state is more stable than the +4 oxidation state due to inert pair effect.

Suggest Corrections

Similar questions

P b^{4 +}

P b^{2 +}

P b^{4 +}

P b^{2 +}

[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^{4 +}

P b^{2 +}

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

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program