Question

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

• A. Due to poor shieding of $6s^2$electron by inner d and f electron
• B. Due to poor shieding of $6p^2$electron by inner s and f electron
• C. Due to poor shieding of $6f^{14}$electron by inner electron
• D. Due to poor shieding of $5d^{10}$electron by inner f electron

Answer 1

The correct option is A Due to poor shieding of $6s^2$electron by inner d and f electron

Lead · Electron configuration
$\left[Xe\right]4f^{14}5d^{10}6s^{2}6p^2$
Due to poor shieding of $6s^2$electron by inner d and f electron it is difficult to remove $6s^2$ 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.