Question

$P{b}^{4+}$ is less stable than $P{b}^{2+}$. This is due to:

• A. Poor shielding of $6f^{14}$ electrons by inner electrons
• B. Poor shielding of $6s^2$ electrons by inner $d$ and $f$ electrons
• C. Poor shielding of $6p^2$ electrons by inner $s$ and $f$ electron
• D. Poor shielding of $5d^{10}$ electrons by inner $f$ electrons

Answer 1

The correct option is B Poor shielding of $6s^2$ electrons by inner $d$ and $f$ electrons

Electronic configuration of $Pb$ is $\left[Xe\right]4f^{14}5d^{10}6s^{2}6p^2$.
Due to poor shielding of $6s^2$ electrons by inner $d$ and $f$ electron it is difficult to remove $6s^2$ electron thus they are inert. This is known as inert pair effect.
Hence, in case of $Pb$, the $+2$ oxidation state is more stable than the $+4$ oxidation state due to inert pair effect.