Question

Due to inert pair effect, the most stable oxidation state
of lead is (Atomic number of Pb = 82)
-2
+2
- 4
+4​

Answer 1

Due to the inert pair effect, the most stable oxidation state of Lead is +2.

• Lead (Pb) is a member of the Carbon family, i.e. it belongs to Group - 14.
• The electronic configuration of Pb is [Xe] 5d¹⁰4f¹⁴6s²6p²
• Due to the electrons in d and f shells, there is a high shielding effect of electrons in it. Hence, the lower oxidation state, i.e. +2 dominates.

Answer 2

To find:

The most stable oxidation state of Lead (Pb)?

Solution:

• You need to remember this as an information because this question is often asked in competitive exam MCQs.

• Pb has +2 oxidation state as it's most stable form.

• Still I am explaining the proper reason !

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• If you see group 14 of the MODERN PERIODIC TABLE , the main elements are Carbon, Silicon, Germanium, Tin, Lead.

• In the upper elements of this group, both s and p orbitals take part in bonding, hence C, Si, Ge show +4 oxidation state.

• Now, see the electronic configuration of Pb:

$\boxed{ \bf Pb = [Xe] \: 4 {f}^{14} \: 5 {d}^{10} \: 6 {s}^{2} \: 6 {p}^{2} }$

• Here , a peculiar effect occurs named INERT PAIR EFFECT !

• The electrons of 6s doesn't take part as bonding electrons and hence don't contribute to positive oxidation state.

• Only the electrons of 6p contribute.

• Hence , the most stable oxidation state is +2 for Pb.

• As a matter of fact, this effect is seen in all the lower elements of the p block groups.