Which of the following statement is true with respect to inert-pair effect?

Options

A. `Sn^(2+)` is oxidising agent while `Pb^(4+)` is reducing agent
B. `Sn^(2+) and Pb^(4+)` are both oxidising agent and reducing agent
C. `Sn^(2+)` is reducing agent while `Pb^(2+)` is oxidising agent
D. `Sn^(2+)` is reducing agent while `Pb^(4+)` is oxidising agent

Answer

To determine which statement is true with respect to the inert-pair effect, let's break down the concept step by step.

### Step-by-Step Solution:

1. **Understanding the Inert-Pair Effect**:
- The inert-pair effect refers to the tendency of the outermost s-electrons (the 'inert pair') to remain non-bonding or unshared in heavier elements of groups 13-16 of the periodic table. This effect becomes more pronounced as we move down a group in the periodic table.

2. **Identifying Relevant Elements**:
- The elements in question are Sn (Tin) and Pb (Lead), which are in group 14 of the periodic table. Sn is in the 4th period, while Pb is in the 6th period.

3. **Oxidation States**:
- Both Sn and Pb can exhibit oxidation states of +2 and +4. The +4 oxidation state is generally less stable for Pb compared to Sn due to the inert-pair effect.

4. **Stability of Oxidation States**:
- For Sn, the +4 oxidation state (Sn^4+) is stable, but for Pb, the +2 oxidation state (Pb^2+) is more stable than the +4 oxidation state (Pb^4+). This is due to the inert-pair effect, where the 6s electrons are less likely to participate in bonding as we move down the group.

5. **Conclusion**:
- The true statement regarding the inert-pair effect is that Pb^2+ is more stable than Pb^4+ due to the inert-pair effect, while Sn^2+ is less stable compared to Sn^4+.

### Final Answer:
The true statement with respect to the inert-pair effect is that Pb^2+ is more stable than Pb^4+.