Correct statbility order of metal cation is /are :

Options

A. `Pb^(2+) lt Sn^(2+)`
B. `Pb^(4+) lt Pb^(2+)`
C. `Sn^(4+) lt Sn^(2+)`
D. `Pb^(4+) lt Sn^(4+)`

Explanation

To determine the stability order of the metal cations Pb²⁺, Pb⁴⁺, Sn²⁺, and Sn⁴⁺, we will analyze the electronic configurations and the inert pair effect.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- Lead (Pb) has an atomic number of 82. Its electronic configuration is:
- Pb: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
- Tin (Sn) has an atomic number of 50. Its electronic configuration is:
- Sn: [Kr] 4d¹⁰ 5s² 5p²

2. **Determine the Oxidation States**:
- For Pb:
- Pb²⁺ is formed by losing 2 electrons from the 6p² subshell.
- Pb⁴⁺ is formed by losing 2 additional electrons from the 6s² subshell.
- For Sn:
- Sn²⁺ is formed by losing 2 electrons from the 5p² subshell.
- Sn⁴⁺ is formed by losing 2 additional electrons from the 5s² subshell.

3. **Apply the Inert Pair Effect**:
- The inert pair effect refers to the tendency of the s-electrons (6s² in Pb and 5s² in Sn) to remain non-bonding and not participate in bonding as we move down the group in the periodic table.
- This means that higher oxidation states (like Pb⁴⁺ and Sn⁴⁺) are less stable compared to their lower oxidation states (like Pb²⁺ and Sn²⁺).

4. **Analyze Stability**:
- Pb²⁺ is more stable than Pb⁴⁺ due to the inert pair effect.
- Similarly, Sn²⁺ is more stable than Sn⁴⁺.
- Between Pb²⁺ and Sn²⁺, Pb²⁺ is generally more stable than Sn²⁺ due to the larger atomic size and the effective nuclear charge.

5. **Establish the Stability Order**:
- Based on the analysis, the stability order from most stable to least stable is:
- Sn²⁺ > Pb²⁺ > Sn⁴⁺ > Pb⁴⁺

### Final Stability Order:
- Sn²⁺ > Pb²⁺ > Sn⁴⁺ > Pb⁴⁺