Which of the following is correct order of stability:

Options

A. `TI^(3+) gt BI^(3+)`
B. `PbO_(2) gt PbO`
C. `BiI_(5) lt BiF_(5)`
D. `Sn^(2+) = Ge^(2+)`

Explanation

To determine the correct order of stability among the given species, we will analyze each pair based on their oxidation states and the inert pair effect.

### Step-by-Step Solution:

1. **Analyzing Thallium (Tl) and Bismuth (Bi)**:
- Thallium (Tl) is in the boron group (Group 13) and can exist in +1 and +3 oxidation states. However, due to the inert pair effect, Tl tends to favor the +1 state, making Tl^3+ less stable.
- Bismuth (Bi), in the nitrogen group (Group 15), primarily exists in the +3 oxidation state. The +3 oxidation state of Bi is more stable than Tl^3+.
- **Conclusion**: Bi^3+ is more stable than Tl^3+.

2. **Analyzing Lead (Pb) in PbO2 and PbO**:
- In PbO2, lead is in the +4 oxidation state, while in PbO, it is in the +2 oxidation state.
- The inert pair effect suggests that the +2 oxidation state is more stable than the +4 state for lead.
- **Conclusion**: PbO is more stable than PbO2.

3. **Analyzing Bismuth (Bi) and Bismuth Pentafluoride (BiF5)**:
- Bismuth can exist in the +3 oxidation state, but it can also form BiF5, where it is in the +5 oxidation state.
- The strong oxidizing nature of fluorine allows BiF5 to be stable despite Bi typically favoring lower oxidation states.
- **Conclusion**: BiF5 is stable.

4. **Analyzing Tin (Sn) and Germanium (Ge)**:
- Tin (Sn) and Germanium (Ge) are in the same group (Group 14). Generally, as we move down a group, the +2 oxidation state becomes more stable due to the inert pair effect.
- Therefore, Ge^2+ should be less stable than Sn^2+.
- **Conclusion**: Sn^2+ is more stable than Ge^2+.

### Final Order of Stability:
Based on the analysis, the correct order of stability is:
1. Bi^3+ > Tl^3+
2. PbO > PbO2
3. BiF5 is stable.
4. Sn^2+ > Ge^2+