In which of the following pairs of ions, the higher oxidation state in aqueous solution is more stable than the other?
I. `TI^(+), TI^(3+)` II. `Cu^(+),Cu^(2+)`
III. `Cr^(2+),Cr^(3+)` IV. `V^(2+),V^4+`

Options

A. I, II
B. II, III
C. II, IV
D. II, III, IV

Answer

To determine which pairs of ions have a higher oxidation state that is more stable in aqueous solution, we will analyze each pair one by one.

### Step 1: Analyze Thallium Ions (TI^(+), TI^(3+))
- **Thallium (Tl)** is a p-block element in group 13.
- It can exist in two oxidation states: +1 and +3.
- The +1 oxidation state (Tl^(+)) is more stable than the +3 oxidation state (Tl^(3+)) due to the **inert pair effect**. As you move down the group, the stability of the +1 state increases.
- **Conclusion**: Tl^(3+) is less stable than Tl^(+). Therefore, this pair does not satisfy the condition.

### Step 2: Analyze Copper Ions (Cu^(+), Cu^(2+))
- **Copper (Cu)** can exist in +1 and +2 oxidation states.
- The +2 oxidation state (Cu^(2+)) is more stable than the +1 oxidation state (Cu^(+)). This is because the hydration energy of Cu^(2+) is higher due to its greater charge, leading to increased stability.
- **Conclusion**: Cu^(2+) is more stable than Cu^(+). Therefore, this pair satisfies the condition.

### Step 3: Analyze Chromium Ions (Cr^(2+), Cr^(3+))
- **Chromium (Cr)** can exist in +2 and +3 oxidation states.
- The +3 oxidation state (Cr^(3+)) is more stable than the +2 oxidation state (Cr^(2+)). The electronic configuration of Cr^(3+) is 3d^3, which is half-filled and therefore more stable compared to the 3d^4 configuration of Cr^(2+).
- **Conclusion**: Cr^(3+) is more stable than Cr^(2+). Therefore, this pair satisfies the condition.

### Step 4: Analyze Vanadium Ions (V^(2+), V^(4+))
- **Vanadium (V)** can exist in +2 and +4 oxidation states.
- The +4 oxidation state (V^(4+)) is more stable than the +2 oxidation state (V^(2+)). The electronic configuration of V^(4+) is 3d^1, which is closer to the noble gas configuration, enhancing its stability.
- **Conclusion**: V^(4+) is more stable than V^(2+). Therefore, this pair satisfies the condition.

### Final Conclusion:
The pairs where the higher oxidation state is more stable than the lower oxidation state are:
- II. Cu^(+), Cu^(2+)
- III. Cr^(2+), Cr^(3+)
- IV. V^(2+), V^(4+)

Thus, the correct answer is **II, III, and IV**.