To determine which of the given ions is the most stable in aqueous solution, we need to analyze the electronic configurations of the ions formed from titanium (Ti), vanadium (V), chromium (Cr), and manganese (Mn).
### Step-by-Step Solution:
1. **Identify the Electronic Configurations:**
- Titanium (Ti, Atomic No. 22):
- Electronic configuration: \( \text{[Ar]} \, 4s^2 \, 3d^2 \)
- Vanadium (V, Atomic No. 23):
- Electronic configuration: \( \text{[Ar]} \, 4s^2 \, 3d^3 \)
- Chromium (Cr, Atomic No. 24):
- Electronic configuration: \( \text{[Ar]} \, 4s^2 \, 3d^5 \) (note that Cr has a unique configuration of \( 4s^1 \, 3d^5 \) for stability)
- Manganese (Mn, Atomic No. 25):
- Electronic configuration: \( \text{[Ar]} \, 4s^2 \, 3d^5 \)
2. **Determine the Ions:**
- \( \text{Ti}^{3+} \): Remove 3 electrons (2 from \( 4s \) and 1 from \( 3d \)):
- Configuration: \( \text{[Ar]} \, 3d^1 \)
- \( \text{V}^{3+} \): Remove 3 electrons (2 from \( 4s \) and 1 from \( 3d \)):
- Configuration: \( \text{[Ar]} \, 3d^2 \)
- \( \text{Cr}^{3+} \): Remove 3 electrons (2 from \( 4s \) and 1 from \( 3d \)):
- Configuration: \( \text{[Ar]} \, 3d^3 \)
- \( \text{Mn}^{3+} \): Remove 3 electrons (2 from \( 4s \) and 1 from \( 3d \)):
- Configuration: \( \text{[Ar]} \, 3d^4 \)
3. **Analyze Stability Based on Electron Configuration:**
- Stability in transition metal ions is often associated with the half-filled and fully filled d-orbitals due to exchange energy and symmetry.
- \( \text{Cr}^{3+} \) has \( 3d^3 \) configuration, which is half-filled in the \( t_{2g} \) level.
- \( \text{Mn}^{3+} \) has \( 3d^4 \) configuration, which has one electron in the \( e_g \) level, reducing its stability compared to \( \text{Cr}^{3+} \).
- \( \text{V}^{3+} \) and \( \text{Ti}^{3+} \) have \( 3d^2 \) and \( 3d^1 \) configurations, respectively, which are less stable than \( \text{Cr}^{3+} \).
4. **Conclusion:**
- Among the ions \( \text{Ti}^{3+} \), \( \text{V}^{3+} \), \( \text{Cr}^{3+} \), and \( \text{Mn}^{3+} \), the most stable ion in aqueous solution is \( \text{Cr}^{3+} \) due to its half-filled \( t_{2g} \) orbitals.
### Final Answer:
The most stable ion in aqueous solution is \( \text{Cr}^{3+} \).
