Calculate the number of unpaired electrons in the following gaseous ions: `Mn^(3+),Cr^(3+),V^(3+)` and `Ti^(3+)`. Which one of these is the most stable in aqueous solution?

Explanation

To determine the number of unpaired electrons in the ions \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{V}^{3+}, \text{Ti}^{3+} \), we first need to establish the electron configurations of the neutral atoms and then adjust for the charge of each ion.

### Step 1: Determine the electron configurations of the neutral atoms.

1. **Manganese (Mn)**: Atomic number = 25
- Electron configuration: \( [Ar] 3d^5 4s^2 \)

2. **Chromium (Cr)**: Atomic number = 24
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