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 will first write their electronic configurations and then identify the number of unpaired electrons in each case.

### Step 1: Write the electronic configuration of each ion.

1. **For \( \text{Mn}^{3+} \)**:
- The atomic number of Manganese (Mn) is 25. Its ground state electronic configuration is:
\[
\text{Mn}: [\text{Ar}] 3d^5 4s^2
...