Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^4` configuration?

Explanation

To understand why \( \text{Cr}^{2+} \) is a reducing agent and \( \text{Mn}^{3+} \) is an oxidizing agent despite both having a \( d^4 \) configuration, we can break down the explanation into clear steps:

### Step 1: Determine the electronic configurations

1. **Chromium (Cr)**:
- Atomic number: 24
- Electronic configuration: \( [\text{Ar}] 3d^5 4s^1 \)
- For \( \text{Cr}^{2+} \), it loses two electrons (one from \( 4s \) and one from \( 3d \)):
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