The stable oxidation states of Mn are

Options

A. `+2, +3`
B. `+2, +7`
C. `+3, +7`
D. `+3, +5`

Answer

To determine the stable oxidation states of manganese (Mn), we can follow these steps:

### Step 1: Determine the Electronic Configuration of Manganese
Manganese has an atomic number of 25. Its electronic configuration is:
\[ \text{Mn: } [\text{Ar}] 4s^2 3d^5 \]

### Step 2: Identify the Valence Electrons
Manganese has a total of 7 valence electrons (2 from the 4s subshell and 5 from the 3d subshell).

### Step 3: Understand Stability in Terms of Electron Configuration
Stable oxidation states often correspond to configurations that are either half-filled or fully filled. For d-block elements, configurations that result in half-filled (d^5) or fully filled (d^10) d orbitals are particularly stable.

### Step 4: Determine the Oxidation States
1. **Manganese in the +2 Oxidation State:**
- When manganese loses 2 electrons, it typically loses the 4s electrons first:
\[ \text{Mn}^{2+}: 3d^5 \]
- This configuration is stable because it is half-filled.

2. **Manganese in the +7 Oxidation State:**
- When manganese loses all 2 electrons from the 4s subshell and all 5 from the 3d subshell:
\[ \text{Mn}^{7+}: 3d^0 \]
- This configuration is also stable as it results in an empty d subshell.

### Conclusion
The stable oxidation states of manganese are +2 and +7.

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