What is the oxidation number of `Mn` is `KMnO_(4)" "K_(2)MNO_(4)" "MnSO_(4)" "MnO_(2)`, and `Mn_(3)O_(4)`?

Explanation

To find the oxidation number of manganese (Mn) in the compounds \( KMnO_4 \), \( K_2MnO_4 \), \( MnSO_4 \), \( MnO_2 \), and \( Mn_3O_4 \), we will follow a systematic approach.

### Step-by-step Solution:

1. **Identify the oxidation states of other elements:**
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Sulfur (S) in sulfate (SO₄²⁻) has an oxidation state of +6.

2. **Set up the equation for each compound:**
- Let the oxidation state of Mn be represented as \( x \).

3. **Calculate the oxidation state for each compound:**

**a. For \( KMnO_4 \):**
\[ +1 + x + 4(-2) = 0 \]
\[ +1 + x - 8 = 0 \]
\[ x - 7 = 0 \implies x = +7 \]
- **Oxidation state of Mn in \( KMnO_4 \) is +7.**

**b. For \( K_2MnO_4 \):**
\[ 2(+1) + x + 4(-2) = 0 \]
\[ +2 + x - 8 = 0 \]
\[ x - 6 = 0 \implies x = +6 \]
- **Oxidation state of Mn in \( K_2MnO_4 \) is +6.**

**c. For \( MnSO_4 \):**
\[ x + (+6) + 4(-2) = 0 \]
\[ x + 6 - 8 = 0 \]
\[ x - 2 = 0 \implies x = +2 \]
- **Oxidation state of Mn in \( MnSO_4 \) is +2.**

**d. For \( MnO_2 \):**
\[ x + 2(-2) = 0 \]
\[ x - 4 = 0 \implies x = +4 \]
- **Oxidation state of Mn in \( MnO_2 \) is +4.**

**e. For \( Mn_3O_4 \):**
\[ 3x + 4(-2) = 0 \]
\[ 3x - 8 = 0 \]
\[ 3x = 8 \implies x = \frac{8}{3} \approx +2.67 \]
- **Average oxidation state of Mn in \( Mn_3O_4 \) is \( +\frac{8}{3} \) or approximately +2.67.**

### Summary of Oxidation States:
- **\( KMnO_4 \): +7**
- **\( K_2MnO_4 \): +6**
- **\( MnSO_4 \): +2**
- **\( MnO_2 \): +4**
- **\( Mn_3O_4 \): \( +\frac{8}{3} \) (approximately +2.67)**