In which of the following compounds, Mn has highest oxidation state?

Options

A. `K_(2)MnO_(4)`
B. `MnO_(2)`
C. `KMnO_(4)`
D. `Mn_(3)O_(4)`

Answer

To determine which compound has manganese (Mn) in the highest oxidation state, we will analyze the oxidation states of Mn in the given compounds step by step.

### Step 1: Analyze K2MnO4
1. Let the oxidation state of Mn be \( x \).
2. The oxidation state of potassium (K) is +1, and there are 2 potassium atoms, contributing \( +2 \).
3. The oxidation state of oxygen (O) is -2, and there are 4 oxygen atoms, contributing \( -8 \).
4. Set up the equation:
\[ 2(+1) + x + 4(-2) = 0 \]
\[ 2 + x - 8 = 0 \]
\[ x - 6 = 0 \implies x = +6 \]

### Step 2: Analyze KMnO4
1. Let the oxidation state of Mn be \( x \).
2. The oxidation state of potassium (K) is +1.
3. The oxidation state of oxygen (O) is -2, and there are 4 oxygen atoms, contributing \( -8 \).
4. Set up the equation:
\[ +1 + x + 4(-2) = 0 \]
\[ 1 + x - 8 = 0 \]
\[ x - 7 = 0 \implies x = +7 \]

### Step 3: Analyze MnO2
1. Let the oxidation state of Mn be \( x \).
2. The oxidation state of oxygen (O) is -2, and there are 2 oxygen atoms, contributing \( -4 \).
3. Set up the equation:
\[ x + 2(-2) = 0 \]
\[ x - 4 = 0 \implies x = +4 \]

### Step 4: Analyze Mn3O4
1. Let the oxidation state of Mn be \( x \).
2. The oxidation state of oxygen (O) is -2, and there are 4 oxygen atoms, contributing \( -8 \).
3. Set up the equation:
\[ 3x + 4(-2) = 0 \]
\[ 3x - 8 = 0 \]
\[ 3x = 8 \implies x = \frac{8}{3} \approx +2.67 \]

### Conclusion
Now we compare the oxidation states of Mn in all compounds:
- In K2MnO4, Mn is +6.
- In KMnO4, Mn is +7.
- In MnO2, Mn is +4.
- In Mn3O4, Mn is approximately +2.67.

The highest oxidation state of Mn is in KMnO4, which is +7.

### Final Answer
The compound in which Mn has the highest oxidation state is **KMnO4**.

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