`K_(2)MnO_(4)+H^(+) to KMnO_(4)+MnO_(2)darr`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To solve the given reaction \( K_2MnO_4 + H^+ \rightarrow KMnO_4 + MnO_2 \), we will follow these steps:

### Step 1: Identify the oxidation states of each element in the reactants and products.

1. **In \( K_2MnO_4 \)**:
- Potassium (K) has an oxidation state of +1.
- Let the oxidation state of manganese (Mn) be \( x \).
- Oxygen (O) has an oxidation state of -2.
- The equation for the compound is:
\[ 2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6 \]
- Therefore, Mn in \( K_2MnO_4 \) is +6.

2. **In \( KMnO_4 \)**:
- Potassium (K) is +1.
- Let the oxidation state of manganese (Mn) be \( y \).
- The equation is:
\[ +1 + y + 4(-2) = 0 \implies 1 + y - 8 = 0 \implies y = +7 \]
- Therefore, Mn in \( KMnO_4 \) is +7.

3. **In \( MnO_2 \)**:
- Let the oxidation state of manganese (Mn) be \( z \).
- The equation is:
\[ z + 2(-2) = 0 \implies z - 4 = 0 \implies z = +4 \]
- Therefore, Mn in \( MnO_2 \) is +4.

### Step 2: Determine the changes in oxidation states.

- In the reaction:
- Mn changes from +6 in \( K_2MnO_4 \) to +7 in \( KMnO_4 \) (oxidation).
- Mn changes from +6 in \( K_2MnO_4 \) to +4 in \( MnO_2 \) (reduction).

### Step 3: Classify the reaction.

- Since one species (Mn) is being oxidized (from +6 to +7) and reduced (from +6 to +4) simultaneously, this type of reaction is known as a **disproportionation reaction**.

### Conclusion

The correct classification of the reaction \( K_2MnO_4 + H^+ \rightarrow KMnO_4 + MnO_2 \) is that it is a **disproportionation reaction**.

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