`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 determine the type of reaction for the given equation \( K_2MnO_4 + H^+ \rightarrow KMnO_4 + MnO_2 \), we will analyze the oxidation states of manganese in the reactants and products.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( K_2MnO_4 \) and \( H^+ \)
- Products: \( KMnO_4 \) and \( MnO_2 \)

2. **Assign Oxidation States**:
- In \( K_2MnO_4 \):
- Potassium (K) = +1 (2 K atoms contribute +2)
- Oxygen (O) = -2 (4 O atoms contribute -8)
- Let the oxidation state of manganese (Mn) be \( x \).
- The equation becomes: \( 2(+1) + x + 4(-2) = 0 \)
- Simplifying: \( 2 + x - 8 = 0 \) → \( x = +6 \)
- In \( KMnO_4 \):
- Potassium (K) = +1
- Oxygen (O) = -2 (4 O atoms contribute -8)
- Let the oxidation state of manganese (Mn) be \( y \).
- The equation becomes: \( 1 + y + 4(-2) = 0 \)
- Simplifying: \( 1 + y - 8 = 0 \) → \( y = +7 \)
- In \( MnO_2 \):
- Oxygen (O) = -2 (2 O atoms contribute -4)
- Let the oxidation state of manganese (Mn) be \( z \).
- The equation becomes: \( z + 2(-2) = 0 \)
- Simplifying: \( z - 4 = 0 \) → \( z = +4 \)

3. **Determine Changes in Oxidation States**:
- In \( K_2MnO_4 \), Mn changes from +6 to +7 in \( KMnO_4 \) (oxidation).
- In \( K_2MnO_4 \), Mn changes from +6 to +4 in \( MnO_2 \) (reduction).

4. **Identify the Type of Reaction**:
- Since the same element (Mn) is both oxidized (to +7) and reduced (to +4) in the reaction, this is a **disproportionation reaction**.

### Conclusion:
The reaction \( K_2MnO_4 + H^+ \rightarrow KMnO_4 + MnO_2 \) is a **disproportionation reaction**.