`H_(2)O_(2)+H_(2)O_(2)rarr2H_(2)O+O_(2)` is an example of dispropotionation because -

Options

A. Oxidation number of oxygen only decreases
B. Oxidation number of oxygen only increases
C. Oxidation number of oxygen decreases as well as increases
D. Oxidation number of oxygen neither decreases nor increases

Answer

To determine why the reaction \( H_2O_2 + H_2O_2 \rightarrow 2H_2O + O_2 \) is an example of disproportionation, we need to analyze the oxidation states of the elements involved, particularly oxygen.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( H_2O_2 \) (hydrogen peroxide)
- Products: \( H_2O \) (water) and \( O_2 \) (oxygen gas)

2. **Assign Oxidation States**:
- In \( H_2O_2 \), the oxidation state of oxygen is \(-1\).
- In \( H_2O \), the oxidation state of oxygen is \(-2\).
- In \( O_2 \), the oxidation state of oxygen is \(0\).

3. **Determine Changes in Oxidation States**:
- For the oxygen in \( H_2O_2 \) to \( H_2O \):
- Change from \(-1\) to \(-2\) (decrease).
- For the oxygen in \( H_2O_2 \) to \( O_2 \):
- Change from \(-1\) to \(0\) (increase).

4. **Conclusion on Disproportionation**:
- In this reaction, the same element (oxygen) undergoes both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) simultaneously.
- Therefore, this reaction is classified as a disproportionation reaction.

5. **Final Answer**:
- The correct reason for this reaction being an example of disproportionation is that the oxidation number of oxygen decreases as well as increases.