`XeF_(4)+H_(2)O to Xe+XeO_(3)+HF+O_(2)`

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 chemical equation \( \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{HF} + \text{O}_2 \), we need to analyze the reaction and identify the type of reaction it represents.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( \text{XeF}_4 \) (Xenon tetrafluoride) and \( \text{H}_2\text{O} \) (water).
- Products: \( \text{Xe} \) (Xenon), \( \text{XeO}_3 \) (Xenon trioxide), \( \text{HF} \) (Hydrofluoric acid), and \( \text{O}_2 \) (Oxygen).

2. **Determine the Oxidation States**:
- In \( \text{XeF}_4 \), the oxidation state of Xenon (Xe) is +4.
- In \( \text{XeO}_3 \), the oxidation state of Xenon is +6.
- In elemental Xenon (\( \text{Xe} \)), the oxidation state is 0.
- In \( \text{HF} \), the oxidation state of Fluorine (F) is -1.

3. **Analyze the Changes in Oxidation States**:
- Xenon goes from +4 in \( \text{XeF}_4 \) to 0 in \( \text{Xe} \) (reduction).
- Xenon also goes from +4 in \( \text{XeF}_4 \) to +6 in \( \text{XeO}_3 \) (oxidation).
- This indicates that one species of Xenon is being both oxidized and reduced in the same reaction.

4. **Classify the Reaction**:
- Since one element (Xenon) is undergoing both oxidation (to +6) and reduction (to 0), this reaction is classified as a **disproportionation reaction**. In a disproportionation reaction, a single reactant is transformed into two different products with different oxidation states.

5. **Conclusion**:
- The reaction \( \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{HF} + \text{O}_2 \) is a **disproportionation reaction**.

### Final Answer:
The reaction is a **disproportionation reaction**.