Xenon trioxide `(XeO_(3))` forms xenate ion in alkaline medium.
`XeO_(3)+NaOH rarr Na[HXeO_(4)]`
But the xenate ions slowly disproportionate in alkaline solution as `Na[HXeO_(4)]+NaOH rarr Z+Xe+O_(2)+H_(2)O`
The compound Z is expexted to be

Options

A. `Na_(2)XeO_(3)`
B. `Na_(2)XeO_(4)`
C. `Na_(4)XeO_(6)`
D. `Na_(4)XeO_(4)`

Answer

To solve the problem, we need to analyze the reactions involving xenon trioxide (XeO₃) and its interaction with sodium hydroxide (NaOH) in an alkaline medium.

### Step-by-Step Solution:

1. **Identify the Initial Reaction**:
- Xenon trioxide reacts with sodium hydroxide to form the xenate ion.
- The reaction can be written as:
\[ \text{XeO}_3 + \text{NaOH} \rightarrow \text{Na[HXeO}_4] \]

2. **Disproportionation Reaction**:
- The xenate ion (Na[HXeO₄]) undergoes disproportionation in the alkaline solution.
- The reaction can be represented as:
\[ \text{Na[HXeO}_4] + \text{NaOH} \rightarrow Z + \text{Xe} + \text{O}_2 + \text{H}_2\text{O} \]

3. **Determine the Product Z**:
- We need to identify the compound Z formed in the reaction.
- The xenate ion (Na[HXeO₄]) has a xenon oxidation state of +6. Upon disproportionation, xenon is released as elemental Xe (oxidation state 0), and oxygen is released as O₂.
- The remaining product Z must have xenon in a lower oxidation state.

4. **Balancing the Reaction**:
- The balanced equation for the disproportionation reaction is:
\[ 2 \text{Na[HXeO}_4] + 2 \text{NaOH} \rightarrow \text{Na}_4\text{XeO}_6 + \text{Xe} + \text{O}_2 + 2 \text{H}_2\text{O} \]
- Here, we have 4 sodium (Na) on the product side, indicating that Z must be Na₄XeO₆.

5. **Conclusion**:
- The compound Z is expected to be Na₄XeO₆.

### Final Answer:
The compound Z is **Na₄XeO₆**.

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