Write down the hydrolysis of `XeF_(6)` in strongly alkaline medium.

Answer

To write down the hydrolysis of \( \text{XeF}_6 \) in a strongly alkaline medium, we can follow these steps:

### Step 1: Identify the reactants
We start with xenon hexafluoride, \( \text{XeF}_6 \), and hydroxide ions \( \text{OH}^- \) from the strong alkaline medium.

### Step 2: Write the balanced chemical equation
The hydrolysis of \( \text{XeF}_6 \) in a strongly alkaline medium is an autoredox process. The balanced equation for this reaction can be written as:

\[ 2 \text{XeF}_6 + 16 \text{OH}^- \rightarrow \text{XeO}_6^{4-} + \text{Xe} + 12 \text{F}^- + 8 \text{H}_2\text{O} \]

### Step 3: Analyze the oxidation states
In \( \text{XeF}_6 \), the oxidation state of xenon (Xe) is +6. During the reaction:
- In \( \text{XeO}_6^{4-} \), the oxidation state of xenon changes to +8.
- The elemental xenon (Xe) has an oxidation state of 0.

### Step 4: Identify the type of reaction
This reaction is a disproportionation reaction, where the same species (Xe in this case) is both oxidized and reduced.

### Final Answer
Thus, the hydrolysis of \( \text{XeF}_6 \) in a strongly alkaline medium can be summarized as follows:

\[ 2 \text{XeF}_6 + 16 \text{OH}^- \rightarrow \text{XeO}_6^{4-} + \text{Xe} + 12 \text{F}^- + 8 \text{H}_2\text{O} \]