Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y' . Compounds 'X' and 'Y' and the oxidation state of Xe are respectively :

Options

A. `XeOF_4 (+6) and XeO_6(+6)`
B. `XeO_2 (+4) and XeO_3 (+6)`
C. `XeOF_4(+6) and XeO_2F_2 (+6)`
D. `XeO_2F_2(+6) and XeO_2 (+4)`

Answer

To solve the question regarding the partial hydrolysis of xenon hexafluoride (XeF6) and the resulting compounds 'X' and 'Y', we can follow these steps:

### Step 1: Understand the Reaction
Xenon hexafluoride (XeF6) undergoes partial hydrolysis when it reacts with water. The products of this reaction are typically xenon oxides and fluorides.

### Step 2: Identify the Products
From the hydrolysis of XeF6, we can identify the two main products:
- Compound 'X': Xenon oxyfluoride (XeOF4)
- Compound 'Y': Xenon dioxide difluoride (XeO2F2)

### Step 3: Write the Reaction
The reaction can be represented as follows:
\[ \text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{XeOF}_4 + \text{HF} \]
\[ \text{XeOF}_4 + \text{H}_2\text{O} \rightarrow \text{XeO}_2\text{F}_2 + \text{HF} \]

### Step 4: Determine the Oxidation State of Xenon in Each Compound
1. **For Compound 'X' (XeOF4)**:
- Let the oxidation state of xenon be \( x \).
- The oxidation states of oxygen and fluorine are -2 and -1, respectively.
- The equation becomes:
\[ x + (-2) + 4(-1) = 0 \\ x - 2 - 4 = 0 \\ x - 6 = 0 \\ x = +6 \]
Thus, the oxidation state of xenon in XeOF4 is +6.

2. **For Compound 'Y' (XeO2F2)**:
- Let the oxidation state of xenon be \( x \).
- The equation becomes:
\[ x + 2(-2) + 2(-1) = 0 \\ x - 4 - 2 = 0 \\ x - 6 = 0 \\ x = +6 \]
Thus, the oxidation state of xenon in XeO2F2 is also +6.

### Step 5: Conclusion
The compounds 'X' and 'Y' are:
- Compound 'X': XeOF4
- Compound 'Y': XeO2F2
- The oxidation states of xenon in both compounds are +6.

### Final Answer
Compounds 'X' and 'Y' are XeOF4 and XeO2F2, respectively, and the oxidation state of xenon in both compounds is +6.

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