`XeF_4` disproportionates in water giving reduced and oxidised products What is the ratio of oxidation states of Xe in the reduced and the oxidised products:

Explanation

To solve the problem, we need to analyze the disproportionation reaction of \( \text{XeF}_4 \) in water and determine the oxidation states of xenon in the reduced and oxidized products. Let's break it down step by step.

### Step 1: Write the Disproportionation Reaction
The reaction of \( \text{XeF}_4 \) with water can be written as:
\[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{XeO}_3 + \text{HF} \]
This indicates that \( \text{XeF}_4 \) is disproportionating into \( \text{XeO}_3 \) (oxidized product) and \( \text{Xe} \) (reduced product).

### Step 2: Determine the Oxidation State of Xenon in \( \text{XeF}_4 \)
Let the oxidation state of xenon in \( \text{XeF}_4 \) be \( x \). The oxidation state of fluorine is \(-1\). Since there are four fluorine atoms, we can set up the equation:
\[ x + 4(-1) = 0 \]
Solving for \( x \):
\[ x - 4 = 0 \implies x = +4 \]
Thus, the oxidation state of xenon in \( \text{XeF}_4 \) is \( +4 \).

### Step 3: Determine the Oxidation State of Xenon in \( \text{XeO}_3 \)
Now, we will find the oxidation state of xenon in \( \text{XeO}_3 \). Let the oxidation state of xenon in \( \text{XeO}_3 \) be \( y \). The oxidation state of oxygen is \(-2\). Since there are three oxygen atoms, we can set up the equation:
\[ y + 3(-2) = 0 \]
Solving for \( y \):
\[ y - 6 = 0 \implies y = +6 \]
Thus, the oxidation state of xenon in \( \text{XeO}_3 \) is \( +6 \).

### Step 4: Determine the Oxidation State of Xenon in the Elemental Form
In the elemental form of xenon (i.e., \( \text{Xe} \)), the oxidation state is \( 0 \).

### Step 5: Identify the Reduced and Oxidized Products
- The oxidized product is \( \text{XeO}_3 \) with an oxidation state of \( +6 \).
- The reduced product is elemental xenon \( \text{Xe} \) with an oxidation state of \( 0 \).

### Step 6: Calculate the Ratio of Oxidation States
Now we can find the ratio of the oxidation states of xenon in the reduced product to the oxidized product:
\[ \text{Ratio} = \text{Oxidation state in reduced product} : \text{Oxidation state in oxidized product} = 0 : 6 \]

### Final Answer
The ratio of oxidation states of xenon in the reduced and oxidized products is:
\[ \boxed{0 : 6} \]