The answer to each of the following is a non-negative integer
In the following reaction, find the difference in oxida- tion state of Xe in the underlined species (numerical value only).
`2[HXeO_4]^-+2OHto[XeO_6]^(4-)+Xe+O_2+2H_2O`

Answer

To find the difference in oxidation state of xenon (Xe) in the given reaction, we will follow these steps:

### Step 1: Identify the oxidation states of xenon in the reactants and products.
We need to determine the oxidation state of xenon in both the reactant \([HXeO_4]^-\) and the product \([XeO_6]^{4-}\).

### Step 2: Calculate the oxidation state of xenon in \([HXeO_4]^-\).
- The formula for \([HXeO_4]^-\) contains:
- 1 hydrogen (H) with an oxidation state of +1.
- 4 oxygen (O) atoms, each with an oxidation state of -2.

Using the formula:
\[ \text{Oxidation state of Xe} + \text{Oxidation state of H} + 4 \times \text{Oxidation state of O} = \text{Charge} \]
Substituting the values:
\[ \text{Oxidation state of Xe} + 1 + 4 \times (-2) = -1 \]
\[ \text{Oxidation state of Xe} + 1 - 8 = -1 \]
\[ \text{Oxidation state of Xe} - 7 = -1 \]
\[ \text{Oxidation state of Xe} = +6 \]

### Step 3: Calculate the oxidation state of xenon in \([XeO_6]^{4-}\).
- The formula for \([XeO_6]^{4-}\) contains:
- 6 oxygen (O) atoms, each with an oxidation state of -2.

Using the formula:
\[ \text{Oxidation state of Xe} + 6 \times \text{Oxidation state of O} = \text{Charge} \]
Substituting the values:
\[ \text{Oxidation state of Xe} + 6 \times (-2) = -4 \]
\[ \text{Oxidation state of Xe} - 12 = -4 \]
\[ \text{Oxidation state of Xe} = +8 \]

### Step 4: Find the difference in oxidation states.
Now we can find the difference in oxidation states of xenon between the two species:
\[ \text{Difference} = \text{Oxidation state in } [XeO_6]^{4-} - \text{Oxidation state in } [HXeO_4]^- \]
\[ \text{Difference} = 8 - 6 = 2 \]

### Final Answer:
The difference in oxidation state of Xe in the underlined species is **2**.

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