The oxidation state of xenon in perxenate ion is +n. Give the value of 'n'.

Answer

To find the oxidation state of xenon in the perxenate ion (XeO6^4-), we can follow these steps:

### Step 1: Identify the formula of the perxenate ion
The perxenate ion is represented as XeO6^4-.

### Step 2: Assign oxidation states
Let the oxidation state of xenon (Xe) be represented as \( x \). The oxidation state of oxygen (O) is typically -2.

### Step 3: Set up the equation
In the perxenate ion, there are 6 oxygen atoms, so the total contribution from oxygen is:
\[ 6 \times (-2) = -12 \]
Since the overall charge of the perxenate ion is -4, we can set up the equation:
\[ x + (-12) = -4 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ x - 12 = -4 \]
Adding 12 to both sides results in:
\[ x = -4 + 12 \]
\[ x = +8 \]

### Conclusion
Thus, the oxidation state of xenon in the perxenate ion is +8. Therefore, the value of \( n \) is 8.

### Final Answer
The value of \( n \) is 8.

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