Determine the oxidation number of Xe atom in `Ba_2XeO_6`

Options

A. `+8`
B. `-8`
C. `+5`
D. `-7`

Answer

To determine the oxidation number of the xenon (Xe) atom in the compound \( \text{Ba}_2\text{XeO}_6 \), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- Barium (Ba) typically has an oxidation state of +2.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write the formula and set up the equation
The formula for the compound is \( \text{Ba}_2\text{XeO}_6 \).
- There are 2 barium atoms, contributing a total of \( 2 \times (+2) = +4 \).
- There are 6 oxygen atoms, contributing a total of \( 6 \times (-2) = -12 \).

### Step 3: Set up the equation for the oxidation number of xenon (Xe)
Let the oxidation number of xenon be \( x \). The overall charge of the compound is neutral (0). Therefore, we can set up the following equation:

\[ \text{Total charge} = \text{Charge from Ba} + \text{Charge from Xe} + \text{Charge from O} \]
\[ 0 = (+4) + (x) + (-12) \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:

\[ x - 12 + 4 = 0 \]
\[ x - 8 = 0 \]
\[ x = +8 \]

### Conclusion
The oxidation number of xenon (Xe) in \( \text{Ba}_2\text{XeO}_6 \) is +8.