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 atom in the compound \( Ba_2XeO_6 \), we can follow these steps:

### Step 1: Identify the oxidation state of barium
Barium (Ba) is an alkaline earth metal located in Group II of the periodic table. The oxidation state of barium in compounds is typically +2. Since there are two barium atoms in the compound, the total contribution from barium is:
\[ 2 \times (+2) = +4 \]

### Step 2: Determine the total charge contributed by oxygen
Oxygen (O) generally has an oxidation state of -2. In the compound \( Ba_2XeO_6 \), there are six oxygen atoms. Therefore, the total contribution from oxygen is:
\[ 6 \times (-2) = -12 \]

### Step 3: Set up the equation for the overall charge
The overall charge of the compound must equal zero since it is neutral. We can express this as:
\[ \text{Total charge} = \text{Charge from Ba} + \text{Charge from Xe} + \text{Charge from O} \]
Substituting the known values:
\[ 0 = (+4) + X + (-12) \]
Where \( X \) is the oxidation state of xenon.

### Step 4: Solve for the oxidation state of xenon
Rearranging the equation gives:
\[ X - 8 = 0 \]
Thus:
\[ X = +8 \]

### Conclusion
The oxidation number of the xenon atom in \( Ba_2XeO_6 \) is +8.

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