A compound of Xe and F is found to have 53.3% Xe (atomic weight = 133) Oxidation number of Xe in this compound is

Options

A. `+2`
B. 0
C. `+4`
D. `+6`

Answer

To determine the oxidation number of xenon (Xe) in the compound with fluorine (F), we can follow these steps:

### Step 1: Determine the percentage of fluorine in the compound.
Given that the compound contains 53.3% xenon, we can find the percentage of fluorine by subtracting the percentage of xenon from 100%.

\[ \text{Percentage of F} = 100\% - 53.3\% = 46.7\% \]

### Step 2: Calculate the moles of xenon and fluorine.
Assuming the total mass of the compound is 100 grams, we can calculate the moles of xenon and fluorine using their respective atomic weights.

- Moles of xenon (Xe):
\[ \text{Moles of Xe} = \frac{\text{mass of Xe}}{\text{atomic weight of Xe}} = \frac{53.3 \text{ g}}{133 \text{ g/mol}} \approx 0.4 \text{ moles} \]

- Moles of fluorine (F):
\[ \text{Moles of F} = \frac{\text{mass of F}}{\text{atomic weight of F}} = \frac{46.7 \text{ g}}{19 \text{ g/mol}} \approx 2.45 \text{ moles} \]

### Step 3: Find the simplest whole number ratio of moles.
To find the simplest ratio, divide both the moles of xenon and fluorine by the smallest number of moles calculated.

- Moles of xenon ratio:
\[ \frac{0.4}{0.4} = 1 \]

- Moles of fluorine ratio:
\[ \frac{2.45}{0.4} \approx 6.125 \approx 6 \]

Thus, the mole ratio of xenon to fluorine is approximately 1:6.

### Step 4: Write the empirical formula of the compound.
Based on the mole ratio, the empirical formula of the compound is:
\[ \text{XeF}_6 \]

### Step 5: Determine the oxidation state of xenon in XeF6.
Let the oxidation state of xenon be \( X \). Since fluorine has an oxidation state of -1 and there are 6 fluorine atoms, we can set up the equation:

\[ X + 6(-1) = 0 \]

This simplifies to:
\[ X - 6 = 0 \]

Thus, solving for \( X \):
\[ X = +6 \]

### Conclusion
The oxidation number of xenon in the compound XeF6 is +6.

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