Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with non metals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different.
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. a. `+2`
B. b. `0`
C. c. `+4`
D. d. `+6`

Answer

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

### Step-by-Step Solution:

1. **Determine the Percentage Composition:**
- The compound contains 53.3% xenon (Xe) and 46.7% fluorine (F) (since 100% - 53.3% = 46.7%).

2. **Find the Molar Mass of Each Element:**
- The atomic weight of xenon (Xe) is given as 133 g/mol.
- The atomic weight of fluorine (F) is 19 g/mol.

3. **Calculate the Number of Moles of Each Element:**
- For xenon:
\[ \text{Moles of Xe} = \frac{\text{mass percentage of Xe}}{\text{molar mass of Xe}} = \frac{53.3}{133} \approx 0.40075 \text{ moles} \]
- For fluorine:
\[ \text{Moles of F} = \frac{\text{mass percentage of F}}{\text{molar mass of F}} = \frac{46.7}{19} \approx 2.457 \text{ moles} \]

4. **Determine the Ratio of Moles:**
- To find the simplest whole number ratio, divide both mole values by the smaller number of moles (0.40075):
- For xenon:
\[ \frac{0.40075}{0.40075} = 1 \]
- For fluorine:
\[ \frac{2.457}{0.40075} \approx 6 \]
- Therefore, the ratio of xenon to fluorine in the compound is 1:6.

5. **Write the Empirical Formula:**
- The empirical formula of the compound is \( \text{XeF}_6 \).

6. **Assign Oxidation States:**
- Let the oxidation number of xenon be \( x \).
- The oxidation number of fluorine is -1 (since fluorine is more electronegative).
- The sum of the oxidation numbers in a neutral compound is zero:
\[ x + 6(-1) = 0 \]
- Simplifying this gives:
\[ x - 6 = 0 \implies x = +6 \]

7. **Conclusion:**
- The oxidation number of xenon (Xe) in this compound is +6.