The oxidation number of `Fe` in `Fe_(0.94)O` is

Options

A. `200`
B. `200//94`
C. `94//200`
D. None

Explanation

To find the oxidation number of iron (Fe) in the compound \( \text{Fe}_{0.94}\text{O} \), we can follow these steps:

### Step 1: Assign oxidation numbers
- Let the oxidation number of iron (Fe) be \( x \).
- The oxidation number of oxygen (O) is known to be \(-2\).

### Step 2: Write the equation for the compound
- The compound is neutral, so the sum of the oxidation numbers must equal zero.
- The formula for the compound indicates that there are 0.94 moles of Fe and 1 mole of O.
- Therefore, the equation can be set up as follows:
\[ 0.94x + (-2) = 0 \]

### Step 3: Solve for \( x \)
- Rearranging the equation gives:
\[ 0.94x - 2 = 0 \]
- Adding 2 to both sides:
\[ 0.94x = 2 \]
- Now, divide both sides by 0.94:
\[ x = \frac{2}{0.94} \]

### Step 4: Simplify the fraction
- To simplify, you can multiply the numerator and denominator by 100:
\[ x = \frac{200}{94} \]

### Step 5: Final answer
- Therefore, the oxidation number of Fe in \( \text{Fe}_{0.94}\text{O} \) is:
\[ x = \frac{200}{94} \approx 2.13 \]

### Conclusion
- The oxidation number of Fe in \( \text{Fe}_{0.94}\text{O} \) is \( \frac{200}{94} \).