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 Fe₀.₉₄O, we can follow these steps:

### Step 1: Identify the known oxidation states
In the compound Fe₀.₉₄O, we know that the oxidation state of oxygen (O) is -2.

### Step 2: Set up the equation for the oxidation state
Let the oxidation state of iron (Fe) be represented as \( x \). Since there are 0.94 moles of Fe in the compound, we can express the total oxidation contribution from iron as \( 0.94x \).

### Step 3: Write the charge balance equation
Since the compound is neutral, the sum of the oxidation states must equal zero. Therefore, we can set up the equation:

\[ 0.94x + (-2) = 0 \]

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

\[ 0.94x - 2 = 0 \]
\[ 0.94x = 2 \]
\[ x = \frac{2}{0.94} \]
Calculating this gives:

\[ x \approx 2.13 \]

### Step 5: Conclusion
The oxidation number of iron (Fe) in Fe₀.₉₄O is approximately +2.13.

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