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 Fe in the compound \( \text{Fe}_{0.94}\text{O} \), we can follow these steps:

### Step 1: Understand the composition of the compound
The compound \( \text{Fe}_{0.94}\text{O} \) indicates that there are 0.94 moles of iron (Fe) for every mole of oxygen (O). In this case, we need to find the oxidation state of iron (Fe).

### Step 2: Assign oxidation states
We know that the oxidation state of oxygen (O) in most compounds is -2. Therefore, we can denote the oxidation state of iron (Fe) as \( x \).

### Step 3: Set up the equation
Since the compound is neutral, the sum of the oxidation states must equal zero. Thus, we can write the equation:
\[ 0.94x + (-2) = 0 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ 0.94x - 2 = 0 \]
\[ 0.94x = 2 \]
\[ x = \frac{2}{0.94} \]

### Step 5: Calculate \( x \)
Now, we can perform the division:
\[ x = \frac{2}{0.94} \approx 2.13 \]

### Step 6: Conclusion
The oxidation number of Fe in \( \text{Fe}_{0.94}\text{O} \) is approximately \( +2.13 \).

### Final Answer
The oxidation number of Fe in \( \text{Fe}_{0.94}\text{O} \) is approximately \( +2.12 \) (rounded to two decimal places).

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