The oxidation sates of metal in the compounds `Fe_(0.94)O` and `[Cr(P Ph_(3))_(3)(CO)_(3)]` respectively are

Options

A. `(200)/(94),0`
B. `0,(94)/(200)`
C. `2,1`
D. `1, (200)/(94)`

Explanation

To find the oxidation states of metals in the compounds \( Fe_{0.94}O \) and \( [Cr(PPh_{3})_{3}(CO)_{3}] \), we will follow these steps:

### Step 1: Determine the oxidation state of iron in \( Fe_{0.94}O \)

1. **Identify the oxidation state of oxygen**: In most compounds, oxygen has an oxidation state of -2.
2. **Let the oxidation state of iron (Fe) be \( x \)**. Since there are 0.94 moles of iron, the contribution of iron to the oxidation state will be \( 0.94x \).
3. **Set up the equation**: The sum of the oxidation states must equal zero because the compound is neutral. Thus, we write the equation:
\[ 0.94x + (-2) = 0 \]
4. **Solve for \( x \)**:
\[ 0.94x - 2 = 0 \implies 0.94x = 2 \implies x = \frac{2}{0.94} \approx 2.13 \]

### Step 2: Determine the oxidation state of chromium in \( [Cr(PPh_{3})_{3}(CO)_{3}] \)

1. **Identify the oxidation states of the ligands**:
- \( PPh_{3} \) (triphenylphosphine) is a neutral ligand, so its oxidation state is 0.
- \( CO \) (carbon monoxide) is also a neutral ligand, so its oxidation state is 0.
2. **Let the oxidation state of chromium (Cr) be \( y \)**. There is one chromium atom in the complex, so the equation becomes:
\[ y + 0 \cdot 3 + 0 \cdot 3 = 0 \]
3. **Solve for \( y \)**:
\[ y = 0 \]

### Final Answer
The oxidation states of the metals in the compounds \( Fe_{0.94}O \) and \( [Cr(PPh_{3})_{3}(CO)_{3}] \) are approximately \( 2.13 \) and \( 0 \), respectively.

### Summary
- Oxidation state of Fe in \( Fe_{0.94}O \): \( \frac{200}{94} \) or approximately \( 2.13 \)
- Oxidation state of Cr in \( [Cr(PPh_{3})_{3}(CO)_{3}] \): \( 0 \)