The formula of an oxide of iron is `Fe_(0.93)O_(1.00)`.  If the compound has hundred `O^(-2)` ions, then it contains 

Options

A. `93Fe^(+2)" ions"`
B. `93Fe^(+3)" ions"`
C. `79Fe^(+2), 14Fe^(+3)`
D. `93Fe^(+2), 14Fe^(+3)`

Answer

To solve the problem, we need to determine how many iron (Fe) ions are present in the compound \( Fe_{0.93}O_{1.00} \) when there are 100 \( O^{2-} \) ions.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The given formula is \( Fe_{0.93}O_{1.00} \). This indicates that for every 1 mole of the compound, there are 0.93 moles of iron and 1 mole of oxygen.

2. **Scaling the Formula**:
Since the problem states there are 100 \( O^{2-} \) ions, we can scale the formula. We multiply both the iron and oxygen by 100.
\[ Fe_{0.93}O_{1.00} \rightarrow Fe_{93}O_{100} \]

3. **Charge Neutralization**:
Each \( O^{2-} \) ion contributes a charge of -2. Therefore, with 100 \( O^{2-} \) ions, the total negative charge is:
\[ \text{Total negative charge} = 100 \times (-2) = -200 \]

4. **Positive Charge from Iron**:
To neutralize the -200 charge from oxygen, the total positive charge from iron must also be +200. We denote the number of \( Fe^{2+} \) ions as \( x \) and the number of \( Fe^{3+} \) ions as \( 93 - x \) (since there are 93 iron atoms in total).

5. **Setting Up the Charge Equation**:
The positive charge contributed by the iron ions can be expressed as:
\[ 2x + 3(93 - x) = 200 \]

6. **Solving the Equation**:
Expanding the equation gives:
\[ 2x + 279 - 3x = 200 \]
Rearranging terms:
\[ -x + 279 = 200 \]
\[ -x = 200 - 279 \]
\[ -x = -79 \quad \Rightarrow \quad x = 79 \]

7. **Finding the Number of Iron Ions**:
Since \( x = 79 \), this means there are 79 \( Fe^{2+} \) ions. The remaining iron ions are \( 93 - 79 = 14 \), which are \( Fe^{3+} \) ions.

### Final Answer:
The compound contains **79 \( Fe^{2+} \) ions** and **14 \( Fe^{3+} \) ions**.