The non-stoichiometric compound `Fe_0.94O` is formed when some `Fe^(+2)` ions are replaced by `Fe^(+3)` ions. What is the percentage of `Fe^(+3)` ions in this ionic lattice ?

Options

A. `18%`
B. `25%`
C. `13%`
D. `6%`

Answer

To find the percentage of \( \text{Fe}^{3+} \) ions in the non-stoichiometric compound \( \text{Fe}_{0.94}\text{O} \), we can follow these steps:

### Step 1: Understand the composition of the compound
We have the compound \( \text{Fe}_{0.94}\text{O} \), which indicates that there are 0.94 moles of iron (Fe) for every mole of oxygen (O).

### Step 2: Calculate the total charge contributed by oxygen
Oxygen typically has a charge of \( -2 \). Therefore, for 100 moles of oxygen:
\[ \text{Charge from O} = 100 \times (-2) = -200 \]

### Step 3: Determine the total positive charge needed for neutrality
Since the compound is neutral, the total positive charge from iron must equal the total negative charge from oxygen:
\[ \text{Total positive charge} = +200 \]

### Step 4: Set up the equation for iron ions
Let \( x \) be the number of \( \text{Fe}^{3+} \) ions. Then, the number of \( \text{Fe}^{2+} \) ions will be:
\[ \text{Number of } \text{Fe}^{2+} \text{ ions} = 94 - x \]

### Step 5: Write the equation for total positive charge
The total positive charge can be expressed as:
\[ \text{Charge from } \text{Fe}^{3+} + \text{Charge from } \text{Fe}^{2+} = 200 \]
This can be formulated as:
\[ 3x + 2(94 - x) = 200 \]

### Step 6: Simplify the equation
Expanding the equation gives:
\[ 3x + 188 - 2x = 200 \]
Combining like terms results in:
\[ x + 188 = 200 \]

### Step 7: Solve for \( x \)
Subtracting 188 from both sides gives:
\[ x = 200 - 188 = 12 \]
So, there are 12 \( \text{Fe}^{3+} \) ions.

### Step 8: Calculate the percentage of \( \text{Fe}^{3+} \) ions
To find the percentage of \( \text{Fe}^{3+} \) ions in the lattice:
\[ \text{Percentage of } \text{Fe}^{3+} = \left( \frac{x}{94} \right) \times 100 = \left( \frac{12}{94} \right) \times 100 \approx 12.76\% \]

### Conclusion
The percentage of \( \text{Fe}^{3+} \) ions in the ionic lattice is approximately **12.76%**.