The spinel structure consists of an array of `O^(2-)` ions in fcc arrangement. Gereral formula of spinel is `AB_(2)O_(4)`. Cations of A occupy 1/8th the tetrahedral voids and cations of B ions occupy half of the octahedral voids. If oxide ions are replaced by `X^(-8//3)` ions then number of an ionic vacancy per unit cell will be

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will analyze the spinel structure and the effect of replacing oxide ions with \( X^{-8/3} \) ions.

### Step 1: Understand the Spinel Structure
The spinel structure has the general formula \( AB_2O_4 \). In this structure:
- \( O^{2-} \) ions are arranged in a face-centered cubic (FCC) lattice.
- Cations of type \( A \) occupy \( \frac{1}{8} \) of the tetrahedral voids.
- Cations of type \( B \) occupy half of the octahedral voids.

### Step 2: Determine the Number of Oxide Ions in the Unit Cell
In an FCC lattice, there are 4 oxide ions per unit cell. This is calculated as follows:
- Each corner contributes \( \frac{1}{8} \) of an ion and there are 8 corners.
- Each face center contributes \( \frac{1}{2} \) of an ion and there are 6 faces.

Thus, the total number of oxide ions is:
\[ \text{Total O}^{2-} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 3: Charge Balance Before Replacement
The total negative charge contributed by the oxide ions in the unit cell is:
\[ \text{Total charge from O}^{2-} = 4 \times (-2) = -8 \]

### Step 4: Replacement of Oxide Ions
We are replacing the \( O^{2-} \) ions with \( X^{-8/3} \) ions. Let \( n \) be the number of \( X^{-8/3} \) ions replacing the oxide ions. The charge contributed by \( n \) ions of \( X^{-8/3} \) is:
\[ \text{Total charge from } X^{-8/3} = n \times \left(-\frac{8}{3}\right) = -\frac{8n}{3} \]

### Step 5: Charge Neutrality Condition
For the crystal to remain neutral, the total negative charge must be equal to the total positive charge. Therefore, we set up the equation:
\[ -8 = -\frac{8n}{3} \]

### Step 6: Solve for \( n \)
To find \( n \), we solve the equation:
\[ -8 = -\frac{8n}{3} \implies 8 = \frac{8n}{3} \implies n = 3 \]

### Step 7: Determine the Number of Ionic Vacancies
Since we started with 4 oxide ions and replaced them with 3 \( X^{-8/3} \) ions, we have:
\[ \text{Number of vacancies} = 4 - n = 4 - 3 = 1 \]

### Conclusion
Thus, the number of ionic vacancies per unit cell is **1**.