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 need to 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 \(A\) occupy \(1/8\) of the tetrahedral voids.
- Cations \(B\) occupy \(1/2\) of the octahedral voids.

### Step 2: Determine the Number of Ions in the Unit Cell
In an FCC arrangement:
- There are 4 oxide ions (\(O^{2-}\)) per unit cell, since each corner contributes \(1/8\) and each face center contributes \(1/2\).

### Step 3: Replacement of Oxide Ions
When \(O^{2-}\) ions are replaced by \(X^{-8/3}\) ions:
- Let \(n\) be the number of \(X^{-8/3}\) ions replacing \(O^{2-}\) ions.
- The total charge contributed by the \(O^{2-}\) ions in the unit cell is \(4 \times (-2) = -8\).
- The total charge contributed by \(n\) \(X^{-8/3}\) ions is \(n \times (-\frac{8}{3})\).

### Step 4: Charge Balance
To maintain charge neutrality:
\[ -8 + n \left(-\frac{8}{3}\right) = 0 \]
Solving for \(n\):
\[ -8 = -\frac{8n}{3} \]
Multiplying both sides by \(-3\):
\[ 24 = 8n \]
\[ n = 3 \]

### Step 5: Calculate Ionic Vacancies
Since \(3\) \(X^{-8/3}\) ions replace \(4\) \(O^{2-}\) ions, we have:
- Initially, there were \(4\) \(O^{2-}\) ions.
- After replacement, \(3\) \(X^{-8/3}\) ions are present.

This means:
\[ \text{Number of vacancies} = 4 - 3 = 1 \]

### Conclusion
Thus, the number of ionic vacancies per unit cell is \(1\).