The spinel structure `AB_(2)O_(4)` consists of an fcc array of `O^(2-)` ions in which the:

Options

A. A cation occupies one-eighth of the tetrahedral holes and B cation occupied one-half of octahedral holes
B. A cation occupies one-fourth of the tetrahedral holes and the B cations the octahedral holes
C. A cation occupies one-eighth of the octahdral hole and the B cation the tetrahdral holes
D. A cation occupies one-fourth of the octahdral holes and the B cations the tetrahedral holes

Answer

To solve the question regarding the spinel structure \( AB_2O_4 \), we will analyze the arrangement of ions in the crystal lattice and identify the correct occupancy of the cation sites.

### Step-by-Step Solution:

1. **Understanding the Spinel Structure**:
- The spinel structure is represented by the formula \( AB_2O_4 \).
- In this structure, \( A \) is a divalent cation (e.g., \( A^{2+} \)) and \( B \) is a trivalent cation (e.g., \( B^{3+} \)).
- The oxide ions \( O^{2-} \) form a face-centered cubic (FCC) lattice.

2. **Counting the Ions in the FCC Lattice**:
- In an FCC lattice, there are 4 oxide ions per unit cell (6 face-centered ions contribute \( \frac{1}{2} \) each and 8 corner ions contribute \( \frac{1}{8} \) each).
- Thus, the total contribution is:
\[ 6 \times \frac{1}{2} + 8 \times \frac{1}{8} = 3 + 1 = 4 \]

3. **Identifying the Types of Voids**:
- In a cubic close-packed structure, the number of octahedral voids is equal to the number of atoms, which is 4.
- The number of tetrahedral voids is twice the number of atoms, which is 8.

4. **Occupancy of Cation Sites**:
- For the normal spinel structure \( AB_2O_4 \):
- The \( A^{2+} \) cation occupies \( \frac{1}{8} \) of the tetrahedral voids.
- The \( B^{3+} \) cation occupies \( \frac{1}{2} \) of the octahedral voids.

5. **Calculating the Number of Cations**:
- Since there are 8 tetrahedral voids, \( A^{2+} \) occupies \( 1 \) tetrahedral void (as \( \frac{1}{8} \times 8 = 1 \)).
- Since there are 4 octahedral voids, \( B^{3+} \) occupies \( 2 \) octahedral voids (as \( \frac{1}{2} \times 4 = 2 \)).
- This gives us the ratio of \( A^{2+} : B^{3+} = 1 : 2 \).

6. **Evaluating the Options**:
- The first option states that \( A \) cation occupies \( \frac{1}{8} \) of the tetrahedral holes and \( B \) cation occupies half of the octahedral holes. This matches our findings.
- The other options do not match the occupancy described above.

7. **Conclusion**:
- The correct answer is the first option.

### Final Answer:
The correct option is: **A cation occupies one-eighth of the tetrahedral holes, and B cation occupies half of the octahedral holes**.