The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in `A_(x)B`. The value of x is _________.

Answer

To solve the problem, we need to determine the value of \( x \) in the empirical formula \( A_xB \) for a compound with a cubic close-packed (ccp) arrangement of anions and cations occupying all the octahedral sites.

### Step-by-Step Solution:

1. **Understanding the Cubic Close Packing (CCP)**:
- In a cubic close-packed structure, the anions are arranged in a close-packed manner. The unit cell of a ccp has a coordination number of 12 for the anions.

2. **Calculating the Number of Anions in the Unit Cell**:
- The number of anions per unit cell (denoted as \( z \)) in a ccp structure is 4. This is because:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom (total contribution from corners = \( 8 \times \frac{1}{8} = 1 \)).
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom (total contribution from faces = \( 6 \times \frac{1}{2} = 3 \)).
- Thus, total \( z = 1 + 3 = 4 \).

3. **Cations in the Octahedral Sites**:
- In a ccp structure, there are octahedral voids where cations can occupy. The number of octahedral voids in a ccp unit cell is also 4.

4. **Determining the Composition**:
- Since all octahedral sites are occupied by cations, we have 4 cations for every 4 anions.
- Therefore, the formula can be represented as \( A_4B_4 \).

5. **Finding the Empirical Formula**:
- The empirical formula is the simplest whole number ratio of the elements in the compound.
- To simplify \( A_4B_4 \), we divide both coefficients by their greatest common divisor, which is 4.
- This gives us \( A_1B_1 \) or simply \( AB \).

6. **Conclusion**:
- The value of \( x \) in the empirical formula \( A_xB \) is 1.

### Final Answer:
The value of \( x \) is **1**.