In the unit cell of KCl (NaCl type), Cl^(-) ions constitute ccp and `K^(+)` ion fall into the octahedral holes. These holes are:

Options

A. One at the centre and 6 at the centres of the faces
B. one at the centre and 12 at the centres of the edges
C. 8 at the centres of 8 small cubes forming the unit cell
D. none of these

Answer

To determine the number of octahedral holes in the unit cell of KCl (which has a NaCl type structure), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
KCl crystallizes in a face-centered cubic (FCC) lattice structure, similar to NaCl. In this structure, Cl⁻ ions form a close-packed arrangement, while K⁺ ions occupy the octahedral holes.

2. **Identify the Octahedral Holes**:
In an FCC unit cell, there are two types of voids: tetrahedral and octahedral. We are interested in the octahedral voids.

3. **Count the Octahedral Voids**:
- **Edge-Centered Voids**: There are 12 edges in a cube, and each edge contributes 1/4 of an octahedral void to the unit cell (since each edge is shared by 4 unit cells). Therefore, the contribution from the edges is:
\[ 12 \text{ edges} \times \frac{1}{4} = 3 \text{ octahedral voids from edges} \]
- **Body-Centered Void**: There is 1 octahedral void located at the center of the unit cell (body-centered).

4. **Total Count of Octahedral Voids**:
Adding the contributions from the edge-centered voids and the body-centered void gives us:
\[ 3 \text{ (from edges)} + 1 \text{ (body-centered)} = 4 \text{ octahedral voids} \]

5. **Conclusion**:
Therefore, in the unit cell of KCl (NaCl type), there are a total of 4 octahedral holes.

### Final Answer:
The octahedral holes in the unit cell of KCl (NaCl type) are 4 (3 from edge centers and 1 from the body center).

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