How many `Na^(o+)` ions sorround each `CI^(Θ)` ion in `NaCI` crystal lattice ?

Options

A. 4
B. 6
C. 8
D. 12

Answer

To determine how many `Na^(o+)` ions surround each `Cl^(Θ)` ion in the NaCl crystal lattice, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Crystal Structure**:
- NaCl crystallizes in a face-centered cubic (FCC) lattice structure. In this arrangement, both sodium (Na) and chloride (Cl) ions are arranged in a cubic geometry.

2. **Visualizing the Unit Cell**:
- The unit cell of NaCl consists of Na and Cl ions. In the FCC structure, the Na ions occupy the corners and the face centers of the cube, while the Cl ions are located in the octahedral voids.

3. **Counting the Surrounding Ions**:
- For each Cl ion in the NaCl lattice, we need to count how many Na ions are in close proximity.
- In the FCC structure, each Cl ion is surrounded by Na ions located at the corners of the cube and the face centers.

4. **Identifying the Coordination Number**:
- In the NaCl crystal lattice, each Cl ion is surrounded by 6 Na ions. This is because:
- There are 4 Na ions in the plane of the Cl ion (one on each corner of the square face) and 2 Na ions directly above and below the Cl ion (one at the top face center and one at the bottom face center).

5. **Conclusion**:
- Therefore, each `Cl^(Θ)` ion in the NaCl crystal lattice is surrounded by 6 `Na^(o+)` ions.

### Final Answer:
Each `Cl^(Θ)` ion in the NaCl crystal lattice is surrounded by 6 `Na^(o+)` ions.

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