One unit cell of NaCl contains

Options

A. `1Na^(+)and6Cl^(-)`
B. `6Na^(+)and1Cl^(-)`
C. `4Na^(+)and4Cl^(-)`
D. `6Na^(+)and6Cl^(-)`

Answer

To determine the composition of one unit cell of NaCl, we need to analyze the arrangement of sodium (Na⁺) and chloride (Cl⁻) ions in the crystal structure of NaCl.

### Step-by-Step Solution:

1. **Understanding the Structure**:
- NaCl crystallizes in a face-centered cubic (FCC) lattice structure. In this structure, the Na⁺ ions occupy the octahedral voids, while the Cl⁻ ions are located at the corners and face centers of the unit cell.

2. **Counting the Chloride Ions (Cl⁻)**:
- In the FCC lattice, the Cl⁻ ions are located at:
- 8 corners of the cube, each contributing 1/8 of a Cl⁻ ion to the unit cell.
- 6 face centers, each contributing 1/2 of a Cl⁻ ion to the unit cell.
- Calculation:
- Contribution from corners: \(8 \times \frac{1}{8} = 1\)
- Contribution from face centers: \(6 \times \frac{1}{2} = 3\)
- Total Cl⁻ ions in the unit cell: \(1 + 3 = 4\)

3. **Counting the Sodium Ions (Na⁺)**:
- The Na⁺ ions occupy the octahedral voids in the FCC structure.
- In a face-centered cubic lattice, there are 4 octahedral voids per unit cell.
- Therefore, there are 4 Na⁺ ions in the unit cell.

4. **Final Composition**:
- From the above calculations, we find that one unit cell of NaCl contains:
- 4 Na⁺ ions
- 4 Cl⁻ ions

### Conclusion:
Thus, one unit cell of NaCl contains 4 Na⁺ and 4 Cl⁻ ions.

### Answer:
4 Na⁺ and 4 Cl⁻ (Option 3)

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