How many effiective `Na^(+)` and `Cl^(-)` ions are present respectively in a uint cell of NaCl solid (Rock salt structure ) if all ions along line connecting opposite face centres are absent ?

Options

A. 3,3
B. `(7)/(2),4`
C. `(7)/(2),(7)/(2)`
D. `4,(7)/(2)`

Answer

To solve the problem of how many effective \( \text{Na}^+ \) and \( \text{Cl}^- \) ions are present in a unit cell of NaCl solid (rock salt structure) when all ions along the line connecting opposite face centers are absent, we can follow these steps:

### Step 1: Understand the Structure of NaCl
In the rock salt structure of NaCl:
- The chloride ions (\( \text{Cl}^- \)) are arranged in a face-centered cubic (FCC) lattice.
- The sodium ions (\( \text{Na}^+ \)) occupy the octahedral voids.

### Step 2: Count the Chloride Ions
In a unit cell of NaCl:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell (since each corner atom is shared by 8 unit cells).
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell (since each face atom is shared by 2 unit cells).

Calculating the total contribution of chloride ions:
- Contribution from corners: \( 8 \times \frac{1}{8} = 1 \)
- Contribution from face centers: \( 6 \times \frac{1}{2} = 3 \)

Thus, the total number of \( \text{Cl}^- \) ions before considering the absence of ions is:
\[ 1 + 3 = 4 \]

### Step 3: Account for the Absence of Ions
According to the problem, the ions along the line connecting opposite face centers are absent. This means:
- 2 face-centered \( \text{Cl}^- \) ions are absent (one from each face).
- The body-centered \( \text{Na}^+ \) ion is also absent.

After removing these ions:
- Remaining \( \text{Cl}^- \) ions: \( 4 - 2 = 2 \)

### Step 4: Count the Sodium Ions
For sodium ions in the unit cell:
- There is 1 body-centered \( \text{Na}^+ \) ion.
- There are 12 edge-centered octahedral voids, each contributing \( \frac{1}{4} \) to the unit cell.

Calculating the total contribution of sodium ions:
- Contribution from body center: \( 1 \)
- Contribution from edges: \( 12 \times \frac{1}{4} = 3 \)

Thus, the total number of \( \text{Na}^+ \) ions before considering the absence of ions is:
\[ 1 + 3 = 4 \]

### Step 5: Account for the Absence of Sodium Ion
Since the body-centered \( \text{Na}^+ \) ion is absent:
- Remaining \( \text{Na}^+ \) ions: \( 4 - 1 = 3 \)

### Conclusion
After considering the absence of the specified ions, we find the effective number of ions in the unit cell:
- Effective \( \text{Na}^+ \) ions: 3
- Effective \( \text{Cl}^- \) ions: 3

Thus, the final answer is:
**Effective \( \text{Na}^+ \) ions: 3 and Effective \( \text{Cl}^- \) ions: 3.**