Find the coordination of `Na^(o+)` in `Na_(2)O`.

Explanation

To find the coordination number of \( \text{Na}^+ \) in \( \text{Na}_2\text{O} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of \( \text{Na}_2\text{O} \)**:
- \( \text{Na}_2\text{O} \) consists of sodium ions (\( \text{Na}^+ \)) and oxide ions (\( \text{O}^{2-} \)).
- In this compound, oxide ions form a face-centered cubic (FCC) lattice.

2. **Counting the Oxide Ions in the FCC Lattice**:
- In an FCC lattice, there are 8 corner atoms and 6 face-centered atoms.
- Each corner atom contributes \( \frac{1}{8} \) to the unit cell, and each face-centered atom contributes \( \frac{1}{2}). - Total contribution from corner atoms: \( 8 \times \frac{1}{8} = 1 \).
- Total contribution from face-centered atoms: \( 6 \times \frac{1}{2} = 3 \).
- Therefore, the total number of oxide ions in one unit cell is \( 1 + 3 = 4 \).

3. **Identifying the Tetrahedral Voids**:
- In a cubic unit cell, there are 8 tetrahedral voids.
- Each body diagonal of the cube contains 2 tetrahedral voids.
- Since there are 4 body diagonals in a cube, the total number of tetrahedral voids is \( 4 \times 2 = 8 \).

4. **Determining the Coordination Number**:
- The coordination number of an ion in a crystal structure is defined as the number of nearest neighboring ions surrounding it.
- For tetrahedral voids, the coordination number is 4.
- Since \( \text{Na}^+ \) occupies the tetrahedral voids in the FCC lattice formed by \( \text{O}^{2-} \), the coordination number of \( \text{Na}^+ \) is also 4.

5. **Final Answer**:
- Therefore, the coordination number of \( \text{Na}^+ \) in \( \text{Na}_2\text{O} \) is **4**.