`Na` and `Mg` crystallize in bcc- and fcc-type crystals, respectively, then the number of atoms of `Na` and `Mg` present in the unit cell of their respective crystal is

Options

A. `4` and `2`
B. `9` and `14`
C. `14` and `9`
D. `2` and `4`

Explanation

To solve the problem, we need to determine the number of atoms present in the unit cells of sodium (Na) and magnesium (Mg) based on their crystal structures: body-centered cubic (BCC) for Na and face-centered cubic (FCC) for Mg.

### Step-by-Step Solution:

1. **Understanding the Crystal Structures**:
- Sodium (Na) crystallizes in a body-centered cubic (BCC) structure.
- Magnesium (Mg) crystallizes in a face-centered cubic (FCC) structure.

2. **Calculating the Number of Atoms in BCC (for Na)**:
- In a BCC unit cell, atoms are located at:
- 8 corners of the cube
- 1 atom at the center of the cube
- Contribution from corner atoms:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- Total contribution from corners = \( 8 \times \frac{1}{8} = 1 \) atom.
- Contribution from the body-centered atom:
- The body-centered atom contributes 1 atom.
- Therefore, the total number of atoms in the BCC unit cell for Na is:
\[ \text{Total atoms} = 1 (\text{from corners}) + 1 (\text{from body center}) = 2 \text{ atoms} \]

3. **Calculating the Number of Atoms in FCC (for Mg)**:
- In an FCC unit cell, atoms are located at:
- 8 corners of the cube
- 6 face centers of the cube
- Contribution from corner atoms:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- Total contribution from corners = \( 8 \times \frac{1}{8} = 1 \) atom.
- Contribution from face-centered atoms:
- Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell.
- Total contribution from faces = \( 6 \times \frac{1}{2} = 3 \) atoms.
- Therefore, the total number of atoms in the FCC unit cell for Mg is:
\[ \text{Total atoms} = 1 (\text{from corners}) + 3 (\text{from faces}) = 4 \text{ atoms} \]

4. **Final Answer**:
- The number of atoms in the unit cell of Na (BCC) is 2.
- The number of atoms in the unit cell of Mg (FCC) is 4.
- Thus, the final answer is:
\[ \text{Na: 2, Mg: 4} \]

### Conclusion:
The number of atoms of Na and Mg present in their respective unit cells is **2 and 4**. Therefore, the correct option is **D: 2 and 4**.