If the edge-length of the unit cell of sodium chloride is 600 pm, and the ionic radius of `CI^(-)` ion is 190 pm, then the ionic radius of `Na^(+)` ion is

Options

A. 310 pm
B. 110 pm
C. 220 pm
D. None

Explanation

To find the ionic radius of the sodium ion (Na⁺) in sodium chloride (NaCl), we can use the relationship between the ionic radii of the cation and anion in a face-centered cubic (FCC) lattice.

### Step-by-Step Solution:

1. **Understand the Structure**: Sodium chloride (NaCl) crystallizes in a face-centered cubic (FCC) lattice. In this structure, the relationship between the ionic radii of the cation (Na⁺) and the anion (Cl⁻) can be expressed as:
\[ r_{Na^+} + r_{Cl^-} = \frac{a}{2} \]
where \( r_{Na^+} \) is the ionic radius of sodium, \( r_{Cl^-} \) is the ionic radius of chloride, and \( a \) is the edge length of the unit cell.

2. **Given Values**:
- Edge length of the unit cell, \( a = 600 \) pm
- Ionic radius of chloride ion, \( r_{Cl^-} = 190 \) pm

3. **Calculate \( \frac{a}{2} \)**:
\[ \frac{a}{2} = \frac{600 \text{ pm}}{2} = 300 \text{ pm} \]

4. **Set Up the Equation**:
Substitute the known values into the equation:
\[ r_{Na^+} + 190 \text{ pm} = 300 \text{ pm} \]

5. **Solve for \( r_{Na^+} \)**:
Rearranging the equation gives:
\[ r_{Na^+} = 300 \text{ pm} - 190 \text{ pm} \]
\[ r_{Na^+} = 110 \text{ pm} \]

6. **Conclusion**: The ionic radius of the sodium ion (Na⁺) is \( 110 \) pm.

### Final Answer:
The ionic radius of Na⁺ is **110 pm**.

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