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⁺) given the edge length of the unit cell of sodium chloride (NaCl) and the ionic radius of the chloride ion (Cl⁻), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of NaCl:**
- Sodium chloride crystallizes in a face-centered cubic (FCC) lattice structure.
- In this structure, Cl⁻ ions occupy the corners and face centers of the cube, while Na⁺ ions occupy the octahedral voids.

2. **Identify the Relationship Between Edge Length and Ionic Radii:**
- The edge length (A) of the unit cell is related to the ionic radii of Na⁺ and Cl⁻ ions.
- The formula that relates the edge length to the ionic radii in NaCl is:
\[ A = 2 \times r_{Cl^-} + 2 \times r_{Na^+} \]
- This can be rearranged to:
\[ r_{Na^+} = \frac{A}{2} - r_{Cl^-} \]

3. **Substitute the Given Values:**
- Given:
- Edge length \( A = 600 \, \text{pm} \)
- Ionic radius of Cl⁻ \( r_{Cl^-} = 190 \, \text{pm} \)
- Substitute these values into the rearranged formula:
\[ r_{Na^+} = \frac{600 \, \text{pm}}{2} - 190 \, \text{pm} \]

4. **Calculate the Ionic Radius of Na⁺:**
- Calculate \( \frac{600 \, \text{pm}}{2} \):
\[ \frac{600}{2} = 300 \, \text{pm} \]
- Now, substitute this back into the equation:
\[ r_{Na^+} = 300 \, \text{pm} - 190 \, \text{pm} = 110 \, \text{pm} \]

5. **Conclusion:**
- The ionic radius of the sodium ion (Na⁺) is \( 110 \, \text{pm} \).

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

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