The radii of `Na^+` and `Cl^-` ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is

Options

A. 276 pm
B. 138 pm
C. 552 pm
D. 415 pm

Explanation

To find the edge length of the NaCl unit cell, we can use the relationship between the ionic radii of the Na⁺ and Cl⁻ ions and the edge length of the cubic unit cell.

### Step-by-Step Solution:

1. **Identify the Ionic Radii:**
- The radius of Na⁺ ion (r₊) = 95 pm
- The radius of Cl⁻ ion (r₋) = 181 pm

2. **Understand the Structure of NaCl:**
- NaCl crystallizes in a face-centered cubic (FCC) lattice structure.
- In this structure, the Na⁺ and Cl⁻ ions touch each other along the edge of the cube.

3. **Determine the Edge Length (A):**
- In the NaCl unit cell, the edge length (A) can be expressed in terms of the ionic radii:
\[ A = r₋ + r₊ + r₋ + r₊ = 2r₋ + 2r₊ \]
- This is because along the edge of the cube, there are two Cl⁻ ions and two Na⁺ ions.

4. **Substitute the Values:**
- Substitute the values of the radii into the equation:
\[ A = 2 \times r₋ + 2 \times r₊ = 2 \times 181 \, \text{pm} + 2 \times 95 \, \text{pm} \]

5. **Calculate the Edge Length:**
- Calculate each term:
\[ 2 \times 181 = 362 \, \text{pm} \]
\[ 2 \times 95 = 190 \, \text{pm} \]
- Now, add these two results:
\[ A = 362 \, \text{pm} + 190 \, \text{pm} = 552 \, \text{pm} \]

6. **Final Result:**
- The edge length of the NaCl unit cell is **552 pm**.