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 follow these steps:

### Step 1: Understand the structure of NaCl
NaCl crystallizes in a face-centered cubic (FCC) lattice structure. In this structure, Na⁺ ions occupy octahedral voids, while Cl⁻ ions are located at the corners and face centers of the cube.

### Step 2: Identify the ionic radii
Given the ionic radii:
- Radius of Na⁺ = 95 pm
- Radius of Cl⁻ = 181 pm

### Step 3: Determine the relationship between ionic radii and edge length
In the FCC lattice of NaCl, the edge length (A) can be related to the ionic radii. The edge length can be expressed as:
\[ A = 2 \times (r_{Na^+} + r_{Cl^-}) \]
where \( r_{Na^+} \) is the radius of the Na⁺ ion and \( r_{Cl^-} \) is the radius of the Cl⁻ ion.

### Step 4: Substitute the values into the formula
Now, substituting the given values into the formula:
\[ A = 2 \times (95 \text{ pm} + 181 \text{ pm}) \]

### Step 5: Calculate the edge length
Calculating the sum of the ionic radii:
\[ A = 2 \times (95 + 181) \]
\[ A = 2 \times 276 \]
\[ A = 552 \text{ pm} \]

### Conclusion
The edge length of the NaCl unit cell is **552 pm**.

---