LiCl has rock salt structure, with an edge length of 6.0Å. If anions are in contact, the ionic radius of chloride ion is

Options

A. 2.12 Å
B. 1.82 Å
C. 4.24 Å
D. 3.68 Å

Explanation

To find the ionic radius of the chloride ion (Cl⁻) in LiCl, which has a rock salt structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- LiCl crystallizes in a face-centered cubic (FCC) structure, also known as the rock salt structure. In this arrangement, the chloride ions (Cl⁻) are located at the corners and the centers of the faces of the cube, while the lithium ions (Li⁺) occupy the octahedral voids.

2. **Given Data**:
- The edge length (a) of the cubic unit cell is given as 6.0 Å.

3. **Face Diagonal Calculation**:
- In an FCC structure, the face diagonal can be calculated using the formula:
\[ \text{Face diagonal} = a\sqrt{2} \]
- Substituting the given edge length:
\[ \text{Face diagonal} = 6.0 \, \text{Å} \times \sqrt{2} \approx 6.0 \, \text{Å} \times 1.414 \approx 8.485 \, \text{Å} \]

4. **Ionic Radii Relationship**:
- In the FCC structure, when the anions (Cl⁻) are in contact along the face diagonal, the distance across the face diagonal is equal to the sum of the radii of the two chloride ions:
\[ \text{Face diagonal} = 4r \]
- Here, \( r \) is the ionic radius of the chloride ion.

5. **Setting Up the Equation**:
- From the face diagonal calculation:
\[ 4r = 8.485 \, \text{Å} \]

6. **Solving for the Ionic Radius \( r \)**:
- Rearranging the equation to find \( r \):
\[ r = \frac{8.485 \, \text{Å}}{4} \approx 2.121 \, \text{Å} \]

### Final Answer:
The ionic radius of the chloride ion (Cl⁻) in LiCl is approximately **2.121 Å**.

---