The edge length of a face-centred cubic unit cell is 508 pm. If the radius of the cation is 110 pm the radius of the anion is

Options

A. 288 pm
B. 398 pm
C. 618 pm
D. 144 pm

Answer

To find the radius of the anion in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Edge length of the FCC unit cell (a) = 508 pm
- Radius of the cation (r⁺) = 110 pm

2. **Understand the Arrangement in FCC**:
- In a face-centered cubic unit cell, the cations occupy the octahedral voids, while the anions are located at the corners and the face centers of the cube.

3. **Relationship Between Edge Length and Ionic Radii**:
- For an FCC unit cell, the relationship between the edge length (a), the radius of the cation (r⁺), and the radius of the anion (r⁻) can be expressed as:
\[ 2r^+ + r^- = a \]
- Here, \(2r^+\) accounts for the two cations that touch along the edge, and \(r^-\) is the radius of the anion.

4. **Substituting Known Values**:
- Substitute the known values into the equation:
\[ 2(110 \text{ pm}) + r^- = 508 \text{ pm} \]
- This simplifies to:
\[ 220 \text{ pm} + r^- = 508 \text{ pm} \]

5. **Solve for the Radius of the Anion**:
- Rearranging the equation to solve for \(r^-\):
\[ r^- = 508 \text{ pm} - 220 \text{ pm} \]
- This gives:
\[ r^- = 288 \text{ pm} \]

6. **Final Answer**:
- The radius of the anion is 288 pm.