The edge length of a face centred cubic cell of an ionic substance 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

Explanation

To find the radius of the anion in a face-centered cubic (FCC) structure of an ionic substance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
In a face-centered cubic (FCC) lattice, the anions are located at the corners and face centers of the cube, while the cations occupy the octahedral voids.

2. **Identify the Given Values**:
- Edge length of the FCC cell (A) = 508 pm
- Radius of the cation (R_c) = 110 pm

3. **Relate the Radii to the Edge Length**:
In an FCC structure, the relationship between the radius of the anion (R_a) and the radius of the cation (R_c) can be expressed as:
\[ R_a + 2R_c = \frac{A}{2} \]
This equation comes from the geometry of the FCC structure, where the diagonal of the face of the cube is equal to the sum of the radii of the anions and cations.

4. **Substitute the Known Values**:
Substitute the known values into the equation:
\[ R_a + 2(110 \text{ pm}) = \frac{508 \text{ pm}}{2} \]
\[ R_a + 220 \text{ pm} = 254 \text{ pm} \]

5. **Solve for the Radius of the Anion**:
Rearranging the equation gives:
\[ R_a = 254 \text{ pm} - 220 \text{ pm} \]
\[ R_a = 34 \text{ pm} \]

6. **Final Answer**:
The radius of the anion (R_a) is **34 pm**.