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

Options

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

Explanation

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

2. **Identify the Relationship**: For an FCC structure, the relationship between the edge length (a) of the unit cell, the radius of the cation (Rc), and the radius of the anion (Ra) can be expressed as:
\[ a = 2(Ra + Rc) \]
This is because the cation and anion touch along the face diagonal.

3. **Given Values**: From the problem, we know:
- Edge length, \( a = 508 \) pm
- Radius of cation, \( Rc = 110 \) pm

4. **Substitute the Known Values**: Plug the known values into the equation:
\[ 508 = 2(Ra + 110) \]

5. **Simplify the Equation**: Divide both sides by 2:
\[ 254 = Ra + 110 \]

6. **Solve for Ra**: Rearranging the equation gives:
\[ Ra = 254 - 110 \]
\[ Ra = 144 \text{ pm} \]

7. **Conclusion**: The radius of the anion is \( 144 \) pm.

### Final Answer:
The radius of the anion (Ra) is **144 pm**.

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