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

Options

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

Explanation

To solve the problem, we need to find the radius of the anion in a face-centered cubic (FCC) unit cell given the edge length and the radius of the cation. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Edge length of the FCC unit cell, \( a = 508 \, \text{pm} \)
- Radius of the cation, \( r_{cation} = 110 \, \text{pm} \)

2. **Understand the Structure of FCC:**
- In a face-centered cubic (FCC) unit cell, the cations and anions are arranged such that they touch along the face diagonal of the cube.
- The face diagonal can be expressed in terms of the edge length \( a \) of the cube.

3. **Calculate the Face Diagonal:**
- The face diagonal \( d \) of the cube can be calculated using the formula:
\[ d = a\sqrt{2} \]
- Substituting the value of \( a \):
\[ d = 508 \, \text{pm} \times \sqrt{2} \approx 508 \, \text{pm} \times 1.414 \approx 718.18 \, \text{pm} \]

4. **Set Up the Relationship Between the Radii:**
- Along the face diagonal, the arrangement of ions can be represented as:
\[ r_{anion} + 2 \cdot r_{cation} + r_{anion} = d \]
- This simplifies to:
\[ 2 \cdot r_{anion} + 2 \cdot r_{cation} = d \]

5. **Substitute Known Values:**
- Substitute \( r_{cation} = 110 \, \text{pm} \) and \( d \approx 718.18 \, \text{pm} \):
\[ 2 \cdot r_{anion} + 2 \cdot 110 = 718.18 \]
- This simplifies to:
\[ 2 \cdot r_{anion} + 220 = 718.18 \]

6. **Solve for \( r_{anion} \):**
- Rearranging the equation gives:
\[ 2 \cdot r_{anion} = 718.18 - 220 \]
\[ 2 \cdot r_{anion} = 498.18 \]
\[ r_{anion} = \frac{498.18}{2} \approx 249.09 \, \text{pm} \]

### Final Answer:
The radius of the anion is approximately \( 249.09 \, \text{pm} \).