An ionic solid `A^(o+)B^(Θ)` crystallizes as an fcc structure. If the edge length of cell is `508` pm and the radius of anion is `144 pm`, the radius of cation is

Options

A. `110` pm
B. `364` pm
C. `220` pm
D. `288` pm

Explanation

To find the radius of the cation in the ionic solid \( A^{+}B^{-} \) that crystallizes in a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**: In a face-centered cubic (FCC) structure, the cations and anions are arranged such that the cations occupy the octahedral sites, and the anions occupy the face-centered positions.

2. **Relation between Edge Length and Ionic Radii**: For an FCC lattice, the relationship between the edge length \( a \), the radius of the cation \( r_{A} \), and the radius of the anion \( r_{B} \) is given by:
\[ a = 2r_{A} + 2r_{B} \]
This equation arises because along the face diagonal of the cube, there are two cation radii and two anion radii.

3. **Given Values**:
- Edge length \( a = 508 \) pm
- Radius of anion \( r_{B} = 144 \) pm

4. **Substituting the Known Values**: We can rearrange the equation to solve for the radius of the cation \( r_{A} \):
\[ 508 = 2r_{A} + 2(144) \]

5. **Calculating**:
- First, calculate \( 2 \times 144 \):
\[ 2 \times 144 = 288 \text{ pm} \]
- Substitute this back into the equation:
\[ 508 = 2r_{A} + 288 \]
- Now, isolate \( 2r_{A} \):
\[ 2r_{A} = 508 - 288 \]
\[ 2r_{A} = 220 \text{ pm} \]
- Finally, divide by 2 to find \( r_{A} \):
\[ r_{A} = \frac{220}{2} = 110 \text{ pm} \]

6. **Final Answer**: The radius of the cation \( A^{+} \) is \( 110 \) pm.

### Summary:
The radius of the cation \( A^{+} \) is \( 110 \) pm.