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 \( A^{o+}B^{Θ} \) that crystallizes in a face-centered cubic (FCC) structure, we can follow these steps:

### Step 1: Understand the FCC Structure
In a face-centered cubic (FCC) structure, the atoms are located at the corners and the centers of each face of the cube. The effective coordination number is 12, and the arrangement allows for the calculation of the relationship between the edge length of the cube and the radii of the cations and anions.

### Step 2: Identify the Given Values
- Edge length of the cube \( a = 508 \, \text{pm} \)
- Radius of the anion \( r_{B} = 144 \, \text{pm} \)

### Step 3: Relate the Edge Length to the Ionic Radii
In an FCC structure, the relationship between the radius of the cation \( r_{A} \) and the radius of the anion \( r_{B} \) can be derived from the geometry of the face diagonal. The face diagonal of the cube can be expressed as:
\[ \text{Face diagonal} = \sqrt{2}a \]
The face diagonal consists of one cation and two anions:
\[ \text{Face diagonal} = r_{A} + 2r_{B} + r_{A} = 2r_{A} + 2r_{B} \]
Thus, we have:
\[ \sqrt{2}a = 2r_{A} + 2r_{B} \]

### Step 4: Rearranging the Equation
We can simplify the equation:
\[ \sqrt{2}a = 2(r_{A} + r_{B}) \]
Dividing both sides by 2 gives:
\[ \frac{\sqrt{2}a}{2} = r_{A} + r_{B} \]

### Step 5: Substitute the Known Values
Substituting the values we have:
\[ \frac{\sqrt{2} \times 508 \, \text{pm}}{2} = r_{A} + 144 \, \text{pm} \]
Calculating \( \frac{\sqrt{2} \times 508}{2} \):
\[ \frac{\sqrt{2} \times 508}{2} \approx \frac{1.414 \times 508}{2} \approx \frac{717.092}{2} \approx 358.546 \, \text{pm} \]
Thus, we have:
\[ 358.546 \, \text{pm} = r_{A} + 144 \, \text{pm} \]

### Step 6: Solve for the Radius of the Cation
Now, we can solve for \( r_{A} \):
\[ r_{A} = 358.546 \, \text{pm} - 144 \, \text{pm} \approx 214.546 \, \text{pm} \]

### Final Answer
The radius of the cation \( r_{A} \) is approximately \( 214.5 \, \text{pm} \).

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