An ionic solid AB crystallizes as a bcc structure . The distance between cation and anion in the lattice is 338 pm . Calculate the edge length of the unit cell .

Options

A. `195.15` pm
B. `97.58` pm
C. `390.3` pm
D. `780.6` pm

Answer

To solve the problem, we need to find the edge length of the unit cell of the ionic solid AB, which crystallizes in a body-centered cubic (BCC) structure. Given that the distance between the cation (A) and the anion (B) is 338 pm, we can use this information to calculate the edge length of the unit cell.

### Step-by-Step Solution:

1. **Understand the BCC Structure**:
- In a BCC structure, there are two types of ions: one cation (A) and one anion (B). The cation is located at the body center, and the anions are located at the corners of the cube.

2. **Determine the Relationship Between Distance and Edge Length**:
- The distance between the cation and anion in the BCC structure can be expressed in terms of the edge length (a) of the unit cell. The relationship is given by:
\[ \text{Distance} = \frac{\sqrt{3}}{2} a \]
- This is because the cation at the center is at (a/2, a/2, a/2) and the anion at a corner is at (0, 0, 0), and the distance can be calculated using the 3D distance formula.

3. **Set Up the Equation**:
- We know the distance between the cation and anion is 338 pm, so we set up the equation:
\[ \frac{\sqrt{3}}{2} a = 338 \text{ pm} \]

4. **Solve for Edge Length (a)**:
- Rearranging the equation to solve for \( a \):
\[ a = \frac{2 \times 338 \text{ pm}}{\sqrt{3}} \]
- Now, calculate the value:
\[ a = \frac{676 \text{ pm}}{\sqrt{3}} \approx \frac{676 \text{ pm}}{1.732} \approx 390.3 \text{ pm} \]

5. **Final Answer**:
- The edge length of the unit cell is approximately **390.3 pm**.