An ionic solid `A^(o+)B^(Θ)` crystallizes as an bcc structure. The distance between cation and anion in the lattice is `338 pm`. The edge length of cell is

Options

A. `338` pm
B. `390.3` pm
C. `292.7` pm
D. `507` pm

Explanation

To find the edge length of the body-centered cubic (BCC) ionic solid \( A^{o+}B^{Θ} \), we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC structure, the cations (A\(^+\)) are located at the body center of the cube, while the anions (B\(^-\)) are located at the corners of the cube.

### Step 2: Identify the Distance Between Cation and Anion
The distance between the cation and anion is given as \( 338 \, \text{pm} \). This distance is equivalent to the distance from a corner atom (anion) to the body-centered atom (cation).

### Step 3: Relate the Distance to the Edge Length
In a BCC structure, the relationship between the edge length \( a \) of the cube and the distance \( d \) between the corner and the body center is given by the formula:
\[ d = \frac{\sqrt{3}}{2} a \]
Where \( d \) is the distance between the corner (anion) and the body center (cation).

### Step 4: Set Up the Equation
We can set up the equation using the given distance:
\[ \frac{\sqrt{3}}{2} a = 338 \, \text{pm} \]

### Step 5: Solve for the Edge Length \( a \)
To find \( a \), we rearrange the equation:
\[ a = \frac{2 \times 338 \, \text{pm}}{\sqrt{3}} \]

### Step 6: Calculate the Edge Length
Now we calculate the value:
\[ a = \frac{676 \, \text{pm}}{\sqrt{3}} \approx \frac{676}{1.732} \approx 390.3 \, \text{pm} \]

### Conclusion
The edge length of the cell is approximately \( 390.3 \, \text{pm} \).

### Final Answer
The edge length of the cell is \( 390.3 \, \text{pm} \) (Option B).

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