`AB` crystallizes in a body centred cubic lattice with edge length `a` equal to `387p m` .The distance between two oppositely charged ions in the lattice is :

Options

A. 335 pm
B. 250 pm
C. 200 pm
D. 300 pm

Answer

To solve the problem of finding the distance between two oppositely charged ions in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Understand the BCC Lattice Structure
In a body-centered cubic lattice, there is one atom at each corner of the cube and one atom at the center of the cube. The distance between the ions can be determined using the relationship between the edge length of the cube and the radius of the ions.

### Step 2: Use the Formula for Radius in BCC
The relationship between the edge length \( a \) and the radius \( R \) of the ions in a BCC lattice is given by the formula:
\[ R = \frac{\sqrt{3}}{4} a \]

### Step 3: Substitute the Given Edge Length
Given that the edge length \( a = 387 \) picometers, we can substitute this value into the formula:
\[ R = \frac{\sqrt{3}}{4} \times 387 \text{ pm} \]

### Step 4: Calculate the Radius
Now, we will calculate the radius:
\[ R = \frac{\sqrt{3}}{4} \times 387 \approx \frac{1.732}{4} \times 387 \approx 0.433 \times 387 \approx 167.57 \text{ pm} \]

### Step 5: Calculate the Distance Between Oppositely Charged Ions
The distance \( d \) between two oppositely charged ions in the lattice is twice the radius:
\[ d = 2R = 2 \times 167.57 \text{ pm} \approx 335.14 \text{ pm} \]

### Step 6: Round the Result
Rounding this to the nearest whole number gives us:
\[ d \approx 335 \text{ pm} \]

### Conclusion
Thus, the distance between two oppositely charged ions in the body-centered cubic lattice is approximately **335 picometers**.