`NH_(4)Cl` crystallizes in a body-centered cubic type lattice with a unit cell edge length of 387 pm. The distance between the oppositively charged ions in the lattice is

Options

A. 335.1 pm
B. 83.77 pm
C. 274.46 pm
D. 137.23 pm

Answer

To find the distance between the oppositely charged ions in the body-centered cubic (BCC) lattice of \( NH_4Cl \), we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC lattice:
- The cations (positive ions) are located at the center of the cube.
- The anions (negative ions) are located at the corners of the cube.

For \( NH_4Cl \):
- The \( NH_4^+ \) ion is the cation and is located at the center of the cube.
- The \( Cl^- \) ions are located at the eight corners of the cube.

### Step 2: Identify the Unit Cell Edge Length
The edge length of the unit cell is given as \( a = 387 \, pm \).

### Step 3: Calculate the Body Diagonal
In a cubic lattice, the body diagonal can be calculated using the formula:
\[ \text{Body Diagonal} = \sqrt{3} \times a \]
Substituting the value of \( a \):
\[ \text{Body Diagonal} = \sqrt{3} \times 387 \, pm \approx 669.1 \, pm \]

### Step 4: Determine the Distance Between Oppositely Charged Ions
The distance between the \( NH_4^+ \) ion at the center and a \( Cl^- \) ion at one of the corners can be found by taking half of the body diagonal:
\[ \text{Distance} = \frac{\text{Body Diagonal}}{2} = \frac{\sqrt{3} \times a}{2} \]
Calculating this gives:
\[ \text{Distance} = \frac{669.1 \, pm}{2} \approx 334.55 \, pm \]

### Step 5: Final Answer
Thus, the distance between the oppositely charged ions \( NH_4^+ \) and \( Cl^- \) in the lattice is approximately:
\[ \text{Distance} \approx 334.55 \, pm \approx 335 \, pm \]