Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is 180 pm, the size of ammonium ion would be

Options

A. 174 pm
B. 158 pm
C. 142 pm
D. 126 pm

Explanation

To solve the problem of finding the size of the ammonium ion (NH4+) in a body-centered cubic (BCC) lattice of ammonium chloride (NH4Cl), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of BCC**:
In a body-centered cubic lattice, there are atoms located at each corner of the cube and one atom at the center of the cube.

2. **Identify the Given Data**:
- Edge length of the unit cell (A) = 390 pm
- Radius of chloride ion (R−) = 180 pm

3. **Relate the Edge Length to the Ionic Radii**:
In a BCC structure, the relationship between the edge length (A) and the ionic radii (R+ for ammonium ion and R− for chloride ion) can be established using the diagonal of the cube. The body diagonal (BD) can be expressed in terms of the edge length:
\[ BD = \sqrt{3} \times A \]
The body diagonal also equals the sum of the diameters of the two ions:
\[ BD = 2R+ + 2R− \]

4. **Set Up the Equation**:
From the above relationships, we can write:
\[ \sqrt{3} \times A = 2R+ + 2R− \]

5. **Substituting the Known Values**:
Substitute A = 390 pm and R− = 180 pm into the equation:
\[ \sqrt{3} \times 390 = 2R+ + 2 \times 180 \]

6. **Calculate the Left Side**:
Calculate \(\sqrt{3} \times 390\):
\[ \sqrt{3} \approx 1.732 \implies 1.732 \times 390 \approx 676.08 \text{ pm} \]

7. **Simplify the Equation**:
Substitute this value into the equation:
\[ 676.08 = 2R+ + 360 \]

8. **Isolate R+**:
Rearranging gives:
\[ 2R+ = 676.08 - 360 \]
\[ 2R+ = 316.08 \]
\[ R+ = \frac{316.08}{2} \approx 158.04 \text{ pm} \]

9. **Final Result**:
The size of the ammonium ion (R+) is approximately 158 pm.

### Summary:
The size of the ammonium ion (NH4+) is approximately 158 pm.