A solid AB has ZnS-type structure. The edge lenth of unit cell is 400 pm abd the radius of `B^-` ion is 0.130 nm. Then the radius of `A^+` ion is

Options

A. 35.8 pm
B. 43.2 pm
C. 60.5 pm
D. 53.2 pm

Answer

To find the radius of the \( A^+ \) ion in a solid \( AB \) with a ZnS-type structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
The ZnS-type structure is a face-centered cubic (FCC) lattice where the \( B^- \) ions occupy the face-centered positions and the \( A^+ \) ions occupy the tetrahedral voids.

2. **Identify Given Values**:
- Edge length of the unit cell (\( a \)) = 400 pm = \( 400 \times 10^{-12} \) m
- Radius of \( B^- \) ion (\( R^- \)) = 0.130 nm = \( 0.130 \times 10^{-9} \) m = \( 130 \) pm

3. **Relate the Radii to the Edge Length**:
In a ZnS-type structure, the relationship between the radii of the ions and the edge length is given by:
\[ R^+ + R^- = \frac{\sqrt{3}}{4} a \]
where \( R^+ \) is the radius of the \( A^+ \) ion.

4. **Substitute the Known Values**:
Substitute \( R^- \) and \( a \) into the equation:
\[ R^+ + 130 \text{ pm} = \frac{\sqrt{3}}{4} \times 400 \text{ pm} \]

5. **Calculate the Right Side**:
Calculate \( \frac{\sqrt{3}}{4} \times 400 \):
\[ \frac{\sqrt{3}}{4} \times 400 \approx 173.2 \text{ pm} \]

6. **Rearrange to Find \( R^+ \)**:
Now rearranging the equation gives:
\[ R^+ = 173.2 \text{ pm} - 130 \text{ pm} \]
\[ R^+ = 43.2 \text{ pm} \]

7. **Final Answer**:
The radius of the \( A^+ \) ion is \( 43.2 \) pm.