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 solve the problem, we need to find the radius of the A^+ ion in a solid AB that has a ZnS-type structure. We are given the edge length of the unit cell and the radius of the B^- ion.

### 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. **Convert Units**:
- The edge length of the unit cell (a) is given as 400 pm (picometers).
- We also need to convert the radius of B^- ion from nm to pm for consistency:
\[ r_B = 0.130 \, \text{nm} = 0.130 \times 10^3 \, \text{pm} = 130 \, \text{pm} \]

3. **Calculate the Distance Between Ions**:
- In an FCC structure, the distance between the A^+ and B^- ions can be calculated using the formula:
\[ d = \frac{a \sqrt{3}}{4} \]
- Substituting the value of a:
\[ d = \frac{400 \, \text{pm} \times \sqrt{3}}{4} = \frac{400 \times 1.732}{4} \approx 173.2 \, \text{pm} \]

4. **Set Up the Equation**:
- The distance \(d\) between the A^+ and B^- ions can also be expressed in terms of their radii:
\[ d = r_A + r_B \]
- Therefore, we can write:
\[ r_A + r_B = 173.2 \, \text{pm} \]

5. **Substitute the Known Value**:
- We know \(r_B = 130 \, \text{pm}\). Substituting this into the equation:
\[ r_A + 130 \, \text{pm} = 173.2 \, \text{pm} \]

6. **Solve for \(r_A\)**:
- Rearranging the equation gives:
\[ r_A = 173.2 \, \text{pm} - 130 \, \text{pm} = 43.2 \, \text{pm} \]

7. **Final Answer**:
- The radius of the A^+ ion is:
\[ r_A = 43.2 \, \text{pm} \]