A solid AB has `NaCl` type structure with edge length 580.4 pm. Then radius of `A^(+)` is 100 p m. What is the radius of `B^(-)` in pm?

Options

A. 190.2
B. 540.13
C. 525
D. 78.12

Answer

To solve the problem, we need to find the radius of the anion \( B^{-} \) in a solid \( AB \) that has an NaCl type structure. The given data includes the edge length of the unit cell and the radius of the cation \( A^{+} \).

### Step-by-Step Solution:

1. **Understand the Structure**:
- The NaCl structure is a face-centered cubic (FCC) lattice where each ion is in contact with its nearest neighbors. In this structure, the cation \( A^{+} \) and anion \( B^{-} \) touch each other along the edge of the cube.

2. **Use the Relationship**:
- The relationship between the edge length \( a \), the radius of the cation \( r_A \), and the radius of the anion \( r_B \) is given by:
\[ r_A + r_B = \frac{a}{2} \]
- Here, \( a \) is the edge length of the unit cell.

3. **Substitute the Given Values**:
- We know:
- Edge length \( a = 580.4 \, \text{pm} \)
- Radius of cation \( r_A = 100 \, \text{pm} \)
- Substitute these values into the equation:
\[ 100 \, \text{pm} + r_B = \frac{580.4 \, \text{pm}}{2} \]

4. **Calculate \( \frac{a}{2} \)**:
- Calculate \( \frac{580.4 \, \text{pm}}{2} \):
\[ \frac{580.4}{2} = 290.2 \, \text{pm} \]

5. **Solve for \( r_B \)**:
- Now, substitute back into the equation:
\[ 100 \, \text{pm} + r_B = 290.2 \, \text{pm} \]
- Rearranging gives:
\[ r_B = 290.2 \, \text{pm} - 100 \, \text{pm} \]
\[ r_B = 190.2 \, \text{pm} \]

6. **Conclusion**:
- The radius of the anion \( B^{-} \) is \( 190.2 \, \text{pm} \).

### Final Answer:
The radius of \( B^{-} \) is \( 190.2 \, \text{pm} \).

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