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 find the radius of the anion \( B^{-} \) in a solid \( AB \) with a NaCl type structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- The NaCl type structure is a face-centered cubic (FCC) lattice where cations and anions are arranged in a specific manner. In this structure, cations occupy the octahedral voids and anions occupy the corners and face centers of the cube.

2. **Given Data**:
- Edge length of the unit cell \( a = 580.4 \, \text{pm} \)
- Radius of cation \( A^{+} = 100 \, \text{pm} \)

3. **Relation Between Edge Length and Ionic Radii**:
- In an NaCl type structure, 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:
\[ a = 2(r_{A} + r_{B}) \]
- This equation arises because the cation and anion touch along the edge of the cube.

4. **Substituting Known Values**:
- We can rearrange the formula to solve for \( r_{B} \):
\[ r_{B} = \frac{a}{2} - r_{A} \]
- Now, substituting the values:
\[ r_{B} = \frac{580.4 \, \text{pm}}{2} - 100 \, \text{pm} \]

5. **Calculating the Radius of Anion**:
- First, calculate \( \frac{580.4 \, \text{pm}}{2} \):
\[ \frac{580.4}{2} = 290.2 \, \text{pm} \]
- Now, substitute this value back into the equation:
\[ r_{B} = 290.2 \, \text{pm} - 100 \, \text{pm} = 190.2 \, \text{pm} \]

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

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

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