A binary solid (AB) has a rock salt structure. If the edge length is 400pm, radius of cation is 80pm the radius of anion is:

Options

A. 100pm
B. 120pm
C. 250pm
D. 325pm

Answer

To solve the problem of finding the radius of the anion in a binary solid (AB) with a rock salt structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- The rock salt structure (NaCl type) consists of cations (B) occupying the octahedral voids and anions (A) forming a face-centered cubic (FCC) lattice.

2. **Identify the Given Values**:
- Edge length of the unit cell (a) = 400 pm
- Radius of the cation (r_B) = 80 pm

3. **Relationship in the Unit Cell**:
- In the rock salt structure, the edge length (a) can be expressed in terms of the radii of the cation and anion. The relationship is given by:
\[ a = 2r_A + 2r_B \]
- Here, \( r_A \) is the radius of the anion.

4. **Rearranging the Equation**:
- We can rearrange the equation to solve for \( r_A \):
\[ r_A = \frac{a}{2} - r_B \]

5. **Substituting the Values**:
- Substitute the known values into the equation:
\[ r_A = \frac{400 \text{ pm}}{2} - 80 \text{ pm} \]
- Calculate \( \frac{400 \text{ pm}}{2} \):
\[ r_A = 200 \text{ pm} - 80 \text{ pm} \]

6. **Final Calculation**:
- Now, calculate \( r_A \):
\[ r_A = 120 \text{ pm} \]

### Conclusion:
The radius of the anion (r_A) is **120 pm**.

---