A binary solid `(A^(+)B^(-))` has a rock salt structure .If the edge length is `300 p m` and radius of cation is 60 pm the radius of anion is

Options

A. 100 pm
B. 125 pm
C. 250 pm
D. 90 pm

Explanation

To solve the problem, we need to find the radius of the anion \( B^{-} \) in a binary solid \( (A^{+}B^{-}) \) with a rock salt structure. Given the edge length of the unit cell and the radius of the cation, we can use the relationship between these parameters in the rock salt structure.

### Step-by-Step Solution:

1. **Understand the Rock Salt Structure**:
- In a rock salt structure (like NaCl), the cations occupy the octahedral voids, and the anions form a face-centered cubic (FCC) lattice.

2. **Identify the Given Values**:
- Edge length of the unit cell, \( a = 300 \, \text{pm} \)
- Radius of the cation \( A^{+} \), \( R_{A} = 60 \, \text{pm} \)

3. **Set Up the Relationship**:
- In the rock salt structure, the relationship between the radius of the cation \( R_{A} \) and the radius of the anion \( R_{B} \) can be expressed as:
\[ 2R_{A} + 2R_{B} = a \]
- This is because the cation and anion touch each other along the edge of the cube.

4. **Simplify the Equation**:
- Dividing the entire equation by 2 gives:
\[ R_{A} + R_{B} = \frac{a}{2} \]
- Substituting the known values:
\[ R_{A} + R_{B} = \frac{300 \, \text{pm}}{2} = 150 \, \text{pm} \]

5. **Substitute the Radius of the Cation**:
- Now, substitute \( R_{A} = 60 \, \text{pm} \) into the equation:
\[ 60 \, \text{pm} + R_{B} = 150 \, \text{pm} \]

6. **Solve for the Radius of the Anion**:
- Rearranging gives:
\[ R_{B} = 150 \, \text{pm} - 60 \, \text{pm} = 90 \, \text{pm} \]

7. **Conclusion**:
- The radius of the anion \( B^{-} \) is \( 90 \, \text{pm} \).

### Final Answer:
The radius of the anion \( B^{-} \) is \( 90 \, \text{pm} \).

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