A binary solid `(A^(+)B^(-))` has a ruck salt structure. If the edge length is 400pm and radius of cation is 75pm the radius of anion is

Options

A. 100 mm
B. 125 pm
C. 250 pm
D. 325 pm

Answer

To find the radius of the anion in a binary solid with a rock salt structure, we can follow these steps:

### Step 1: Understand the Structure
In a rock salt structure (like NaCl), the cations and anions are arranged in a face-centered cubic lattice. The distance between the cation and anion is equal to half the edge length of the unit cell.

### Step 2: Given Data
- Edge length (a) = 400 pm
- Radius of cation (rₐ) = 75 pm

### Step 3: Calculate the Distance Between Cation and Anion
The distance between the cation and anion in the rock salt structure is given by:
\[ \text{Distance} = \frac{a}{2} \]
Substituting the given edge length:
\[ \text{Distance} = \frac{400 \, \text{pm}}{2} = 200 \, \text{pm} \]

### Step 4: Relate the Distances to the Radii
In the rock salt structure, the distance between the cation and anion can also be expressed as the sum of their radii:
\[ \text{Distance} = rₐ + rᵦ \]
Where:
- \( rᵦ \) is the radius of the anion.

### Step 5: Set Up the Equation
From the previous steps, we have:
\[ rₐ + rᵦ = 200 \, \text{pm} \]
Substituting the known value of the cation radius:
\[ 75 \, \text{pm} + rᵦ = 200 \, \text{pm} \]

### Step 6: Solve for the Radius of the Anion
Rearranging the equation to find \( rᵦ \):
\[ rᵦ = 200 \, \text{pm} - 75 \, \text{pm} \]
\[ rᵦ = 125 \, \text{pm} \]

### Conclusion
The radius of the anion is 125 pm.