A solid AB has rock salt structure. If the edge length is 520 pm and radius of `A^(+)` is 80 pm, the radius of anion `B^(–)` would be -

Options

A. 440 pm
B. 220 pm
C. 360 pm
D. 180 pm

Answer

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

### Step 1: Understand the Rock Salt Structure
In a rock salt structure, cations (\( A^{+} \)) occupy the octahedral voids, while anions (\( B^{-} \)) are located at the corners and face centers of the cubic lattice.

### Step 2: Identify Given Values
- Edge length \( a = 520 \) pm
- Radius of cation \( r_{A^{+}} = 80 \) pm

### Step 3: Relate Edge Length to Ionic Radii
In the rock salt structure, the edge length of the cube can be expressed in terms of the radii of the cation and anion. The relationship is given by:
\[ a = 2r_{A^{+}} + 2r_{B^{-}} \]
This equation accounts for the diameter of the cation and the diameter of the anion along the edge of the cube.

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 520 = 2(80) + 2r_{B^{-}} \]

### Step 5: Simplify the Equation
Calculate \( 2(80) \):
\[ 520 = 160 + 2r_{B^{-}} \]

### Step 6: Isolate \( r_{B^{-}} \)
Subtract 160 from both sides:
\[ 520 - 160 = 2r_{B^{-}} \]
\[ 360 = 2r_{B^{-}} \]

### Step 7: Solve for \( r_{B^{-}} \)
Divide both sides by 2:
\[ r_{B^{-}} = \frac{360}{2} = 180 \text{ pm} \]

### Conclusion
The radius of the anion \( B^{-} \) is \( 180 \) pm.

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