A solid `AB` has `NaCl` structure. If the radius of the cation `A` is `100` pm, what is the radius of anion `B`?

Explanation

To find the radius of anion `B` in the solid `AB` with NaCl structure, we can follow these steps:

### Step 1: Understand the NaCl Structure
In the NaCl structure, the cation (A) occupies octahedral voids in a face-centered cubic (FCC) lattice formed by the anions (B).

### Step 2: Identify the Given Data
We are given:
- Radius of cation A (rₐ) = 100 pm.

### Step 3: Use the Radius Ratio for NaCl Structure
For a stable NaCl structure, there is a specific range for the radius ratio (rₐ/rᵦ) of the cation to the anion. The radius ratio must satisfy the following condition:
\[ 0.414 \leq \frac{rₐ}{rᵦ} \leq 0.732 \]

### Step 4: Rearrange the Radius Ratio Equation
We can rearrange the equation to find the radius of the anion (rᵦ):
\[ \frac{rₐ}{rᵦ} \geq 0.414 \]
This implies:
\[ rᵦ \leq \frac{rₐ}{0.414} \]

### Step 5: Substitute the Known Values
Now, substitute the value of rₐ into the equation:
\[ rᵦ \leq \frac{100 \text{ pm}}{0.414} \]

### Step 6: Calculate the Radius of Anion B
Now, perform the calculation:
\[ rᵦ \leq 241.54 \text{ pm} \]

### Conclusion
Thus, the radius of anion B is approximately **241.54 pm**.

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