A solid AB has the NaCl structure. If the radius of the cation `A^(+)` is 150 pm, calculate the maximum possible value of the radius of the anion `B^(-)`.

Explanation

To solve the problem of finding the maximum possible value of the radius of the anion \( B^{-} \) in a solid \( AB \) with the NaCl structure, follow these steps:

### Step 1: Understand the Structure
The NaCl structure consists of cations and anions arranged in a face-centered cubic lattice. In this structure, each cation is surrounded by six anions and vice versa.

### Step 2: Identify Given Information
- The radius of the cation \( A^{+} \) is given as \( r_{A^{+}} = 150 \) pm.

### Step 3: Use the Radius Ratio
For stable 6-fold coordination in ionic solids, the radius ratio \( \frac{r_{A^{+}}}{r_{B^{-}}} \) must fall within a specific range. For NaCl structure, this range is:
\[ 0.414 \leq \frac{r_{A^{+}}}{r_{B^{-}}} \leq 0.732 \]

### Step 4: Calculate Maximum Radius of Anion
To find the maximum possible radius of the anion \( B^{-} \), we will use the lower limit of the radius ratio:
\[ \frac{r_{A^{+}}}{r_{B^{-}}} = 0.414 \]

Rearranging the equation gives:
\[ r_{B^{-}} = \frac{r_{A^{+}}}{0.414} \]

Substituting the value of \( r_{A^{+}} \):
\[ r_{B^{-}} = \frac{150 \text{ pm}}{0.414} \]

### Step 5: Perform the Calculation
Now, calculate \( r_{B^{-}} \):
\[ r_{B^{-}} \approx \frac{150}{0.414} \approx 362.32 \text{ pm} \]

### Final Answer
The maximum possible value of the radius of the anion \( B^{-} \) is approximately \( 362.32 \) pm.

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