solid AB crystallises as NaCl structure and the radius of the cation is 0.100nm. The maximum radius of the anion can be :

Options

A. 0.137 nm
B. 0.241 nm
C. 0.274 nm
D. 0.482 nm

Explanation

To determine the maximum radius of the anion in the solid AB that crystallizes in the NaCl structure, we can follow these steps:

### Step 1: Understand the NaCl Structure
The NaCl structure consists of cations (Na⁺) and anions (Cl⁻) arranged in a face-centered cubic (FCC) lattice. The coordination number for both ions in this structure is 6.

### Step 2: Use the Radius Ratio Rule
In ionic solids, the stability of the structure is influenced by the radius ratio of the cation to the anion. For a coordination number of 6 (as in NaCl), the radius ratio (r⁺/r⁻) is approximately 0.414.

### Step 3: Identify the Given Values
From the problem, we know:
- The radius of the cation (r⁺) = 0.100 nm
- The radius ratio for coordination number 6 = 0.414

### Step 4: Set Up the Equation
Using the radius ratio formula:
\[ \frac{r^+}{r^-} = 0.414 \]
We can rearrange this to find the radius of the anion (r⁻):
\[ r^- = \frac{r^+}{0.414} \]

### Step 5: Substitute the Known Values
Substituting the value of r⁺ into the equation:
\[ r^- = \frac{0.100 \, \text{nm}}{0.414} \]

### Step 6: Calculate the Radius of the Anion
Now, performing the calculation:
\[ r^- \approx \frac{0.100}{0.414} \approx 0.241 \, \text{nm} \]

### Conclusion
The maximum radius of the anion (r⁻) can be approximately 0.241 nm.

### Final Answer
The maximum radius of the anion can be **0.241 nm**.

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