Given that the radius of `Na^(+)` ion is .0 95 Å and that of `Cl^(-)` ion is .1 81 Å, hence in the close packed lattice of `Cl^(-)` ions `Na^(+)` ions prefer to occupy.

Options

A. Tetrahedral site
B. Octahedral site
C. Cubic site
D. Trigonal site

Explanation

To determine the position of Na⁺ ions in the close-packed lattice of Cl⁻ ions, we can use the radius ratio rule. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Radius of Na⁺ ion (r⁺) = 0.95 Å
- Radius of Cl⁻ ion (r⁻) = 1.81 Å

### Step 2: Calculate the radius ratio
The radius ratio (r⁺/r⁻) is calculated as follows:
\[ \text{Radius Ratio} = \frac{r^+}{r^-} = \frac{0.95 \, \text{Å}}{1.81 \, \text{Å}} \]

### Step 3: Perform the calculation
Calculating the above expression:
\[ \text{Radius Ratio} = \frac{0.95}{1.81} \approx 0.524 \]

### Step 4: Analyze the radius ratio
The radius ratio of 0.524 falls between the critical values for different coordination numbers:
- For octahedral voids, the radius ratio range is approximately 0.414 to 0.732.

### Step 5: Conclusion
Since the radius ratio (0.524) lies within the range for octahedral coordination, we can conclude that Na⁺ ions will occupy octahedral voids in the close-packed lattice of Cl⁻ ions.

### Final Answer
Na⁺ ions prefer to occupy octahedral voids in the close-packed lattice of Cl⁻ ions.

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