The ionic radii of `Rb^(+)` and `I^(-)` are 1.46 and `2.16Å`. The most probable type of structure exhibited by it is:

Options

A. CsCl type
B. NaCl type
C. ZnS type
D. `CaF_2` type

Answer

To determine the most probable type of structure exhibited by the ionic compound formed from `Rb^(+)` and `I^(-)`, we can follow these steps:

### Step 1: Identify the Ionic Radii
The ionic radii given are:
- Radius of `Rb^(+)` (Rubidium cation) = 1.46 Å
- Radius of `I^(-)` (Iodine anion) = 2.16 Å

### Step 2: Calculate the Radius Ratio
The radius ratio is calculated using the formula:
\[ \text{Radius Ratio} = \frac{\text{Radius of Cation}}{\text{Radius of Anion}} = \frac{1.46}{2.16} \]
Calculating this gives:
\[ \text{Radius Ratio} = 0.676 \]

### Step 3: Determine Coordination Number
Based on the radius ratio, we can determine the coordination number. The range for coordination number 6 (octahedral voids) is between 0.414 and 0.732. Since our calculated radius ratio (0.676) falls within this range, we conclude:
- Coordination Number = 6
- Type of voids = Octahedral

### Step 4: Identify the Structure Type
Now, we need to identify which of the given structure types corresponds to a coordination number of 6 and octahedral voids:
1. **CaCl2 type structure**: Coordination number is 8:8 (not suitable).
2. **NaCl type structure**: Coordination number is 6:6 (suitable).
3. **ZnS type structure**: Coordination number is 4:4 (not suitable).
4. **CaF2 type structure**: Coordination number is 8:4 (not suitable).

### Conclusion
The most probable type of structure exhibited by the ionic compound formed from `Rb^(+)` and `I^(-)` is the **NaCl type structure**.

### Final Answer
The correct option is **NaCl type structure**.

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