An ionic compound AB has fluorite type structures. If the radius `B^(-) is 200`pm, then the ideal radius of `A^(+)` would be:

Options

A. (a) 82.8pm
B. (b) 146.4pm
C. (c) 40pm
D. (d) 45pm

Answer

To solve the problem, we need to determine the ideal radius of the cation \( A^{+} \) given the radius of the anion \( B^{-} \) in an ionic compound \( AB \) with a fluorite-type structure.

### Step-by-Step Solution:

1. **Understand the Structure**:
- The fluorite structure is characterized by the formula \( AB_2 \), where the cation \( A^{+} \) is in a cubic close-packed (CCP) arrangement and the anions \( B^{-} \) occupy the tetrahedral voids.

2. **Identify the Radius Ratio**:
- For a fluorite structure, the radius ratio \( r_A/r_B \) (where \( r_A \) is the radius of the cation \( A^{+} \) and \( r_B \) is the radius of the anion \( B^{-} \)) should fall within the range of 0.225 to 0.414 for stable structures.

3. **Use the Lower Limit of the Radius Ratio**:
- To find the ideal radius of the cation, we will use the lower limit of the radius ratio, which is approximately 0.225.

4. **Given Data**:
- The radius of the anion \( B^{-} \) is given as \( 200 \) pm.

5. **Set Up the Equation**:
- We can express the relationship using the radius ratio:
\[ \frac{r_A}{r_B} = 0.225 \]
- Rearranging gives:
\[ r_A = 0.225 \times r_B \]

6. **Substitute the Value of \( r_B \)**:
- Substitute \( r_B = 200 \) pm into the equation:
\[ r_A = 0.225 \times 200 \text{ pm} \]

7. **Calculate \( r_A \)**:
- Performing the multiplication:
\[ r_A = 0.225 \times 200 = 45 \text{ pm} \]

8. **Conclusion**:
- The ideal radius of the cation \( A^{+} \) is \( 45 \) pm.

### Final Answer:
The ideal radius of \( A^{+} \) is **45 pm**.

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