` M^(+)X^(–)` has fcc structure with octahedral voids. Anion has radius of 300 pm. The ideal radius of cation will be :

Options

A. `0.414xx300`
B. `0732xx300`
C. `0.155xx300`
D. `0.225xx300`

Answer

To find the ideal radius of the cation \( M^{+} \) in the compound \( M^{+}X^{-} \) with a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- The compound \( M^{+}X^{-} \) has a face-centered cubic (FCC) structure. In this structure, the anions (X) occupy the face-centered positions and the octahedral voids.

2. **Identify the Given Data**:
- The radius of the anion \( r_{-} \) is given as 300 pm (picometers).

3. **Determine the Radius Ratio**:
- For an FCC structure, the radius ratio of the cation to the anion in octahedral voids is given by:
\[ \frac{r_{+}}{r_{-}} = 0.732 \]
- This means that the radius of the cation \( r_{+} \) can be expressed in terms of the radius of the anion \( r_{-} \).

4. **Calculate the Radius of the Cation**:
- Rearranging the formula gives us:
\[ r_{+} = 0.732 \times r_{-} \]
- Substituting the value of \( r_{-} \):
\[ r_{+} = 0.732 \times 300 \, \text{pm} \]
- Performing the multiplication:
\[ r_{+} = 219.6 \, \text{pm} \]

5. **Conclusion**:
- The ideal radius of the cation \( M^{+} \) is approximately 219.6 pm.

### Final Answer:
The ideal radius of the cation \( M^{+} \) is approximately **220 pm** (rounded to three significant figures).