MgO crystallizes in a cubic type crystal system. The ionic radii for `Mg^(2+) and O^(2-)` are 0.066 abd 0.140nm respectively One can conclude that the `Mg^(2+)` ions occypy:

Options

A. a cubic hole in a simple structure
B. every tetrahedral hole in a close packed structure
C. an octahedral hole in a cubic packed structure
D. every other tetrahedral hole in a close paked structure

Answer

To determine where the `Mg^(2+)` ions occupy in the cubic crystal system of magnesium oxide (MgO), we will use the radius ratio rule. Here’s a step-by-step solution:

### Step 1: Identify the ionic radii
We are given the ionic radii:
- Radius of `Mg^(2+)` = 0.066 nm
- Radius of `O^(2-)` = 0.140 nm

### Step 2: Calculate the radius ratio
The radius ratio (r) is calculated using the formula:
\[ r = \frac{r_{cation}}{r_{anion}} \]
Substituting the given values:
\[ r = \frac{0.066 \, \text{nm}}{0.140 \, \text{nm}} \]

### Step 3: Perform the calculation
Calculating the radius ratio:
\[ r = \frac{0.066}{0.140} \approx 0.471 \]

### Step 4: Determine the type of void occupied
Now we will use the radius ratio to determine which type of void the `Mg^(2+)` ions will occupy:
- If \( 0.225 < r < 0.414 \), the cation occupies tetrahedral holes.
- If \( 0.414 < r < 0.732 \), the cation occupies octahedral holes.
- If \( r > 0.732 \), the cation occupies cubic holes.

Since our calculated radius ratio \( r \approx 0.471 \) falls in the range \( 0.414 < r < 0.732 \), we conclude that:
- `Mg^(2+)` ions will occupy the octahedral holes.

### Step 5: Conclusion
Thus, we conclude that the `Mg^(2+)` ions occupy the octahedral holes in the cubic crystal structure of magnesium oxide (MgO).

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