Ionic radii of `Mg^(2+)` and `O^(2-)` ions are 66pm and 140pm respectively. The type of intersitital void and coordination number of `Mg^(2+)` ion respectively are

Options

A. tetrahedral , 6
B. octahedral , 6
C. tetrahedral , 4
D. octahedral , 8

Answer

To determine the type of interstitial void and the coordination number of the `Mg^(2+)` ion, we can follow these steps:

### Step 1: Identify the ionic radii
We are given the ionic radii:
- Radius of `Mg^(2+)` = 66 pm
- Radius of `O^(2-)` = 140 pm

### Step 2: Calculate the radius ratio
The radius ratio (R) is calculated using the formula:
\[ R = \frac{R_{cation}}{R_{anion}} \]
Substituting the values:
\[ R = \frac{66 \, \text{pm}}{140 \, \text{pm}} \]
\[ R \approx 0.4714 \]

### Step 3: Determine the type of void
To determine the type of void, we refer to the radius ratio ranges:
- If \( 0.225 < R < 0.414 \), the void is tetrahedral.
- If \( 0.414 < R < 0.732 \), the void is octahedral.
- If \( R > 0.732 \), the void is cubic.

Since our calculated radius ratio \( R \approx 0.4714 \) falls within the range \( 0.414 < R < 0.732 \), we conclude that the type of void is **octahedral**.

### Step 4: Determine the coordination number
The coordination number for an octahedral void is always 6, as it is surrounded by 6 spheres of atoms.

### Final Answer
Thus, the type of interstitial void is **octahedral void** and the coordination number of `Mg^(2+)` ion is **6**.

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