A compound formed by Mg, Al and O, is found to have cubic close array of oxide ions in which `Mg^(2+)` occupying `(1^(th))/(8)` of tetrahedral voids and `Al^(3+)` ions occupying `1/2` of the octahedral voids. The formula for the compound is:

Options

A. `MgAlO`
B. `MgAl_4O_2`
C. `Mg_2AL_3O_2`
D. `MgAl_2O_4`

Explanation

To find the formula of the compound formed by Mg, Al, and O, we need to analyze the information given about the positions of the ions in the cubic close-packed structure of oxide ions.

### Step-by-Step Solution:

1. **Identify the Number of Oxide Ions:**
In a cubic close-packed (ccp) structure, there are 4 oxide ions (O^2-) per unit cell.

2. **Determine the Number of Octahedral and Tetrahedral Voids:**
- The number of octahedral voids in a ccp structure is equal to the number of atoms, which is 4 in this case.
- The number of tetrahedral voids is twice the number of atoms, which gives us 8 tetrahedral voids.

3. **Calculate the Number of Mg^2+ Ions:**
- It is given that Mg^2+ occupies \( \frac{1}{8} \) of the tetrahedral voids.
- Since there are 8 tetrahedral voids, the number of Mg^2+ ions is:
\[ \text{Number of Mg}^{2+} = \frac{1}{8} \times 8 = 1 \]

4. **Calculate the Number of Al^3+ Ions:**
- It is given that Al^3+ occupies \( \frac{1}{2} \) of the octahedral voids.
- Since there are 4 octahedral voids, the number of Al^3+ ions is:
\[ \text{Number of Al}^{3+} = \frac{1}{2} \times 4 = 2 \]

5. **Determine the Overall Charge Balance:**
- The total positive charge contributed by the ions is:
\[ \text{Charge from Mg}^{2+} = 1 \times (+2) = +2 \]
\[ \text{Charge from Al}^{3+} = 2 \times (+3) = +6 \]
\[ \text{Total positive charge} = +2 + +6 = +8 \]
- The total negative charge from the oxide ions (O^2-) is:
\[ \text{Charge from O}^{2-} = 4 \times (-2) = -8 \]

6. **Write the Formula:**
- The overall charge is balanced (total positive charge = total negative charge).
- The formula of the compound can be written as:
\[ \text{Mg}^{2+} \text{Al}_2^{3+} \text{O}_4^{2-} \]
- Simplifying this gives the empirical formula:
\[ \text{MgAl}_2\text{O}_4 \]

### Final Answer:
The formula for the compound is **MgAl₂O₄**.