In spinel, `Mg^(2+)` is present in one-eighth of `TVs` in an fcc lattice of oxide ions and `Al^(3+)` ions are present in half of the `OVs`. The Formula of spinel is

Options

A. `MgAl_(3)O_(3)`
B. `MgAl_(2)O_(3)`
C. `MgAl_(2)O_(4)`
D. `MgAl_(3)O_(4)`

Answer

To determine the formula of spinel based on the information provided, we can follow these steps:

### Step 1: Understand the FCC Lattice of Oxide Ions
In a face-centered cubic (FCC) lattice, oxide ions (O²⁻) are arranged. The number of oxide ions in an FCC lattice can be calculated as follows:
- There are 8 corner atoms, each contributing 1/8 to the unit cell.
- There are 6 face-centered atoms, each contributing 1/2 to the unit cell.

Calculating the total:
\[ \text{Total O²⁻ ions} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 oxide ions in the unit cell.

### Step 2: Calculate the Number of Tetrahedral and Octahedral Voids
In an FCC lattice:
- The number of octahedral voids is equal to the number of atoms in the lattice, which is 4 (the same as the number of oxide ions).
- The number of tetrahedral voids is twice the number of atoms in the lattice, which is \(2 \times 4 = 8\).

### Step 3: Determine the Number of Mg²⁺ and Al³⁺ Ions
According to the question:
- Mg²⁺ ions occupy one-eighth of the tetrahedral voids.
- The number of tetrahedral voids is 8, so the number of Mg²⁺ ions is:
\[ \text{Number of Mg²⁺} = \frac{1}{8} \times 8 = 1 \]

- Al³⁺ ions occupy half of the octahedral voids.
- The number of octahedral voids is 4, so the number of Al³⁺ ions is:
\[ \text{Number of Al³⁺} = \frac{1}{2} \times 4 = 2 \]

### Step 4: Write the Formula of Spinel
Now we can summarize the ions present:
- 1 Mg²⁺ ion
- 2 Al³⁺ ions
- 4 O²⁻ ions

The formula can be written as:
\[ \text{Formula} = \text{Mg}^{2+} \text{Al}_{2}^{3+} \text{O}_{4}^{2-} \]
This simplifies to:
\[ \text{MgAl}_2\text{O}_4 \]

### Final Answer
The formula of spinel is:
\[ \text{MgAl}_2\text{O}_4 \]