In the spinel structur, oxides ions are cubical-closet packed whereas 1/8th of tetrahedral voids are occupied by `A^(2+)` cation and 1/2 of octahedral voids are occupied by `B^(+)` cations. The general formula of the compound having spinel structure is:

Options

A. `A_(2)B_(2)O_(4)`
B. `AB_(2)O_(4)`
C. `A_(2)B_(2)O_(2)`
D. `A_(4)B_(2)O_(2)`

Answer

To derive the general formula of a compound with a spinel structure, we will follow these steps:

### Step 1: Understand the structure
In the spinel structure, oxide ions (O²⁻) form a cubic close-packed (CCP) arrangement. This means that the oxide ions are arranged in a face-centered cubic (FCC) lattice.

### Step 2: Calculate the number of oxide ions
In a face-centered cubic (FCC) unit cell:
- There are 8 corner atoms, contributing \( \frac{1}{8} \) each, giving a total contribution of \( 8 \times \frac{1}{8} = 1 \).
- There are 6 face-centered atoms, contributing \( \frac{1}{2} \) each, giving a total contribution of \( 6 \times \frac{1}{2} = 3 \).

Thus, the total number of oxide ions in one unit cell is:
\[ 1 + 3 = 4 \]
So, there are 4 oxide ions (O²⁻) in the unit cell.

### Step 3: Calculate the number of tetrahedral and octahedral voids
- The total number of tetrahedral voids in a CCP structure is \( 2 \times \) the number of atoms (or ions) in the structure. Since we have 4 oxide ions, the total number of tetrahedral voids is:
\[ 2 \times 4 = 8 \]

- The total number of octahedral voids is equal to the number of atoms (or ions) in the structure, which is:
\[ 4 \]

### Step 4: Determine the number of cations in the voids
According to the problem:
- \( \frac{1}{8} \) of the tetrahedral voids are occupied by \( A^{2+} \) cations.
- \( \frac{1}{2} \) of the octahedral voids are occupied by \( B^{+} \) cations.

Calculating the number of \( A^{2+} \) cations:
- Total tetrahedral voids = 8
- Number of \( A^{2+} \) cations = \( \frac{1}{8} \times 8 = 1 \)

Calculating the number of \( B^{+} \) cations:
- Total octahedral voids = 4
- Number of \( B^{+} \) cations = \( \frac{1}{2} \times 4 = 2 \)

### Step 5: Write the general formula
Now we can summarize the number of each type of ion:
- Number of \( A^{2+} \) cations = 1
- Number of \( B^{+} \) cations = 2
- Number of oxide ions (O²⁻) = 4

Thus, the general formula for the compound is:
\[ \text{AB}_2\text{O}_4 \]

### Conclusion
The general formula of the compound having a spinel structure is \( \text{AB}_2\text{O}_4 \).

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