In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,
R (radius of tetrahedral void) = 0.225 R
In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,
R (radius of octahedral void = 0.414 R).
In the spinel structure, oxide ions are cubic close packed whereas 1/8th of tetrahedral voids are occupied by `A^(2+)` cations and 1/2 of octahedral voids are occupied by `B^(3+)` 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_(4)O_(2)`
D. `A_(4)B_(2)O_(2)`

Answer

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

### Step 1: Determine the number of oxide ions in the unit cell
In a cubic close-packed (CCP) structure, which is the same as face-centered cubic (FCC), the number of atoms (or ions) per unit cell (Z) can be calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.

Calculating the contributions:
\[ Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]
Thus, the number of oxide ions (O\(^{2-}\)) per unit cell is 4.

### Step 2: Determine the number of tetrahedral voids
In a close-packed structure, the number of tetrahedral voids is twice the number of atoms in the unit cell:
\[ \text{Number of tetrahedral voids} = 2 \times Z = 2 \times 4 = 8 \]
Since \( \frac{1}{8} \) of the tetrahedral voids are occupied by \( A^{2+} \) cations:
\[ \text{Number of } A^{2+} \text{ cations} = \frac{1}{8} \times 8 = 1 \]

### Step 3: Determine the number of octahedral voids
The number of octahedral voids is equal to the number of atoms in the unit cell:
\[ \text{Number of octahedral voids} = Z = 4 \]
Since \( \frac{1}{2} \) of the octahedral voids are occupied by \( B^{3+} \) cations:
\[ \text{Number of } B^{3+} \text{ cations} = \frac{1}{2} \times 4 = 2 \]

### Step 4: Write the general formula
Now, we can combine the contributions of the ions to write the general formula:
- Number of \( A^{2+} \) cations = 1
- Number of \( B^{3+} \) cations = 2
- Number of oxide ions = 4

Thus, the general formula of the compound is:
\[ \text{General formula} = A_1B_2O_4 \text{ or } AB_2O_4 \]

### Final Answer
The general formula of the compound having a spinel structure is \( AB_2O_4 \).

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