In a solid ,oxide `(O^(2-))` ions are arranged in ccp, cations `(A^(3+))` occupy one -fourth of tetrahedral void and cations `(B^(3+))` occupy half of the octahedral voids . What is the formula of the compound?

Explanation

To determine the formula of the compound based on the arrangement of ions in a solid oxide, we will follow these steps:

### Step 1: Identify the arrangement of oxide ions
The oxide ions (O²⁻) are arranged in a cubic close-packed (CCP) structure. In a CCP structure, the number of oxide ions (Z) is 4.

### Step 2: Determine the number of tetrahedral and octahedral voids
In a CCP structure:
- The number of tetrahedral voids is twice the number of atoms, which is \(2Z\). Therefore, the number of tetrahedral voids is \(2 \times 4 = 8\).
- The number of octahedral voids is equal to the number of atoms, which is \(Z\). Therefore, the number of octahedral voids is \(4\).

### Step 3: Calculate the number of cations in the voids
- Cation A³⁺ occupies one-fourth of the tetrahedral voids:
\[ \text{Number of A³⁺ ions} = \frac{1}{4} \times 8 = 2 \]

- Cation B³⁺ occupies half of the octahedral voids:
\[ \text{Number of B³⁺ ions} = \frac{1}{2} \times 4 = 2 \]

### Step 4: Establish the ratio of cations to oxide ions
Now, we have:
- Number of A³⁺ ions = 2
- Number of B³⁺ ions = 2
- Number of O²⁻ ions = 4 (since Z = 4)

The ratio of A³⁺ : B³⁺ : O²⁻ is:
\[ 2 : 2 : 4 \]
This can be simplified to:
\[ 1 : 1 : 2 \]

### Step 5: Write the empirical formula
From the ratio, we can write the empirical formula of the compound as:
\[ \text{ABO}_2 \]

### Final Answer
The formula of the compound is **ABO₂**.

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