Question

In a solid, oxide ions are arranged in ccp. Cation A occupies \(1/8\)th of the tetrahedral voids and cation B occupies \(1/4\)th of the octahedral voids. What is the formula of the compound?

• A. \(\text{ABO}_{4}\)
• B. \(\text{AB}_{2}\text{O}_{3}\)
• C. \(\text{A}_{2}\text{BO}_{4}\)
• D. \(\text{AB}_{2}\text{O}_{4}\)

Hint:

A ccp lattice has 4 atoms, 8 tetrahedral voids and 4 octahedral voids per unit cell.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a ccp arrangement with \(N\) oxide ions there are \(2N\) tetrahedral voids and \(N\) octahedral voids.

Step 2: Count the Ions
Take 4 oxide ions per unit cell: 8 tetrahedral and 4 octahedral voids.
Cation A fills \(\dfrac{1}{8}\times 8=1\).
Cation B fills \(\dfrac{1}{4}\times 4=1\).

Step 3: Write the Formula
Ratio \(\text{A}:\text{B}:\text{O}=1:1:4\), so the formula is \(\text{ABO}_{4}\).

Step 4: Compare the options
This is option (A).

Final Answer:
The formula of the compound is ABO4. \[ \boxed{\text{(A) }\text{ABO}_{4}} \]