Question

A compound is formed by \(A\) (cations), \(B\) (cations) and \(O\) (anions). Atoms of \(O\) form a ccp lattice. Atoms of \(A\) occupy \(50\%\) of octahedral voids and atoms of \(B\) occupy \(25\%\) of tetrahedral voids. What is the molecular formula of the compound?

• A. \(AB_2O_4\)
• B. \(AB_2O_2\)
• C. \(ABO_3\)
• D. \(ABO_2\)

Hint:

For a ccp (fcc) lattice containing \(N\) particles: \[ \text{Octahedral voids}=N \] \[ \text{Tetrahedral voids}=2N \] Always calculate the number of ions occupying the voids and then reduce the ratio to obtain the empirical formula.

Answer 1

The Correct Option is D

Concept: If \(N\) anions form a ccp (fcc) lattice, then \[ \text{Number of octahedral voids}=N \] and \[ \text{Number of tetrahedral voids}=2N. \]

Step 1: Assume \(N\) oxide ions are present. \[ O=N \] Since oxygen forms ccp arrangement, \[ \text{Octahedral voids}=N \] \[ \text{Tetrahedral voids}=2N. \]

Step 2: Calculate number of \(A\) ions. \(A\) occupies \(50\%\) of octahedral voids. \[ A=\frac{50}{100}\times N \] \[ A=\frac{N}{2}. \]

Step 3: Calculate number of \(B\) ions. \(B\) occupies \(25\%\) of tetrahedral voids. \[ B=\frac{25}{100}\times 2N \] \[ B=\frac{N}{2}. \]

Step 4: Find the simplest ratio. \[ A:B:O = \frac{N}{2}:\frac{N}{2}:N \] Dividing by \[ \frac{N}{2}, \] \[ A:B:O = 1:1:2. \] Therefore, \[ \boxed{ABO_2} \]

Final Answer: \[ \boxed{ABO_2} \] \[ \boxed{\text{Answer = (D)}} \]