Question

In a compound, oxides ions form fcc lattice, whereas $\frac{1}{4}th$ of tetrahedral voids are occupied by cations $A^{2+}$ and $\frac{1}{8}th$ of octahedral voids are occupied by cations $B^{3+}$ ions. The simplest formula of the compound is

• A. $A_{4}B{O}_8$
• B. $A{B}_2{O}_4$
• C. $A{B}_4{O}_8$
• D. $A_{2}B{O}_4$

Answer 1

The correct option is A $A_{4}B{O}_8$

In the compound , since oxide ions forms the ccp arrangements, it occupies the corners and face centres of the cubic unit cell.
$\text{No. of}{O}^{2-}\text{ions in unit cell}=N=4$
$\therefore$
$\text{Number of tetrahedral voids}=2N=8$
$\text{Number of octahedral voids}=N=4$

$A^{2+}$ ions occupies $\frac{1}{4}th$ of the tetrahedral voids
$\therefore$
$\text{Number of}{A}^{2+}\text{ions in an unit cell}=\frac{1}{4}\times 8=2$

$B^{3+}$ ions occupies $\frac{1}{8}th$ of the octahedral voids
$\therefore$
$\text{Number of}{B}^{3+}\text{ions in an unit cell}=\frac{1}{8}\times 4=\frac{1}{2}$

$\text{The ratio of}{A}^{2+}:B^{3+}:O^{2-}\text{is}=2:\frac{1}{2}:4$
Formula of compound is $A_2{B}_{\frac{1}{2}}{O}_4$
Simplest formula :
$A_{4}B{O}_8$