Question

In a cubic closed packed structure of mixed oxides, the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions $\left(A^{2+}\right)$ while one half of the octahedral voids occupied trivalent ions$\left(B^{3+}\right)$. What is the formula of the oxide?

• A. $A{B}_2{O}_3$
• B. $A_{2}B{O}_4$
• C. $A{B}_2{O}_4$
• D. None of the above

Answer 1

The correct option is C $A{B}_2{O}_4$

Number of oxide ions $=4$ (because of FCC lattice)
Number of ions A $=\frac{1}{8}\times 8=1$
Number of ions B $=\frac{1}{2}\times 4=2$
Therefore the formula of the compound is $A{B}_2{O}_4$