Question

In a solid, oxide ions are arranged in CCP, cations A occupy $\left(\frac{1}{8}\right)^{\text {th }}$ of the tetrahedral voids and cations B occupy $\left(\frac{1}{4}\right)^{\text {th }}$ of the octahedral voids. The formula of the compound is :-

• A. $ABO _{4}$
• B. $AB _{2} O _{3}$
• C. $A _{2} BO _{4}$
• D. $AB _{4} O _{4}$

Answer 1

Answer: (A)

No. of $O ^{2-}=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
No. of $A =\frac{1}{8} \times 8=1$
No. of $8=\frac{1}{4} \times 4=1$
Formula $\rightarrow ABO _{4}$