In a solid, oxide ions are arranged in CCP, cations A occupy ((1/8)) text th of the tetrahedral voids and cations B occupy ((1/4)) text th of the octahedral voids. The formula of the compound is :-

Question

In a solid, oxide ions are arranged in CCP, cations A occupy ((1/8)) text th of the tetrahedral voids and cations B occupy ((1/4)) 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

Tardigrade Admin · Accepted Answer

No. of O 2-=8 × (1/8)+6 × (1/2)=4 No. of A =(1/8) × 8=1 No. of 8=(1/4) × 4=1 Formula arrow ABO 4

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

$A B O_{4}$

$A B_{2} O_{3}$

$A_{2} B O_{4}$

$A B_{4} O_{4}$

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 $\to A B O_{4}$

Q. In a solid, oxide ions are arranged in CCP, cations A occupy (81​)th of the tetrahedral voids and cations B occupy (41​)th of the octahedral voids. The formula of the compound is :-

ABO4​

AB2​O3​

A2​BO4​

AB4​O4​