JEE PYQ: Solid State Question 14
Question 14 - 2019 (08 Apr Shift 1)
Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:
(1) $\text{A}_2\text{BO}_4$
(2) $\text{AB}_2\text{O}_4$
(3) $\text{A}_2\text{B}_2\text{O}$
(4) $\text{A}_4\text{B}_2\text{O}$
Show Answer
Answer: (2)
Solution
No. of lattice points = No. of B atoms = 4. No. of atoms of A $= \frac{1}{2} \times$ No. of Oh voids $= \frac{1}{2} \times 4 = 2$. No. of atoms of O = No. of all Td voids $= 2 \times 4 = 8$. Hence, A : B : O = 1 : 2 : 4. Therefore, the formula of the compound is $\text{AB}_2\text{O}_4$.
BETA
