First slide

Crystal lattices and unit cells

Question

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:

Moderate

A

${AB}_{2} O_{4}$
B

$A_{4} B_{2} O$
C

$A_{2} B_{2} O$
D

$A_{2} {BO}_{4}$

Solution

Number of atoms per unit cell in ccp = N =4
Number of octahedral voids = N =4
A occupies half of the octahedral voids, thus A atoms per unit cell = 4/2 =2
Tetrahedral voids = 2 × N =8
So number of oxygen atoms = 8
A:B:0:: 2:4:8
Thus, structure of bimetallic oxide is AB₂O₄ .

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