Element 'B' forms ccp structure and 'A' occup | vedclass

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(B) In a $ccp$ (cubic close-packed) structure,the number of atoms per unit cell is $Z = 4$.$1$. Number of $B$ atoms $= 4$.$2$. Number of octahedral vo

Vedclass · Accepted Answer

$AB_2O_4$

Vedclass · Answer

(B) In a $ccp$ (cubic close-packed) structure,the number of atoms per unit cell is $Z = 4$.$1$. Number of $B$ atoms $= 4$.$2$. Number of octahedral voids $= 4$. Since '$A$' occupies half of the octahedral voids,the number of $A$ atoms $= 4 	imes \frac{1}{2} = 2$.$3$. Number of tetrahedral voids $= 2 	imes Z = 2 	imes 4 = 8$. Since oxygen atoms occupy all the tetrahedral voids,the number of oxygen atoms $= 8$.The ratio of atoms $A:B:O$ is $2:4:8$,which simplifies to $1:2:4$.Therefore,the formula of the bimetallic oxide is $AB_2O_4$.

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

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