The structure of a mixed oxide is cubic close | vedclass

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(D) In a $ccp$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.For a $ccp$ structure,the number of tetrahedral voids is $2 	imes 4 =

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$AB_2O_2$

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(D) In a $ccp$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.For a $ccp$ structure,the number of tetrahedral voids is $2 	imes 4 = 8$ and the number of octahedral voids is $4$.Given that $A$ ions occupy $\frac{1}{4}$ of the tetrahedral voids,the number of $A$ ions $= \frac{1}{4} 	imes 8 = 2$.Given that $B$ ions occupy all the octahedral voids,the number of $B$ ions $= 4$.The ratio of $A : B : O$ is $2 : 4 : 4$,which simplifies to $1 : 2 : 2$.Therefore,the formula of the oxide is $AB_2O_2$.

The structure of a mixed oxide is cubic close packed $(ccp)$. The cubic unit cell of the mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal $A$ and the octahedral voids are occupied by a monovalent metal $B$. The formula of the oxide is:

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