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(A) In a close-packed structure of oxide ions $(O^{2-})$,let the number of $O^{2-}$ ions be $N = 4$. Number of octahedral voids $= N = 4$. Number of t

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$A_3 B_4 O_3$

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(A) In a close-packed structure of oxide ions $(O^{2-})$,let the number of $O^{2-}$ ions be $N = 4$. Number of octahedral voids $= N = 4$. Number of tetrahedral voids $= 2N = 8$. $A$ occupies all octahedral voids,so number of $A$ atoms $= 4$. $B$ occupies $(2/3)$rd of tetrahedral voids,so number of $B$ atoms $= 8 	imes (2/3) = 16/3$. The ratio of $A : B : O$ is $4 : (16/3) : 4$. Multiplying by $3$ to get the simplest integer ratio: $12 : 16 : 12$. Dividing by $4$: $3 : 4 : 3$. Thus,the molecular formula is $A_3 B_4 O_3$.

In a mixed oxide of $A$ and $B$,$A$ occupies all the octahedral voids while $B$ occupies $(2/3)$rd of the tetrahedral voids. The molecular formula of this oxide is

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