In a solid compound,the oxide ions are arrang | vedclass

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(A) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.The number of tetrahedral voids is $2N = 2 	imes 4 = 8$.The number of oct

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$ABO_3$

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(A) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.The number of tetrahedral voids is $2N = 2 	imes 4 = 8$.The number of octahedral voids is $N = 4$.Cations $A$ occupy $1/6^{th}$ of the tetrahedral voids: $A = \frac{1}{6} 	imes 8 = \frac{4}{3}$.Cations $B$ occupy $1/3^{rd}$ of the octahedral voids: $B = \frac{1}{3} 	imes 4 = \frac{4}{3}$.The ratio of $A : B : O$ is $\frac{4}{3} : \frac{4}{3} : 4$.Multiplying by $3/4$,we get the ratio $1 : 1 : 3$.Therefore,the formula of the compound is $ABO_3$.

In a solid compound,the oxide ions are arranged in $CCP$. Cations $A$ occupy $1/6^{th}$ of the tetrahedral voids and cations $B$ occupy one third of the octahedral voids. The probable formula of the compound is

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