In a solid,oxide ions are arranged in HCP. Ca | vedclass

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(C) Let the number of oxide ions $(O^{2-})$ in the $HCP$ lattice be $N = 6$.The number of $OHV$ (Octahedral Voids) is equal to the number of ions,so $

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$ABO_3$

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(C) Let the number of oxide ions $(O^{2-})$ in the $HCP$ lattice be $N = 6$.The number of $OHV$ (Octahedral Voids) is equal to the number of ions,so $OHV = 6$.The number of $THV$ (Tetrahedral Voids) is twice the number of ions,so $THV = 2 	imes 6 = 12$.Cations $A$ occupy $\frac{1}{6}$ of $THV = \frac{1}{6} 	imes 12 = 2$.Cations $B$ occupy $\frac{1}{3}$ of $OHV = \frac{1}{3} 	imes 6 = 2$.The ratio of atoms $A : B : O$ is $2 : 2 : 6$,which simplifies to $1 : 1 : 3$.Therefore,the empirical formula is $ABO_3$.

In a solid,oxide ions are arranged in $HCP$. Cations $A$ occupy one-sixth of $THV$ and cations $B$ occupy one-third of the $OHV$. The empirical formula of the compound is

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