In a crystal lattice,anions A form an hcp arr | vedclass

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(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$. Since anions $A$ form the $hcp$ array,the number of anions $A = 6$. The number of te

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$C_4 A_3$

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(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$. Since anions $A$ form the $hcp$ array,the number of anions $A = 6$. The number of tetrahedral voids is equal to $2 	imes$ (number of atoms in the lattice) $= 2 	imes 6 = 12$. Given that $2/3$ of the tetrahedral voids are occupied by cations $C$,the number of cations $C = \frac{2}{3} 	imes 12 = 8$. The ratio of $C : A = 8 : 6$,which simplifies to $4 : 3$. Therefore,the formula of the crystal is $C_4 A_3$.

In a crystal lattice,anions $A$ form an $hcp$ array,and $2/3$ of the tetrahedral voids are occupied by cations $C$. What is the formula of the crystal?

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