An ionic solid is HCP of B^{2-} ions and A^{x | vedclass

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(D) In an $HCP$ lattice,the number of $B^{2-}$ ions per unit cell is $N = 6$. The number of tetrahedral voids is $2N = 2 	imes 6 = 12$. The $A^{x+}$

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$4$

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(D) In an $HCP$ lattice,the number of $B^{2-}$ ions per unit cell is $N = 6$. The number of tetrahedral voids is $2N = 2 	imes 6 = 12$. The $A^{x+}$ ions occupy one-fourth of the tetrahedral voids,so the number of $A^{x+}$ ions is $\frac{1}{4} 	imes 12 = 3$. The formula of the compound is $A_3B_6$,which simplifies to $AB_2$. For the compound to be electrically neutral,the total positive charge must equal the total negative charge: $3 	imes (+x) + 6 	imes (-2) = 0$. $3x - 12 = 0$,which gives $3x = 12$,so $x = 4$.

An ionic solid is $HCP$ of $B^{2-}$ ions and $A^{x+}$ ions are in one-fourth of tetrahedral voids. The value of $x$ should be

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