In a mineral MX_2,M^{+2} ions are arranged in | vedclass

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(A) In a $ccp$ structure,the number of $M^{+2}$ ions per unit cell is $4$. Since $X^{-}$ ions occupy the tetrahedral voids,and there are $2$ tetrahedr

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$4, 8, 8, 100\%$

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(A) In a $ccp$ structure,the number of $M^{+2}$ ions per unit cell is $4$. Since $X^{-}$ ions occupy the tetrahedral voids,and there are $2$ tetrahedral voids per atom in a $ccp$ lattice,the number of $X^{-}$ ions per unit cell is $2 	imes 4 = 8$. Thus,the number of cations $(M^{+2})$ is $4$ and the number of anions $(X^{-})$ is $8$. In this structure (fluorite-like),the coordination number of the cation $(M^{+2})$ is $8$. Since all tetrahedral voids are occupied by $X^{-}$ ions,the percentage of occupied tetrahedral voids is $100\%$. Therefore,the values are $4, 8, 8, 100\%$.

In a mineral $MX_2$,$M^{+2}$ ions are arranged in a $ccp$ structure and $X^{-}$ ions occupy the tetrahedral voids. What are the number of cations and anions per unit cell,the coordination number of the cation,and the percentage of occupied tetrahedral voids,respectively?

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