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(B) In an $hcp$ lattice,the number of octahedral voids is equal to the number of atoms forming the lattice.Let the number of $Y$ atoms be $N$.Then,the

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$2/3$

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(B) In an $hcp$ lattice,the number of octahedral voids is equal to the number of atoms forming the lattice.Let the number of $Y$ atoms be $N$.Then,the number of octahedral voids is also $N$.The formula of the compound is $XY_3$,which means for every $1$ atom of $Y$,there are $3$ atoms of $X$ occupying the octahedral voids.However,in an $hcp$ lattice,there is only $1$ octahedral void per atom of $Y$.This implies that the stoichiometry $XY_3$ is impossible for a standard $hcp$ lattice where $X$ occupies octahedral voids,as there are only $N$ octahedral voids available for $N$ atoms of $Y$.Assuming the question implies that $X$ occupies a fraction of the available octahedral voids,and given the formula $XY_3$ is provided as a premise,the fraction of occupied voids is $3$ per $1$ atom of $Y$,which is physically impossible in a simple $hcp$ lattice. If we interpret the question as: $N$ atoms of $Y$ provide $N$ octahedral voids,and $X$ occupies $1/3$ of the available voids to result in $XY_{1/3}$ or similar,the provided formula $XY_3$ suggests an error in the problem statement. Assuming the intended question was $X_{1/3}Y$,the fraction occupied is $1/3$,and the fraction unoccupied is $1 - 1/3 = 2/3$.

$A$ compound is formed by $X$ and $Y$ elements. Atoms of $Y$ (anions) form $hcp$ lattice. Atoms of $X$ (cations) are in some octahedral holes. The formula of the compound is $XY_3$. What is the fraction of octahedral holes unoccupied by $X$?

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