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Atom of element $\ce{B}$ forms HCP lattice and those of the element $\ce{A}$ occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements $\ce{A}$ and $\ce{B}$?

My doubt is that since the element $\ce{B}$ is in HCP, the number of octahedral and tetrahedral voids will be 6 and 12, respectively, which means that $2/3 \times 12 = 8$ tetrahedral voids occupied by $\ce{A}$.

Since $\ce{B}$ is in HCP lattice, there must be 12 voids, i.e. the ratio will be $8 : 12 \implies 2 : 3$, so $\ce{A2B3}$ must be the answer.

But according to NCERT textbook (example 1.2) the answer is $\ce{A4B3}$.

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Since B forms an HCP lattice, it's z=6 (z is the number of atoms power unit cell of the lattice) then the ratio comes out to be 8:6 that gives us A4B3.

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Let no. of atoms of B used in hcp lattice =n Number of tetrahedralvoids=2n Number of atoms of A=(2/3)2n=(4/3)n

A:B=(4/3)n(n)

formulaof the compound is A4B3

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  • $\begingroup$ NCERT answer is correct $\endgroup$ – user69960 Nov 8 '18 at 4:26

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