A compound of formula A_2B_3 has the hcp latt | vedclass

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Question

(A) In an $hcp$ lattice,the number of tetrahedral voids is twice the number of atoms forming the lattice.Let the number of $B$ atoms forming the $hcp$

Vedclass · Accepted Answer

$hcp$ lattice $- B$,$\frac{1}{3}$ tetrahedral voids $- A$

Vedclass · Answer

(A) In an $hcp$ lattice,the number of tetrahedral voids is twice the number of atoms forming the lattice.Let the number of $B$ atoms forming the $hcp$ lattice be $N$.Then,the number of tetrahedral voids $= 2N$.Given the formula is $A_2B_3$,the ratio of atoms is $A:B = 2:3$.If $B$ forms the $hcp$ lattice ($N$ atoms),then $A$ atoms occupy the tetrahedral voids.Fraction of tetrahedral voids occupied by $A = \frac{	ext{Number of } A 	ext{ atoms}}{	ext{Number of tetrahedral voids}} = \frac{2N/3}{2N} = \frac{1}{3}$.Thus,$B$ forms the $hcp$ lattice and $A$ occupies $\frac{1}{3}$ of the tetrahedral voids.

$A$ compound of formula $A_2B_3$ has the $hcp$ lattice. Which atom forms the $hcp$ lattice and what fraction of tetrahedral voids is occupied by the other atoms?

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