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A compound of formula $A_{2} B_{3}$ 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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hcp lattice – B , $\frac{2}{3}$ tetrahedral voids – A

hcp lattice – A , $\frac{2}{3}$ tetrahedral voids – B

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

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

answer is D.

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Detailed Solution

Here can also be written as $A_{4} B_{6}$

Thus , hcp has six atoms so, ‘B’ forms hcp lattice and ‘A’ is present in tetrahedral void.

Total tetrahedral voids =12

Fraction of tetrahedral voids occupied by $A = \frac{4}{12} = \frac{1}{3}$

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