JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

A compound of formula A₂ B₃ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms:

A
hcp lattice - B, $\frac{1}{3}$ Tetrahedral voids - A
B
hcp lattice - A, $\frac{2}{3}$ Tetrahedral voids - B
C
hcp lattice - B, $\frac{2}{3}$ Tetrahedral voids - A
D
hcp lattice - A, $\frac{1}{3}$ Tetrahedral voids - B

Solution

Total effective atoms in HCP unit cell = 6

Total no. of tetrahedral voids = 12

If B is placed at HCP lattice points then $\frac{1}{3}$ of tetrahedral voids will be occupied by A.

So the general formulae becomes A2B3.

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