JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Solid State

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In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:

(A) $M_{2} A_{3}$
(B) $MA_{3}$
(C) $M_{4} A$
(D) $M_{4} A_{3}$ ✓ Correct

Answer: (D)

Solution

Take

n

atoms of A in the hcp lattice. An hcp array of

n

atoms has

2 n

tetrahedral voids, and M fills two-thirds of them, so the number of M atoms is

\frac{2}{3} \times 2 n = \frac{4 n}{3}

. The ratio is therefore

M : A = \frac{4 n}{3} : n = 4 : 3

so the binary compound is

M_{4} A_{3}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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