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JEE Main 202118 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 18 March
Shift: Morning Shift
Subject: Chemistry
Chapter: Solid State

JEE Main 2021 · 18 March · Morning Shift · Chemistry · Solid State

In a binary compound, atoms of element A form a hcp structure and...

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:

Options

A ${M}_{2}{A}_{3}$
B ${M}_{4}{A}_{3}$
C ${M}_{4}A$
D ${MA}_{3}$

Correct answer

B. ${M}_{4}{A}_{3}$

Step-by-step solution

In HCP unit cell number of effective number of atoms is $6$ ,

so $A = 6$ .

Number of tetrahedral voids

= 2 x number of atoms forming the lattice

= $2X6$

${M}_{12 \times \dfrac{2}{3}}{A}_{6}$
${M}_{8}{A}_{6}$
${M}_{4}{A}_{3}$

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