JEE PYQ: Solid State Question 4
Question 4 - 2021 (18 Mar Shift 1)
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:
(1) $\text{M}_2\text{A}_3$
(2) $\text{M}_4\text{A}_3$
(3) $\text{M}_4\text{A}$
(4) $\text{MA}_3$
Official Ans. by NTA (2)
Show Answer
Answer: (2)
Solution
In hcp, for 6 atoms of A (one unit cell of hcp has 6 atoms), there are $12 \times \frac{2}{3} = 8$ atoms of M occupying tetrahedral voids. So formula: $\text{M}_8\text{A}_6 = \text{M}_4\text{A}_3$.
BETA
