JEE PYQ: Solid State Question 21
Question 21 - 2019 (10 Jan Shift 2)
A compound of formula $\text{A}_2\text{B}_3$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms:
(1) hcp lattice $-$ A, $\dfrac{2}{3}$ Tetrahedral voids $-$ B
(2) hcp lattice $-$ A, $\dfrac{1}{3}$ Tetrahedral voids $-$ B
(3) hcp lattice $-$ B, $\dfrac{2}{3}$ Tetrahedral voids $-$ A
(4) hcp lattice $-$ B, $\dfrac{1}{3}$ Tetrahedral voids $-$ A
Show Answer
Answer: (4)
Solution
Here, $\text{A}_2\text{B}_3$ can also be written as $\text{A}_4\text{B}_6$. 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 $= 4/12 = \dfrac{1}{3}$.
BETA
