Solid State - Result Question 8
8. A compound of formula $A _2 B _3$ has the hep lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms ?
(2019 Main, 10 Jan II)
(a) hcp lattice- $A, \dfrac{2}{3}$ tetrahedral voids- $B$
(b) hcp lattice- $A, \dfrac{1}{3}$ tetrahedral voids- $B$
(c) hcp lattice- $B, \dfrac{1}{3}$ tetrahedral voids- $A$
(d) hcp lattice- $B, \dfrac{2}{3}$ tetrahedral voids- $A$
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Answer:
Correct Answer: 8. (c)
Solution:
- Total effective number of atoms in hcp unit lattice $=$ Number of octahedral voids in $h c p=6$
$\therefore$ Number of tetrahedral voids (TV) in hcp
$ \begin{aligned} & =2 \times \text { Number of atoms in hep lattice } \\ & =2 \times 6=12 \end{aligned} $
As, formula of the lattice is $A _2 B _3$.
| Suppose, | $A$ | $B$ |
|---|---|---|
| $\left(\dfrac{1}{3} \times TV\right)$ | (hcp) | |
| $\Rightarrow$ | $\dfrac{1}{3} \times 12$ | 6 |
| $\Rightarrow$ | $\dfrac{2}{3}$ | 1 |
| $\Rightarrow$ | 2 | 3 |
So, $A=\dfrac{1}{3}$ tetrahedral voids, $B=$ hcp lattice
BETA
