Question

If the anions A form hexagonal closest packing and cations C occupy and 2/3rd of the octahedral voids in it, then the general formula for the compound is:
 A.$\text{ CA}$
B.$\text{C}{{A}_{2}} $
C.$\text{ }{{\text{C}}_{2}}{{A}_{3}}$
D.$\text{ }{{\text{C}}_{3}}{{A}_{2}}$

Answer 1

Hint:In hcp packing, number of octahedral voids are equal to the number of atoms in the closed packing structure. For hcp packing, there are 6 atoms per unit cell. The elements are found in a variety of crystal packing arrangements. The most common lattice structures for metals are those obtained by stacking the atom spheres into the most compact arrangement

Complete step by step answer:
We are given that anion A forms the hexagonal close packing and we know that there are 6 atoms per unit cell in it.
So, A= 6. Now this is also equal to the number of octahedral voids as there is one octahedral void corresponding to one atom.
So number of octahedral voids = 6.
 Cations C occupy only 2/3 of the octahedral voids. So, number of cations,
\[C=\dfrac{2}{3}\times \]octahedral voids
\[\text{= }\dfrac{2}{3}\times 6=4\]
So, ratio \[\text{C:A=4:6 = 2:3}\] the general formula of compound is \[{{C}_{2}}{{A}_{3}}\]

Thus, option C is correct.
Additional information:
 Basic concept in crystal structure is the unit cell. It is the smallest unit of volume that permits identical cells to be stacked together to fill the space. By repeating the pattern of the unit cell over and over in all directions, the entire crystal lattice can be constructed. An important characteristic of a unit cell is the number of atoms it contains. The total number of atoms in the entire crystal is the number in each cell multiplied by the number of unit cells.

Note:
Hexagonal close packed structure (hcp):- The first layer has the atoms packed into a plane- triangular lattice in which every atom has 6 immediate neighbours. This arrangement for the atom is labelled as A. For the second layer B, there is some plane triangle structure, the atoms sit in the holes formed by the first layer. The first layer has two equivalent sets of holes but the atoms of the second layer can occupy only one set. The third layer is labelled C. It can be placed over the atoms of the first layer and the structure is repeated as ABABABA……… and the hexagonal closest packed arrangement is obtained.
Also study the concepts of Crystal structures, the different types of arrangements in the structures like octahedral and tetrahedral voids and their number present in different types of structures.