Question

In a cubic lattice, the closed packed structure of mixed oxides of the lattice is made up of oxide ions, one eight of the tetrahedral voids by divalent ions $\left( {{A^{2 + }}} \right)$ while one half of the octahedral voids are occupied by trivalent ions $\left( {{B^{3 + }}} \right)$. What is the formula of the oxides?

Answer 1

Hint: We can calculate the formula of the oxides by taking the ratios of A,B, and O. The ratio of A,B, and O are calculated using the number of tetrahedral and octahedral voids per oxide ion present in the lattice.

Complete step by step answer:
Given data contains,
One eight of the tetrahedral voids by divalent ions $\left( {{A^{2 + }}} \right)$.
One half of the octahedral voids are made up by trivalent ions $\left( {{B^{3 + }}} \right)$.
The lattice contains compounds of oxide ion. We can now calculate the number of oxide ions in the lattice.
Oxide ions in the lattice=$\dfrac{8}{8} = 1$
The number of oxide ions present in the lattice is one.
Let us now calculate the number of ions of A. The number of ions of A is calculated by multiplying 1/8 with the number of tetrahedral voids and number of oxide ions.
The number of tetrahedral voids is two.
Number of ions of A$ = \dfrac{1}{8} \times {\text{Number of tetrahedral voids}} \times {\text{Number of oxide ion in lattice}}$
$\Rightarrow$ Number of ions of A = $\dfrac{1}{8} \times 2 \times 1$
$\Rightarrow$ Number of ions of A = $\dfrac{1}{4}$
$\therefore$ the number of ions of A is $\dfrac{1}{4}$.
Let us now calculate the number of ions of B. The number of ions of B is calculated by multiplying 1/2 with the number of octahedral voids and number of oxide ions.
The number of octahedral voids is one.
Number of ions of B$ = \dfrac{1}{2} \times {\text{Number of octahedral voids}} \times {\text{Number of oxide ion in lattice}}$
$\Rightarrow$ Number of ions of B = $\dfrac{1}{2} \times 1 \times 1$
$\Rightarrow$Number of ions of B = $\dfrac{1}{2}$
$\therefore$ the number of ions of B is $\dfrac{1}{2}$.
Let us now take the ratios of A,B, and O
$A:B:O = \dfrac{1}{4}:\dfrac{1}{2}:1$
Now the formula of the compound will be ${A_{1/4}}{B_{1/2}}O$.
Let us take the whole number ratio to get the formula of the oxide.
$A:B:O = 1:2:4$

$\therefore$ The formula of the oxide is $A{B_2}{O_4}$.

Note:
We must know about that cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic, and rhombohedral are seven simple crystal structures. They vary in their arrangement of crystallographic axes and angles. We can relate hexagonal closed packed systems to face-centered cubic cells. An example of a compound that has face-centered cubic lattice is sodium chloride. Examples of other compounds that have face-centered cubic structures are lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc.