Question: Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms oc...

Question

Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:
A. ${A_2}B{O_4}$
B. ${A_2}{B_2}O$
C. ${A_4}{B_2}O$
D. $A{B_2}{O_4}$

Answer 1

Solution

In a ccp structure, there are a total eight number of atoms present at the corners of a unit cell and a total of 6 numbers of atoms occupy the center of the faces of the unit cell. Thus, overall there are a total number of 4 atoms per unit cell present. In the ccp structure, the type of close packing is ABCABC…

Complete step by step answer: In the given question, it has been said that the element ‘B’ forms a ccp structure. Now, in the case of a ccp structure, the contribution of the atoms is as follows:
Total contribution of atoms at the corners + Total contribution of atoms at the face centers = Total number of atoms
$\Rightarrow \dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 = 1 + 3 = 4atoms/unit cell$ = Total number of atoms
In a ccp lattice structure, the total number of octahedral voids is 4 and the total number of tetrahedral voids is 8. The octahedral void is formed when the atom is at the edge or at the body center. The tetrahedral void is formed at the body diagonal of the unit cell.
As it has been said that the atom ‘A’ occupies half of the octahedral voids, this means that the number of atoms of ‘A’ are:
$A = \dfrac{1}{2} \times 4 = 2$
Again, oxygen atoms occupy all the tetrahedral voids. This means that the number of oxygen atoms present in the lattice structure is:
$O = 8$
Thus, the molecular formula of the compound is ${A_2}{B_4}{O_8}$ or in more simpler terms, $A{B_2}{O_4}$ .
So, the correct option is D. $A{B_2}{O_4}$ .

Note: In ccp, one corner and its three face centers form a tetrahedral void and two tetrahedral voids are obtained along one cube diagonal. Thus, overall there are 8 tetrahedral voids in ccp. In ccp, six face centers form an octahedral void. On superimposing the two triangles on one another, an octahedral site is created. Thus, in ccp, the total number of octahedral voids is equal to four (one-fourth at each edge centre and there are a total of 12 edge centers and one at the body centre).