Question

Structure of a mixed oxide is cubic-close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is-
[A] ${{A}_{2}}B{{O}_{2}}$
[B] ${{A}_{2}}{{B}_{3}}{{O}_{4}}$
[C] $A{{B}_{2}}{{O}_{2}}$
[D] $AB{{O}_{2}}$

Answer 1

To solve this you need to remember the number of octahedral voids and tetrahedral voids in a ccp cell. The number of tetrahedral voids is 8 and octahedral voids is half the number of tetrahedral voids. Use this to find the formula of the oxide.

Complete answer:
We know that the term "closest packed structures" refers to the most tightly packed or space-efficient composition of crystal structures.
In the question it is given to us that the divalent metal A takes up the one-fourth of the tetrahedral voids and the monovalent metal B takes up the octahedral voids.
The void surrounded by four spheres sitting at the corners of a regular tetrahedron is called a tetrahedral void. When two tetrahedral voids from two different layers are aligned, together they form an octahedral void. This kind of void is surrounded by 6 atoms.
We already know that the number of tetrahedral and octahedral voids in CCP unit cells is 8 and 4 respectively. The number of particles in ccp is 4. Hence, the number of tetrahedral and octahedral voids is 8 and 4 respectively.
Therefore, we can write that the number of oxide ions i.e. ${{O}^{2-}}$ is 4.
A occupies one-fourth of the tetrahedral voids i.e. $\frac{1}{4}\times 8$ = 2
And B takes up the octahedral voids i.e. 4.
Therefore, the ratio A: B: O = 2:4:4 = 1:2:2.
Therefore A = 1, B = 2 and O = 2.
So, the formula of the oxide is $A{{B}_{2}}{{O}_{2}}$

Therefore, the correct answer is option [C] $A{{B}_{2}}{{O}_{2}}$ .

Note:
Face Centered Cubic i.e. fcc and cubic close packed i.e. ccp, these are two different names for the same lattice. We can think of this cell as being made by inserting another atom into each face of the simple cubic lattice - hence the "face-centred cubic" name.
If we divide an FCC unit cell into 8 small cubes, then each small cube has 1 tetrahedral void at its own body centre. Thus, there are 8 tetrahedral voids in total in one unit cell.