In an antifluorite structure, cations occupy
A) octahedral voids
B) centre of cube
C) tetrahedral voids
D) corners of cube

Hint:Let us find the basic difference between fluorite and antifluorite structure. In these two structures, the positions of the cations and anions are reversed.

Complete answer:
There are two types of structures, fluorite structure and antifluorite structure.We will observe that the fluorite structure is formed by compounds having general formula \[{\text{M}}{{\text{X}}_2}\] . Calcium fluoride with general formula \[{\text{Ca}}{{\text{F}}_2}\] has fluorite structure. In the fluorite structure, the cations form face centered cubic (fcc) arrangement. The cations are present at eight corners of a regular cube. The cations are also present at six face centers of the regular cube. The anions are present in the tetrahedral voids. There are eight tetrahedral voids in one fcc unit cell.

In the antifluorite structure, the positions of cations and anions are reversed.
 We will observe that the antifluorite structure is formed by compounds having general formula \[{{\text{A}}_2}{\text{X}}\] . Potassium oxide, lithium oxide, sodium oxide with general formula \[{{\text{K}}_2}{\text{O, L}}{{\text{i}}_2}{\text{O, N}}{{\text{a}}_2}{\text{O}}\] respectively have antifluorite structure. In the antifluorite structure, the anions form face centered cubic (fcc) arrangement. The anions are present at eight corners of a regular cube. The anions are also present at six face centers of the regular cube. The cations are present in the tetrahedral voids. There are eight tetrahedral voids in one fcc unit cell. Thus, in an antifluorite structure, cations occupy tetrahedral voids.

Hence, the correct answer is the option (C).

Note:In the face centred cubic unit cell, each corner ion contributes one eight to the unit cell and there are eight corner ions per unit cell. Each face centered ion contributes one half to the unit cell and there are six face centered ions per unit cell. Thus, total number of anions in one unit cell of antifluorite structure is \[\left( {8 \times \dfrac{1}{8}} \right) + \left( {6 \times \dfrac{1}{2}} \right) = 1 + 3 = 4\] . The number of cations and the number of tetrahedral voids is twice or \[2 \times 4 = 8\] .