In fluorite structure (CaF2), the cations (Ca2+) ion forms ccp arrangement and anions $(F^{2-}) ion occupy all tetrahedral voids.
So, along face diagonal, cations will be touching each other. If rf is radius of cations and a is the edge length of the cube, then,
√2a=4r+⇒a=2√2r+
So, a3=16√2r3+
Now, if r− is the radius of the anion. So, for tetrahedral voids, we have
Radius of cation (r+)=10.225 (For this crystal type)
Radius of anion $(r_-)
So, packing fraction =4×43×(r+)3+8×43π(r−)316√2r3+
=π3√2+2π3√2(r−r+)3
=π3√2[1+2×(0.225)3]=0.749
So, void fraction =1−0.749=0.251
Void % =25.1%