An ionic solid MX crystallises in the rock-salt structure: the anions form a cubic close-packed lattice, and the cations occupy all the octahedral…

Let the anion radius be $r_-$ and the cation radius $r_+$. Anions in a cubic close-packed lattice touch along the face diagonal, so $\sqrt2\,a=4r_-$ and $a=2\sqrt2\,r_-$. The cation sits at the centre of an octahedral void along the cell edge, where it touches two anions, so $2(r_++r_-)=a$, which gives $r_+/r_-=\sqrt2-1=0.414$. This is the limiting radius ratio for octahedral coordination. The unit cell contains $4$ anions and $4$ cations. The occupied volume is $4\times\tfrac43\pi(r_+^3+r_-^3)=\tfrac{16\pi}{3}r_-^3\,[1+(\sqrt2-1)^3]$ and the cell volume is $a^3=16\sqrt2\,r_-^3$. Packing fraction $=\dfrac{\pi\,[1+0.07107]}{3\sqrt2}=0.7931$, i.e. $79.31\%$. Counting only the anions would give the familiar $74\%$, so the filled octahedral voids raise the packing efficiency by about five percentage points.