Question

Find the void space in fluorite structure per unit volume of unit cell.

Answer 1

In fluorite structure $\left(Ca{F}_2\right)$, the cations $\left(C{a}^{2+}\right)$ ion forms ccp arrangement and anions $(F^{2-}) ion occupy all tetrahedral voids.

So, along face diagonal, cations will be touching each other. If $r_f$ is radius of cations and $a$ is the edge length of the cube, then,

$\sqrt{2}a=4r_{+}\Rightarrow a=2\sqrt{2}{r}_{+}$

So, $a^3=16\sqrt{2}{r}_{+}^3$

Now, if $r_{-}$ is the radius of the anion. So, for tetrahedral voids, we have

Radius of cation $\left(r_{+}\right)=\frac{1}{0.225}$ (For this crystal type)

Radius of anion $(r_-)

So, packing fraction $=\frac{4\times \frac{4}{3}\times (r_{+}{)}^3+8\times \frac{4}{3}\pi (r_{-}{)}^3}{16\sqrt{2}{r}_{+}^3}$

$=\frac{\pi }{3\sqrt{2}}+\frac{2\pi }{3\sqrt{2}}{\left(\frac{r_{-}}{r_{+}}\right)}^3$

$=\frac{\pi }{3\sqrt{2}}[1+2\times (0.225{)}^3]=0.749$

So, void fraction $=1-0.749=0.251$

Void % $=25.1$%