Question

Compute the percentage void space per unit volume of unit cell in zinc-fluride structure. In zinc fluoride, anions occupy

fcc positions and half of the tetrahedral holes are occupied by cations.

• A. 25%
• B. 30%
• C. 40%
• D. 20%

Answer 1

The correct option is A

25%

Since anions occupy fcc positions and half of the tetrahedral holes are occupied by cations. Since there are four anions and 8 tetrahedral holes per

unit cell, the fraction of volume occupied by spheres/unit volume of the unit cell is

$=\frac{4\times \left(\frac{4}{3}\pi r_a^3\right)+\frac{1}{2}\times 8\times \left(\frac{4}{3}\pi r_c^3\right)}{16\sqrt{2}{r}_a^3}$ = $\frac{\pi }{3\sqrt{2}}\{1+{\left(\frac{r_c}{r_a}\right)}^3\}$

$\ast \ast fortetrahedralholes,$
$=\frac{r_c}{r_a}=0.225$
$=\frac{\pi }{3\sqrt{2}}\{1+{\left(0.225\right)}^3\}=0.7496$
$\therefore \text{Void volume = 1-0.7496 = 0.2504/unit volume of unit cell \% void space = 25.04\%}$