Chemistry - Solid State Question with Solution | TestHub

In bcc structure are $\mathrm{r}++\mathrm{r}=\frac{\sqrt{3}}{2}\mathrm{a}$

Hence, for $\mathrm{CsCl},\mathrm{Rc}+\mathrm{ra}=\frac{\sqrt{3}}{2}\mathrm{a}$

Since, NaCl crystalise in fcc structure

$\therefore 2\mathrm{rC}+2\mathrm{ra}=$ edge length of the unit cell

hence, $\mathrm{rC}+\mathrm{ra}=\frac{1}{2}$

Since packing fraction of a bcc unit cell is 0.68

∴ void space $=1-0.68=0.32$

In fcc unit cell $\mathrm{PF}=74\%\therefore \mathrm{VF}=100-74=26\%$