Question

I a crystal, oxide ions are arranged at fcc and A²⁺ions occupy 1/8th of the tetrahedral voids and ions B³⁺ occupy 1/2 of the octahedral voids. Calculate the packing fraction of the crystal if oxide ions of the crystals are removed from alternate corners and A²⁺ions are placed at 2 of the corners.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{If}\mathrm{alternate}\mathrm{corner}\mathrm{atoms}\mathrm{are}\mathrm{removed} \\ \mathrm{Number}\mathrm{of}\mathrm{O}\mathrm{ions}=6\times \frac{1}{2}+4\times \frac{1}{8}=\frac{7}{2} \\ \mathrm{One}\mathrm{eighth}\mathrm{of}\mathrm{the}\mathrm{tetrahedral}\mathrm{voids}\mathrm{are}\mathrm{occupied}\mathrm{by}\mathrm{divalent}\mathrm{cations}\left({\mathrm{A}}^{2+}\right) \\ \mathrm{Therefore},\mathrm{Number}\mathrm{of}\mathrm{divalent}\mathrm{cations}\mathrm{in}\mathrm{tetrahedral}\mathrm{voids}=2\times \frac{1}{8}+\frac{1}{8}\times 8=\frac{1}{4}+1=\frac{5}{4} \\ \mathrm{Half}\mathrm{of}\mathrm{the}\mathrm{octahedral}\mathrm{voids}\mathrm{are}\mathrm{occupied}\mathrm{by}\mathrm{trivalent}\mathrm{cations}=4\times \frac{1}{2}=2 \\ \mathrm{Packing}\mathrm{fraction}=\frac{7}{2}\times \frac{4}{3}{\mathrm{\pi r}}^3+\frac{5}{4}\times \frac{4}{3}{{\mathrm{\pi r}}^3}_{{\mathrm{A}}^{2+}}+2\times \frac{4}{3}{{\mathrm{\pi r}}^3}_{{\mathrm{B}}^{3+}}...\left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{2}} \\ \frac{{\mathrm{r}}_{{\mathrm{A}}^{2+}}}{{\mathrm{r}}^{-}}=0.225 \\ \mathrm{Therefore},{\mathrm{r}}_{{\mathrm{A}}^{2+}}=0.225{\mathrm{r}}^{-} \\ \frac{{\mathrm{r}}_{{\mathrm{B}}^{3+}}}{{\mathrm{r}}^{-}}=0.414 \\ \mathrm{Therefore},{\mathrm{r}}_{{\mathrm{B}}^{3+}}=0.414{\mathrm{r}}^{-} \\ \mathrm{Putting}\mathrm{all}\mathrm{the}\mathrm{values}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ \mathrm{Packing}\mathrm{fraction}=0.676 \end{gathered}$$