Question

Select the right expression for determining the Packing fraction (PF) of NaCl unit cell (assume an ideal crystal), if the ions along an edge diagonal are absent: (Where $r_{+}$ represents the radius of the cation and $r_{-}$ represents the radius of the anion)

• A. $PF=\frac{d\frac{4}{3}[r_{+}^3+r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$
• B. $PF=\frac{\frac{4}{3}\pi [\frac{5}{2}{r}_{+}^3+4r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$
• C. $PF=\frac{\frac{4}{3}\pi [\frac{5}{2}{r}_{+}^3+r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$
• D. $PF=\frac{\frac{4}{3}\pi [\frac{7}{2}{r}_{+}^3+r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$

Answer 1

The correct option is B $PF=\frac{\frac{4}{3}\pi [\frac{5}{2}{r}_{+}^3+4r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$

An edge diagonal runs from the middle of one edge - from the edge centre to the diagonally opposite edge centre.
$N{a}^{+}$ ions occupy the edge centres and $C{l}^{-}$ ions occupy the FCC lattice.
Removing atoms along an edge diagonal, means that two edge centres and one body centre get removed. So, out of 4, $2\times \frac{1}{4}$ (edge centres) and $1$ (body centre) are not present anymore, giving us $4-\frac{3}{2}=\frac{5}{2}N{a}^{+}$ atoms.
$N{a}^{+}=5/2,C{l}^{-}=4$
For FCC, $2\sqrt{2}a=4R$ and $R=r_{-}$ here.
Volume of unit cell = $a^3=16\sqrt{2}{r}_{-}^3$
Occupied volume = $\frac{4\pi r^3\times \frac{5}{2}\times r_{+}^3}{3}+\frac{4\pi r^3\times 4\times r_{-}^3}{3}=\frac{4\pi }{3}\times [\frac{5r_{+}^3}{2}+4r_{-}^3]$
Hence, packing fraction = Occupied volume / Total volume
= $\frac{\frac{4\pi }{3}\times [\frac{5r_{+}^3}{2}+4r_{-}^3]}{16\sqrt{2}{r}_{-}^3}$