Question

For a NaCl unit cell, the $N{a}^{+}$ occupies the edge centres and body centred positions of cube while $C{l}^{-}$ occupies the corners and face centres of the cube.

Which is the correct expression for determining packing fraction (P.F.) of NaCl unit cell, if ions along an edge diagonal are absent?

Answer 1

Given, in NaCl

$N{a}^{+}$ occupies the 12 edge centres and 1 body centre
$C{l}^{-}$ occupies 8 corners and 6 face centers in the cube which is an fcc lattice arrangement.
Along an edge diagonal, one body centred atom and two opposite edge centre atoms are present.
Therefore, if the ions present along edge diagonal is removed
$\text{Effective number of}N{a}^{+}\text{ions in an unit cell}=10\times \frac{1}{4}$
(Here $2N{a}^{+}$ from edges and $N{a}^{+}$ from body centred is removed.)
$\therefore$
$Z_{N{a}^{+}}=\frac{5}{2}$

No $C{l}^{-}$ ion numbers are affected during the removal of edge diagonal. So,
$\text{Effective no. of}C{l}^{-}\text{ions in one unit cell}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
$Z_{C{l}^{-}}=4$

For fcc lattice arrangement,
We know,
$\sqrt{2}a=4r^{-}$
$a=\frac{4r^{-}}{\sqrt{2}}$
$\text{Volume of unit cell}=a^3=16\sqrt{2}{r}_{-}^3$
$\text{Packing fraction}=\frac{(Z_{N{a}^{+}}\times \frac{4}{3}\pi r_{+}^3)+Z_{C{l}^{-}}\times \frac{4}{3}\pi r_{-}^3}{a^3}$

$\text{Packing fraction}=\frac{\frac{4}{3}\pi (\frac{5}{2}{r}_{+}^3+4r_{-}^3)}{16\sqrt{2}{r}_{-}^3}$