Question

Which 3-D packing has the maximum empty space in its crystal lattice?

Answer 1

Packing Efficiency of Hexagonal Close Packing :

$\text{Total no. of spheres in a hcp unit cell}=(12\times \frac{1}{6})+(2\times \frac{1}{2})+3=6$
where,
$(12\times \frac{1}{6})\text{is for corners atoms of two hexagonal layers}$
$(2\times \frac{1}{2})\text{is for the centre atoms of two hexagonal layers}$

$3\text{is for the atoms of middle layer}$
$\text{Total volume occupied}=6\times \frac{4}{3}\times \pi \times r^3=8\pi r^3$
$\text{Volume of hcp unit cell (hexagon)}=\text{Base area}\times \text{Height}$


$\text{Base area of a regular hexagon}=6\times \frac{\sqrt{3}}{4}(2r{)}^2=6\times \sqrt{3}{r}^2$

$\text{Height of Unit Cell (h)}=4r\times \sqrt{\frac{2}{3}}$

$\text{Volume of hexagon (V)}=(6\times \sqrt{3}{r}^2)\times (4r\times \sqrt{\frac{2}{3}})V=24\sqrt{2}{r}^3$
$\text{Packing fraction}=\frac{8\pi r^3}{24\sqrt{2}{r}^3}=\frac{\pi }{3\sqrt{2}}P.F\approx 0.74$

$\text{Fraction empty space}=1-0.74=0.26$


For body centred cubic arrangement, $Z_{eff}=2$

$4r=\sqrt{3}a$
where,
r is radius of atom
a is edge length of cube
Thus, $a=\frac{4r}{\sqrt{3}}$
$\text{Volume of cell}=a^3$
$\text{Volume of cell}=(\frac{4r}{\sqrt{3}}{)}^3$
$\text{Packing fraction}=\frac{\text{Total volume occupied by sphere}}{\text{Volume of unit cell}}$
$\text{Packing fraction}=\frac{Z\times \frac{4}{3}\pi r^3}{a^3}$
$=\frac{2\times \frac{4}{3}\pi r^3}{{\left(\frac{4r}{\sqrt{3}}\right)}^3}=\frac{\sqrt{3}\pi }{8}=0.68$
$\text{Fraction empty space}=1-0.68=0.32$

For simple cubic (sc):


$Z_{eff}\text{for sc =1}$
So,

$\text{The total volume occupied by sphere}V=1\times \frac{4}{3}\times \pi r^3$
$\text{Volume of the cube}=a^3=(2r{)}^3$
$\text{Packing fraction (P.F)}=\frac{1\times \frac{4}{3}\times \pi r^3}{(2r{)}^3}P.F=\frac{\pi }{6}=0.52$
$\text{Fraction empty space}=1-0.52=0.48$

In ccp lattice:


Packing fraction (P.F) of ccp:

$P.F=\frac{\text{No of spheres in one ccp unit cell}\times \text{Volume of each sphere}}{\text{Total Volume of the Unit cell}}$

ccp unit cell have FCC arrangement



Here,
$\sqrt{2}\times a=4ra=2\sqrt{2}\times r$

$Z_{eff}\text{for FCC}=(8\times \frac{1}{8})+(6\times \frac{1}{2})=4$
So,

$\text{The total volume occupied by sphere}V=4\times \frac{4}{3}\times \pi r^3$
$\text{Volume of the cube}=a^3=(2\sqrt{2}r{)}^3$

$\text{Packing fraction (P.F)}=\frac{4\times \frac{4}{3}\times \pi r^3}{(2\sqrt{2}r{)}^3}P.F=\frac{\pi }{3\sqrt{2}}\approx 0.74$
$\text{Fraction empty space}=1-0.74=0.26$

Packing	Empty space (fraction)
HCP	$\approx 0.26$
CCP	$\approx 0.26$
BCC	$\approx 0.32$
SCP	$\approx 0.48$

Option (c) is correct