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

Packing Efficiency of Hexagonal Close Packing :

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

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


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

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

\(\text{Volume of hexagon (V)}=(6\times \sqrt{3}r^2)\times\left(4r\times \sqrt{\dfrac{2}{3}}\right)\\V=24\sqrt{2}r^3\)
\(\text{Packing fraction}=\dfrac{8\pi r^3}{24\sqrt{2}r^3}=\dfrac{\pi}{3\sqrt{2}}\\P.F\approx0.74\)

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


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, \(a = \dfrac {4~r}{\sqrt{3}}\)
\(\text{Volume of cell }= a^3\)
\(\text{Volume of cell }= (\dfrac {4~r}{\sqrt{3}})^3\)
\(\text{Packing fraction }= \dfrac{\text{Total volume occupied by sphere}}{\text{Volume of unit cell}}\)
\(\text{Packing fraction }= \dfrac{Z \times \dfrac{4}{3}\pi r^3}{a^3} \)
\(= \dfrac{2 \times \dfrac{4}{3}πr^3}{\left(\dfrac{4r}{\sqrt{3}} \right)^3}\\
=\dfrac {\sqrt{3}~\pi }{8}\\
=0.68 \)
\(\text{Fraction empty space }= 1 - 0.68 = 0.32\)

For simple cubic (sc):


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

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

In ccp lattice:


Packing fraction (P.F) of ccp:

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

ccp unit cell have FCC arrangement



Here,
\(\sqrt{2}\times a=4r\\a=2\sqrt{2}\times r\)

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

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

\(\text{Packing fraction (P.F)}=\dfrac{4\times \dfrac{4}{3}\times \pi r^3}{(2\sqrt{2}r)^3}\\P.F=\dfrac{\pi}{3\sqrt{2}}\approx0.74\)
\(\text{Fraction empty space} =1-0.74=0.26\)

Packing	Empty space (fraction)
HCP	\(\approx0.26\)
CCP	\(\approx0.26\)
BCC	\(\approx 0.32\)
SCP	\(\approx0.48\)

Option (c) is correct