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Question

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

A
Hexagonal closed packing (hcp)
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B
Body centred cubic close packing (bcc)
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C
Simple cubic close packing (scp)
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D
Cubic close packing (ccp)
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Solution

The correct option is C Simple cubic close packing (scp)
Packing Efficiency of Hexagonal Close Packing :

Total no. of spheres in a hcp unit cell =(12×16)+(2×12)+3=6
where,
(12×16) is for corners atoms of two hexagonal layers
(2×12) is for the centre atoms of two hexagonal layers

3 is for the atoms of middle layer
Total volume occupied=6×43×π×r3=8πr3
Volume of hcp unit cell (hexagon)=Base area×Height


Base area of a regular hexagon=6×34(2r)2=6×3r2

Height of Unit Cell (h)=4r×23

Volume of hexagon (V)=(6×3r2)×(4r×23)V=242r3
Packing fraction=8πr3242r3=π32P.F0.74

Fraction empty space=10.74=0.26


For body centred cubic arrangement, Zeff=2

4r=3 a
where,
r is radius of atom
a is edge length of cube
Thus, a=4 r3
Volume of cell =a3
Volume of cell =(4 r3)3
Packing fraction =Total volume occupied by sphereVolume of unit cell
Packing fraction =Z×43πr3a3
=2×434r3(4r2)3=3 π8=0.68
Fraction empty space =10.68=0.32

For simple cubic (sc):


Zeff for sc =1
So,

The total volume occupied by sphere V=1×43×πr3
Volume of the cube=a3=(2r)3
Packing fraction (P.F)=1×43×πr3(2r)3P.F=π6=0.52
Fraction empty space =10.52=0.48

In ccp lattice:


Packing efficiency (P.F) of ccp:

P.F=No of spheres in one ccp unit cell ×Volume of each sphereTotal Volume of the Unit cell

ccp unit cell have FCC arrangement



Here,
2×a=4Ra=22×R

Zeff for FCC=(8×18)+(6×12)=4
So,

The total volume occupied by sphere V=4×43×πr3
Volume of the cube=a3=(22R)3

Packing efficiency (P.F)=4×43×πr3(22R)3P.F=π320.74
Fraction empty space=10.74=0.26
Packing Empty space (fraction)
HCP 0.26
CCP 0.26
BCC 0.32
SCP 0.48

Option (c) is correct

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