As the name suggest the atoms are closely packed. But what does closed packed mean? It simply means that the atoms are arranged in such a manner that the space between them is as small as possible. Snooker balls in a triangle are closely packed, with each one sitting in the cavity made by the two adjacent ones.
If we extend this idea to a 3d array, there will be a cavity made by three atoms on the plane below. So the second atomic plane is slightly offset from the first plane. We call the first plane A and the second B, but where does the next plane, C, go? There are two place it can sit, straight above the 'A' plane,
or in line with neither of the two previous planes. The picture below illustrates the two possible positions
In the Cubic close packed structure the 3rd plane in fact is position inline with neither of the previous two, and so adopts the c position. The resulting stacking is shown below. From the packing it is not immediately clear that there is a cubic unit cell but if wee look along a (xyz???) plane,
the cube can then be seen. The following picture shows the cubic arrangement.
The unit cell for the ccp arrangement has an atom at each of the vertices, and one in the centre of the faces. the atomic coordinates for the positions are:
(0,0,0) (0,0,1) (0,1,0) (1,0,0) (1,1,0) (1,0,1) (1,1,1) (0,1,1)
and then the face centres which are at (1/2,0,1/2) (1/2,1/2,0) (0,1/2,1/2) (1,1/2,1/2) (1/2,1,1/2) (1/2,1/2,1)
The coordination number of the ccp structure is..(continue)
Metals which exhibit such packing include the coinage metals- Cu, Ag, Au the transition metals in group 10- Ni, Pd, and Pt.
These metals are more dense than metals which adopt the simple cubic or body centred cubic packing, this is due to the percentage of the unit cell volume occupied. The volume occupied by the ions is clearly proportional to the density of the metal.
To calculate the volume occupied:
Define radius of an atom as r
The length of the unit cell, a, is found using Pythagoras(assume the facial atoms are touching, although the unit cell projection doesn't show this its can be seen on closer inspection of the array).
Where hyp=4r, hence a=2√2r
the volume of the unit cell is
=a3
=(2√2r)3
=16√2r3
Having established the volume in terms of r we need to know the volume occupied by the atoms or ions. There are 6 atoms on the faces of the cube, only half of each is in the unit cell(since each is shared with the next unit cell). There are also 8 atoms at the vertices, of which 1/8 is in the unit cell.
So the total number of atoms
= 6*1/2 + 8*1/8=3+1=4
the volume of a sphere is given by
4/3πr3
So to find the total occupied volume we simply multiply by 4
4*4/3πr3=16/3πr3
To then calculate the percentage occupied we need to find the occupied volume as a fraction of the unit cell volume.
(16/3πr3) / (16√2r3)
=π/(3√)
=0.74048=74.01%
