Metallic Crystal Structures Study Notes
Overview of material to know:
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Structure: FCC, BCC, SC (be able to draw)
Spheres per unit cell: FCC, BCC, SC
Atomic radius change: hex, FCC, BCC, SC
Number of nearest neighbours: hex, FCC, BCC, SC.
Packing Efficiency: hex, FCC, BCC, SC.
Should know that there are 7 crystal systems and 14 Bravais lattices, but only
need to reproduce the cubic ones.
Density  atomic radius etc.. calculations
Note that FCC (face centered cubic) and CCP (cubic close packing) are the same.
Questions from Assignments
Assignment 2 Q10-14
Readings from text
6.1-6.4
Web resources
These resources offer 3-D images of crystal structures and are recommended for
visualising the structures of metallic and salt crystal structures.
http://www.dawgsdk.org/crystal/en/library/bcc#0003 metal and salt crystals (no simple
cubic)
cubic lattice
http://www.youtube.com/watch?v=T3VGfk0wVos&feature=related
http://www.youtube.com/watch?v=bAMYoAOHx5U&feature=related
cubic close packing
http://www.youtube.com/watch?v=fs_5-KA3iiE&feature=related
calcium fluorite
http://www.youtube.com/watch?v=YIyGoAOEIew&feature=related
salt crystals
http://www.neubert.net/Crystals/CRYStruc.html
Structure: Be able to draw BCC (body centered cubic), SC (simple cubic) and FCC
(face centered cubic). Note that FCC is also called CCP (cubic close packing).
Spheres per unit cell: FCC, BCC, SC
Within the unit cells there are partial spheres. This is most easily viewed in the following
images. There are 3 positions of spheres, each with different percentages of themselves
within the given unit cell. These are center (1), face (1/2) and corner (1/8).
Simple cubic
8 corner×1/8 =1
1 sphere total
BCC
8 corner × 1/8 =1
1 center × 1 =1
2 spheres total
Number of nearest neighbours: hex, FCC, BCC, SC.
FCC
8 corner × 1/8 =1
6 face × 1/2 =3
4 spheres total
All spheres in a given metallic crystal system have the same number of nearest
neighbours, as they are in identical environments. Nearest neighbours are the atoms that
are directly in contact with the atom.
BCC
The number of nearest neighbours is easiest to see with the BCC
The center sphere is directly touching 8 spheres around it and therefore the number of
nearest neighbours is 8.
SC
For SC, it may be easier to visualise the number of nearest neighbours if one considers
how the stacking works to build up the crystal structure. The center atom has 4 nearest
neighbours in the plane and one directly above and one directly below for a total of 6.
Recall that in stacking a simple cubic structure, each sphere sits directly on top of the one
below.
1
4
2
3
FCC
If we look at the number of nearest neighbours for the atom labelled A, there are 4
(labelled 1-4) in the plane with the face centered atom. Then there are 4 more that atom A
is touching (labelled 5-8). As the atom A is only half a sphere, you can visualise the
mirror image. The rest of the sphere is still touching atoms 1-4 and then another set of 4
spheres, analogous to 5-8 for a total of 12 nearest neighbours.
5
1
2
8
6
A
3
7
4
Hex
Hexagonal also has 12 nearest neighbours-do not need to know structure, but should
know that both FCC and hex are close packed structures and have the same number
of nearest neighbours and the same packing efficiency.
Packing Efficiency: BCC, SC, hex, FCC
SC
radius of each sphere = r
edge length of unit cell (a) = 2r
volume of cell =a3= 8r3
volume of sphere = 4/3πr3 = 4.19r3
% occupancy = 1 sphere/unit cell
= (4.19r3/8r3) x 100%
= 52%
BCC
d
2 atoms/cell
d=4rFe =√3aBCC
radius of each sphere = r
diagonal length of unit cell = r+2r+r = 4r=d
d=√3a= 4r
edge length (a) = 4r/√3 = 4r/1.73 = 2.31r
volume of cell = a3= 12.33r3
% occupancy = 2 sphere/unit cell
= 2(4.19r3/12.33r3) x 100%
= 68%
FCC
c
4 atoms/cell
radius of each sphere = r
diagonal length of face = r+2r+r = 4r=c
c=√2a= 4r
edge length (a) = 4r/√2 = 4r/1.41 = 2.83r
volume of cell = a3= 22.63r3
% occupancy =4 sphere/unit cell
= 4(4.19r3/22.63r3) x 100%
= 74%
Table of values associated with lattice structures
Crystal
structure
Spheres per
unit cell
SC
BCC
FCC
hexagonal
1
2
4
n/a
Number of
nearest
neighbours
6
8
12
12
packing
efficiency
reproduce
structure
52%
68%
74%
74%
yes
yes
yes
no
Atomic radius change: hex, FCC, BCC, SC
The metallic radius decreases with decreasing coordination number (aka number of
nearest neighbours):
coordination number 12
relative radius
1.00
8
0.97
6
0.96
4
0.88
Atoms in BCC structures are eight-coordinate. This gives packing efficiency of 68%,
lower than close packing efficiencies of 74% (12 coordinate) for FCC and hex. The six
coordinate simple cubic has the smallest radius with a packing efficiency of 54%. The
coordination number of 4 is included for completeness (ie tetrahedral in salt structures)
only.
There are 7 crystal systems and 14 Bravais lattices (only need to reproduce the cubic
ones).
Density  atomic radius etc.. calculations
In the cases to be described, the value “a” is always the edge length of the unit cell. Note
that the edge length is specific to both the type of unit cell (FCC, BCC, SC) and to the
elemental composition of the spheres (Fe, Co, Mn).
A review of the geometry of triangles
c
a
b
For a right angle triangle, recall Pythagoras
a2 +b2 = c2
if a and b are equal then a=b
a2 +a2 = c2 and therefore
2×a2 =c2
or
c=√2×a, which is generally written as √2a
so the triangle can be shown as
c=√2 a
a
a
Putting this on a 3-D cube, we can see
a
c
a
a
or
c
a
a
a
Triangle and lattice structures for FCC
If this is placed on a FCC, the following figure is obtained, where c=4r, with r being the
radius of the spheres. It is important to note that the line MUST go through touching
spheres.
c
In the following figure, all the red lines are correct for c=4r, and the green lines are
correct for a. Note that the “a” values could not be equal to 4r or 2r because the spheres
are not touching. Also, consider how many nearest neighbours one can identify within in
this cube for a face centered sphere (6) and how many would exist in total (12).
c
Example One (FCC)
What is the density of solid iron at temperatures above 910 °C, assuming the iron is in
FCC (formed at elevated temperatures) and the radius of Fe of 124 Picometers?
ρ= ? g/cm2 (FCC) aw=55.847 g/mole
rFe =1.24*10-8 cm
ρ=(grams per cell)/a3 (volume of cell)
4rFe =√2aFCC
aFCC= 4rFe
4*(1.24*10-8 cm)
=
√2
√2
aFCC =3.51*10-8 cm
(4 atoms/cell)(55.847 grams/mol)
6.022*1023 atoms/mol
ρ=
(3.51*10-8
cm)3
= 8.57 grams/cm3
Example 2 (FCC)
What is the density of FCC Ni, assuming a sphere radius of 125 Picometers?
Triangle and lattice structures for BCC
Recall Pythagoras
a2 +a2 = c2 and therefore
c=√2×a ( generally written as √2a)
In this example below a2 +c2 = d2 and as a2 +a2 = c2 therefore
a2 + a2 +a2 = d2 or d=√3×a, which is generally written as √3a
a
d
a
c
If this is placed on a BCC crystal structure, the following figure is obtained, where d=4r,
with r being the radius of the spheres. The line goes through the two corner spheres (1r
each) and the entire center sphere (2r). It is important to note that the line MUST go
through touching spheres.
d
Example One (BCC)
What is the atomic radius for BCC iron , given a density of 7.86 g/cm3 ?
ρ= 7.86 g/cm2 (BCC) aw=55.847 g/mole
(atoms/cell)(atomic weight)
ρ=
(grams per cell)
=
NA
(volume of cell)
a3
(2 atoms/cell)(55.847 grams/mol)
a3=g/ρ =
6.022*1023 atoms/mol
7.86
=2.36*10-23 cm3
grams/cm3
aBCC = 2.86* 10-8 cm
4rFe =√3aBCC
rFe =√3aBCC
4
rFe =1.24*10-8 cm
Example Two (BCC)
What is the density for BCC nickel , given an atomic radius of 125 picometers ?
The actual density of Nickel is 8.90 g/cm3. Is nickel more likely FCC or BCC?
We calculated a density of 8.78 for FCC and 8.09 for BCC, so nickel would be FCC.
(Note that hexagonal would have the same density as FCC, as the packing efficiency is
the same. Also note that we have ignored the change in atomic radii with change in
coordination number for the above examples)