CRYSTAL STRUCTURE
Crystal Structure
Objectives
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Relationships between structures-bonding-properties of engineering
materials.
Arrangements in crystalline solids
Give examples of each: Lattice, Crystal Structure, Unit Cell and
Coordination Numbers
Describe hard-sphere packing and identify cell symmetry.
Define directions and planes (Miller indices) for crystals
Outline
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Structure of the Atom and Atomic Bonding
Electronic Structure of the Atom
Lattice, Unit Cells, Basis, and Crystal Structures
Points, Directions, and Planes in the Unit Cell
Crystal Structures of Ionic Materials
Covalent Structures
Crystal Structure
Crystal Structure
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Lattice- A collection of points that divide space into smaller equally sized
segments.
Unit cell - A subdivision of the lattice that still retains the overall
characteristics of the entire lattice.
Atomic radius - The apparent radius of an atom, typically calculated
from the dimensions of the unit cell, using close-packed directions
(depends upon coordination number).
Packing factor - The fraction of space in a unit cell occupied by atoms.
Types of Crystal Structure
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Body centered cubic (BCC)
Face centered cubic (FCC)
Hexagonal close packed (HCP)
Crystal Structure
A number of metals are shown below with their room temperature crystal
structure indicated. There are substances without crystalline structure at
room temperature; for example, glass and silicone. All metals and alloys
are crystalline solids, and most metals assume one of three different lattice,
or crystalline, structures as they form: body-centered cubic (BCC), facecentered cubic (FCC), or hexagonal close-packed (HCP).
Aluminum (FCC)
Chromium
(BCC)
Iron (gamma) (BCC)
Iron
(BCC)
Silver (FCC)
Titanium
(HCP)
Copper (FCC)
(delta) Lead (FCC)
Tungsten
(BCC)
Iron
(FCC)
(alpha)
Nickel (FCC)
Zinc (HCP)
Crystal Structure
Number of Lattice Points in Cubic Crystal Systems

In the SC unit cell: point / unit cell
= (8 corners)1/8 = 1

In BCC unit cells: point / unit cell
= (8 corners)1/8 + (1 center)(1) = 2

In FCC unit cells: point / unit cell
= (8 corners)1/8 + (6 faces)(1/2) = 4
Relationship between Atomic Radius and Lattice Parameters
In SC, BCC, and FCC structures when one atom is located at each lattice
point.
Crystal Structure
Packing Factor

In a FCC cell, there are four lattice points per cell; if there is one atom per
lattice point, there are also four atoms per cell. The volume of one atom is
4πr3/3 and the volume of the unit cell is a0 3
Crystal Structure
Density
Density of BCC iron, which has a lattice parameter of 0.2866 nm.
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Atoms/cell = 2, a0 = 0.2866 nm = 2.866  10-8 cm
Atomic mass = 55.847 g/mol
Volume of unit cell = = (2.866  10-8 cm)3 = 23.54  10-24 cm3/cell
Avogadro’s number NA = 6.02  1023 atoms/mol
(number of atoms/cell )(atomic mass of iron)
Density  
(volume of unit cell)(Avog adro' s number)
(2)(55.847)
3


7
.
882
g
/
cm
(23.54 1024 )(6.02 1023 )
Crystal Structure
Points, Directions, and Planes in the Unit Cell
Geometry
Unit Cell: The basic structural unit of a crystal
structure. A unit cell is the smallest component
of the crystal that reproduces the whole crystal
when stacked together with purely
translational repetition.
R
R
R
R
a
Atomic configuration in
Face-Centered-Cubic
Atomic configuration in Simple Cubic
Crystal Structure
Bravais Lattices
Crystal Structure
Unit Cells Types
A unit cell is the smallest component of the crystal that reproduces the
whole crystal when stacked together with purely translational repetition.
• Primitive (P) unit cells contain only a single lattice point.
• Internal (I) unit cell contains an atom in the body center.
• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.
• Centered (C) unit cell contains atoms centered on the sides of the unit cell.
Primitive
Body-Centered
Face-Centered
End-Centered
Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monclinic,
triclinic, trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only
14 types of 3-D lattice.
Crystal Structure
Counting Number of Atoms Per Unit Cell
Counting Atoms in 3D Cells
Atoms in different positions are shared by differing numbers of unit cells.
• Vertex atom shared by 8 cells => 1/8 atom per cell.
• Edge atom shared by 4 cells => 1/4 atom per cell.
• Face atom shared by 2 cells => 1/2 atom per cell.
• Body unique to 1 cell => 1 atom per cell.
Simple Cubic
8 atoms but shared by 8 unit cells. So,
8 atoms/8 cells = 1 atom/unit cell
How many atoms/cell for
Body-Centered Cubic?
And, Face-Centered Cubic?
Crystal Structure
Atomic Packing Fraction for FCC
APF = vol. of atomic spheres in unit cell
total unit cell vol.
No. of atoms per unit cell =
Volume of one atom=
Volume of cubic cell =
“R” related to “a” by
Face-Centered-Cubic
Arrangement
4/cell
4R3/3
a3
2a  4R
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell

= 0.74
Crystal Structure
APF for BCC
Again, For BCC
Vatoms
APF 

Vcell


3
4 a 3  
 
2  
3 
 4 
 


a3
3

 0.68
8

• BCC: a = b = c = a and angles a = b =g= 90°.
• 2 atoms in the cubic cell: (0, 0, 0) and (1/2, 1/2, 1/2).
Crystal Structure
FCC Stacking
A
B
C
Highlighting
the stacking
Highlighting
the faces
Crystal Structure
FCC Stacking
ABCABC.... repeat along <111> direction gives Cubic Close-Packing (CCP)
• Face-Centered-Cubic (FCC) is the most efficient packing of hard-spheres of any lattice.
• Unit cell showing the full symmetry of the FCC arrangement : a = b =c, angles all 90°
• 4 atoms in the unit cell: (0, 0, 0) (0, 1/2, 1/2) (1/2, 0, 1/2) (1/2, 1/2, 0)
HCP Stacking
Crystal Structure
A
B
Highlighting
the stacking
A
Highlighting
the cell
Layer A
Layer B
Layer A
Crystal Structure
HCP Stacking
ABABAB.... repeat along <111> direction gives Hexagonal Close-Packing (HCP)
• Unit cell showing the full symmetry of the HCP arrangement is hexagonal
• Hexagonal: a = b, c = 1.633a and angles a = b = 90°, g = 120°
• 2 atoms in the smallest cell: (0, 0, 0) and (2/3, 1/3, 1/2).
Crystal Structure
Crystallographic Points, Directions, and Planes.
To define a point within a unit cell….
Express the coordinates uvw as fractions of unit cell vectors a, b, and c
(so that the axes x, y, and z do not have to be orthogonal).
pt. coord.
pt.

c
origin 
b

a
x (a)
y (b)
z (c)
0
0
0
1
0
0
1
1
1
1/2
0
1/2
Crystal Structure
Crystallographic Points, Directions, and Planes.
Crystallographic
coordinates.
directions
and Direction B
1. Two points are 1, 1, 1 and 0, 0, 0
2. 1, 1, 1, -0, 0, 0 = 1, 1, 1
3. No fractions to clear or integers
to reduce
4. [111]
Direction A
1. Two points are 1, 0, 0, and 0, 0, 0
2. 1, 0, 0, -0, 0, 0 = 1, 0, 0
3. No fractions to clear or integers to
reduce
4. [100]
Direction C
1. Two points are 0, 0, 1 and 1/2, 1,
0
2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1
3. 2(-1/2, -1, 1) = -1, -2, 2
4. [ 1 22]
Crystal Structure
Crystallographic Points, Directions, and Planes.
Procedure:
1. Any line (or vector direction) is specified by 2 points.
•
The first point is, typically, at the origin (000).
c
b
2.
Determine length of vector projection in each of 3 axes in
units (or fractions) of a, b, and c.
•
X (a), Y(b), Z(c)
1
1
0
3.
Multiply or divide by a common factor to reduce the lengths
to the smallest integer values, u v w.
4.
Enclose in square brackets: [u v w]: [110] direction.
a
[1 1 0]
5. Designate negative numbers by a bar
•
Pronounced “bar 1”, “bar 1”, “zero” direction.
6.
“Family” of [110] directions is designated as <110>.

DIRECTIONS will help define PLANES (Miller Indices or plane normal).
Crystal Structure
Crystallographic Points, Directions, and Planes.
Figure 2.9 Crystallographic planes
and intercepts
Plane B
1. The plane never intercepts the z
axis, so x = 1, y = 2, and z =
2.1/x = 1, 1/y =1/2, 1/z = 0
3. Clear fractions:
1/x = 2, 1/y = 1, 1/z = 0
4. (210)
Plane A
1. x = 1, y = 1, z = 1
2.1/x = 1, 1/y = 1,1 /z = 1
3. No fractions to clear
4. (111)
Plane C
1. We must move the origin, since
the plane passes through 0, 0, 0.
Let’s move the origin one lattice
parameter in the y-direction.
Then, x = ∞ , y = -1, and z = ∞
2.1/x = 0, 1/y = 1, 1/z = 0
3. No fractions to clear.
4 (o1-o)
Crystal Structure
Miller Indices for HCP Planes
4-index notation is more important for planes in HCP, in order to
distinguish similar planes rotated by 120o.
t
As soon as you see [1100], you will know that it
is HCP, and not [110] cubic!
Find Miller Indices for HCP:
1.
r
s
2.
3.
4.
5.
Find the intercepts, r and s, of the plane with any two of
the basal plane axes (a1, a2, or a3), as well as the
intercept, t, with the c axes.
Get reciprocals 1/r, 1/s, and 1/t.
Convert reciprocals to smallest integers in same ratios.
Get h, k, i , l via relation i = - (h+k), where h is associated
with a1, k with a3, i with a2, and l with c.
Enclose 4-indices in parenthesis: (h k i l) .
Crystal Structure
Miller Indices for HCP Planes
What is the Miller Index of the pink plane?
1.
The plane’s intercept a1, a3 and c at r=1, s=1 and t=
, respectively.
2.
The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.
3.
They are already smallest integers.
4.
We can write (h k i l) = (1 ? 1 0).
5.
Using i = - (h+k) relation, k=–2.
6.
Miller Index is
(12 11)
Crystal Structure
Linear Density in FCC
LD =
Number of atoms centered on a direction vector
Length of the direction vector
Example: Calculate the linear density of an FCC crystal along [1 1 0].
ASK
a. How many spheres along blue line?
b. What is length of blue line?
ANSWER
a. 2 atoms along [1 1 0]
in the cube.
b. Length = 4R
LD110 

XZ = 1i + 1j + 0k = [110]
2atoms
1

4R
2R
Crystal Structure
Planar Packing Density in FCC
PD =
Area of atoms centered on a given plane
Area of the plane
Example: Calculate the PPD on (1 1 0) plane of an FCC crystal.
• Find area filled by atoms in plane: 2R2
• Find Area of Plane: 8√2 R2
a  2 2R
Hence,
PPD 
4R
2R 2
8 2R 2


4 2
 0.555
Always independent of R!
Crystal Structure
Theoretical Density, 
• crystal structure = FCC: 4 atoms/unit cell
• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)
• atomic radius R = 0.128 nm
Result: theoretical Cu = 8.89 g/cm3
Compare to actual: Cu = 8.94 g/cm3
Crystal Structure
Crystal Structures of Ionic Materials
Factors need to be considered in order to understand crystal structures of
ionically bonded solids are:
• Ionic Radii
• Electrical Neutrality
• Connection between Anion Polyhedra
• Visualization of Crystal Structures Using Computers
The cesium chloride structure, a SC unit cell with two ions (Cs+ and CI-) per
lattice point. (b) The sodium chloride structure, a FCC unit cell with two ions
(Na+ + CI-) per lattice point
Crystal Structure
Crystal Structures of Ionic Materials
The zinc blende unit cell, (b) plan
view.
Fluorite unit cell, (b) plan view.
Crystal Structure
Crystal Structures of Ionic Materials
The perovskite unit cell showing the A and B site cations and
oxygen ions occupying the face-center positions of the unit cell.
Crystal Structure
Covalent Structures
The silicon-oxygen tetrahedron and
the resultant β-cristobalite form of
silica
Tetrahedron and (b) the diamond cubic (DC)
unit cell. This open structure is produced
because of the requirements of covalent
bonding
Crystal Structure
Covalent Structures
Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is
produced because of the requirements of covalent bonding