1. CRYSTAL STRUCTURE

2. Crystal Structure Objectives
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
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

3. Crystal Structure Crystal Structure
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
Body centered cubic (BCC)
Face centered cubic (FCC)
Hexagonal close packed (HCP)

4. 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), face-centered cubic (FCC), or hexagonal close-packed (HCP).
Aluminum (FCC)
Chromium (BCC)
Copper (FCC)
Iron (alpha) (FCC)
Iron (gamma) (BCC)
Iron (delta) (BCC)
Lead (FCC)
Nickel (FCC)
Silver (FCC)
Titanium (HCP)
Tungsten (BCC)
Zinc (HCP)

5. Number of Lattice Points in Cubic Crystal Systems
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.

6. 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

7. Crystal Structure Density
Density of BCC iron, which has a lattice parameter of nm.
Atoms/cell = 2, a0 = nm =  10-8 cm
Atomic mass = g/mol
Volume of unit cell = = (2.866  10-8 cm)3 =  cm3/cell
Avogadro’s number NA = 6.02  1023 atoms/mol

8. 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

9. Crystal Structure
Bravais Lattices

10. 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.
Face-Centered
Primitive
Body-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.

11. 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?

12. Crystal Structure Atomic Packing Fraction for FCC Face-Centered-Cubic
Arrangement
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
4/cell
4R3/3 a3
= 0.74

13. Crystal Structure APF for BCC Again, For BCC
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).

14. Crystal Structure A B C FCC Stacking Highlighting the stacking
the faces

15. 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)

16. Crystal Structure A B A HCP Stacking Highlighting the stacking
the cell
Layer A
Layer B
Layer A

17. 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).

18. 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.
x (a) y (b) z (c)
0 0 0
origin
1 0 0
1 1 1
1/2 0 1/2

19. Crystal Structure Crystallographic Points, Directions, and Planes.
Crystallographic directions and coordinates.
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
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

20. Crystal Structure Crystallographic Points, Directions, and Planes.
Procedure:
Any line (or vector direction) is specified by 2 points.
The first point is, typically, at the origin (000).
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)
Multiply or divide by a common factor to reduce the lengths to the smallest integer values, u v w.
Enclose in square brackets: [u v w]: [110] direction. a b c
DIRECTIONS will help define PLANES (Miller Indices or plane normal).
5. Designate negative numbers by a bar
Pronounced “bar 1”, “bar 1”, “zero” direction.
6. “Family” of [110] directions is designated as <110>.

21. 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)

22. Crystal Structure t r s 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:
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) . r s

23. Crystal Structure Miller Indices for HCP Planes
What is the Miller Index of the pink plane?
The plane’s intercept a1, a3 and c at r=1, s=1 and t= , respectively.
The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.
They are already smallest integers.
We can write (h k i l) = (1 ? 1 0).
Using i = - (h+k) relation, k=–2.
Miller Index is

24. 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
2 atoms along [1 1 0] in the cube.
Length = 4R
XZ = 1i + 1j + 0k = [110]

25. Planar Packing Density in FCC
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
Hence,
Always independent of R!

26. Crystal Structure Theoretical Density, 
• crystal structure = FCC: 4 atoms/unit cell
• atomic weight = g/mol (1 amu = 1 g/mol)
• atomic radius R = nm

27. 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

28. Crystal Structures of Ionic Materials
The zinc blende unit cell, (b) plan view.
Fluorite unit cell, (b) plan view.

29. 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.

30. 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
The silicon-oxygen tetrahedron and the resultant β-cristobalite form of silica

31. 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