1. Crystal and Amorphous Structure
Engineering Materials
Crystal and Amorphous Structure
in Materials

2. Engineering Materials
Types of Atomic Bonds
Atomic Bonding
Strong Primary Bonds
Weak Secondary Bonds
Ionic Bond
Covalent Bond
Metallic Bond
Fluctuating Dipoles
Permanent Dipoles

3. Ionic Bond A primary bond formed by the transfer of one or
Engineering Materials
Types of Atomic Bonds
Ionic Bond
A primary bond formed by the transfer of one or
more electron from an electropositive atom to an
electronegative one.
The ions are bonded together in a solid crystal by electrostatic forces.
Example: NaCl crystal (see animation NaCl4 & 7)

4. Covalent Bond A primary bond resulting from the sharing of electrons.
Engineering Materials
Types of Atomic Bonds
Covalent Bond
A primary bond resulting from the sharing of
electrons.
Its involves the overlapping of half-filled orbitals
of two atoms. I
Example: Diamond, H2, H2O, methane (animation Covalent… 7)

5. .. . + F .. . O + .. . N + Covalent Bond Types of Atomic Bonds
Engineering Materials
Types of Atomic Bonds
Covalent Bond F .. . +
Single bond
Double bond . O .. +
Triple bond . N .. +

6. Metallic Bond A primary bond resulting from the sharing of
Engineering Materials
Types of Atomic Bonds
Metallic Bond
A primary bond resulting from the sharing of
delocalized outer electrons in the form of an
electron charged cloud by an aggregate of metal
atoms.
Example elemental sodium

7. Metallic Bond Types of Atomic Bonds Engineering Materials Positive ion
Valance electrons in the form of
electron charge clouds +

8. Engineering Materials
Crystal and Amorphous Structure in Materials
The physical structure of solid materials mainly depends on the arrangements of
the atoms, ions, or molecules that make up the solid and the bonding forces between
them.
IF the atoms or ions of solid are arranged in pattern that repeats itself in three
dimensions. Forming solid that has long range order (LRO)  crystalline solid or
crystalline materials
CRYSTAL a solid composed of atoms, ions, or molecules arranged in a pattern that
is repeats in three dimensions.
Examples metals, alloys and some ceramic materials
IF the atoms or ions of solid are NOT arranged in a long range and repeatable manner
Then forming solid that has short range order (SRO) Amorphous or non crystalline
materials.
Example liquid water has a short range order

9. Engineering Materials
Crystal and Amorphous Structure in Materials
Space lattice
A three dimensional array of points each of which has identical surroundings
Unit cell
Is a repeating unit of a space lattice. The axial lengths (a, b, c) and the axial angles
(alfa, beta, gamma) are the lattice constants of the unit cell

10. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
There are 7 different crystal classes according to the values of the axial lengths
and the axial Angles (lattice constants, a, b, c, alfa, beta, and gamma).
Cubic
Tetragonal
Orthorhombic
Rhombohedral
Hexagonal
Monoclinic
Triclinic
Also there are 4 types of unit cell
Simple
Body centered
Faced centered
Side centered b c a

11. Materials Engineering

12. Engineering Materials
Principle metallic crystal structure
1. Simple Cubic crystal structure
2. Body- Centered Cubic (BCC) crystal structure
3. Faced- centered Cubic (FCC) crystal structure
4. Hexagonal Close-Packed (HCP) crystal structure

13. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Body- centered cubic (BCC) crystal structure
The atoms in the BCC unit cell contact each other across the cube diagonal, so the relationship between the length of the cube side (a) and the atomic radius (R) is;
The central atom in the unit cell is surrounded by 8 nearest neighbor, therefore, the BCC unit cell has a coordination number of 8.

14. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Body- centered cubic (BCC) crystal structure
Atomic packing factor (APF);
The APF for the BCC unit cell is 68%; which means that 68% of the volume of the BCC unit cell is occupied by atoms and the remaining 32% is empty space.
Many metals (iron, chromium, vanadium) have the BCC crystal structure at room temperature.

15. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Body- centered cubic (BCC) crystal structure
The number of atoms in the BCC unit cell = 2
1 (at the center) + 8 * (1/8) = 2 atoms per unit cell
Example 1
Iron at 20 C is BCC with atoms of atomic radius nm. Calculate the lattice
constant (a) for the cube edge of the iron unit cell.
Example 2
Calculate the atomic packing factor (APF) for the BCC unit cell, assuming the atoms
to be hard spheres.

16. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Faced- centered cubic (FCC) crystal structure
The atoms in the FCC unit cell contact each other across the cube face diagonal, so the relationship between the length of the cube side (a) and the atomic radius (R) is;
X2=a2+a2 a x
90 deg a
Each atom is surrounded by 12 other atoms, therefore, the FCC unit cell has a coordination number of 12.

17. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Faced- centered cubic (FCC) crystal structure
Atomic packing factor (APF);
The APF for the FCC unit cell is 78%; which means that 78% of the volume of the FCC unit cell is occupied by atoms and the remaining 22% is empty space.
Many metals (copper, lead, nickel) have the FCC crystal structure at elevated temperatures (912 to 1394 C).

18. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Faced- centered cubic (FCC) crystal structure
The number of atoms in the FCC unit cell = 4
8 * (1/8) + 6 * (1/2) = 4 atoms per unit cell
Example 3
Calculate the atomic packing factor (APF) for the FCC unit cell, assuming the atoms
to be hard spheres.

19. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Hexagonal close- packed (HCP) crystal structure
60 deg c a
120 deg
Each atom is surrounded by 12 other atoms, therefore, the HCP unit cell has a coordination number of 12.

20. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Hexagonal close- packed (HCP) crystal structure
Atomic packing factor (APF);
The APF for the HCP unit cell is 78%; and equal to that of FCC.

21. Crystal and Amorphous Structure in Materials
Engineering Materials
Crystal and Amorphous Structure in Materials
Hexagonal close- packed (HCP) crystal structure
The number of atoms in the HCP unit cell = 2
4 * (1/6) + 4 * (1/12) + 1 (center) = 2 atoms per unit cell c a
Example 4
Calculate the volume of the zinc crystal structure unit cell by using the following
data: Pure zinc has the HCP crystal structure with lattice constants a= nm
and c= nm.

22. Engineering Materials
Volume, planar, and linear density unit-cell calculation
Example 5:
Copper has an FCC crystal structure and an atomic radius of nm. Assuming
the atoms to be hard sphere that touch each other along the face diagonals of the FCC
unit cell, calculate a theoretical value for the density of copper in megagrams per cubic
meter. The atomic mass of copper is g/mol.
1g = 10 -6Mg

23. Engineering Materials
Volume, planar, and linear density unit-cell calculation
Example
Calculate the planar atomic density on the (110) plane of the α iron BBC
lattice in atoms per square milimeters. The lattice constant of α is nm.

24. Engineering Materials
Volume, planar, and linear density unit-cell calculation
Example
Calculate the planar atomic density on the (110) plane of the α iron BBC
lattice in atoms per square milimeters. The lattice constant of α is nm.
(110) a

25. Engineering Materials
Volume, planar, and linear density unit-cell calculation
Example 6:
Calculate the linear atomic density in the (110) direction in the copper crystal
lattice in atoms milimeters. Copper is FCC and has a lattice constant of nm

26. Engineering Materials
Volume, planar, and linear density unit-cell calculation
No. of atomic diameters intersected by the
Length of line are =2 atoms. a
(110)
Length of line= length of face diagonal=
Example 6:
Calculate the linear atomic density in the (110) direction in the copper crystal
lattice in atoms milimeters. Copper is FCC and has a lattice constant of nm

27. Crystal and Amorphous Structure
Engineering Materials
Crystal and Amorphous Structure
in Materials
H W1
Molybdenum at 20 C is BCC and has an atomic radius of nm. Calculate a value
for its lattice constant a in nanometers.
Lithium at 20 C is BCC and has a lattice constant of nm. Calculate a value
for the atomic radius of a lithium atom in nanometers.
What is the coordination number for the atoms in the FCC crystal structure?
Gold is FCC and has a lattice constant of nm. Calculate a value for the
atomic radius of a gold atom in nanometers.
5. Calculate the atomic packing factor for the FCC structure.