Atomic Structure and Bonding CHAPTER 2 Atomic Structure and Bonding 1

Why Study Atomic Structure and Interatomic Bonding?

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

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)

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

Covalent Bonding - Examples F F F + F F F Bond Energy=160KJ/mol O + O O O O = O Bond Energy=28KJ/mol N N N N N + N Bond Energy=54KJ/mol 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

Metallic Bond of Zinc

Metallic Bond Types of Atomic Bonds Positive Ion Engineering Materials Types of Atomic Bonds Metallic Bond Positive Ion Valence electron charge cloud

Metallic bonding, particularly the outer electrons of the atom, accounts for many physical properties of metals, such as strength, malleability, ductility, thermal and electrical conductivity, opacity and luster As the number of bonding electrons increases, the more attraction thus the bonding energies and melting points of the metals also increases

The atoms in the bonded state are in a more stable energy condition than when they are unbonded Ionic bonds form between oppositely charged ions because there is a net decrease in potential energy of the ions after bonding The lattice energies and melting points of ionically bonded solids are relatively high

The high thermal and electrical conductivities of metals support the theory that some electrons are free to move through the metal crystal lattice. Most metals can be deformed a considerable amount without fracturing because the metal atoms can slide past each other without completely disrupting the metallically bonded structure

The bond energies and melting points of different metals vary greatly. In general, the fewer valence electrons per atom involve in metallic bonding, the more metallic the bonding is. That is, the valence electrons are freer to move ( bec few only). Metals with fewer delocalized electrons has low bonding energies and melting points ( bec electrons are less attracted in the ion so less amount of energy needed to break away the bond of ions )

Physical Properties explained through the Bonding Melting points and boiling points Metals tend to have high melting and boiling points because of the strength of the metallic bond. The strength of the bond varies from metal to metal and depends on the number of electrons which each atom delocalises into the sea of electrons, and on the packing.

Electrical conductivity Metals conduct electricity. The delocalised electrons are free to move throughout the structure in 3-dimensions. Thermal conductivity Metals are good conductors of heat. Heat energy is picked up by the electrons as additional kinetic energy (it makes them move faster). The energy is transferred throughout the rest of the metal by the moving electrons (free electrons )

Crystal and Amorphous Structure in Materials CHAPTER 3 Crystal and Amorphous Structure in Materials

Why study the structures of metals and ceramics?

A crystal's structure and symmetry play a role in determining many of its physical properties (opaque and translucent) thus the behavior of a material 20

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 STRUCTURE 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

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 Unit

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 constant of the unit cell

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 basic types of unit cell Simple Body centered Faced centered Side centered b c a

Materials Engineering

Engineering Materials Principal metallic crystal structures 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

Coordination number - the number of closes neighbors to which the atom is bonded

Body- centered cubic (BCC) crystal structure Examples are lithium, sodium, potassium, chromium, barium, vanadium and tungsten . Metals which have a bcc structure are usually harder and less malleable than close-packed metals such as gold. When the metal is deformed, the planes of atoms must slip over each other, and this is more difficult in the bcc structure

Body- centered cubic (BCC) crystal structure

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.

Body- centered cubic (BCC) crystal structure

Table 3.2

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.

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 0.124 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.

Faced- Centered Cubic (FCC) crystal structure Examples : aluminum, copper, gold, iridium, lead, nickel, platinum and silver.

Figure 3.6 Faced- centered cubic (FCC) crystal structure

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.

Figure 3.7 Faced- centered cubic (FCC) crystal structure

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 74%; which means that 78% of the volume of the FCC unit cell is occupied by atoms and the remaining 26% is empty space. Many metals (copper, lead, nickel) have the FCC crystal structure at elevated temperatures (912 to 1394 C).

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.

Hexagonal close- packed (HCP) crystal structure beryllium, cadmium, magnesium, titanium, zinc and zirconium.  

Hexagonal close- packed (HCP) crystal structure

Hexagonal close- packed (HCP) crystal structure

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.

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=0.2665 nm and c=0.4947 nm.

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 74%; and equal to that of FCC.

Hexagonal close- packed (HCP) crystal structure Question For an HCP unit cell (consider the primitive cell), (a) how many atoms are there inside the unit cell, (b)What is the coordination number for the atoms, (c ) what is the atomic packing factor (d) what is the ideal c/a ratio for HCP metals. And (e) repeat a through consïdering the “larger” cell.

Engineering Materials Volume, planar, and linear density unit-cell calculation Example 5: Copper has an FCC crystal structure and an atomic radius of 0.1278 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 63.54 g/mol. 1g = 1x10 -6 Mg 1 nm = 1x10-9 m

Engineering Materials Volume, planar, and linear density unit-cell calculation

Engineering Materials Volume, planar, and linear density unit-cell calculation Example Calculate the planar atomic density on the (110) plane of the α iron BCC lattice in atoms per square milimeters. The lattice constant of α is 0.287 nm. (110) a

Figure 3.20

Engineering Materials Volume, planar, and linear density unit-cell calculation Example 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 0.361 nm

Engineering Materials Volume, planar, and linear density unit-cell calculation No. of atomic diameters intersected by the Length of line are 0.5+1+0.5=2 atoms. a (110) Length of line= length of face diagonal=

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 0.361 nm