1. Elements of Materials Science and Engineering
ChE 210
Chapter 2

2. Chapter Outline
Individual atoms and ions
Molecules
Macromolecules
Three dimensional bonding
Interatomic distances
Generalization based on atomic bonding
Understanding of interatomic bonding is the first step towards understanding/explaining materials properties

3. Individual atoms and ions
Atoms = nucleus (protons and neutrons) + electrons
Charges:
Electrons and protons have negative and positive charges of the same magnitude, 1.6 × Coulombs. Neutrons are electrically neutral.
Masses:
Protons and Neutrons have the same mass, 1.67 × kg. Mass of an electron is much smaller, 9.11 × kg and can be neglected in calculation of atomic mass.
# protons gives chemical identification of the element
# protons = atomic number (Z)
# neutrons defines isotope number
The atomic mass (A) = mass of protons + mass of neutrons

4. Atomic mass units. Atomic weight.
The atomic mass unit (amu) is often used to express atomic weight. 1 amu is defined as 1/12 of the atomic mass of the most common isotope of carbon atom that has 6 protons (Z=6) and six neutrons (N=6).
Mproton ≈ Mneutron = 1.66 x g = 1 amu.
The atomic mass of the 12C atom is 12 amu.
The atomic weight of an element = weighted average of
the atomic masses of the atoms naturally occurring isotopes. Atomic weight of carbon is amu.
The atomic weight is often specified in mass/mole.

5. Atomic mass units. Atomic weight.
A mole is the amount of matter that has a mass in grams equal to the atomic mass in amu of the atoms (A mole of carbon has a mass of 12 grams).
The number of atoms in a mole is called the Avogadro number, Nav = × 1023.
Nav = 1 gram/1 amu.
Example:
Atomic weight of iron = amu/atom = g/mol

6. Calculations Diamond (carbon):  = 3.5 g/cm3, M = 12 g/mol
The number of atoms per cm3, n, for material of density  (g/cm3) and atomic mass M (g/mol):
n = Nav x  / M
Graphite (carbon):  = 2.3 g/cm3, M = 12 g/mol
n = 6×1023 atoms/mol × 2.3 g/cm3/12 g/mol = 11.5 x 1022 atoms/cm3
Diamond (carbon):  = 3.5 g/cm3, M = 12 g/mol
n = 6×1023 atoms/mol x 3.5 g/cm3/12 g/mol = 17.5 x1022 atoms/cm3
Water (H2O) d = 1 g/cm3, M = 18 g/mol
n = 6×1023 molecules/mol x 1 g/cm3/18 g/mol = 3.3 x 1022 molecules/cm3

7. Calculations
For material with n = 6 × 1022 atoms/cm3 we can calculate the mean distance between atoms L = (1/n)1/3 = 0.25 nm.
! the scale of atomic structures in solids – a fraction of 1 nm or a few A.

9. The Periodic Table
Elements in the same column (Elemental Group) share similar properties. Group number indicates the number of electrons available for bonding.
0: Inert gases (He, Ne, Ar...) have filled subshells: chem. Inactive
IA: Alkali metals (Li, Na, K…) have one electron in outermost occupied s subshell - eager to give up electron – chem. Active
VIIA: Halogens (F, Br, Cl...) missing one electron in outermost occupied p shell - want to gain electron - chem. active

10. Periodic Table - Electronegativity
Electronegativity - a measure of how willing atoms are to accept electrons Subshells with one electron - low electronegativity.
Subshells with one missing electron –high electronegativity.
Electronegativity increases from left to right
Metals are electropositive – they can give up their few valence electrons to become positively charged ions

11. Bonding Energies and Forces

12. Bonding Energies and Forces
The repulsion between atoms, when they are brought close to each other, is related to the Pauli principle: when the electronic clouds surrounding the atoms starts to overlap, the energy of the system increases abruptly.
The origin of the attractive part, dominating at large distances, depends on the particular type of bonding.

13. Bond Angles
Hybridized orbitals are formed by linear combination of pure atomic orbitals
The type of hybridization determines the bond angle in the molecule:
s + 3p 4 Sp BA = o
s + 2p  3 sp BA = 120o
s + p  2 sp BA = 180o

15. In symmetrical molecules, e.g. CH4
Atoms are equally spaced around the central atom.
~ placing the carbon at the cube centre and pointing the orbitals towards four of the 8 corners.
Distortion occurs if some of electrons are present as lone pair, e.g. H2O and NH3

16. Delocalized electrons
 bond is a single covalent bond and is formed by sharing of two electrons   
Not all the valence electrons are localized e.g. benzene

17. Example 2-2.1, p 30
How much energy is required or released if 2.6 Kg of acetylene C2H2 react with hydrogen to form C2H4?
One triple and one H—H bond are eliminated
Two C—H and one double bonds are formed
H—H KJ/mol C—H KJ/mol
C=C KJ/mol C‬≡C KJ/mol

18. Example p 31
How many delocalized electrons in naphthalene molecule
How many wave patterns (electron states
Formula: C10H8
Total valence electrons = (4/C)(10 C) + (1/H)(8 H) = 48
Electrons in bonds: C—H 8x2 = 16
C—C 11x2 = 22
Reminder = 48-(16+22)= 10 delocalized electrons
Electron states = number of carbon atoms

19. Macromolecules (polymers)
Large number of small repeated units called mers
High molecular weight compounds
Natural and synthetic
Linear or branched
Thermoplastic or thermosetting
Melting range and not melting points
No boiling points

20. Linear Molecules
Linear polymeric chains such as polyvinyl chloride (PVC)
n = degree of polymerization: the number of mers per molecule