Jif 419 webex1

JIF 419

Materials Science

Course Manager: Assoc Prof Dr Saw Kim Guan

Textbook: Materials Science and Engineering (4th

Ed) by Callister & Rethwisch

Academic Planner

• 2 assignments: 30 Nov 2015 (1st assignment)

15 Feb 2016 (2nd assignment)

• Web-conference sessions: 3 before Intensive

• Intensive course (19 Jan 2016 - 7 Feb 2016)

• Continuous Examination: types of materials, atomic structure, bonding in solids, crystal structures, crystallographic points, directions and planes, point defects

• The contributions of course work and examinations:

Final Exam: 70%

Assignments: 10%

Continuous exam: 20%

2

Classification of materials(metals)

• Atoms in metals are very orderly and dense

• Metallic materials have large numbers of non-localized electrons (electrons NOT bound to particular atoms)

• These delocalized electrons make metals to be good electrical & heat conductor and NOT transparent to visible light.

Classification of materials(ceramics)

• Ceramics are compounds b/w metallic & non-metallic elements

• Mostly oxides, nitrides and carbides

• e.g. aluminium oxide, silicon dioxide, silicon carbide, silicon nitride, porcelain

• Hard, brittle, strong, able to stand heat

• Modern usage – engine parts, cookware, cutlery

Classification of materials(polymers)

• Many polymers are organic compounds that are chemically based on carbon, hydrogen and non-metallic elements

• Consist of very large molecular structures, often chainlike and have a backbone of carbon atoms

• e.g. nylon, polyethylene, silicone rubber

• Ductile and pliable, relatively inert chemically

• Low electrical conductivity and non-magnetic

Classification of materials(composites)

• Composites consists of two or more individual materials

• Designed to achieve a combination of properties that is not present in any single material

• designed to incorporate the best characteristics of each of the component material

• E.g. fibreglass – strong, low density

• Modern usage – sport equipment, engine parts

7

Atomic Structure

• Some of the following properties

1) Electrical

2) Thermal

3) Optical

are determined by electronic structure

8

Electron Configurations• Valence electrons – electrons in unfilled shells

• Filled shells more stable

• Valence electrons are most available for bonding and tend to control the chemical properties

– example: C (atomic number = 6)

1s2 2s2 2p2

valence electrons

9

The Periodic Table• Columns: Similar Valence Structure

Adapted from

Fig. 2.8,

Callister &

Rethwisch 9e.

Electropositive elements:

Readily give up electrons

to become + ions.

Electronegative elements:

Readily acquire electrons

to become - ions.

giv

e u

p 1

e-

giv

e u

p 2

e-

giv

e u

p 3

e- in

ert

gases

accept 1e

-

accept 2e

-

O

Se

Te

Po At

I

Br

He

Ne

Ar

Kr

Xe

Rn

F

ClS

Li Be

H

Na Mg

BaCs

RaFr

CaK Sc

SrRb Y

10

• Ranges from 0.9 to 4.1,

Smaller electronegativity Larger electronegativity

• Large values: tendency to acquire electrons.

Electronegativity

11

Each H: has 1 valence e-,

needs 1 more

Electronegativities

are the same.

Fig. 2.12, Callister & Rethwisch 9e.

Covalent Bonding• similar electronegativity share electrons

• Example: H2

shared 1s electronfrom 2nd hydrogen

atom

H

H2

shared 1s electronfrom 1st hydrogen

atom

H

12

Covalent Bonding: Carbon sp3

• Example: CH4

C: has 4 valence e-,

needs 4 more

H: has 1 valence e-,

needs 1 more

Electronegativities of C and H

are comparable so electrons

are shared in covalent bonds.Fig. 2.15, Callister & Rethwisch 9e.(Adapted from J.E. Brady and F. Senese, Chemistry:

Matter and Its Changes, 4th edition. Reprinted with

permission of John Wiley and Sons, Inc.)

13

Ionic bond – metal + nonmetal

donates accepts

electrons electrons

Dissimilar electronegativities

ex: MgO Mg 1s2 2s2 2p6 3s2 O 1s2 2s2 2p4

[Ne] 3s2

Mg2+ 1s2 2s2 2p6 O2- 1s2 2s2 2p6

[Ne] [Ne]

14

• Occurs between + and - ions.

• Requires electron transfer.

• Large difference in electronegativity required.

• Example: NaCl

Ionic Bonding

Na (metal)

unstable

Cl (non-metal)

unstable

electron

+ -Coulombic

Attraction

Na (cation)

stable

Cl (anion)

stable

15

Primary Bonding• Metallic Bond -- delocalized as electron cloud

• Ionic-Covalent Mixed Bonding

% ionic character =

where XA & XB are Pauling electronegativities

%)100(x

Ex: MgO XMg = 1.3XO = 3.5

16

Arises from interaction between dipoles

• Permanent dipoles-molecule induced

• Fluctuating dipoles

-general case:

-ex: liquid HCl

-ex: polymer

Adapted from Fig. 2.20,

Callister & Rethwisch 9e.

Adapted from Fig. 2.22,


Secondary Bonding

asymmetric electronclouds

+ - + -secondary

bonding

HH HH

H 2 H 2

secondary bonding

ex: liquid H 2

H Cl H Clsecondary bonding

secondary bonding

+ - + -

secondary bonding

17

Type

Ionic

Covalent

Metallic

Secondary

Bond Energy

Large!

Variable

large-Diamond

small-Bismuth

Variable

large-Tungsten

small-Mercury

smallest

Comments

Nondirectional (ceramics)

Directional

(semiconductors, ceramics

polymer chains)

Nondirectional (metals)

Directional

inter-chain (polymer)

inter-molecular

Summary: Bonding

18

• atoms pack in periodic, 3D arrays

Crystalline materials...

-metals

-many ceramics

-some polymers

• atoms have no periodic packing

Noncrystalline materials...

-complex structures

-rapid cooling

crystalline SiO2

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.11(b),


Adapted from Fig. 3.11(a),


Materials and Packing

Si Oxygen

• typical of:

• occurs for:

19

Crystal Systems

7 crystal systems

14 crystal lattices

Unit cell: smallest repetitive volume which

contains the complete lattice pattern of a crystal.

a, b, and c are the lattice constants

Credit: www.alientravelguide.com

21

Crystallographic Directions

1. Determine coordinates of vector tail, pt. 1:

x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2.

2. Tail point coordinates subtracted from head

point coordinates.

3. Normalize coordinate differences in terms

of lattice parameters a, b, and c:

4. Adjust to smallest integer values

5. Enclose in square brackets, no commas

[uvw]ex:

pt. 1 x1 = 0, y1 = 0, z1 = 0

=> 1, 0, 1/2

=> [ 201 ]

z

x

Algorithm

y

=> 2, 0, 1

pt. 2

headpt. 1:

tail

pt. 2 x2 = a, y2 = 0, z2 = c/2

22

Crystallographic Directions

-4, 1, 2

families of directions <uvw>

z

x

where the overbar represents a negative

index

[ 412 ]=>

y

Example 2:

pt. 1 x1 = a, y1 = b/2, z1 = 0

pt. 2 x2 = -a, y2 = b, z2 = c

=> -2, 1/2, 1

pt. 2

head

pt. 1:

tail

Multiplying by 2 to eliminate the fraction

Crystal Structures

• A lattice is a 3D array of points coinciding with atomic positions

• The atomic order in crystalline solids indicates that small groups of atoms form a repetitivepattern

• The repeat entities are called unit cells

Metallic Crystal Structures

• The atomic bonding is metallic and non-directional in nature

• 3 crystal structures for most common metals –face-centered cubic structure (FCC), body-centered cubic structure & hexagonal close-packed structure

Face-centered cubic structure (FCC)

• Has a unit cell of cubic geometry

• Atoms located at each corner and the centers of all the cubic faces

• The coordination number (the no. of nearest neighbour atoms) is 12

• The atomic packing factor (sum of the sphere volumes of all atoms within a unit cell) is 0.74

FCC unit cell

Hard sphere unit cell representation; the unit cell contains 4 atoms

Reduced-sphere unit cell representation

FCC coordination number

Hard sphere unit cell representation shows that:

The front face atom X has four corner nearest neighbouratoms surrounding it (indicated as 1, 2, 3, 4), four face atoms that are in contact from behind (two of these are indicated as I and II), and four other face atoms residing in the next unit cell to the front, which is not shown.

FCC

Hard sphere unit cell representation

Using Pythagoras’ theorem we have

a2 + a2 = (4R)2

Thus by simplifying we have

a (unit cell length) = 2R 2 = 2.83R

FCC unit cell volume

Hard sphere unit cell representation

V= a3 = (2R 2)3 =8×2×R3 2=16 R3 2

Atomic packing factor for FCC

The atomic packing factor is the fraction of the solid sphere volume in a unit cell

APF = vol of atoms in unit cell/total unit cell vol

Note that there are 4 atoms per FCC unit cell

Vol of atoms = 4 × =

Total unit cell vol = 16 R3 2

34

3R 316

3R

APF = = 0.74 3

3

16

3

16 2

R

R

Body centered cubic structure

• The BCC structure has a cubic unit cell with atoms located at all eight corners and one atom at the cube center

• The single atom at the center is wholly contained within the unit cell

• The coordination number is 8

• The atomic packing factor is 0.68

BCC unit cell

Hard sphere unit cell representation; the unit cell contains 2 atoms

Reduced-sphere unit cell representation

The unit cell length a is 2.31 R

BCC coordination number

Hard sphere unit cell representation shows that:

Each center atom X has the eight corner atoms as it nearest neighbours (indicated by 1, 2, …7. The last atom 8 is hidden behind X and not shown)

Therefore the coordination no. is 8 for BCC structure

34

Atomic Packing Factor: BCC

APF =

4

3

π ( 3 a/4 ) 32

atoms

unit cell atom

volume

a 3

unit cell

volume

length = 4R =

Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

aRAdapted from

Fig. 4.1(a), Callister &

Rethwisch 9e.

a

a2

a3

Source: www. chemprofessor.com

Source: practicalmaintenance.net

36

• Coordination # = 12

• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection

Adapted from Fig. 4.3(a),


Hexagonal Close-Packed Structure

(HCP)

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

• c/a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

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