chapter 2author.uthm.edu.my/uthm/www/content/lessons/4781... · • molecules with metals...

CHAPTER 2

har/jkbr/fkmp/2016 Materials Science 1

2.1 Atomic structure and bonding

2.2 Crystal system and Bravais lattice

2.3 Metallic crystal structure

2.4 Crystallographic directions and planes

2.5 Crystal structure analysis (X-Ray Diffraction)


ISSUES TO ADDRESS...

• How do atoms assemble into solid structures?(for now, focus on metals)

• How does the density of a material depend onits structure?

• When do material properties vary with thesample (i.e., part) orientation?


ATOMBasic Unit of an Element

Diameter : 10 –10 m.Neutrally Charged

NucleusDiameter : 10 –14 m

Accounts for almost all massPositive Charge

Electron CloudMass : 9.109 x 10 –28 g

Charge : -1.602 x 10 –9 CAccounts for all volume

ProtonMass : 1.673 x 10 –24 g

Charge : 1.602 x 10 –19 C

NeutronMass : 1.675 x 10 –24 g

Neutral Charge

STRUCTURE OF ATOMS

2.1 ATOMIC AND BONDING STRUCTURE

Fundamental concepts → atomic number, atomic weight, atomic mass unit (amu), mole etc.

Electron in atoms → atomic models, quantum numbers, electron configurations, periodic table.

Primary interatomic bonds → ionic bonding, covalent bonding, metallic bonding.

Secondary bonding (Van Der Waals) → hydrogen bonding, polar-molecule-induced dipole bonds, permanent dipole bonds.


Nucleus: Z = # protons

2

orbital electrons: n = principal quantum number

n=3 2 1

= 1 for hydrogen to 94 for plutonium

N = # neutrons

Atomic mass A ≈ Z + N

Adapted from Fig. 2.1, Callister 6e.

BOHR ATOM

• have discrete energy states• tend to occupy lowest available energy state.

3

Inc

rea

sin

g e

ne

rgy

n=1

n=2

n=3

n=4

1s

2s

3s

2p

3p

4s

4p3d

Electrons...


ELECTRON ENERGY STATES

4

• have complete s and p subshells• tend to be unreactive.

Stable electron configurations...

Z Element Configuration

2 He 1s2

10 Ne 1s22s22p6

18 Ar 1s22s22p63s23p6

36 Kr 1s22s22p63s23p63d104s24p6

Adapted from Table 2.2, Callister 6e.

STABLE ELECTRON CONFIGURATIONS

5

• Why? Valence (outer) shell usually not filled completely.

• Most elements: Electron configuration not stable.

Element Hydrogen Helium Lithium Beryllium Boron Carbon ... Neon Sodium Magnesium Aluminum ... Argon ... Krypton

Atomic # 1 2 3 4 5 6

10 11 12 13

18 ... 36

Electron configuration 1s1

1s2 (stable) 1s22s1 1s22s2 1s22s22p1 1s22s22p2 ...

1s22s22p6 (stable) 1s22s22p63s1 1s22s22p63s2 1s22s22p63s23p1 ...

1s22s22p63s23p6 (stable) ...

1s22s22p63s23p63d104s246 (stable)

Adapted from Table 2.2, Callister 6e.

SURVEY OF ELEMENTS

6

• Columns: Similar Valence Structure

Electropositive elements:Readily give up electronsto become + ions.

Electronegative elements:Readily acquire electronsto become - ions.

He

Ne

Ar

Kr

Xe

Rn

ine

rt g

as

es

ac

ce

pt

1e

a

cc

ep

t 2

e

giv

e u

p 1

e

giv

e u

p 2

e

giv

e u

p 3

e

F Li Be

Metal

Nonmetal

Intermediate

H

Na Cl

Br

I

At

O

S Mg

Ca

Sr

Ba

Ra

K

Rb

Cs

Fr

Sc

Y

Se

Te

Po


THE PERIODIC TABLE

7

• Ranges from 0.7 to 4.0,

Smaller electronegativity Larger electronegativity

He -

Ne -

Ar -

Kr -

Xe -

Rn -

F 4.0

Cl 3.0

Br 2.8

I 2.5

At 2.2

Li 1.0

Na 0.9

K 0.8

Rb 0.8

Cs 0.7

Fr 0.7

H 2.1

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

• Large values: tendency to acquire electrons.

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by CornellUniversity.

ELECTRONEGATIVITY

Na (metal) unstable

Cl (nonmetal) unstable

electron

+ - Coulombic Attraction

Na (cation) stable

Cl (anion) stable

8

• Occurs between + and - ions.

• Requires electron transfer.

• Large difference in electronegativity required.

• Example: NaCl

IONIC BONDING

9

• Predominant bonding in Ceramics

Give up electrons Acquire electrons

He -

Ne -

Ar -

Kr -

Xe -

Rn -

F 4.0

Cl 3.0

Br 2.8

I 2.5

At 2.2

Li 1.0

Na 0.9

K 0.8

Rb 0.8

Cs 0.7

Fr 0.7

H 2.1

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

CsCl

MgO

CaF2

NaCl

O 3.5

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by CornellUniversity.

EXAMPLES: IONIC BONDING

10

• Requires shared electrons

• Example: CH4

C: has 4 valence e,needs 4 more

H: has 1 valence e,needs 1 more

Electronegativitiesare comparable.

shared electrons from carbon atom

shared electrons from hydrogen atoms

H

H

H

H

C

CH4


COVALENT BONDING

11

• Molecules with nonmetals

• Molecules with metals and nonmetals

• Elemental solids (RHS of Periodic Table)

• Compound solids (about column IVA)

He -

Ne -

Ar -

Kr -

Xe -

Rn -

F 4.0

Cl 3.0

Br 2.8

I 2.5

At 2.2

Li 1.0

Na 0.9

K 0.8

Rb 0.8

Cs 0.7

Fr 0.7

H 2.1

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

SiC

C(diamond)

H2O

C 2.5

H2

Cl2

F2

Si 1.8

Ga 1.6

GaAs

Ge 1.8

O 2.0

co

lum

n I

VA

Sn 1.8

Pb 1.8

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 isadapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University.

EXAMPLES: COVALENT BONDING

12

• Arises from a sea of donated valence electrons

(1, 2, or 3 from each atom).

• Primary bond for metals and their alloys

+ + +

+ + +

+ + + Adapted from Fig. 2.11, Callister 6e.

METALLIC BONDING

13

Arises from interaction between dipoles

• Permanent dipoles-molecule induced

• Fluctuating dipoles

+ - secondary bonding

+ -

H Cl H Clsecondary bonding

secondary bonding

HH HH

H2 H2

secondary bonding

ex: liquid H2asymmetric electron clouds

+ - + -secondary

bonding

-general case:

-ex: liquid HCl

-ex: polymer


Adapted from Fig. 2.14,Callister 6e.

Adapted from Fig. 2.14,Callister 6e.

SECONDARY BONDING

14

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

15

• Bond length, r

• Bond energy, Eo

F F

r

• Melting Temperature, Tm

Eo=

“bond energy”

Energy (r)

ro r

unstretched length

r

larger Tm

smaller Tm

Energy (r)

ro

Tm is larger if Eo is larger.

PROPERTIES FROM BONDING: TM

16

• Elastic modulus, E

• E ~ curvature at ro

cross sectional area Ao

L

length, Lo

F

undeformed

deformed

L F Ao

= E Lo

Elastic modulus

r

larger Elastic Modulus

smaller Elastic Modulus

Energy

ro unstretched length E is larger if Eo is larger.

PROPERTIES FROM BONDING: E

17

• Coefficient of thermal expansion,

• ~ symmetry at ro

is larger if Eo is smaller.

L

length, Lo

unheated, T1

heated, T2

= (T2-T1) L

Lo

coeff. thermal expansion

r

smaller

larger

Energy

ro

PROPERTIES FROM BONDING:

18

Ceramics

(Ionic & covalent bonding):

Metals

(Metallic bonding):

Polymers

(Covalent & Secondary):

secondary bonding

Large bond energy

large Tm

large Esmall

Variable bond energy

moderate Tm

moderate Emoderate

Directional Properties

Secondary bonding dominatessmall Tsmall Elarge

SUMMARY: PRIMARY BONDS

2.2 CRSYTAL SYSTEMS & BRAVAIS LATTICE SYSTEM

Crystal – a solid composed of atoms, ions or molecules arranged in a pattern that is repeated in three dimensions.

Solid materials – classified according to the regularity with which atoms or ions are arranged.

Atomic arrangement – determine the solid materials microstructure and properties.

Examples : ductility and strength


SOLID

MATERIALS


- Atoms are positioned/situated in an orderly and repeated pattern, 3D arrays

- Examples : all metals, many ceramic and some polymers

- Atoms are distributed randomly and disordered.

-Occurs for– complex structure

- rapid cooling

- Examples : polymer & a few ceramics

MATERIALS AND PACKING


• have the simplest crystal structures.• tend to be densely packed.• have several reasons for dense packing:• Typically, only one element is present, so allatomic radii are the same.• Metallic bonding is not directional.• Nearest neighbor distances tend to be small inorder to lower bond energy.

METALLIC CRYSTALS…

CRYSTAL SYSTEM

Crystal structure – a regular three-dimensional pattern of atoms or ions in space.

Space lattice – a three-dimensional array of points coinciding with atom positions (or sphere centers).

Lattice point – atom position in the crystal lattice

Unit cell – small groups of atoms form a repetitive pattern (Figure 1)


Crystal Structure


Geometry of a unit cell

Space lattice

7 systems of unit cell for metallic crystal structures

(Table 1 & Figure 2) :

◦ Ciubic

◦ Tetragonal

◦ Orthorhombic

◦ Rhombohedral

◦ Hexagonal

◦ Monoclinic

◦ Triclinic

These systems can be divided into 14 types of standard unit cell geometries – Bravais Unit Cell System


Table 1 : Lattice parameter for the 7 Crystal Systems



Figure 2 :

14 unit cell geometries (BravaisUnit Cell)

2.3 METALLIC CRYSTAL STRUCTURE

Most of the metallic materials - 3 types of crystal structure :

Face-centered cubic (FCC) example : γ – Fe, Cu, Al, Arg, Au

Body-centered cubic (BCC)example : α – Fe, Cr, W

Hexagonal close-packed (HCP)example: Mg, Ti, Zn, Cd.


Atomic packing factor (APF);

Vs = nVa

Vc Vc

Coordination number;

number of nearest-neighbor or touching atoms for each atom in the crystal structure


(Total sphere volume)

THE SIMPLE CUBIC STRUCTURE


• Rare due to poor packing (only Po has this structure)

THE BODY-CENTERED CUBIC STRUCTURE


34Ra

where

a = unit cell length

R = atomic radius

Number of atoms per unit cell = 2

Total sphere volume, Vs

Unit cell volume, Vc

Atomic Packing Factor, (APF)

68% of the unit cell is occupied with atoms.

3

34)2( RVs

3aVc


3)34( R

3

3

)3

4(

)3

4(2

R

R

V

VAPF

c

s 68.0

THE FACE-CENTERED CUBIC STRUCTURE


222

4

42

RRa

Ra

where

a = unit cell length

R = atomic radius

Atoms located at each of the corners and the centers of all the cube facesExample : aluminium, silver, gold, copper etcNumber of atoms per unit cell = 4Total sphere volume, Vs

Unit cell volume, Vc

Atomic Packing Factor, (APF)

74% of the unit cell space is packed with atoms.


3

34)4( RVs

3aVc 3)24( R

3

3

)2

4(

)3

4(4

R

R

V

VAPF

c

s 74.0

HEXAGONAL CLOSE-PACKED


Theoretical density, ρ = n x A Vc x NA

wheren = number of atoms associated with

each unit cell A = atomic weightVc = volume of the unit cell

(NA) = Avogadro’s number(6.023x 1023

atoms/mol)


Example :

Copper has the FCC crystal structure with atomic radius, R =

0.1278nm. Assume the atoms to be hard spheres and packed as

close together as possible along the FCC unit cell cross-section.

Calculate the theoretical volume density of copper in mg/m3.

(Atomic mass of Cu = 63.54 g/mol)


SOLUTION :

For FCC,

24

42

Ra

Ra

nm

nm

361.0

2

)1278.0(4

Theoretical volume density,

6.02 x 1023 atoms → 63.54 g


unitcellVolume

unitcellMassPv

/

/

)10(1002.6

)4(54.63 3

23mg

mg281022.4

4 atoms →

Mass of unit cell, m =

393 )10361.0( a

3291070.4 m

329

28

1070.4

1022.4

m

mg

v

mPv

3/98.8 mmg 3/98.8 cmg

Volume of unit cell, V =

@

2.4 CRYSTALLOGRAPHIC PLANES AND DIRECTIONS

Deformation in metal such as forging, drawing etc. – moves according to certain planes and directions in the crystal structure.

Example : Ferum (Fe) – Magnetic effect is strong in [100] direction compare to [111] direction.

Miller Indices – used to explain the planes and directions position in crystal.



Atomic position in BCC unit cellAtomic position in cubic unit cell

CRYSTALLOGRAPHIC DIRECTIONS

A line or vector between 2(two) points.

Steps in determination of the three directional indices :

1. A vector of convenient length is positioned such that it passes through the origin of the coordinate system. Any vector may be translated throughout the crystal lattice without alteration, if parallelism is maintained.

2. The length of the vector projection on each of the three axes is determined; they are measured in terms of the unit cell dimensions a, b and c.

3. These three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values.

4. The three direction indices are enclosed by square brackets with no separating commas, thus : [ ].

→ +ve

→ -ve


uvw

wvu

All parallel direction vectors have the same direction indices.

Crystallographically equivalent directions – the atom spacing along each direction is the same.

other directions of a family <110> and <111>

- INDICES OF A FAMILY

]100[],010[],001[],001[],010[],100[100



111

X

0

Z

Y

011

0

X

Y

Z

212

2

1

0 120

0

X

Y

Z

2

1


320

3

2

0

X

Z

Y

221

2

1

0Y

X

Z

210

2

1

0

x

y

Z

0

2

1

Z

Y

X

201

CRYSTALLOGRAPHIC PLANES

The reciprocals of the fractional intercepts (with fractions cleared) which the plane makes with the crystallographic x, y and z axes of the three nonparallel edges of the cubic unit cell.

Procedure for determining the Miller indices for a cubic crystal plane is as follows :

1. Choose the plane that does not pass through the origin at

(0,0,0).

2. Determine the intercepts of the plane in terms of the crystallograhic x, y and z axes for a unit cube. These intercepts may be fractions.

3. Form the reciprocals of these intercepts. A plane that parallela an axis may be considered to have an infinite intercept, and therefore, a zero index.

4. If necessary, these three numbers are changed to the set of smallest integers by multiplication or division by a common factor.

5. Finally, the integer indices, not separated by commas, are enclosed with parentheses, thus : (hkl)



Z

X

Y

(1 0 1)

(111)

Nota :

= ( _ _ _ )

= ( 0 _ _ )

= ( _ 0 _ )

= ( _ _ 0 )

x

z

y

Problems

Draw the following crystallograhic planes in cubic unit cells :

)234)((

)221)((

)110)((

)101)((

d

c

b

a


zz

y

(a)

x

z

x

y(b)

PLANES OF A FAMILY – sets of equivalent lattice planes are related by the symmetry of the crystal system.

An important relationships for the cubic system (only)

→ the direction indices of a direction perpendicular to a crystal plane are the same as the Miller indices of that plane.

Example : the [100] direction is perpendicular to the (100) crystal plane.

001,010,100}100{



Determine the planes indices of the cubic direction shown in figures below.

(a) (b) (c)

(d) (e) (f)

CRYTALLOGRAPHIC PLANES IN HEXAGONAL UNIT CELLS


)(hkil

The four coordinate axes (a1, a2, a3

and c) of he HCP crystal structure unit cell

The HCP crystal planes indices, are donated by the letters h, k, i and l.

Enclosed in parentheses as

a1, a2 and a3 – basal axes which make 120 with each other.

c axis – the vertical axis located at the center of the unit cell.

PLANAR ATOMIC DENSITY (P)

plane of Area

plane selectedby dintersecte are centers whoseatoms of no. Equiv.

P


• The units of planar density = m-2 , nm-2

Example Problem

Calculate the planar atomic density p on the plane of the iron BCC lattice in atoms per square millimeter. The lattice constant, a of iron is 0.287 nm.

Figure : (a) A BCC atomic-site unit cell showing a shaded (110) plane

(b) Areas of atoms in BCC unit cell cut by the (110) plane.


Solution :

Equivalent number of atoms intersected by the (110) plane ;

1 atom at the center + (4 x ¼ atoms at four corners of plane) = 2 atoms

2atoms/mm

1317.2x10

2nm17.2atoms/

P

P

2nm) (0.287 2

atoms 2


Area of the plane (110) ;

(√2 a)(a) = √2 a2

The planar atomic density is

LINEAR ATOMIC DENSITY (l)

vector direction of Length

vector direction on centered atoms ofNumber

l


• The units of linear density = m-1 , nm-1

Example Problem

Calculate the linear atomic density l in the [110] direction in the copper crystal lattice in atoms per millimeter. Copper is FCC and has a lattice constant of 0.361 nm.

Figure : Diagram for calculating the atomic linear density in the [110] direction in an FCC unit cell


Solution :

The number of atomic diameters intersected by this length of line are

½ + 1 + ½ = 2 atoms.

atoms/mm 6

10 x 3.92atoms/nm 3.92

l

l

nm) (0.361 2

atoms 2


Length of the direction vector = √2 a

The linear atomic density is

2.5 Crystal Structure Analysis (X-Ray Diffraction)

Spacing between planes → distance between 2 parallel planes having similar miller indices, given by (hkl is miller indices for plane).

The analysis is done by using the X-Ray Diffractometer Machine.

hkd


XRD Profile


Carbon monoxidemolecules arrangedon a platinum (111)surface.

Iron atoms arrangedon a copper (111)surface. These Kanjicharacters representthe word “atom”.

Atoms can be arranged and imaged!!!

SCANNING TUNNELING MICROSCOPY

Polymorph materials – example: Iron (Fe)

SUMMARY


Atoms may assemble into crystalline or amorphousstructures.

We can predict the density of a material, providedwe know the atomic weight, atomic radius, andcrystal structure (e.g., FCC, BCC, HCP).

Material properties generally vary with singlecrystal orientation (i.e., they are anisotropic), butproperties are generally non-directional (i.e., theyare isotropic) in polycrystals with randomlyoriented grains.

chapter 2author.uthm.edu.my/uthm/www/content/lessons/4781... · • molecules with metals...

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