ME 210: Material Science
Atomic Bonding
Dr. Aaron L. Adams,
Assistant Professor
Mechanical Engineering Department
Alabama A&M University
Fall 2014, Lecture 2
Atomic Bonding
Learning Objectives
• What promotes bonding?
• Describe ionic, covalent, and metallic, secondary
(i.e., hydrogen and van der Waals) bonds.
• Which materials exhibit each of these bonding
types?
• Describe interatomic forces and how they influence
structure and properties.
READING: Chapter 2, pages 17-37
Structure Determines Properties
It all starts with atoms (structure & bonding)
Macrostructure
Decreasing size …..powers of ten Microstructure
~103 m
More
Fe
More
C
Atomic
structure
2 3
54 nm
~10-6 m
(1 µm)
~10-10 m
(1 Å)
52 n
m
122 nm
Atoms Bond
► Atoms are the smallest unit of matter that retains the
identity of the substance.
What does this image tell us? HRTEM image of a crystalline material A quartz crystal In materials,
atoms
assemble in
specific
ways to build
up the
structure of a
material.
1 4
Atomic Structure
►  Some properties are determined by the
electronic structure of an atom.
1.  Chemical
2.  Electrical
3.  Thermal
4.  Optical
5.  Mechanical behavior
5
Atomic Structure
► Atoms are composed of three kinds of subatomic
particles:
•  Electrons:
–  Negatively charged (–)
–  Mass = 9.11 x 10-31 kg
•  Protons:
–  Positively charged (+)
–  Mass = 1.67 x 10-27 kg
•  Neutrons
–  No charge (i.e., electrically neutral)
–  Mass = 1.67 x 10-27 kg
6
Atomic Structure
► Atomic number (Z):
•  Characterizes each element
•  This is the # of protons in nucleus of an atom.
(= # of electrons in a neutral atom)
► Atomic mass (A):
•  Summed masses of protons and neutrons in the
nucleus of an atom.
•  Also equals the mass in grams of Avogadro’s
number (i.e., 6.022 × 1023) of atoms or molecules.
•  Units = g/mol
7
Atomic Structure
► Atomic mass - continued
•  Atomic mass unit (amu):
–  Alternative unit of measure for atomic mass.
–  Equal to 1/12 the mass of carbon 12.
–  1 amu/atom = 1 g/mol
C: 12.011 amu/atom OR AC =12.011 g/mol
H: 1.008 amu/atom OR AH = 1.008 g/mol
etc…
8
In neutral atoms:
► # protons = # electrons
(+1)
(-1)
Consider Ca (calcium, Z = 20):
•  If 20 protons are present in an atom then there must be
20 electrons to balance the overall charge — this
makes the atom neutral.
•  The neutrons have no charge; therefore the # does not
need to equal the # protons or # electrons.
9
How many subatomic particles are
there in an atom?
►  Atomic number (Z): this number
indicates the number of protons
in an atom.
•  Ex: Hydrogen’s atomic number
is 1. # protons?
–  So hydrogen has 1 proton
•  Ex: Carbon’s atomic number is
6. # protons?
–  So carbon has 6 protons
**The number of protons identifies
the atom.
Increasing # of electrons (protons) Atomic Structure
Bohr Model of the atom
“Atoms are treated like a solar system”
-­‐ The 3rd orbital
can hold up to
18 e-
-­‐ -­‐ -­‐ K -­‐ The 4th orbital
and any after
can hold up to
32 e-
-­‐ -­‐ L N M “Orbitals/Shells”
Nucleus consists of
protons and neutrons
-­‐ -­‐ The 1st orbital can
hold up to 2 e-
-­‐ The 2nd orbital can
hold up to 8 e-
INSIDE the NUCLEUS • Protons have (+) charge • Neutrons have (0) charge Have same mass (~10-­‐27 kg) # depends on atom type Mass of atom dep. on # proton & neutrons. OUTSIDE the NUCLEUS • Electrons circle the nucleus in specific orbitals • Electrons have (–) charge • Have very liFle mass (~10-­‐31 kg) 11
Electronic Structure
► Deficiencies of Bohr model led to development of
wave-mechanical model → electrons have wavelike
and particulate properties.
► Some of the wavelike characteristics are :
•  Electron positions are in orbitals defined by a probability distribution.
•  Size, shape, orientation of probability distribution (i.e. electron cloud)
are characterized by quantum numbers.
•  Energy levels separate into electron subshells designated by QNs.
Quantum #
Designation
n = principal (energy level-shell)
 = subsidiary (orbitals)
ml = magnetic
K, L, M, N, O (1, 2, 3, etc.)
s, p, d, f (0, 1, 2, 3,…, n -1)
1, 3, 5, 7 (- to +)
ms = spin
½, -½
12
Bohr Atom vs. Wave Mechanics
Versus “Electron orbitals have different shapes”
(Depends on l)
3d 3p Energy 3s 2p 4p 4s Further from the nucleus (higher energy state) 2s 1s 13
Filling of shells is based on orbital type (Au?au principle) Electron Energy States
Electrons...
• have discrete energy states
• tend to occupy lowest available energy state first.
Max # e-
Subshell
Shell / Principal Q.N.
10
6
4d
4p
N-shell n = 4
10
2
Adapted from Fig.
2.6, Callister &
Rethwisch 9e.
3d
4s
3p
3s
M-shell n = 3
2
2
2p
2s
L-shell n = 2
2
1s
K-shell n = 1
Energy
(From K. M. Ralls, T. H.
Courtney, and J. Wulff,
Introduction to Materials
Science and
Engineering, p. 22.
Copyright © 1976 by
John Wiley & Sons, New
York. Reprinted by
permission of John
Wiley & Sons, Inc.)
14
Electronic Configurations
ex: Fe: atomic #, Z=26 1s2 2s2 2p6 3s2 3p6 3d 6 4s2
4d
4p
N-shell n = 4
valence
electrons
3d
4s
Energy
Adapted from Fig. 2.6,
Callister & Rethwisch 9e.
(From K. M. Ralls, T. H. Courtney,
and J. Wulff, Introduction to
Materials Science and Engineering,
p. 22. Copyright © 1976 by John
Wiley & Sons, New York. Reprinted
by permission of John Wiley &
Sons, Inc.)
3p
3s
M-shell n = 3
2p
2s
L-shell n = 2
1s
K-shell n = 1
15
SURVEY OF ELEMENTS
• Most elements: Electron configuration not stable.
Element
Atomic #
Hydrogen
1
Helium
2
Lithium
3
Beryllium
4
Boron
5
Carbon
6
...
Neon
10
Sodium
11
Magnesium
12
Aluminum
13
...
Argon
18
...
...
Krypton
36
Electron configuration
1s 1
(stable)
1s 2
1s 2 2s 1
1s 2 2s 2
1s 2 2s 2 2p1
1s 2 2s 2 2p2
...
1s 2 2s 2 2p6
(stable)
1s 2 2s 2 2p6 3s 1
1s 2 2s 2 2p6 3s 2
1s 2 2s 2 2p6 3s 2 3p1
...
Adapted from Table 2.2,
Callister & Rethwisch 4e.
(stable)
1s 2 2s 2 2p6 3s 2 3p6
...
1s 2 2s 2 2p6 3s 2 3p6 3d 10 4s 2 4p 6 (stable)
• Why? Valence (outer) shell usually not filled completely.
16
Electron Configurations
► Valence electrons – those in outermost shells.
► Atoms with filled shells are more stable.
► Valence electrons readily available for bonding
and tend to control the chemical properties.
•  example: C (atomic number = 6)
K
L
2 2s2 2p2
1s
e config.:
Shell →
valence electrons [He] 2s2 2p2 Short hand means of wriCng electronic configuraCon 17
Check out hFp://www.periodictable.com. It’s very informaRve! 18
Why atoms bond? Atoms want to be stable (i.e., fill their orbitals) This results in electronic (electron) transfer or sharing This leads to chemical reacRons and/or the formaRon of atomic bonds ex. H2, O2, NaCl, LiF, etc… More on this in a moment 19
How to read and use a Periodic Table Outer Shells have 1 electron Increasing # of electrons (protons) Outer Shells missing 2 electrons Outer Shells missing 1 electron All Shells filled (noble) Atomic # = # of protons (or # electrons in a neutral atom) Atomic Mass Unit (amu): 1 amu = 1.66 x 10-­‐27 kg (based on the unit of mass of 1/12 of 12C) 63.55 g Cu Atomic mass of Cu = 63.55 amu or Avogadro’s number (6.02 x 1023) of Cu atoms “…or 1 mole of Cu atoms weighs 63.55 20
g” The Periodic Table
give up 1egive up 2egive up 3e-
accept 2eaccept 1einert gases
• Columns: Similar Valence Structure
K Ca Sc
Se Br Kr
H
He
Li Be
O
F Ne
Na Mg
S
Cl Ar
Rb Sr
Y
Cs Ba
Te
I
Adapted from
Fig. 2.6,
Callister &
Rethwisch 4e.
Xe
Po At Rn
Fr Ra
Electropositive elements:
Readily give up electrons
to become + ions.
Electronegative elements:
Readily acquire electrons
to become - ions.
21
Electronegativity
• Ranges from 0.9 to 4.1,
• Large values: tendency to acquire electrons or share their electrons w/
other atoms.
ity v
R
a
g
rone
lect
e
g
asin
e
r
c
In
Smaller electronegativity
Larger electronegativity
22
Primary Types of Atomic Bonds
► Covalent:
•  Shared outer shell electrons. Directional.
► Metallic:
•  Shared sea of electrons. Non-directional.
► Ionic:
•  Donation of valence electron (e-) to balance charge.
Non-directional.
► van der Waals / Secondary:
•  Attraction between (+) and (-) charged regions.
What kind of bonds will you see in different types of materials? Table 2.3
Bond energies and
melting
temperatures for
various
substances
Bond type
influences
structure
and
properties!
Stronger bonds Weaker bonds Ionic Bonding
•
•
•
•
Occurs between (+) and (-) ions.
Requires electron transfer.
Large difference in electronegativity required.
Example: NaCl
Na (metal)
unstable
1
extra
e−
Missing
1 e−
Cl (nonmetal)
unstable
electron
Na (cation)
stable
+
Coulombic
Attraction
Cl (anion)
stable
25
The Ionic Bond
Electron
Transfer
(donation)
Ionic bonding between sodium
(Na) and chlorine (Cl) atoms.
Transfer of e− from Na to Cl
creates a Na+ cation and a Cl−
anion.
The ionic bond is due to the
coulombic attraction between
ions of opposite charge.
Na
Cl
Ionic Bond
Ionic Bonds are non-directional
Na+
Cl-
Cation (Na+) becomes smaller than the neutral Na atom.
It gives up the extra electron.
26
Anion (Cl−) becomes larger than the Cl neutral atom.
It accepts the extra electron. Outer shell has only one electron Outer shell missing one electron Electron donaRon makes both ions “happy” Both have filled valence shells. Formation of an ionic bond between sodium and chlorine in which the effect of ionization on
atomic radius is illustrated. The cation (Na+) becomes smaller than the neutral atom (Na), while
the anion (Cl−) becomes larger than the neutral atom (Cl).
27
Lower
electronegativity
Ionic bond = metal
+
Higher
electronegativity
nonmetal
donates
electrons
accepts
electrons
Dissimilar electronegativities
ex: MgO
Mg: 1s2 2s2 2p6 3s2
[Ne]+3s2
Recall the shorthand Mg2+: 1s2 2s2 2p6
[Ne]
(stable)
O2-­‐ Mg2+ O: 1s2 2s2 2p4
[Ne]-3s2
O2–: 1s2 2s2 2p6
[Ne]
(stable)
28
Yet another example of the ionic bond
CaCon Has extra e-­‐ − − Electron
Transfer
(donation)
3+ − − Lithium atom
Li
− 3+ − − − − − 9+ − − Lithium ion
Li+
− Fluorine atom
F
Fl-­‐ − − − − Li+ − 9+ − − − Anion Needs an e-­‐ − − − Fluorine atom
F-
Figure SchemaCc illustraCon of ionic bonding between lithium and fluorine. As before, the effect of ionizaCon on atomic radius is illustrated. The caCon (Li+) becomes smaller than the neutral atom (Li), while the anion (F−) becomes larger than the neutral atom (F). 29
IONIC BONDING & STRUCTURE
• Charge Neutrality:
--Net charge in the
structure should
be zero.
CaF2
--General form:
AmXp
m, p determined by charge neutrality
• Stable structures:
--maximum # of nearest neighbors
with opposite charge.
--∴ ionic bonds are not directional
30
There are forces between atoms! •  AFracRon (draws together) •  Repulsion (pushes apart) Na+
Cl-
rNa1+ = 0.102 nm
rCl1− = 0.181 nm
Net bonding force curve for
a Na+−Cl− pair showing an
equilibrium bond length of
ro = 0.28 nm. This curve is the difference between the aFracRve and repulsive curves (FN = FA + FR) At equilibrium bond length, FA and FR must balance 31
(aka, FN = 0) FA (coulombic force of attraction)
FN (net bonding force)
+ r
0 − repulsion
Force
attraction
ro
FR (coulombic force of repulsion)
Repulsive b/c of overlapping electric fields Similar to Figure 2.8(a) from text • The bonding energy, E, is related to
the bonding force, F, through:
P. 27 E = ∫ Fdr ∴ F =
such that:
Na+
Cl-
dE
dr
∞
EN = ∫ FN dr
Equilibrium bond length (ro) occurs
where F = 0 and E is a minimum Force
r
r0
+ r
0 − Externally applied compressive force
is required to push ions closer
together than ro.
Externally applied tensile force is
required to pull ions further apart
than ro.
Mechanical behavior depends on
bonding! See Chap. 7!
32
Energy
bonding force = 0 + r
0 − E0
minimum bonding energy Figure 2.8 from text Note these general curve shapes applied to other bonding types too Coulombic force of attraction for a Na+ – Cl− pair: dE A d ( − A / r )
FA =
=
dr
dr
A
⎛ − A ⎞
= − ⎜ 2 ⎟ = 2
⎝ r ⎠ r
Na+
Clr
e = single electron charge (1.602x10-­‐19 C) P. 29 A = A0 ( Z1 e)( Z 2 e)
ProporRonality constant (9x109V/C) Z = valence of the charged ion, i.e. +1 for Na+ and -­‐1 for Cl-­‐ FA (attractive force)
AXracCve force between two opposite charges These equations apply to ionic bonds
but NOT covalent 33
Closer the charged species, the greater the aFracRon, up to a point! Interatomic separation r
Ao =
1
4πε 0
From P. 29
NaCl Crystal Structure Regular stacking of Na+ and Cl− ions in solid NaCl, which is
indicative of the non-directional nature of ionic bonding Maximize # of nearest neighbors of opposite charge Na+ Cl-­‐ 34
CLASS ROOM EXAMPLE: So far, we have concentrated on the coulombic force of attraction
between ions. But like ions repel each other. A nearest-neighbor pair of
Na+ ions (shown below) are separated by a distance of 2ro , where ro is
defined as the distance between Cl– and Na+ ions. Calculate the
coulombic force of repulsion between such a pair of like ions.
35
CLASS ROOM EXAMPLE: So far, we have concentrated on the coulombic force of attraction
between ions. But like ions repel each other. A nearest-neighbor pair of
Na+ ions (shown below) are separated by a distance of 2ro , where ro is
defined as the distance between Cl– and Na+ ions. Calculate the
coulombic force of repulsion between such a pair of like ions.
Given:
ro = 0.102 + 0.181 (nm) = 0.28 nm = 0.28 ×10−9 m
r = 2ro
FR ≅ −FA = −
2 Ao ( Z1 e)( Z 2 e)
r2
r = distance between two charges
36
To solve for force of repulsion, you need to find the distance between adjacent Na1+ OR Cl1-­‐ ions. This is r.
How do we calculate r? Geometry. 2
o
2
o
r = r + r = ro 2
r
ro
Just some simple geometry 3D ro = rNa1+ + rCl1−
= 0.102 + 0.181 (nm)
= 0.283 nm
ro
r = ro2 + ro2 = ro 2 = ( rNa1+ + rCl1− ) 2 = 0.28 2 nm
This interatomic distance (r) will be the one that you subsRtute into the equaRon for coulombic force. 37
CLASS ROOM EXAMPLE: So far, we have concentrated on the coulombic force of attraction
between ions. But like ions repel each other. A nearest-neighbor pair of
Na+ ions (shown below) are separated by a distance of 2ro , where ro is
defined as the distance between Cl– and Na+ ions. Calculate the
coulombic force of repulsion between such a pair of like ions.
Given:
ro = 0.102 + 0.181 (nm) = 0.28 nm = 0.28 ×10−9 m
r = 2ro
At equilibrium:
Ao ( Z1 e)( Z 2 e)
For like atoms, FR ≅ − FA = −
r2
where Ao = 9 ×109 V ⋅ m / C ;
Z1 , Z 2 = valence ;
2 (1V ⋅ C / m = 1 N )
e = 0.16 ×10−18 C
(9 ×109 V ⋅ m / C )(+1)(0.16 ×10−18 C )(+1)(0.16 ×10−18 C )
−9
∴ FR = −
=
−
1.49
×
10
N
−9
2
2(0.28 ×10 m)
38
CLASS ROOM EXAMPLE -­‐ SOLUTION Calculate the coulombic force of attraction between Ca2+ and O2- in
CaO, which has the NaCl-type structure.
From front cover of text:
Ionic Radii for Ca 2+ and O 2−
rCa2+ = 0.100nm = 0.100 × 10−9 m
rO2− = 0.140 nm = 0.140 × 10−9 m
O2Ca2+
Interatomic Separation:
ao = rCa 2 + + rO2 − = (0.100 ×10−9 m) + (0.140 ×10−9 m) = 0.240 ×10 −9 m
(9 ×109 V ⋅ m / C )(+2)(0.16 ×10−18 C )(+2)(0.16 ×10−18 C )
∴ FA =
(0.240 ×10−9 m) 2
= +16.0 × 10−9 N
2 39
Summary…
► There are five types of atomic bonds: Covalent,
Ionic, Metallic, Hydrogen, and Van der Waals.
► The bond type influences the structure and
properties of a material.
► Ionic bonds result from electron transfer btw. atoms
A
F
=
► Coulombic force of repulsion and coulombic
force
A
r2
of attraction can be calculated using
► The equation relating coulombic force of repulsion
and coulombic force of attraction is: FA = −FR
40
Topics for next time…
► Atomic Bonding – continued
► Coordination numbers
► Start reading chapter 3 in your textbook!
► See you next time!