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Lecture #02
Chapter 2
Atomic Bonding
Learning Objectives
• Describe ionic, covalent, and metallic, hydrogen, and
van der Waals bonds.
• Which materials exhibit each of these bonding
t e ?
types?
• What is coulombic force of repulsion and coulombic
force of attraction and how to calculate both.
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It all starts with atoms
Decreasing size
…..powers of ten
~103 m
~10‐6 m
2
~10‐10 m
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Atoms Bond
• Atoms are the smallest unit of matter that retains
the identity of the substance
Atoms self assemble in
different ways to build
up the structure of a
material.
What do we learn in this image?
2
3
Bohr Model of the atom
“Kind of like the solar system”
‐
3rd
The
ring
can hold
h ld up
to 18 e-
‐
‐
‐
‐
The 4th ring
and any after
can hold up to
32 e-
Nucleus consists of
protons and neutrons
‐
‐
‐
‐
The 1st ring can
hold up to 2 e-
‐
The 2nd ring can
hold up to 8 e-
Electrons have (–) charge
little mass (~10‐29g)
IN the NUCLEUS
Protons have (+) charge
Neutrons have 0 charge
Same mass
depends on atom
type (~10‐27g)
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Bohr Model of the atom
“Orbitals/Shells”
‐
3rd
The
ring
can hold
h ld up
to 18 e-
‐
‐
‐
K
‐
L
M
Nucleus consists of
protons and neutrons
N
‐
The 1st ring can
hold up to 2 e-
‐
The 4th ring
and any after
can hold up to
32 e-
‐
‐
‐
The 2nd ring can
hold up to 8 e-
Electrons have (–) charge
little mass (~10‐29g)
IN the NUCLEUS
Same mass
depends on atom
type (~10‐27g)
Protons have (+) charge
Neutrons have 0 charge
5
Bohr Atom vs. Orbital Description
z
x
y
s Orbital
Versus
“Orbitals adopt certain configurations”
4p
3d
4s
3p
Energy
3s
2p
Further from the nucleus
(higher energy state)
2s
6
1s
Filling of shells is based on orbital type
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• # protons = # electrons
(+1)
(-1)
( 1)
– If 20 protons are present in an atom then 20
electrons are there to balance the overall charge
of the atom—atoms are neutral.
– The neutrons have no charge; therefore they do
not have to equal the # protons or # electrons.
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7
How do we know the number of
subatomic particles in an atom?
• Atomic number: this number indicates the
number
b of
f protons
t
iin an atom.
t
– Ex: Hydrogen’s atomic number is 1. # protons?
• So hydrogen has 1 proton
– Ex: Carbon
Carbon’ss atomic number is 6.
6 # protons?
• So carbon has 6 protons
**The number of protons identifies the atom.
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Check out http://www.periodictable.com. It’s very cool!
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)
Atomic Mass Unit (amu): 1 amu = 1.66 x 10‐27 kg
(based on the unit of mass of 1/12 of 12C)
Atomic mass of Cu = 63.55 amu or
63.55 g Cu
Avogadro’s number (6.02 x 23 mol‐1)
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Atoms want to fill (complete) their shells – results in electronic
(electron) transfer or bonding with each other, ex. H2 → BONDING
K
L
M
N
4p
3d
4s
3p
Energy
3s
2p
2s
1s
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Primary Atomic Bonds
• Covalent:
– Shared outer shell electrons. Directional
• Ionic:
– Donation of valence electron (e‐) to balance
charge.
• Metallic:
– Sea of electrons. Non
Non‐directional.
directional.
• Hydrogen:
– Attraction between H+ to adjacent molecule (‐)
• van der Waals:
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– Attraction between (+) and (‐) charged regions.
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Bond
Bond
Strength
(GPa)
Example of
Bond
Covalent
1,000
Diamond
Ionic
30 - 100
Salt and Ceramics
Metallic
30 - 150
Metals
Hydrogen
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Ice
Van der
d Waals
l
2
Polymers
l
Strongest
Weakest
Bond type influences structure and properties!
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The Ionic Bond
Electron
Transfer
(donation)
Ionic bonding between sodium (Na)
and chlorine (Cl) atoms.
Electron transfer from Na to Cl
creates a cation (Na+) and an anion
(Cl−).
)
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(Na+)
Cation
becomes smaller
than the neutral Na atom.
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Anion (Cl−) becomes larger than
the Cl neutral atom.
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Outer shell
has only one
electron
Outer shell
missing one
electron
Electron donation
makes both ions
“happy”
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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).
Yet another example of the ionic bond
−
−
Electron
Transfer
(donation)
3+
−
−
−
−
−
Lithium atom
Li
Fluorine atom
F
Fl‐
−
−
3+
−
−
−
−
−
9+
−
−
−
−
Lithium ion
Li+
Fluorine atom
F-
−
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−
−
−
Li+
−
9+
−
Figure Schematic illustration of ionic bonding between lithium and fluorine. As before, the effect
of ionization on atomic radius is illustrated. The cation (Li+) becomes smaller than the neutral
atom (Li), while the anion (F−) becomes larger than the neutral atom (F).
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IONIC BONDING & STRUCTURE
• Charge Neutrality:
--Net charge in the
structure should
be zero.
CaF2:
Ca2+ +
cation
Fanions
F-
--General form:
AmXp
m p determined by charge neutrality
m,
• Stable structures:
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--maximize the # of nearest
oppositely charged
neighbors.
Na+
Net bonding force curve for
a Na
N +−Cl
Cl− pair
i showing
h i
an
equilibrium bond length of
ro = 0.28 nm.
Cl-
ro
Same as curve as before
FA (coulombic force of attraction)
F
Force
FN (net bonding force)
This curve is the difference
between the attractive and
repulsive curves
(FN = FA + FR)
0
r
FR (coulombic force of repulsion)
Repulsive b/c of
overlapping electric fields
Similar to Figure 2.8(a) from text
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• Bonding distance is a balanced
distance
Na+
Cl-
• The bonding energy, E, is related
to the bonding force, F, through:
E   Fdr  F 
dE
dr
Force
P. 27
Equilibrium bond length (ao) occurs
where F = 0 and E is a minimum
r0
+
0
r
−
bonding force = 0
Enerrgy
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.
+
0
r
−
E0
minimum bonding energy
Mechanical behavior depends on
bonding! See Chap. 7!
Figure 2.8 from text
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Note these general curve shapes applied to other bonding types too
Coulombic force for a Na+ – Cl− pair.
FA 
dE  A
 2
dr
r
Na+
Clr
e = single electron charge (1.602x10‐19 C)
P. 29
A  A0 ( Z1e)( Z 2 e)
Proportionality
constant
(9x109V/C)
Z valence of the
Z=
charged ion, i.e. +1 for
Na+ and ‐1 for Cl‐
FA (attractive force)
Attractive force between two opposite charges
These equations apply to ionic bonds
but NOT covalent
Closer the charged
species, the
greater the
attraction,
up to a point!
Interatomic separation r
Ao 
1
4 0
From P. 29
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Regular stacking of Na+ and Cl− ions in solid NaCl, which is
indicative of the non-directional nature of ionic bonding
Na+
Cl‐
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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.
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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  0.28 nm  0.28  109 m
r  2ro
(See the next page)
At equilibrium:
A ( Z e)( Z e)
For like atoms, FA  FR   o 1 2 2
r
9
where Ao  9 10 V  m / C ;
2
(1 V  C / m  1 N )
Z1 , Z 2  valence ;
e  0.16 1018 C
 FR  
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(9  109 V  m / C )( 1)(0.16 1018 C )( 1)(0.16 1018 C )
 1.49  109 N
2(0.28  109 m) 2
To solve this problem you need to treat ro as
the distance between adjacent Na1+ and Cl1‐
ions.
Just some simple geometry
Use geometry to determine the distance
between adjacent Na1+ ions or adjacent Cl1‐
ions. This is shown below.
This new interatomic
Thi
i t t i distance
di t
(r) will
ill b
be th
the
one that you substitute into the equation for
coulombic force.
In this problem
ro  rNa1  rCl1
r
ro
ro
From geometry, you can see that:
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r  ro2  ro2  ro 2   rNa1  rCl1  2  0.28 nm for this problem.
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CLASS ROOM EXAMPLE
Calculate the coulombic force of attraction between Ca2+ and O2- in
CaO, which has the NaCl-type structure.
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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
Interatomic Separation:
ao  rCa2   rO2   (0.100 109 m)  (0.140 109 m)  0.240 109 m
(9 109 V  m / C )(2)(0.16
2)(0 16 1018 C )(2)(0
2)(0.16
16  1018 C )
(0.240  109 m) 2
 16.0  109 N
 FR  
2
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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.
• Coulombic force of repulsion and coulombic force
of attraction can be calculated using
A
FA 
r2
• The equation relating coulombic force of repulsion
and coulombic force of attraction is: FA =-FR
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