Crystals of Gallium
1
Structure Changes Everything!
Glass
Aluminum
Rubber
2
Bonding Forces and Energies
• At interactions between
atoms are negligible.
• As the atoms approach each
other, they start to interact
(ie. exert forces on one
another):
– Attractive force, FA ; repulsive
force, FR
– Magnitude of force is a function
of interatomic distance
– The bond-length corresponds to
the distance at which the
attractive and repulsive forces
exactly balance.
3
Bonding Forces and Energies
• Attractive force - depends on specific type of bonding
between atoms for example we have Coulomb
attraction between ions in ionic compounds like NaCl.
• Repulsive force - when outer shell electrons overlap
(electrostatic repulsion)
• Net Force:
FNET  FA  FR
• Equilibrium occurs when attractive force and
repulsive force balance
– FNET=0
– The “sweet spot”
– r0⋍0.3nm for most metals
4
Bonding Forces and Energies
• Energy
– minimum energy is called the
bond energy, E0 (energy that is
required to completely
separate two atoms)
EN   FN dr   FAdr   FR dr  EA  ER
r
r
r



– bond length:
dE N
 0  r0
dr
5
Characteristics of Bonding Forces
and Energies
• Material properties are related to the following
features of the interatomic potential:
– Bond energy
– Bond length
– curvature
• High bonding energy  High melting T
• High curvature  high elastic modulus, E stiff
material; (chap 6)
• Symmetric energy well near r0 low coefficient of
thermal expansion, relatively small dimensional
alterations (chap 19)
6
Ionic Bond
• When the difference
in electronegativity
is large, one
element loses
electrons and the
other gains
electrons. The
resulting ions attract
and form a strong
bond.
• NaCl bond energy
640 kJ/mol
• MgO bond energy
1000 kJ/mol
Covalent Bond
• Stable electron configuration
is achieved by sharing/
overlap of outer orbitals.
• Strong bond.
• Si bond energy 450 kJ/mol
• C bond energy 713 kJ/mol
H
H
H
H
Metallic Bond
• Ion cores in a “sea of
electrons”.
• Electrons are free to drift.
• Excellent conductors of
electricity and heat.
• Electrons act to glue the ion
cores together.
• Al bond energy 324 kJ/mol
• Fe bond energy 406 kJ/mol
-
Mg2+
-
Mg2+
-
-
-
Mg2+
Mg2+
Mg2+
-
Mg2+
Mg2+
Mg2+
-
-
Mg2+
-
-
Secondary Bonding
+
•
•
•
•
-
+
-
Weak bonds.
Also known as van der Waals bonding
Secondary bonding exists between virtually all atoms or molecules.
Due to the presence of dipoles
– Fluctuation induced dipoles
– Permanent dipoles.
• Hydrogen bonding is a special type of secondary bonding.
Strong dipoles are present when H bonds with F, Cl or O.
Properties Derived from Bonding: E
11
Bonding Types
• Primary
– Ionic
– Covalent
– Metallic
• Secondary
– Van der Waals
– Dipole
– Hydrogen
Example: Ionic Bonding
EA   A r
ER  B
r
n
(Z1e)(Z 2e)
A
4 0
12
Chapter 2
Practice Problems
Practice Problems
1.
Two ions are separated by an infinite amount of space. When the two ions are
brought close to one another, they begin to interact with attractive and
repulsive forces. When in an equilibrium position, which of the following
statements most accurately describes the “bond length.”
i.
ii.
iii.
iv.
The bond length is the distance between the two ions where the net
force is zero.
The bond length is the distance between the two ions where the net
force is maximum.
The bond length is the distance between the two ions where the net
energy is zero.
The bond length is the distance between the two ions where the net
energy is minimum.
a) i. and iii.
b) i. and iv.
c) ii. and iii.
d) ii. and iv.
Practice Problems
Answer:
Bond length corresponds to zero net force
or minimum energy. So, i) and iv) are
correct, that is b).
Practice Problems
2. Rank the following bond types in terms of
increasing strength. (ie. From weakest to
strongest)
a) Hydrogen Bonding, Van der Waals Bonding, Metallic
Bonding, Covalent Bonding
b) Van der Waals Bonding, Hydrogen Bonding, Covalent
Bonding, Metallic Bonding
c) Van der Waals Bonding, Hydrogen Bonding, Metallic
Bonding, Covalent Bonding
d) Van der Waals Bonding, Covalent Bonding, Metallic
Bonding, Hydrogen Bonding
Practice Problems
Answer:
The correct answer is c):
Van der Waals Bonding, Hydrogen Bonding,
Metallic Bonding, Covalent Bonding
Practice Problems
3. Consider the bonding in hypothetical
molecule. The force of attraction between two
atoms is given by the equation:
FAttraction
0.90eV.nm2

r3
where, r is the distance between the two atoms.
The force of repulsion is given by the equation:
 5.9 105 eV.nm10
FRepulsion 
r11
Practice Problems
a) Calculate the bond length for this molecule.
b) Calculate the bond energy.
Practice Problems
Answer:
a) Bond length is the separation between atoms
when the net force is zero. This means the magnitude
of the forces (attractive and repulsive) are the same
and the two equations can be set equal to solve for r.
The bond length is 0.30 nm
Practice Problems
b) The bond energy is the energy of the system at the bond
length. So, we have to integrate the force to get the energy
and then substitute the bond length we solved for in a) to
solve.
Evaluate both at r = 0.3nm to get:
E Attraction  5eV
E Repulsion  1eV
E Total  E Attraction  E Repulsion  5eV  1eV  4eV
The bondenergy is 4 eV