Crystals of Gallium 1
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Crystals of Gallium1
Structure Changes Everything!
AluminumGlass
Rubber2
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:
• Equilibrium occurs when attractive force and repulsive force balance – FNET=0– The “sweet spot”– r0⋍0.3nm for most metals
4
RANET FFF
Bonding Forces and Energies• Energy
– minimum energy is called thebond energy, E0 (energy that is required to completely separate two atoms)
– bond length:
r
RAR
r r
ANN EEdrFdrFdrFE
0N r0
dr
dE
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, Estiff
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
H H
H H
• 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
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
rAEA
rBE nR
0
21
4
))((
eZeZA
12
Example: Ionic Bonding
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. The bond length is the distance between the two ions where the net force is zero.
ii. The bond length is the distance between the two ions where the net force is maximum.
iii. The bond length is the distance between the two ions where the net energy is zero.
iv. 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:
where, r is the distance between the two atoms. The force of repulsion is given by the equation:
3
2
Attraction
.90.0F
r
nmeV
11
105
Repulsion
.109.5F
r
nmeV
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
Evaluate both at r = 0.3nm to get:
eV 4 is energy bond The
eVeVeV
eV
eV
415EEE
1E
5E
RepulsionAttractionTotal
Repulsion
Attraction
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.
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