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CHAPTER 10
Molecules and Solids
Molecular Bonds
Molecular Spectra
Johannes Diderik van der Waals
(1837 – 1923)
“You little molecule!”
- Fred Flintstone (insulting his friend Barney Rubble
Molecular
Bonds
The Coulomb force is the only one
to bind atoms.
The combination of attractive and
repulsive forces creates a stable
molecular structure.
Force is related to potential energy F = −dV / dr, where r is the
distance separation.
It is useful to look at molecular binding using potential energy V.
Negative slope (dV / dr < 0) with repulsive force.
Positive slope (dV / dr > 0) with attractive force.
Molecular Bonds
An approximation of the force felt by one
atom in the vicinity of another atom is
where A and B are positive constants.
Because of the complicated shielding effects of the various electron
shells, n and m are not equal to 1.
A stable equilibrium exists with
total energy E < 0. The shape
of the curve depends on the
parameters A, B, n, and m.
Also n > m.
Molecular Bonds
Vibrations are excited thermally, so the exact level of E depends on
temperature.
A pair of atoms is joined.
One would have to supply energy to
raise the total energy of the system to
zero in order to separate the molecule
into two neutral atoms.
The corresponding value of r of a
minimum value is an equilibrium
separation. The amount of energy to
separate the two atoms completely is
the binding energy which is roughly
equal to the depth of the potential well.
Molecular Bonds
Ionic bonds:
The simplest bonding mechanisms.
Periodic Table: Alkali metals (1st Column) bond with halogens (7th
column)
Ex: Sodium (1s22s22p63s1) readily gives up its 3s electron to become
Na+, while chlorine (1s22s22p63s23p5) readily gains an electron to
become Cl−. That forms the NaCl molecule.
Ex: K gives up a 4s electron to 3p shell in Cl.
K->K+ ion 4.34 eV of energy input
electron is transferred
Cl -> Cl- ion 3.62 eV energy given back. 0.72 eV is the difference
K+ and Cl- combine to give neutral KCl. The dissociation energy is 4.42
eV.
Example
10.1 Find b if the PE between the two ions is
ke2 b
PE 
 9
r
r
10.2 Calculate the PE for KCL at its equilibrium distance of
ro  2.79 A
Molecular Bonds
Covalent bonds:
The atoms are not as easily ionized.
Ex: Diatomic molecules formed by the combination of two identical
atoms tend to be covalent.
Larger molecules are formed with covalent bonds.
H2 molecule: two electrons moving the electromagnetic field of two
protons.
From Q.M the two electrons can either have spin zero or one. S=0
state has lower energy
The most probable location of the two electrons are in the middle of
the two protons.
Molecular Bonds
Van der Waals bond:
Weak bond found mostly in liquids and solids at low temperature.
Ex: in graphite, the van der Waals bond holds together adjacent sheets
of carbon atoms. As a result, one layer of atoms slides over the next
layer with little friction. The graphite in a pencil slides easily over
paper.
Two atoms that have zero dipole moment can induce dipole moment
into each other. That is positive and negative charges can be
separated due to proximity of another atom. The weak attractive
force between the induced dipoles is called van der Waals bond.
Molecular Bonds
Hydrogen bond:
Holds many organic molecules together.
Metallic bond:
Free valence electrons may be shared by a number of atoms.
Example
10.3 Calculate the dipole moment of KCl for the equilibrium distance of
ro  2.79 A
Rotational States
Molecular spectroscopy:
We can learn about molecules by studying how molecules absorb,
emit, and scatter electromagnetic radiation.
From the equipartition theorem, the N2 molecule may be thought of as
two N atoms held together with a massless, rigid rod (rigid rotator
model).
In a purely rotational system, the kinetic energy is expressed in terms
of the angular momentum L and rotational inertia I.
Erot
1 2
1 ( I )2
 I rot 
2
2 I
Rotational States
L is quantized.
The energy levels are
Erot varies only as a function of the quantum
number ℓ.
Example
10.4 Calculate the order of magnitude of the rotational energy levels for
H2. Let the equilibrium spacing be 0.74 angstorm.
Vibrational States
A vibrational energy mode can also be excited.
Thermal excitation of a vibrational mode can occur.
It is also possible to stimulate vibrations in molecules using
electromagnetic radiation.
Assume that the two atoms are
point masses connected by a
massless spring with simple
harmonic motion.
Vibrational States
The energy levels are those of
a quantum-mechanical oscillator.
The frequency of a
two-particle oscillator is:
where the reduced mass is μ = m1m2 / (m1 + m2) and the spring
constant is κ.
If it is a purely ionic bond, we can compute κ by assuming that the force
holding the masses together is Coulomb.
and
Vibration and Rotation Combined
It’s possible to excite rotational and vibrational modes simultaneously.
Total energy of simple vibration-rotation system:
Vibrational energies are spaced at regular intervals.
Transition from ℓ + 1 to ℓ:
Photons will have
an energies at
regular intervals:
Vibration and Rotation Combined
An emission-spectrum spacing that varies with ℓ.
The higher the starting energy level, the greater the photon
energy.
Vibrational energies are greater than rotational energies. This energy
difference results in the band spectrum.
Vibration and Rotation Combined
The positions and intensities of the observed bands are ruled by
quantum mechanics. Note two features in particular:
1) The relative intensities of the bands are due to different
transition probabilities.
The probabilities of transitions from an initial state to final state
are not necessarily the same.
2) Some transitions are forbidden by the selection rule that
requires Δℓ = ±1.
Absorption spectra:
Within Δℓ = ±1 rotational state changes, molecules can absorb
photons and make transitions to a higher vibrational state when
electromagnetic radiation is incident upon a collection of a particular
kind of molecule.
Vibration
and
Rotation
Combined
ΔE increases
linearly with ℓ.
Vibration and Rotation Combined
In the absorption spectrum of HCl, the spacing between the peaks
can be used to compute the rotational inertia I. The missing peak in
the center corresponds to the forbidden Δℓ = 0 transition.
The central frequency
Example
10.5 Estimate the vibration energy levels of H2 (r=0.74 angstrom)