Chemistry 2000 Lecture 4:
Molecular spectroscopy
Marc R. Roussel
Evidence for MO theory
How do we know that MO theory is correct?
Bond lengths:
I
I
X-ray or neutron diffraction for solids
Rotational (microwave) or vibrational (infrared)
spectroscopy for gases
Potential energy curve/surface:
I Vibrational (infrared) spectroscopy
Orbital energy diagram:
I Photoelectron spectroscopy
I Electronic absorption (UV/visible) spectroscopy
I Fluorescence spectroscopy
Vibrational levels
E
vibrational levels
R
Stronger bond ←→ narrower potential well
←→ larger vibrational spacing
Infrared (vibrational) spectroscopy
I
Under some circumstances (to be discussed later), molecules
can absorb a photon and be excited from (usually) the ground
vibrational state to (usually) the first excited vibrational state.
E
∆ Evib = h ν
R
I
Vibrational energy spacings correspond to the infrared region
of the electromagnetic spectrum.
I
In the gas phase, we get a combination of rotational and
vibrational excitation, so we get bond length information as
well.
Example: Gas-phase IR spectrum of HCl:
0.4
0.35
0.3
0.25
A
0.2
0.15
0.1
0.05
0
-0.05
2600
2700
2800
2900
3000
3100
~
!/cm-1
Note: ν̃ is the wavenumber, i.e. the inverse of the
wavelength, so it’s proportional to the photon energy.
Solution-phase IR spectroscopy
I
In solution, we don’t see rotational structure, and the
vibrational absorption bands are broadened due to the
deformation of molecules during collisions.
(Recall that the orbital energies, and thus the effective
potential, depend on the molecular geometry.)
Example: IR spectrum of liquid ethanol
Source: Spectral Database of Organic Compounds,
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi, Jan. 16, 2013
Note: The wavenumber axis often runs backward, as shown here.
Infrared spectroscopy and the identification of
compounds
I
One important application of spectroscopy (in general) is for
the identification of unknown compounds.
I
Certain bonds in organic molecules are reproducibly found in
certain spectral regions:
Bond
Spectral region/cm−1
C
2800–3000
H
H
C
C
(including aromatic CH)
O H (non-hydrogen-bonded)
O H (hydrogen-bonded)
3000–3200
3500–3700 (sharp)
3200–3500 (broad)
Example: The IR spectrum of ethanol
C−H stretches
hydrogen−bonded OH
Alkene and alkyne carbon-carbon bond stretches
Bond
C C
C C
Spectral region/cm−1
1640–1675 (sometimes)
1950–2300 (sometimes)
C=C stretch
Example: IR spectrum of liquid cis-3-hexene
CH 3
CH 2
C
alkene CH
alkane CH
Spectrum source: Spectral Database of Organic Compounds,
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi, Jan. 20, 2013
H
CH 2
C
H
CH 3
C=C stretch missing
Example: IR spectrum of liquid trans-3-hexene
CH3
C
alkene CH
alkane CH
H
CH2
C
H
Spectrum source: Spectral Database of Organic Compounds,
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi, Jan. 20, 2013
CH2
CH3
The fingerprint region of the spectrum
I
The region from 900 to 1300 cm−1 is called the fingerprint
region of the IR spectrum.
I
In this region, we typically find many peaks arising from
various low-energy stretching and bending motions of the
molecules.
I
Very difficult to assign peaks in this region but they are very
different even for closely related compounds
I
Used for confirmation that a particular (known) compound
has been isolated
Example: Fingerprint regions of
cis- and trans-3-hexene compared
trans
Transmittance (%)
cis
1200
1200
wavenumber (cm −1)
Review: Molecular dipole moments
I
A bond dipole is a slight separation of charge between two
non-identical atoms connected by a bond.
I
The size of the bond dipole is proportional to the amount of
charge separation and to the bond length.
I
The dipole moment of a molecule is the vector sum of the
bond dipoles.
I
A polar molecule has a non-zero dipole moment.
I
Examples: CO2 , H2 O
Normal modes
I
Except in diatomics, molecular vibrations always involve
motions of multiple atoms, i.e. more than one bond is
deformed at a time.
I
The vibrational modes must conserve overall molecular
momentum.
I
We can choose vibrational modes which, to a good
approximation, are independent motions, called normal modes.
I
A linear molecule has 3N − 5 normal modes, where N is the
number of atoms in the molecule.
I
A nonlinear molecule has 3N − 6 normal modes.
Normal modes of H2 O
O
H
O
O
H
Symmetric stretch
H
H
Asymmetric stretch
H
H
Bend
Normal modes of CO2
O
C
O
O
Symmetric stretch
O
C
O
Asymmetric stretch
C
Bend (×2)
O
Selection rule
I
A selection rule is a rule that tells us when a particular kind of
spectroscopic event can occur.
I
In IR absorption spectroscopy, the key selection rule is that
the dipole moment of the molecule has to change during the
vibration.
I
A normal mode that can absorb an IR photon is said to be
IR active.
Normal modes of H2 O in IR spectroscopy
Which of these modes are IR active?
O
H
O
O
H
Symmetric stretch
H
H
Asymmetric stretch
H
H
Bend
Normal modes of CO2 in IR spectroscopy
Which of these modes are IR active?
O
C
O
O
Symmetric stretch
O
C
O
Asymmetric stretch
C
Bend (×2)
O
Greenhouse gases
I
I
I
I
I
I
When a gaseous molecule becomes vibrationally excited by
absorbing infrared radiation, the excess vibrational energy can
be converted to translational kinetic energy during collisions.
We will see later that there is a direct link between kinetic
energy and temperature.
Gases that absorb in the infrared therefore result in warming
of the atmosphere, i.e. they are greenhouse gases.
Atmospheric temperature is set by a balance of energy coming
in from the Sun and radiated into space by the Earth.
Greenhouse gases keep the Earth warm, but shifting that
balance by slowing the escape of radiation to space necessarily
warms the atmosphere.
N2 , O2 and Ar, the major components of the atmosphere,
don’t absorb in the IR. (Why?)
Greenhouse gases (continued)
I
I
I
I
I
I
The next two most common components of the atmosphere,
water and carbon dioxide, are greenhouse gases.
The atmospheric water content is set by the balance of
evaporation and precipitation and is beyond our control.
We worry a lot about CO2 because we are adding a lot of it to
the atmosphere, thus tipping the energy balance.
From 1959 to 2012, the CO2 concentration in the atmosphere
measured at the Mauna Loa observatory has risen from an
annual average value of 316 ppm to 394 ppm, an increase of
25%.
The rate of increase in the CO2 concentration is also rising,
from about 0.7 ppm y−1 in the early 1960s to about 2 ppm y−1
now.
Warming the atmosphere also results in more water vapor,
worsening the problem.
Photoelectron spectroscopy
I
I
I
I
How do we know that the orbital occupancies predicted by
MO theory are correct?
Photoelectron spectroscopy is similar in principle to the
analysis of the photoelectric effect.
An atom or molecule is ionized using a photon of energy hν.
The kinetic energy of the ejected electron is then
K = hν − Ii
I
I
where Ii is the ionization energy of an electron in orbital i.
The ionization energy of an electron in a particular orbital is
the negative of its orbital energy (εi ).
We measure K and calculate the orbital energy of occupied
orbitals:
−Ii = εi = K − hν
Photoelectron spectroscopy (continued)
I
Removing a valence electron typically requires a photon in the
ultraviolet range.
I
Removing a core electron typically requires an x-ray photon.
Example: Photoelectron spectrum of Ne
K/eV
60
50
40
30
20
10
0
0.7
hν = 60 eV
0.6
2p
Intensity
0.5
0.4
0.3
0.2
2s
0.1
0
0
10
20
30
40
I/eV
MO energy level diagram? Rotate clockwise!
50
60
A complication
I
For molecules, the ion formed also has vibrational levels.
I
As a result, the photoelectron spectrum typically has
vibrational substructure:
E
+
X2
e−
X2
R
I
Instead of one ionization energy, the photoelectron spectrum
gives us a band of several lines corresponding to the ionization
of an electron from a particular orbital.
+
Intensity
∆ Evib(X 2)
Ii
"ionization energy":
X 2 produced in its
vibrational ground state
I
The photoelectron spectrum thus allows us to recover the
vibrational spectrum of the ion formed.
I
We compare the vibrational spectrum of the molecule to that
of the ion.
I
The way in which the vibrational spectrum changed tells us
how the potential energy curve changed, and thus how the
bonding changed.
I
This can be correlated to the MO diagram:
I
I
I
Removing an electron from an orbital not directly involved in
bonding (e.g. the 1π orbital in HF) won’t change the
vibrational spectrum much.
Removing an electron from a bonding orbital will lead to a
weaker bond in the ion, thus to lower vibrational frequencies
for the associated normal mode(s) than in the parent molecule.
Removing an electron from an antibonding orbital. . .
Example: UV photoelectron spectrum of N2
40
3σ
2191 cm-1
35
30
Intensity
25
20
1π
1850 cm-1
15
10
2σ*
2397 cm-1
5
0
15
16
17
18
19
20
I/eV
Note: The vibrational “frequency” of N2 is 2358 cm−1 .