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Lecture 17 Computational Spectroscopy
Predicting real observables
©2013, Jordan, Schmidt & Kable
Lecture 17
Copyright Notice
Some images used in these lectures are taken, with permission, from
“Physical Chemistry”, T. Engel and P. Reid, (Pearson, Sydney, 2006);
denoted “ER” throughout the lectures
and other sources as indicated, in accordance with
the Australian copyright regulations.
©2013, Jordan, Schmidt & Kable
Lecture 17
Learning outcomes
17.1 Calculating the geometry tells you something about the
rotational spectrum
17.2 Calculating force constants allows calculation of the
normal modes of vibration
17.3 HF overestimates frequencies due to incorrect
dissociation
17.4 DFT can get band positions pretty well by multiplication
by a factor of 0.97
17.5 The normal modes of vibration can be mixtures of the
various coordinates
17.6 Most accurate way is to solve Schrödinger equation on
an accurate potential energy surface!
©2013, Jordan, Schmidt & Kable
Lecture 17
How to be chemically accurate?
For chemistry, we would like energies to be accurate to within about 4
kJ/mol. So we need to use a very good method, but it would be very
expensive to do everything with configuration interaction at the complete
basis set limit.
Geometry optimizations and frequency calculations (for zero point energies)
take a lot of effort, and cannot be done with high levels of theory.
We could calculate geometries at one level, and energies at another, e.g.
CCSD/6-31G(d)//B3LYP/6-31G(d). Here we get the density function theory
geometry and then use a so-called coupled-cluster calculation to get the
energy.
But, smart people (Curtiss, Pople) have developed combinations of methods
which end up being chemically accurate, with more modest cost.
©2013, Jordan, Schmidt & Kable
Lecture 17
Level of correlation (# of configurations)
The Quantum Chemistry Landscape
The answer
pointless
Sensible
compromise
Hartree-Fock limit
1st year chemistry
Quality of basis set
©2013, Jordan, Schmidt & Kable
Lecture 17
Level of correlation (# of configurations)
The Quantum Chemistry Landscape
pointless
QCISD/6-311G(d,p)
DE(QCI)
The answer
Diffuse DE(+)
MP4/6-311G(d,p)//
MP2/6-31G(d)
more polarizations
DE(2df)
MP2/6-31G(d)
Hartree-Fock limit
1st year chemistry
Quality of basis set
©2013, Jordan, Schmidt & Kable
Lecture 17
Level of correlation (# of configurations)
G1 Theory
pointless
QCISD/6-311G(d,p)
DE(QCI)
DE(+)
The answer
DE(2df)
G1
(chemically accurate)
MP4/6-311G(d,p)//
MP2/6-31G(d)
The corrections are added to
approximate performing
QCISD calculation with large
basis
MP2/6-31G(d)
Hartree-Fock limit
1st year chemistry
Quality of basis set
©2013, Jordan, Schmidt & Kable
Lecture 17
Predicting Molecular Spectra
Spectra represent transitions between energy levels of a molecule.
Generally speaking, we can separate motions due to translations, rotations
and vibrations, with electronic transitions being separated again.
Etrans<Erot<Evib<Eelec
Generally, we do not consider quantized translational energy levels. Only
in exceptional circumstances would one bother (e.g. confined molecule).
If we know the equilibrium structure of a molecule, we can predict the
rotational energy levels.
©2013, Jordan, Schmidt & Kable
Lecture 17
Predicting Rotational Spectra
If we know the equilibrium structure of a molecule, we can predict the
rotational energy levels. This is easy for a diatomic. The energy levels are
given by
(
)
E = BJ J +1
Where J is the angular momentum quantum number and B is the
rotational constant. This energy level expression comes from solving the
Schrödinger equation. m is the reduced mass.
B=
2
2m R2
©2013, Jordan, Schmidt & Kable
m1m2
m=
m1 + m2
Lecture 17
Predicting Rotational Spectra – carbon monoxide
OK, so I need to know R, the equilibrium bond length. Do an energy
minimization. Do MP2/6-311G(d)
1
***** EQUILIBRIUM GEOMETRY LOCATED *****
comment (A single descriptive title card)
COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
ATOM CHARGE
X
Y
Z
-----------------------------------------------------------O
8.0 0.0000000000 0.0000000000 -0.0190160600
C
6.0 0.0000000000 0.0000000000 1.1190160600
COORDINATES OF ALL ATOMS ARE (ANGS)
ATOM CHARGE
X
Y
Z
-----------------------------------------------------------O
8.0 0.0000000000 0.0000000000 -0.0190160600
C
6.0 0.0000000000 0.0000000000 1.1190160600
INTERNUCLEAR DISTANCES (ANGS.)
-----------------------------O
1 O
2 C
C
0.0000000
1.1380321 *
1.1380321 *
0.0000000
©2013, Jordan, Schmidt & Kable
m1m2 12 ´16
m=
=
= 6.857amu
m1 + m2 12 +16
B=
2
2m R2
= 3.772 ´10-23 J
= 1.899cm-1
R
Lecture 17
Predicting Rotational Spectra – carbon monoxide
OK, so I need to know R, the equilibrium bond length. Do an energy
minimization.
-1
B = 1.899cm = 56.9 GHz
2B = 114 GHz
So, there should be radiation at about
114 GHz, and multiples thereof
emanating from CO in interstellar space.
In this case, we actually have an
experimental transition of 115 GHz.
Let’s go and look for CO in space…
©2013, Jordan, Schmidt & Kable
(
)
E = BJ J +1
6B
J=3
Transitions with DJ=±1
4B
J=2
2B
J=1
0
J=0
Lecture 17
Predicting Rotational Spectra – carbon monoxide
©2013, Jordan, Schmidt & Kable
2B
J=1
0
J=0
Lecture 17
CO maps the universe
©2013, Jordan, Schmidt & Kable
Lecture 17
Spectroscopy in the Daily Mail June 2011
©2013, Jordan, Schmidt & Kable
Lecture 17
Predicting Rotational Spectra – larger species
For larger species we have three rotational constants, A, B and C which
are obtained by finding the principle inertial axes of the molecule.
These axes are such that the moment of
inertia tensor is diagonal.
c
E  ωT Iω
  m y 2  z 2 

I    mxy
  mxz

 mxy
 mx  z 
 myz
2
2
 mxz
 myz



2
2 


m
x

y

b
a
©2013, Jordan, Schmidt & Kable
Lecture 17
Vibrational Spectra
For diatomics, there is only one vibrational mode, so calculation of the
vibrational energy levels is simple.
Harmonic Approximation: The oscillator is assumed to be parabolic, and
the energy levels are given by
1
2
V  k R  Re 
2
E  e v  1 / 2
1
e 
2c
k
m
v=3
7we/2
v=2
5we/2
v=1
3we/2
we/2
©2013, Jordan, Schmidt & Kable
v=0
Re
Lecture 17
Vibrational Spectra – carbon dioxide
We can calculate the force constant, k, by looking for the second
derivative of the potential energy curve.
E  e v  1 / 2
-112.56
1
e 
2c
-112.60
m
Nm-1
m = 6.857 amu
e = 2510
cm-1
RHF/6-311G(d)
-112.58
-112.62
Energy (hartree)
k = 2544
k
-112.64
-112.66
-112.68
-112.70
-112.72
-112.74
-112.76
-112.78
0.8
experiment is 2169 cm-1
©2013, Jordan, Schmidt & Kable
1.0
1.2
1.4
C-O distance (Å)
Lecture 17
Vibrational Spectra – carbon dioxide
Even when we do the most careful differentiation, Hartree-Fock is giving a
frequency that is quite high. DFT does better.
Method
e (cm-1)
RHF/6-311G(d)
2435
B3LYP/6-311G(d)
2210
Experiment
2169
The reason that Hartree-Fock overestimates the frequency is due to
neglect of correlation. HF cannot describe dissociation, so the well is too
steep, and the frequency is too high. If you use HF frequencies, better
think about scaling by about 0.89!
©2013, Jordan, Schmidt & Kable
Lecture 17
Vibrational Spectra – carbon dioxide
Normally, we want to compare to band positions, which also need to
account for anharmonicity which is missing in the harmonic
approximation. As it turns out, if you multiply DFT frequencies by the
magic number 0.97, you get great results.
J=3
0.97×2210 = 2143cm-1.
v=1
R
P
J=2
J=1
J=0
J=3
v=0
J=2
J=1
J=0
©2013, Jordan, Schmidt & Kable
Lecture 17
How do complicated molecules vibrate?
More complicated molecules than diatomics have many vibrational modes
(3N-6 for a non-linear molecule).
q1
q2
In this orientation, we can see that if we roll a ball vertically or
horizontally, these vibrations will repeat forever. These are the normal
modes of the molecule.
©2013, Jordan, Schmidt & Kable
Lecture 17
How do complicated molecules vibrate?
How about in this orientation?
q1
q2
Here, the coordinates q1 and q2 are not the normal modes. Vibrating in q1
will cause movements in q2.
©2013, Jordan, Schmidt & Kable
Lecture 17
How do complicated molecules vibrate?
The normal modes are a rotation of the original coordinates.
q1
q’1
q’2
q2
The mathematical procedure which calculates the normal modes of the
system is known as normal mode analysis.
©2013, Jordan, Schmidt & Kable
Lecture 17
E.g. Carbon dioxide
The normal modes of CO2
are not the individual CO
stretches, but linear
combinations which are
called symmetric and
antisymmetric (also
asymmetric)
©2013, Jordan, Schmidt & Kable
Lecture 17
Doing the job properly
To calculate spectra accurately, you just can’t beat solving the nuclear
Schrödinger equation on the various electronic potential energy surfaces.
How many quanta of the
“normal modes” are in
each wavefunction?
©2013, Jordan, Schmidt & Kable
Lecture 17
Crazy 2d vibrational wavefunctions
©2013, Jordan, Schmidt & Kable
Lecture 17