Transcript Components of the Atom

Chapter 8
Diatomic Molecules
Slide 1
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 2
Hydrogen Molecular Ion:
Born-Oppenheimer Approximation
The simplest molecule is not H2. Rather, it is H2+, which has two
hydrogen nuclei and one electron.
e
ra
a
rb
Rab
b
The H2+ Hamiltonian (in au)
H
1
1
1
1
1
1
a2 
b2  e2 


2Ma
2Mb
2
ra
rb
Rab
KE
Nuc a
KE
Nuc b
KE
Elect
PE
e-N
Attr
PE
PE
e-N
N-N
Attr Repuls
Slide 3
Born-Oppenheimer Approximation
Electrons are thousands of times lighter than nuclei.
Therefore, they move many times faster
The Born-Oppenheimer Approximation states that since nuclei
move so slowly, as the nuclei move, the electrons rearrange almost
instantaneously.
With this approximation, it can be shown that one can separate
nuclear coordinates (R) and electronic coordinates (r), and get
separate Schrödinger Equations for each type of motion.
Nuclear Equation


1
1
1
2
2
a 
b 
 Eel     Rab   Enuc    Rab 

2
M
2
M
R
a
b
ab


Eel is the effective potential energy exerted by the electron(s) on
the nuclei as they whirl around (virtually instantaneously on the
time scale of nuclear motion)
Slide 4
Electronic Equation
 1 2 1 1
       Eelect  E
ra rb 
 2
Because H2+ has only one electron, there are no electron-electron
repulsion terms.
In a multielectron molecule, one would have the following terms:
1. Kinetic energy of each electron.
2. Attractive Potential energy terms of each electron
to each nucleus.
3. Repulsive Potential energy terms between each
pair of electrons
Slide 5
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 6
Systems of Linear Equations: Cramer’s Rule
a11x1  a12 x2  c1
a21x1  a22 x2  c2
In these equations, the aij and ci are constants.
We want to solve these two equations for the values
of the variables, x1 and x2
Cramer’s Rule
c1
c2
x1 
a11
a21
a12
a22
Det of " Augmented " matrix

a12
Det of original matrix of coefficients
a22
a11 c1
a
c2
Det of " Augmented " matrix
x2  12

a11 a12 Det of original matrix of coefficients
a21 a22
Slide 7
A Numerical Example
a11x1  a12 x2  c1
2x1  5 x2  1
a21x1  a22 x2  c2
3 x1  2x2  11
c1
c
x1  2
a11
a21
a12
1 5
a22
11 2 1( 2)  5(11) 57


3

2 5
a12
2( 2)  5(3) 19
3 2
a22
a11 c1
2 1
2(11)  1(3) 19
a
c2
3 11


 1
x2  12

2(

2)

5(3)

19
2 5
a11 a12
3 2
a21 a22
Slide 8
Homogeneous Linear Equations
a11x1  a12 x2  0
a21x1  a22 x2  0
Question: What are the solutions, x1 and x2?
Answer: I got it!! I got it!! x1=0 and x2=0.
Question: That’s brilliant, Corky!!
But that’s the “trivial” solution.
Ya got any other solutions for me?
Answer: Lunch Time!! Gotta go!!
Slide 9
a11x1  a12 x2  0
a21x1  a22 x2  0
Let’s try Cramer’s rule.
0
0
x1 
a11
a21
a12
a22
a12
a22

a11
a12
x2 
a11
a21
0
0
a12
a22

0
a11 a12
a21 a22
0
a11 a12
a21 a22
a11 a12
 0 unless
0
a21 a22
 0 unless
a11 a12
0
a21 a22
a11x1  a12 x2  0
a21x1  a22 x2  0
x1  x2  0 unless
a11 a12
a21 a22
0
If one has a set of N linear homogeneous equations with N unknowns,
there is a non-trivial solution only if the determinant of coefficients is zero.
This occurs often in Quantum Chemistry.
The determinant of coefficients is called the “Secular Determinant”.
Slide 11
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 12
LCAO Treatment of H2+
Molecular Orbitals
When we dealt with multielectron atoms, we assumed that the total
wavefunction is the product of 1 electron wavefunctions (1 for each
electron), and that one could put two electrons into each orbital, one
with spin  and the second with spin .
In analogy with this, when we have a molecule with multiple electrons,
we assume that the total electron wavefunction is product of
1 electron wavefunctions (“Molecular Orbitals”), and that we can put
two electrons into each orbital.
 1,2,3,...N    1MO (1)1    1MO (2) 2    2MO (3) 3    2MO (4) 4  ...
Actually, that’s not completely correct. We really use a
Slater Determinant of product functions to get an Antisymmetrized
total wavefunctions (just like with atoms).
Slide 13
Linear Combination of Atomic Orbitals (LCAO)
Usually, we take each Molecular Orbital (MO) to be a
Linear Combination of Atomic Orbitals (LCAO), where each atomic
orbital is centered on one of the nuclei of the molecule.
For the H2+ ion, there is only 1 electron, and therefore we need only
1 Molecular Orbital.
The simplest LCAO is one where the MO is a combination of
hydrogen atom 1s orbitals on each atom:
  ca1sa  cb1sb  ca1sa  cb1sb
shorthand
Assume that 1sa and 1sb
are each normalized.
a
b
1sa
1sb
Slide 14
Expectation Value of the Energy
  ca1sa  cb1sb
Our goal is to first develop an expression relating the expectation value
of the energy to ca and cb.
Then we will use the Variational Principle to find the best set of
coefficients; i.e. the values of ca and cb that minimize the energy.
E
 * H d


 * d

 H
 

Num
Denom
Slide 15
  ca1sa  cb1sb
Denom     ca1sa  cb1sb ca1sa  cb1sb
Denom  ca2 1sa 1sa  cacb 1sa 1sb  cbca 1sb 1sa  cb2 1sb 1sb
Remember that 1sa 1sa  1sb 1sb  1 because 1sa and 1sb are normalized.
Define: Sab  1sa 1sb  1sb 1sa , where Sab is the overlap integral.
Denom  ca2  2cacbSab  cb2
Slide 16
  ca1sa  cb1sb
Num   H   ca1sa  cb1sb H ca1sa  cb1sb
Num  ca2 1sa H 1sa  cacb 1sa H 1sb  cbca 1sb H 1sa  cb2 1sb H 1sb
Haa  1sa H 1sa
H bb  1sb H 1sb
Hab  1sa H 1sb
H ba  1sb H 1sa  Hab because H is Hermitian
Num  ca2Haa  2cacbHab  cb2H bb
Note: For this particular problem,
Hbb=Haa by symmetry.
However, this is not true
in general.
Slide 17
Num  ca2Haa  2cacbHab  cb2H bb
Denom  ca2  2cacbSab  cb2
2
 H
ca2Haa
 2cacbHab  cb2Hbb
Num

E 

ca2  2cacbSab  cb2
 
Denom
Minimizing <E>: The Secular Determinant
In order to find the values of ca and cb which minimize <E>, we
want:
 E
ca
 0 and
 E
cb
0
It would seem relatively straightforward to take the derivatives
of the above expression for <E> and set them equal to 0.
However, the algebra to get where we want is extremely messy.
If, instead, one uses “implicit differentiation”, the algebra is only
relatively messy.
I’ll show it to you, but you are not responsible for the details.
You are responsibly only for the concept of how we get to
the “Secular Determinant”
Slide 18
ca2Haa  2cacbHab  cb2H bb
E 
ca2  2cacbSab  cb2
c
2
a

 2cacbSab  cb2 E  ca2Haa  2cacbHab  cb2H bb
Differentiate both sides w.r.t. ca: Use product rule on left side


  ca2  2cacbSab  cb2 E   ca2Haa  2cacbHab  cb2Hbb 

ca
ca

ca2  2cacbSab  cb2
Set
 E
ca

 E
ca
 E  2ca  2cbSab  0    2caHaa  2cbHab  0 
 0 and group coefficients of ca and cb
Slide 19
E  2ca  2cbSab    2caHaa  2cbHab 
E ca  E cbSab  caHaa  cbHab
After dividing both
sides by 2
0   Haa  E  ca   Hab  E Sab  cb
or
H
aa
 E  ca   Hab  E Sab  cb  0
This is one equation relating the two coefficients, ca and cb.
We get a second equation if we repeat the procedure, except
differentiate w.r.t. cb and set the derivative =0.
The second
equation is:
H
ab
 E Sab  ca   H bb  E  cb  0
Slide 20
H
aa
 E  ca   Hab  E Sab  cb  0
H
ab
 E Sab  ca   H bb  E  cb  0
Hey!!! Now we have two equations with two unknowns, ca and cb.
All we have to do is use Cramer’s Rule to solve for them.
Not so fast, Corky!! Those are homogeneous equations. The
only way we can get a solution other than the trivial one, ca=cb=0,
is if the determinant of coefficients of ca and cb is zero.
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
The Secular Determinant
Slide 21
Extension to Larger Systems
The 2x2 Secular Determinant resulted from using a wavefunction
consisting of a linear combination of atomic orbitals.
If, instead, you use a linear combination of N orbitals, then you get
an NxN Secular Determinant
A simple way to remember how to build a Secular Determinant is
to use the “generic” formula:
H ij  E Sij  0
After you have made the Secular Determinant, set the diagonal
overlaps, Sii = 1.
Slide 22
Hij  E Sij  0
For example, if
  caa  cbb  ccc
Then the Secular Determinant is:
Haa  E Saa
HaB  E Sab
Hac  E Sac
Hab  E Sab
Hbb  E Sbb
Hbc  E Sbc  0
Hac  E Sac
Hbc  E Sbc
Hcc  E Scc
Setting diagonal Sii = 1
Haa  E
HaB  E Sab
Hab  E Sab
Hbb  E
Hac  E Sac
Hbc  E Sbc
Hac  E Sac
Hbc  E Sbc  0
Hcc  E
Slide 23
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 24
H2+ Energies
Linear Equations
H
aa
 E  ca   Hab  E Sab  cb  0
H
ab
 E Sab  ca   H bb  E  cb  0
Secular Determinant
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Outline: 1. We will expand the Secular Determinant.
This will give us a quadratic equation in <E>.
2. We will solve for the two values of <E> as a function
of Haa, Hab, Sab.
3. We will explain how the matrix elements are evaluated and
show the energies as a function of R
4. For each value of <E>, we will calculate the MO;
i.e. the coefficients, ca and cb.
Slide 25
Expansion of the Secular Determinant
Haa  E
Hab  E Sab
or
Hab  E Sab
0
Hbb  E
E
H
H
aa  E
bb  E
  H
ab  E Sab   0
2
 Haa  E  Hbb    Hab  E Sab   0
2
This expression can be expanded, yielding a quadratic equation
in <E>. This equation can be solved easily using the quadratic
formula.
However, let’s remember that for this problem:
Haa  1sa H 1sa
Hbb  1sb H 1sb
Therefore, Hbb=Haa (by symmetry)
The equation then simplifies to:

E  Haa    Hab  E Sab 
2
2
Slide 26
Solving for the Energies
E
 Haa    Hab  E Sab 
2
2
E  Haa    Hab  E Sab   Hab
E Sab
E 1 Sab   Haa  Hab
E  E Sab  Haa  Hab
Therefore:
or
E 
E

Haa  Hab
1  Sab
H  Hab
 aa
1  Sab
and
E


Haa  Hab
1  Sab
Slide 27
E

H  Hab
 aa
1  Sab
and
E


Haa  Hab
1  Sab
Evaluating the Matrix Elements and Determining <E>+ and <E>This is the easiest part because we won’t do it.
These are very specialized integrals. For Hab and Sab, they involve
two-center integrals. That’s because 1sa is centered on nucleus a,
whereas 1sb is centered on nucleus b.
They can either be evaluated numerically, or analytically using a
special “confocal elliptic” coordinate system. We will just present
the results. They are functions of the internuclear distance, R.
Haa
1 1 
1
     1   e2R
2 R  R
Sab  e
R
 R3


R

1


3


1
Hab   Sab   R  1 e R
2
Slide 28
E
E


Haa  Hab
1  Sab
V
H  Hab
 aa
1  Sab
V



 E


1 Haa  Hab 1


R
1  Sab
R
 E


1 Haa  Hab 1


R
1  Sab
R
<E>+ and <E>- represent the electronic energy of the H2+ ion.
The total energy, <V>+ and <V>- , also includes the
internuclear repulsion, 1/R
Slide 29

 E


1 Haa  Hab 1


R
1  Sab
R
V

 E


1 Haa  Hab 1


R
1  Sab
R
-0.1
-0.2
E (au)
V
<V>-
<V>+
-0.3
Antibonding Orbital
-0.4
Asymptotic limit
of EH as R
V() = -0.50 au
Bonding Orbital
-0.5
1
2
3
4
5
R/a0
( R Calculated
Vp lu s Vmi nu Minimum
s Li ne)
Energy
Emin(cal) = -0.565 au at Rmin(cal) = 2.49 a0 = 1.32 Å
Slide 30
Comparison with Experiment
Emin(cal) = -0.565 au
Rmin(cal) = 1.32 Å
-0.1
E (au)
-0.2
<V>-
<V>+
Dissociation Energy
-0.3
Antibonding Orbital
De(cal) = EH – Emin(cal)
= -0.5 au – (-0.565 au)
-0.4
Bonding Orbital
= +0.065 au•27.21 eV/au
-0.5
1
2
3
4
R/a0
Calculated
( R Vp
lu s Vmi nu sMinimum
Li ne)
Energy
5
= 1.77 eV
Rmin
De
Cal. 1.32 Å
1.77 eV
Expt. 1.06
2.79
The calculated results aren’t great, but it’s a start.
We’ll discuss improvements after looking at the wavefunctions.
Slide 31
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 32
H2+ Wavefunctions (aka Molecular Orbitals)
The LCAO Wavefunction:   ca1sa  cb1sb
Remember that by using the Variational Principle on the expression
for <E>, we developed two homogeneous linear equations
relating ca and cb.
H
aa
 E  ca   Hab  E Sab  cb  0
H
ab
 E Sab  ca   H bb  E  cb  0
We then solved the Secular Determinant of the matrix coefficients
to get two values for <E>
E


Haa  Hab
1  Sab
E


Haa  Hab
1  Sab
We can now plug one of the energies (either <E>+ or <E>-)
into either of the linear equations to get a relationship between
ca and cb for that value of the energy.
Slide 33
Bonding Wavefunction
 Haa  E  ca   Hab  E Sab  cb  0
Plug in
E


Haa  Hab
1  Sab



Haa  Hab 
Haa  Hab
H

c

H

S
 aa
 a  ab
ab  cb  0
1

S
1

S
ab
ab




 H 1  S   H
aa
ab
aa
 Hab  ca   Hab 1  Sab   HaaSab  HabSab  cb  0
Haa  HaaSab  Haa  Hab  ca  Hab  HabSab  HaaSab  HabSab  cb  0
HaaSab  Hab  ca  Hab  HaaSab  cb  0
 HaaSab  Hab 
 Hab  HaaSab 
c


 a 
 cb  0
 HaaSab  Hab 
 HaaSab  Hab 
ca  cb  0
cb  ca
Note: Plugging into the second of the two linear equations
gets you the same result.
Slide 34
   ca 1sa  1sb 
cb  ca +   ca1sa  cb1sb
or
   N 1sa  1sb 
N+ (=ca) is determined by normalizing +
Normalization:
1   * d     
1  N 1sa  1sb  N 1sa  1sb   N2  1sa 1sa  1sa 1sb  1sb 1sa  1sb 1sb

1  N2 1 Sab  Sab  1  N2  2  2Sab 
N 
1
2  2Sab
   N 1sa  1sb 

1
2  2Sab
1sa  1sb 
Slide 35
Antibonding Wavefunction
 Haa  E  ca   Hab  E Sab  cb  0
Plug in
E


Haa  Hab
1  Sab



Haa  Hab 
Haa  Hab
H

c

H

S
 aa
 a  ab
ab  cb  0
1

S
1

S
ab
ab




HW
ca  cb  0
   ca 1sa  1sb   N 1sa  1sb  
cb  ca
1
2  2Sab
1sa  1sb 
HW
Note: Plugging into the second of the two linear equations
gets you the same result.
Slide 36
Plotting the Wavefunctions


 N

1 s a
 1sb 


 N

1 s a
 1sb 
0
1
2

0
Nuc
a
1
2
3
Nuc
b
Nuc
a
2
3
Nuc
b
Note that the bonding MO, +, has significant electron density in
the region between the two nuclei.
Note that the antibonding MO, - , has a node (zero electron density in
the region between the two nuclei.
Slide 37
Improving the Results
One way to improve the results is to add more versatility to the
atomic orbitals used to define the wavefunction.
We used hydrogen atom 1s orbitals:
1sa  1sa 
1

e ra and 1sb  1sb 
1

e  rb
(in atomic units)
Instead of assuming that each nucleus has a charge, Z=1, we can use
an effective nuclear charge, Z’, as a variational parameter.
a 
Z '3

e
 Z ' ra
and b 
Z '3

eZ ' rb
The expectation value for the energy, <E>, is now a function
of both Z’ and R.
Slide 38
  caa  cbb  ca
Z '3

e
 Z ' ra
 cb
Z '3

eZ ' rb
This expression for the wavefunction can be plugged into the
equation for <E>. The values of Z’ and R which minimize <E>
can then be calculated. The best Z’ is 1.24.
Rmin
De
Cal.(Z=1)
1.32 Å
1.77 eV
Cal.(Z’=1.24)
1.06
2.35
Expt.
1.06
2.79
Slide 39
An Even Better Improvement: More Atomic Orbitals
Z-Direction
a
b
Instead of expanding the wavefunction as a linear combination of just
one orbital on each atom, put in more atomic orbitals. e.g.
  c11sa  c2 2sa  c3 2pza  c41sb  c5 2sb  c6 2pza
Note: A completely general rule is that if you assume that a Molecular
Orbital is an LCAO of N Atomic Orbitals, then you will get an
NxN Secular Determinant and N Molecular Orbitals.
Slide 40
Z-Direction
a
b
We ran a calculation using: 4 s orbitals, 2 pz orbitals and 1 dz2 orbital
on each atom.
The calculation took 12 seconds. We’ll call it Cal.(Big)
Rmin
De
Cal.(Z=1)
1.32 Å
1.77 eV
Cal.(Z’=1.24)
1.06
2.35
Cal.(Big)
1.06
2.78
Expt.
1.06
2.79
Slide 41
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 42
MO Treatment of the H2 Molecule
The H2 Electronic Hamiltonian
1
r1a
r1b
a
b
R
r2a
r2b
2
1
1
1
1
1
1
1
H   12  22 




2
2
r1a
r1b
r2a
r2b
r12
KE
e1
KE
e2
PE
e-N
Attr
PE
e-N
Attr
PE PE
PE
e-N e-N
e-e
Attr Attr Repuls
Slide 43
The LCAO Molecular Orbitals
Energy
E 
Haa  Hab
1  Sab
EH  Haa
EH  Haa
1sa
E 
Haa  1sa H 1sa
   N (1sa  1sb )
Haa  Hab
1  Sab
Antibonding Orbital
1sb
   N (1sa  1sb )
Bonding Orbital
is the energy of an electron in a hydrogen
1s orbital.
We can put both electrons in H2 into the bonding orbital,
+, one with  spin and one with  spin.
Slide 44
Notation
Antibonding Orbital
   N 1sa  1sb    u* 1s
Bonding Orbital
   N 1sa  1sb    g 1s
 g 1s
Combin. of
1s orbitals
e- density max. on
symmetric w.r.t.
internuclear axis
inversion
 1s
*
u
antibonding
antisymmetric w.r.t.
inversion
Slide 45
The Molecular Wavefunction
Put 1 electron in g1s with  spin: g1s(1) 1
Put 1 electron in g1s with  spin: g1s(2)2
Form the antisymmetrized product using a Slater Determinant.
 MO
 MO 
1  g 1s(1)1  g 1s(1)1

2!  g 1s(2) 2  g 1s(2)2
1
 g 1s(1)1   g 1s(2)2   g 1s(1)1   g 1s(2) 2 
2
1
 MO   g 1s(1) g 1s(2) 
12  12    spat  spin
2
spat
spin
Slide 46
1
 MO   g 1s(1) g 1s(2) 
12  12 
2
 spin 
1
2
12  1 2 
The spin wavefunction is already normalizeD
(see Chap. 8 PowerPoint for He).
Because the Hamiltonian doesn’t operate on the spin, the spin
wavefunction has no effect on the energy of H2.
This independence is only because we were able to write the total
wavefunction as a product of spatial and spin functions.
This cannot be done for most molecules.
 spat   g 1s(1) g 1s(2)  N 1sa  1sb   N 1sa  1sb  
Slide 47
The MO Energy of H2
 spat   g 1s(1) g 1s(2)  N 1sa  1sb   N 1sa  1sb  
1
1
1
1
1
1
1
H   12  22 




2
2
r1a
r1b
r2a
r2b
r12
The expectation value for the ground state H2 electronic energy
is given by:
E   spat H  spat
using the wavefunction and Hamiltonian above.
The (multicenter) integrals are very messy to integrate, but can
be integrated analytically using confocal elliptic coordinates, to
get E as a function of R (the internuclear distance)
The total energy is then: V (R )  E (R ) 
1
R
Slide 48
Energy (au)
2•EH
-1.0
De: Dissociation Energy
0.0
0.5
1.0
1.5
2.0
2.5
R (Angstroms)
Emin(cal)=-1.099 au
R,min(cal)= 0.85 Å
De(cal)= 2•EH – Emin(cal)
De(cal)= +0.099 au = 2.69 eV
Rmin
De
Cal.(Z=1)
0.85 Å
2.69 eV
Expt.
0.74
4.73
Slide 49
Improving the Results
As for H2+, one can add a variational parameter to the atomic orbitals
used in g1s.
a 
Z '3

e
 Z ' ra
and b 
Z '3

eZ ' rb
(in atomic units)
 spat   g 1s(1) g 1s(2)  N a  b   N a  b  
The energy is now a function of both Z’ and R.
One can find the values of both that minimize the energy.
Rmin
De
Cal.(Z=1)
0.85 Å
2.70 eV
Cal.(Var. Z’)
0.73
3.49
Expt.
0.74
4.73
Slide 50
An Even Better Improvement: More Atomic Orbitals
Z-Direction
a
b
As for H2+, one can make the bonding orbital a Linear Combination
of more than two atomic orbitals; e.g.
 g 1s  c11sa  c2 2sa  c3 2pza  c41sb  c5 2sb  c6 2pza
We performed a Hartree-Fock calculation on H2 using an LCAO
that included 4 s orbitals, 2 pz orbitals and 1 dz2 orbitals on each hydrogen.
Rmin
De
Cal.(Z=1)
0.85 Å
2.70 eV
Cal.(Var. Z’)
0.73
3.49
Cal.(HF-Big)
0.74
3.62
Expt.
0.74
4.73
Question: Hey!! What went wrong??
Slide 51
Question: Hey!! What went wrong??
When we performed this level calculation on H2+, we nailed
the Dissociation Energy almost exactly.
But on H2 the calculated De is almost 25% too low.
Answer: The problem, Corky, is that unlike H2+, H2 has
2 (two, dos, zwei) electrons, whose motions are correlated.
Hartree-Fock calculations don’t account for the electron
correlation energy.
Don’t you remember anything from Chapter 7??
Question: Does your watch also say 3:00 ?
Tee Time – Gotta go!!!
Slide 52
Inclusion of the Correlation Energy
There are methods to calculate the Correlation Energy correction
to the Hartree-Fock results. We’ll discuss these methods in
Chapter 9.
We used a form of “Configuration Interaction”, called QCISD(T),
to calculate the Correlation Energy and, thus, a new value for De
Rmin
De
Cal.(Z=1)
0.85 Å
2.70 eV
Cal.(Var. Z’)
0.73
3.49
Cal.(HF-Big)
0.74
3.62
Cal.(QCISD(T)) 0.74
4.69
Expt.
4.73
0.74
That’s Better!!
Slide 53
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 54
Homonuclear Diatomic Molecules
We showed that the Linear Combination of 1s orbitals on two
hydrogen atoms form 2 Molecular Orbitals, which we used to describe
the bonding in H2+ and H2.
These same orbitals may be used to describe the bonding in
He2+ and lack of bonding in He2.
Linear Combinations of 2s and 2p orbitals can be used to
create Molecular Orbitals, which can be used to describe
the bonding of second row diatomic molecules (e.g. Li2).
We can place two electrons into each Molecular Orbital.
Definition: Bond Order – BO = ½(nB – nA)
nB = number of electrons in Bonding Orbitals
nA = number of electrons in Antibonding Orbitals
Slide 55
Bonding in He2+
He2+ has 3 electrons
Electron Configuration
u* 1s     N 1sa  1sb 

Config   g 1s   u* 1s
2

1
Energy
Antibonding Orbital
1sa
 g 1s     N 1sa  1sb 
1sb
BO = ½(2-1)
= 1/2
Bonding Orbital
Slide 56
Slater Determinant: He2+

Config   g 1s   u* 1s
2
 MO

1
 g 1s(1)1  g 1s(1)1  u* 1s(1)1
1

 g 1s(2) 2  g 1s(2) 2  u* 1s(2) 2
3!
 g 1s(3) 3  g 1s(3)3  u* 1s(3) 3
 MO 
1
3!
 g 1s(1)1  g 1s(2)2  u* 1s(3)3
Shorthand Notation
Slide 57
He2
He2 has 4 electrons
Electron Configuration
 1s     N 1sa  1sb 
*
u

Config   g 1s   1s
2
*
u

2
Energy
Antibonding Orbital
1sa
 g 1s     N 1sa  1sb 
1sb
BO = ½(2-2)
=0
Bonding Orbital
Actually, He2 forms an extremely weak “van der Waal’s complex”,
with Rmin  3 Å and De  0.001 eV [it can be observed at T = 10-3 K.
Slide 58
Second Row Homonuclear Diatomic Molecules
We need more Molecular Orbitals to describe diatomic molecules
with more than 4 electrons.
and
+
Sigma () MO’s
Max. e- density along
internuclear axis
+
2sb
2sa
-
2pza
+
+
and
2pzb
+
Sigma () MO’s
Max. e- density along
internuclear axis
+
and
-
-
2pya
2pyb
Pi () MO’s
Max. e- density above/below
internuclear axis
2pxa and 2pxb also combine to give  MO’s
Slide 59
-
Sigma-2s Orbitals
 u* 2s  N  2sa  2sb 
Energy
Antibonding Orbital
+
2sa
2sb
+
 g 2s  N  2sa  2sb 
Bonding Orbital
Slide 60
Sigma-2p Orbitals
 u* 2 p  N  2 pza  2 pzb 
-
2pza
+
+
-
Energy
Antibonding Orbital
2pzb
 g 2 p  N  2 pza  2 pzb 
Bonding Orbital
Note sign reversal of 2p from 2s and 1s orbitals.
Slide 61
Pi-2p Orbitals
 g* 2 p  N  2 pya  2 pyb 
Energy
Antibonding Orbital
+
+
-
-
2pya
2pyb
 u 2 p  N  2 pya  2 pyb 
Bonding Orbital
There is a degenerate u2p orbital and a degenerate g*2p
orbital arising from analogous combinations of 2pxa and 2pxb
Slide 62
Homonuclear Diatomic Orbital Energy Diagram
 u* 2p
 g* 2p  g* 2p
2pxa
2pya
2pza
 g 2p
 u 2p
2pxb
2pyb
2pzb
 u 2p
 u* 2s
2sa
 g 2s
2sb
 u* 1s
1sa
 g 1s
1sb
Slide 63
 u* 2p
 g* 2p
Consider Li2
(a) What is the electron configuration?
(b) What is the Bond Order?
(c) What is the spin multiplicity?
(Singlet, Doublet or Triplet)
 g 2p
 u 2p
6 Electrons
 1s   1s  
2
 u* 2s
g
*
u
2
g
2s 
 g 2s
BO = ½(4-2) = 1
 u* 1s
S = 0 : Singlet
2
 g 1s
Slide 64
Consider F2
(a) What is the electron configuration?
(b) What is the Bond Order?
(c) What is the spin multiplicity?
(Singlet, Doublet or Triplet)
 u* 2p
 g* 2p
 g 2p
 u 2p
18 Electrons
 1s   1s  
2
 u* 2s
g
*
u
2

*
g 2s   u 2s
2
  u 2p   g 2p   g* 2p 
2
 g 2s
BO = ½(10-8) = 1
 u* 1s
S = 0 : Singlet
4
2
4
 g 1s
Slide 65
Consider O2
(a) What is the electron configuration?
(b) What is the Bond Order?
(c) What is the spin multiplicity?
(Singlet, Doublet or Triplet)
 u* 2p
 g* 2p
 g 2p
 u 2p
16 Electrons
 1s   1s  
2
 u* 2s
g
*
u
2

*
g 2s   u 2s
2
  u 2p   g 2p   g* 2p 
2
 g 2s
BO = ½(10-6) = 2
 u* 1s
S = 1 : Triplet
4
2
2
 g 1s
Slide 66
Consider O2 , O2+ , O2(a) Which has the longest bond?
(b) Which has the highest vibrational frequency?
(c) Which has the highest Dissociation Energy?
 u* 2p
 g* 2p
O2: 16 Electrons – BO = 2
 g 2p
O2+: 15 Electrons – BO = 2.5
 u 2p
O2-: 17 Electrons – BO = 1.5
 u* 2s
O2- has the longest bond.
 g 2s
O2+ has the highest vibrational frequency.
 u* 1s
O2+ has the highest Dissociation Energy.
 g 1s
O2
Slide 67
A More General Picture of Sigma Orbital Combinations
-
+
-
+
2pza
2pzb
+
+
2sa
2sb
+
+
1sa
1sb
The assumption in the past section that
only identical orbitals on the two atoms
combine to form MO’s is actually a bit
simplistic.
In actuality, each of the 6  MO’s is really
a combination of all 6 AO’s.
 MO  c11sa  c2 2sa  c3 2pza  c41sb  c5 2sb  c6 2pzb
Slide 68
Approximate vs. Accurate MO’s in C2
 MO  c11sa  c2 2sa  c3 2pza  c41sb  c5 2sb  c6 2pzb
2p MO (E  -15 eV)
 Approx  0.70  2pza  0.70  2pzb
 Accur   0.07  1sa  0.40  2sa  0.60  2 pza    0.07  1sb  0.40  2sb  0.60  2 pzb 
2s MO (E  -40 eV)
 Approx  0.70  2sa  0.70  2sb
 Accur   0.17  1sa  0.50  2sa  0.23  2 pza    0.17  1sb  0.50  2sb  0.23  2 pzb 
1s MO (E  -420 eV)
 Approx  0.70  1sa  0.70  1sb
 Accur  0.70  1sa  0.01 2sa   0.70  1sb  0.01 2sb 
Slide 69
Outline
• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.
• Math Prelim.: Systems of Linear Equations – Cramer’s Rule
• LCAO Treatment of H2+
• H2+ Energies
• H2+ Wavefunctions
• MO Treatment of the H2 Molecule
• Homonuclear Diatomic Molecules
• Heteronuclear Diatomic Molecules
Slide 70
Heteronuclear Diatomic Molecules
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Hbb  Haa
Therefore, the energies are not symmetrically displaced,
and the magnitudes of the coefficients are no longer equal.
c b  ca
Antibonding (A)
 A  ca' a  cb' b
a E  Haa
E  Hbb
Bonding (B)
b
 B  caa  cbb
Slide 71
Interpretation of Secular Determinant Parameters
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Sab  a b   a*bd
Overlap Integral
+
+
Large Sab
+
+
Small Sab
Generally, Sab  0.1 – 0.2
Commonly, to simplify the calculations,
it is approximated that Sab  0
Slide 72
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Haa  a H a   a*Had
Energy of an electron in atomic orbital,
a, in an unbonded atom.
H bb  b H b   b*Hbd
Energy of an electron in atomic orbital,
b, in an unbonded atom.
Haa , Hbb < 0
Traditionally, Haa and Hbb are called “Coulomb Integrals”
Commonly, Haa is estimated as –IE, where IE is the Ionization Energy
of an electron in the atomic orbital, a.
C+
IE(2p)=11.3 eV
2p
H2s,2s (C )  20.8 eV
H2 p,2 p (C )  11.3 eV
IE(2s)=20.8 eV
2s
Carbon
Slide 73
Haa  E
Hab  E Sab
Hab  a H b   a*Hbd
Hab  E Sab
0
Hbb  E
Interaction energy between atomic orbitals,
a and b .
Traditionally, Hab is called the “Resonance Integral”.
Hab is approximately proportional to: (1) the orbital overlap
(2) the average of Haa and Hab
 H  Hbb 
Hab  K  aa
 Sab
2


K  1.75
Wolfsberg-Helmholtz Formula
(used in Extended Hückel Model)
Hab < 0
Slide 74
Interpretation of Orbital Coefficients
Let’s assume that an MO is a linear combination of 2 normalized AO’s:
 MO  N caa  cbb 
Normalization:
2
1   MO
d   MO  MO
1  N  caa  cbb  N  caa  cbb 
1  N 2 ca2 a a  cb2 b b  2cacb a b 
1  N 2 ca2  cb2  2cacbSab 
N
1
ca2  cb2  2cacbSab
where
Sab  a b
Orbital
Overlap
Slide 75
 MO 
If Sab  0:
 MO 
1
c  c  2cacbSab
2
a
2
b
ca
c c
2
a
2
b
a 
 caa  cbb 
cb
c c
2
a
2
b
b
ca2
fa  2
ca  cb2
Fraction of electron density in orbital a
cb2
fb  2
ca  cb2
Fraction of electron density in orbital b
General:  MO  N cii
ci2
fi 
 ci2
Fraction of electron density in orbital i
Slide 76
ca2
fa  2
ca  cb2
cb2
fb  2
ca  cb2
Homonuclear Diatomic Molecules:
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Hbb  Haa
cb  ca
fa  fb  0.50
Heteronuclear Diatomic Molecules:
Hbb  Haa
cb  ca
fb  fa
Slide 77
 A  ca' a  cb' b
Antibonding (A)
a
E  Haa
E  Hbb
 B  caa  cbb
Bonding (B)
Haa  E
Hab  E Sab
H
aa
 E
 H
bb
 E
b
  H
Hab  E Sab
0
Hbb  E
ab
 E Sab  Hab  E Sab   0
One has a quadratic equation, which can be solved to yield
two values for the energy, <E>.
One can then determine cb/ca for both the bonding
and antibonding orbitals.
Slide 78
A Numerical Example: Hydrogen Fluoride (HF)
1sa(H)
Haa  E
Hab  E Sab
-
+
+
2pzb(F)
Hab  E Sab
0
Hbb  E
 MO  N caa  cbb   N ca1sa (H )  cb 2pzb (F )
Matrix Elements
Haa  a H a  1sa (H ) H 1sa (H)  13.6 eV
Hbb  b H b  2pzb (F ) H 2pzb (F )  17.4 eV
Hab  a H b  1sa (F ) H 2pzb (F )  2.0 eV
Sab  a b  1sa (H ) 2pzb (F )  0
Slide 79
Haa  E
Hab  E Sab
Hab  E Sab
0
Hbb  E
Haa  13.6 eV
Hbb  17.4 eV
Hab  2.0 eV
13.6  E
2
 13.6 
E
E
2
Sab  0
2
0
17.4  E
 17.4 
E
   2  2   0
 31.0 E  232.64  0
31.0  (31.0)2  4(1)(232.64)
E 
2
E
B
31.0  30.44

 18.26 eV
2
E
A

31.0  30.44
 12.74 eV
2
Slide 80
13.6  E
2
2
0
17.4  E
 13.6  E  c  2c  0
2c   17.4  E  c  0
a
a
Bonding MO
 13.6   18.25   ca  2cb  0
cb
 2.33  cb  2.33ca
ca
 B   caa  cbb 
b
b
Antibonding MO
 13.6   12.74   ca  2cb  0
cb
 0.430  cb  0.430ca
ca
 A   caa  cbb 
 ca a  2.33b 
 ca a  0.430b 
 N a  2.33b 
 N a  0.430b 
B 
1
1   2.33 
a  2.33b 
2
 B  0.394a  0.919b
 0.394  1sa (H )  0.919  2 pzb (F )
A 
1
1   0.430 
2
a  0.430b 
 A  0.919a  0.394b
 0.919  1sa (H )  0.394  2 pzb (F )
Slide 81
Electron Densities in Hydrogen Fluoride
Bonding Orbital
 B  0.394a  0.919b  0.394  1sa (H )  0.919  2 pzb (F )
ca2
2
fa  fH  2

0.394
 0.16


ca  cb2
cb2
2
fb  fF  2

0.919
 0.84


ca  cb2
Over 80% of the electron density of the two electrons in the
bonding MO resides on the Fluorine atom in HF.
Antibonding Orbital
 A  0.919a  0.394b  0.919  1sa (H )  0.394  2 pzb (F )
ca2
2
fa  fH  2

0.919
 0.84


ca  cb2
cb2
2
fb  fF  2

0.394
 0.16


ca  cb2
The situation is reversed in the Antibonding MO.
However, remember that there are no electrons in this orbital.
Slide 82
Statistical Thermodynamics: Electronic Contributions
to Thermodynamic Properties of Gases
Here they are again.
  ln Q 
U  kT 2 

 T V ,N
 U 
CV  

 T V ,N
  ln Q 
  ln Q 
H  kT 2 

kT

  lnV 
 T V ,N

T ,N
 H 
CP  

 T P,N
S  k ln Q 
U
T
A  U  TS  kT ln Q
  ln Q 
G  H  TS  kT ln Q  kT 

  lnV T ,N
Slide 83
The Electronic Partition Function

q
elect
  gi e

i
kT
i 0
or
q
elect
e

0
kT
 g 0e

  gi e

 i  0
kT

0
 g1e
kT
e

i 0
q
elect
e

0
kT


 i
kT

ei
T
  gi e
e

0
kT
i 0
q
elect
e

0
kT

  gi e
i 0
hc ei

ei  i 
k
k
e

0
kT
0
kT

1
kT
 g 2e

2
kT
 ...
 
 


 1 0
 2 0
 g 0  g1e kT  g 2e kT  ...






 1
 2
 g 0  g1e kT  g 2e kT  ...






 e1
 e2
 g 0  g1e T  g 2e T  ...


I is the energy of the i'th electronic level above
the ground state, 0
~ is the absorption frequency (in cm-1) to this state.
Slide 84
q elect  e

0
kT

  gi e

ei
T
e

0
kT
i 0




 e1
 e2
T
T
 g 0  g1e
 g 2e
 ...


ei 
 i hc ei

k
k
The ground state energy, 0, is arbitrary (depending upon the point of
reference. It is sometimes taken to be zero.
Excited state energies above ~5,000-10,000 cm-1 don't contribute
significantly to qelect (or to thermodynamic properties). For example,
 ei  20,000 cm 1
T  3,000 K
hc ei (6.63 x1034 )(3.00 x1010 )(20,000)
4
ei 


2.88
x
10
K
23
k
1.38 x10
e

 ei
T
e

2.88 x 104
3,000
At room temperature (298 K): e

 e 9.60  7 x10 5
ei
k
 1x1042
Thus, one can ignore all but very low-lying electronic states, except
at elevated temperatures.
Slide 85
For convenience, we will consider only systems






elect
 e kT g 0  g1e T 
with a maximum of one "accessible" excited state: q
0
e1


Internal Energy and Enthalpy
U
elect
  ln Q elect 
 kT 


T

V ,N
2
Qelect independent of V
H
H
elect
U
elect
elect
  ln Q elect 
  ln Q elect 
elect
 kT 
  kT 
 U
 T
V ,N
  lnV T ,N
2
elect
  ln q elect  N



ln
Q


 kT 2 
  kT 2 

T
 T
V ,N

elect


2   ln q
  nRT 


T


V ,N
V ,N
Slide 86
q
elect
e

0
kT


 e1 
g 0  g1e T 


ln q
 d

2
 nRT 
 dT
V ,N
elect
2   ln q
 nRT 
 T
H elect  U elect
  
 nRT  0 2  
 kT 
 g1e1  
 T 2   e


nRT 2
2
g 0  g1e

e 1
T
elect


 e1 

 ln g0  g1e T 
kT


 0
  kT


 e1
T
0
 d
  dT


H
elect
U
elect
 nE0 
e1
T

nRg1e1e
  nNA 0 

 e1

g ge T
R
NA 
k
H0elect

 
 e1   
 ln g0  g1e T  


 
 
0
1
Hthermelecte1
nRg1e1e
g0  g1e
elect
H elect  U elect  H0elect  Htherm


T
e1
T
E0  NA 0 GS Electronic Energy
per mole
e1 
 1 hc e1

k
k
Slide 87
H elect  nE0 
nRg1e1e
g0  g1e


e1
T
e1 
e1
T
 1 hc e1

k
k
For any numerical examples in this section, we'll assume that
0 = 0 which leads to E0 = NA 0 = 0.
i.e. we're setting the arbitrary reference energy to zero in the electronic
ground state.
In this case, we could have started with:
q elect  e

0
kT



 e1 
 e1
g 0  g1e T   g 0  g1e T


which would have led to:
H elect 
nRg1e1e
g0  g1e


e1
T
e1
T
Slide 88
H elect  U elect 

nRg1e1e
g0  g1e

e1
T
e1
T
e1 
 1 hc e1

k
k
Numerical Example #1
The electronic ground state of CO is a singlet, with no accessible
excited electronic levels. Calculate Helect at 298 K and 3000 K
If there are no "accessible" excited states, that means that
e1/T >> 1. This leads to:
H
elect

nRg1e1e
g0  g1e


e1
T
e1
T
nRg1e1e
0


g0  g1e
By the same reasoning, Helect = Uelect = 0 for molecules with
ground states of any multiplicity if the excited states are inaccessible.
Slide 89
H elect  U elect 
nRg1e1e
g0  g1e


e1
T
e1 
e1
T
 1 hc e1

k
k
Numerical Example #2
The electronic ground state of O2 is a triplet (refer back to orbital diagram
for O2 earlier in the chapter.
There is an excited electronic state doublet 7882 cm-1 above the GS .
Calculate Helect for one mole of O2 at 3000 K and 298 K.




6.63 x1034 J  s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1 

 11,360 K
k
1.38 x1023 J / K
H elect 
nRg1e1e
g0  g1e


e1
T
e1
T

1mol    8.31J / mol  K  (2) 11360 K  e
3  2e

11360
3000

11360
3000
 1400 J  1.40 kJ
At 298 K: H elect  2x1015 kJ  0
Slide 90
Entropy
S
elect
S
 k ln Q
elect
q
elect
elect
 nR ln q
e

0
kT
U elect

 k ln q elect
T
elect

U elect

T

N
U elect
U elect
elect

 Nk ln q

T
T
because Nk = nNAk = nR



 e1 
 e1
g 0  g1e T   g 0  g1e T


because we're assuming
that 0 = 0.
Slide 91
S
elect
 nR ln q
elect
U elect

T
q
elect
 g0  g1e

e1
T
Numerical Example: O2 again
The electronic ground state of O2 is a triplet.
There is an excited electronic state doublet 7882 cm-1 above the GS .




6.63 x1034 J  s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1 

 11,360 K
k
1.38 x1023 J / K
Let's calculate Select for one mole at 3000 K.
Uelect = Helect = 1400 J (from earlier calculation)
q
elect
 g0  g1e

e1
T
 3  2e

11360
3000
 3  0.045  3.045
Slide 92
S
elect
 nR ln q
elect
U elect

T
q
elect
 g0  g1e

e1
T
 3.045
Uelect = 1400 J
S
elect
 nR ln q
elect
U elect
1400 J

 1mol  8.31J / mol  K  ln(3.045) 
T
3000 K
S elect  9.25  0.47  9.72 J / K at 3000 K
At 25 oC = 298 K:
Uelect = 0.00 J
qelect = g0 = 3
Select = Rln(3) + 0/298 = 9.13 J/K at 298 K
Slide 93
S
elect
 nR ln q
elect
At 25 oC = 298 K:
U elect

T
q
elect
 g0  g1e

e1
T
Select = Rln(3) + 0/298 = 9.13 J/K
Recall from above that, if there are no accessible excited states,
then Helect = Uelect = 0, independent of ground-state multiplicity.
In contrast, if the electronic ground-state of a molecule has a
multiplicity, g0 > 1, then Select >0, even if there are no
accessible excited states.
Slide 94
O2 Entropy: Comparison with experiment
O2: Smol(exp) = 205.1 J/mol-K at 298.15 K
Stran  S rot  Svib  151.9  43.8  0.035  195.7
Chap. 3
Chap. 4
Chap. 5
We were about 5% too low. Let's add in the electronic contribution.
Stran  S rot  Svib  Select  151.9  43.8  0.035  9.1  204.8
Note that the agreement with experiment is virtually perfect.
Note: If a molecule has a singlet electronic ground-state and no
accessible excited electronic states, then there is
no electronic contribution to S.
Slide 95
Helmholtz and Gibbs Energy
Aelect  U elect  TSelect  kT lnQelect
Qelect independent of V
G
elect
H
elect
 TS
elect
 kT ln Q
elect
  ln Q elect 
elect
 kT 
 A
  lnV T ,N
Numerical Example: O2 Once More
The electronic ground state of O2 is a triplet.
There is an excited electronic state doublet 7882 cm-1 above the GS .




6.63 x1034 J  s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1 

 11,360 K
k
1.38 x1023 J / K
Let’s calculate Gelect ( = Aelect) for one mole of O2 at 3000 K.
Slide 96
elect
G
q
elect
A
elect
 kT lnQ
 g0  g1e

e1
T
elect
 3  2e
q

11360
3000

Gelect  kT ln Qelect  kT ln q elect
elect
 g0  g1e

e1
T
e1 = 11,360 K
g0 = 2
g1 = 3
 3.045

N
 NkT ln q elect  nRT ln q elect
Nk = nNAk = nR
Gelect  nRT ln qelect   1mol 8.31J / mol  K  (3000 K )ln(3.045)
Gelect  Aelect  27,760 J  27.8 kJ
Slide 97
O2: Entropy
300
S [J/mol-K]
250
200
Expt
T
TR
TRV
TRVE
150
0
1000
2000
3000
4000
5000
Temperature [K]
The agreement of the calculated entropy including electronic
contributions with experiment is close to perfect.
Slide 98
O2: Enthalpy
200
H [kJ/mol]
150
100
Expt
T
TR
TRV
TRVE
50
0
0
1000
2000
3000
4000
5000
Temperature [K]
The agreement of the calculated enthalpy including electronic
contributions with experiment is close to perfect.
Slide 99
O2: Heat Capacity
40
CP [J/mol-K]
35
30
25
Expt
T
TR
TRV
TRVE
20
15
0
1000
2000
3000
4000
5000
Temperature [K]
The calculated heat capacity is the most sensitive function of any
neglect in the calculations.
We have neglected several subtle factors, including vibrational
anharmonicity, centrifugal distortion and vibration-rotation coupling
Slide 100