Components of the Atom
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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:
ca1sa cb1sb 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
caa cbb ccc
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 e2R
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 N2 1sa 1sa 1sa 1sb 1sb 1sa 1sb 1sb
1 N2 1 Sab Sab 1 N2 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
eZ ' rb
The expectation value for the energy, <E>, is now a function
of both Z’ and R.
Slide 38
caa cbb ca
Z '3
e
Z ' ra
cb
Z '3
eZ ' 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)
12 12 spat spin
2
spat
spin
Slide 46
1
MO g 1s(1) g 1s(2)
12 12
2
spin
1
2
12 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
eZ ' 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 caa cbb
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*Had
Energy of an electron in atomic orbital,
a, in an unbonded atom.
H bb b H b b*Hbd
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*Hbd
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 caa cbb
Normalization:
2
1 MO
d MO MO
1 N caa cbb N caa cbb
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
caa cbb
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 cii
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 caa cbb
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 caa cbb 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 caa cbb
b
b
Antibonding MO
13.6 12.74 ca 2cb 0
cb
0.430 cb 0.430ca
ca
A caa cbb
ca a 2.33b
ca a 0.430b
N a 2.33b
N a 0.430b
B
1
1 2.33
a 2.33b
2
B 0.394a 0.919b
0.394 1sa (H ) 0.919 2 pzb (F )
A
1
1 0.430
2
a 0.430b
A 0.919a 0.394b
0.919 1sa (H ) 0.394 2 pzb (F )
Slide 81
Electron Densities in Hydrogen Fluoride
Bonding Orbital
B 0.394a 0.919b 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 0.919a 0.394b 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 x1034 )(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
1x1042
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
g1e1
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
nRg1e1e
nNA 0
e1
g ge T
R
NA
k
H0elect
e1
ln g0 g1e T
0
1
Hthermelecte1
nRg1e1e
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
nRg1e1e
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
nRg1e1e
g0 g1e
e1
T
e1
T
Slide 88
H elect U elect
nRg1e1e
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
nRg1e1e
g0 g1e
e1
T
e1
T
nRg1e1e
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
nRg1e1e
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 x1034 J s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1
11,360 K
k
1.38 x1023 J / K
H elect
nRg1e1e
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 2x1015 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 x1034 J s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1
11,360 K
k
1.38 x1023 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 x1034 J s 3.00 x1010 cm / s 7882 cm 1
hc e1
e1
11,360 K
k
1.38 x1023 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