10 Chapter Homonuclear Diatomic Molecules
Chapter 10: 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
Chapter 10: 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 rb Rab
a
b
The H2+ Hamiltonian (in au)
H =- 1 1 1 2 1 1 1 2 2 - b - e - - + a 2M a 2M b 2 ra rb Rab
KE Nuc a KE Nuc b KE Elect PE e-N Attr PE PE e-N N-N Attr Repuls
Chapter 10: 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 Schrdinger Equations for each type of motion. Nuclear Equation
(c.f. Eq. 4.1 in text)
1 1 1 2 - 2 - b + + Eel ( Rab ) = Enuc ( Rab ) 2M a 2M Rab a b
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)
Chapter 10: Slide 4
Electronic Equation
1 2 1 1 - - - = Eelect = E 2 ra rb
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. 1. 1. Attractive Potential energy terms of each electron to each nucleus. Repulsive Potential energy terms between each pair of electrons
Chapter 10: 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
Chapter 10: Slide 6
Systems of Linear Equations: Cramer's Rule
a11 x1 + a12 x2 = c1 a21 x1 + 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 a12 Det of " Augmented" matrix c2 a22 = x1 = Det of original matrix of coefficients a11 a12 a21 a22 a11 c1
Det of " Augmented" matrix a12 c2 = x2 = Det of original matrix of coefficients a11 a12 a21 a22
Chapter 10: Slide 7
A Numerical Example
a11 x1 + a12 x2 = c1 a21 x1 + a22 x2 = c2 c1 x1 = a12 1 5 2 x1 + 5 x2 = 1 3 x1 - 2 x2 = 11
c2 a22 11 - 2 1 (-2) - 5 11 - 57 = = +3 = = a11 a12 2 5 - 19 2 (-2) - 5 3 a21 a22 3 -2 a11 c1 = 2 1
x2 =
a21 c2 a11 a12 a21 a22
3 11 + 19 2 11 - 1 3 = = -1 = 2 5 - 19 2 (-2) - 5 3 3 -2
Chapter 10: Slide 8
Homogeneous Linear Equations
a11 x1 + a12 x2 = 0 a21 x1 + a22 x2 = 0
Question: What are the solutions, x1 and x2? Answer: x1=0 and x2=0.... But there's more!
Chapter 10: Slide 9
a11 x1 + a12 x2 = 0 a21 x1 + a22 x2 = 0
Let's try Cramer's rule.
0 a21 x1 = 0 a22 a11 a12 a21 a22 a11 0 a12 =0 a22 = 0 a11 a21 a12 a22 = 0 unless a11 a21 a12 a22 =0
a21 0 a 0 x2 = = 0 unless 11 = a11 a12 a11 a12 a21 a21 a22 a21 a22
Chapter 10: Slide 10
a11 x1 + a12 x2 = 0 a21 x1 + a22 x2 = 0 a11 x1 = x2 = 0 unless a21 a12 =0 a22
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".
Chapter 10: Slide 11
Secular Equations
A number of problems in the physical sciences give rise to systems of equations of the form:
a11 x1 + a12 x2 + a13 x3 + ...a1n xn = x1 a21 x1 + a22 x2 + a23 x3 + ...a2 n xn = x2 a31 x1 + a32 x2 + a33 x3 + ...a3n xn = x3 (a11 - ) x1 + a12 x2 + a13 x3 + ...a1n xn = 0 a21 x1 + (a22 - ) x2 + a23 x3 + ...a2 n xn = 0
ion lut so y onl if:
?
an1 x1 + an 2 x2 + an 3 x3 + ...ann xn = xn
a31 x1 + a32 x2 + (a33 - ) x3 + ...a3n xn = 0 an1 x1 + an 2 x2 + an 3 x3 + ...(ann - ) xn = 0
nNo
ial riv t
(a11 - ) a21 a31 an1
a12 (a22 - ) a32 an 2
a13 a23
... ...
a1n a2 n =0 (ann - )
(a33 - ) ... a3n an 3 ...
Expansion is a polynomial of degree n in , the required values of are the n roots of the polynomial.
Chapter 10: Slide 12
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
Chapter 10: Slide 13
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).
Chapter 10: Slide 14
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 = ca 1sa + cb1sb
shorthand
a
b
Assume that 1sa and 1sb are each normalized.
1sa
1sb
Chapter 10: Slide 15
Expectation Value of the Energy
= ca 1sa + 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
* Hd = *d
=
H
=
Num Denom
Chapter 10: Slide 16
= ca 1sa + cb1sb
Denom = = ca 1sa + cb1sb ca 1sa + cb 1sb
2 2 Denom = ca 1sa 1sa + ca cb 1sa 1sb + cb ca 1sb 1sa + cb 1sb 1sb
Remember that
1sa 1sa = 1sb 1sb = 1 because 1sa and 1sb are normalized.
Define: S ab = 1sa 1sb = 1sb 1sa , where Sab is the overlap integral.
2 2 Denom = ca + 2ca cb S ab + cb
Chapter 10: Slide 17
= ca 1sa + cb1sb
Num = H = ca 1sa + cb 1sb H ca 1sa + cb 1sb
2 2 Num = ca 1sa H 1sa + ca cb 1sa H 1sb + cb ca 1sb H 1sa + cb 1sb H 1sb
Define: H aa = 1sa H 1sa
H bb = 1sb H 1sb H ab = 1sa H 1sb H ba = 1sb H 1sa = H ab because H is Hermitian
2 2 Num = ca H aa + 2ca cb H ab + cb H bb Note: For this particular problem, Hbb=Haa by symmetry.
However, this is not true Chapter 10: in general.
Slide 18
2 2 Num = ca H aa + 2ca cb H ab + cb H bb
2 2 Denom = ca + 2ca cb + cb
H E =
2 2 Num ca H aa + 2ca cb H ab + cb H bb = = 2 2 Denom ca + 2ca cb S ab + cb
Minimizing <E>: The Secular Determinant
In order to find the values of ca and cb which minimize <E>, we E E = 0 and =0 want: ca cb It would seem relatively straightforward to take the derivatives of the above expression for <E>. 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're not responsible for the details, only the concept of how we get to the "Secular Determinant"
Chapter 10: Slide 19
2 2 ca H aa + 2ca cb H ab + cb H bb E = 2 2 ca + 2ca cb S ab + cb
(c
2 a
2 2 2 + 2ca cb S ab + cb E = ca H aa + 2ca cb H ab + cb H bb
)
Differentiate both sides w.r.t. ca: Use product rule on left side
2 2 ca + 2ca cb S ab + cb E
[(
ca
) ] = [c H
2 a
2 + 2ca cb H ab + cb H bb aa ca
]
(c
2 a
+ 2ca cb S ab + c
2 b
)
E ca
+ E ( 2ca + 2cb S ab + 0 ) = ( 2ca H aa + 2cb H ab + 0 )
Set
E ca
= 0 and group coefficients of ca and cb
Chapter 10: Slide 20
E ( 2ca + 2cb S ab ) = ( 2ca H aa + 2cb H ab ) E ca + E cb S ab = ca H aa + cb H ab 0 = ( H aa - E ) ca + ( H ab - E S ab ) cb
or
(H
aa
- E ) ca + ( H ab - E S ab ) 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 S ab ) ca + ( H bb - E ) cb = 0
Chapter 10: Slide 21
(H (H
aa
- E ) ca + ( H ab - E S ab ) cb = 0 - E S ab ) ca + ( H bb - E ) cb = 0
ab
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. But wait!! 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.
H aa - E H ab - E S ab H ab - E S ab =0 H bb - E
The Secular Determinant
Chapter 10: Slide 22
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.
Chapter 10: Slide 23
H ij - E Sij = 0
For example, if
= caa + cbb + ccc
Then the Secular Determinant is:
H aa - E S aa H ab - E S ab H ac - E S ac H ab - E S ab H bb - E S bb H bc - E S bc H ac - E S ac H bc - E Sbc = 0 H cc - E S cc
H aa - E H ab - E S ab H ac - E S ac
H ab - E S ab H bb - E H bc - E Sbc
H ac - E S ac H bc - E Sbc = 0 H cc - E
Chapter 10: Slide 24
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
Chapter 10: Slide 25
H2+ Energies
Linear Equations ( H aa - E ) ca + ( H ab - E S ab ) cb = 0 Secular Determinant
H aa - E H ab - E S ab H ab - E S ab =0 H bb - E
(H
ab
- E S ab ) ca + ( H bb - E ) cb = 0
Outline: We will expand the Secular Determinant. 1. 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, caand cb.
Chapter 10: Slide 26
Expansion of the Secular Determinant
H aa - E H ab - E S ab H ab - E S ab =0 H bb - E
(H
aa - E )( H bb - E ) - ( H ab - E S ab ) = 0 2
or
(E
- H aa )( E - H bb ) - ( H ab - E S ab ) = 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: Therefore, Hbb=Haa (by symmetry)
H aa = 1sa H 1sa H bb = 1sb H 1sb
2
(E
- H aa ) = ( H ab - E S ab )
2
Chapter 10: Slide 27
Solving for the Energies
(E
- H aa ) = ( H ab - E S ab )
2
2
E - H aa = ( H ab - E S ab ) = H ab E S ab E E S ab = H aa H ab E (1 S ab ) = H aa H ab
Therefore:
E =
H aa H ab 1 S ab H aa + H ab 1 + S ab
or
E
+
=
and
E
-
=
H aa - H ab 1 - S ab
Chapter 10: Slide 28
E
+
=
H aa + H ab 1 + S ab
and
E
-
=
H aa - H ab 1 - S ab
Evaluating the Matrix Elements and Determining <E>+ and <E>This is the easiest part because we (and the text) 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.
H aa H ab 1 1 1 = - - + 1 + e - 2 R 2 R R 1 = - S ab - ( R + 1)e - R 2
Chapter 10: Slide 29
R3 S ab = e + R + 1 3
-R
E
+
H + H ab = aa 1 + S ab H aa - H ab = 1 - S ab
V
+
= E
+ + + -
1 H aa + H ab 1 = + R 1 + S ab R 1 H aa - H ab 1 = + R 1 - S ab R
E
-
V
-
= E
<E>+ and <E>- represent the electronic energy of the H2+ ion. The total energy, <V>+ and <V>- , also includes the internuclear repulsion, 1/R
Chapter 10: Slide 30
V
+
= E
H 1 Vplus + H ab 1 +( R = aa Vminus Line ) + + R 1 + S ab R
V
-
= E
+ -
1 H aa - H ab 1 = + R 1 - S ab R
-0.1 -0.2
E (au)
<V>+
<V>Antibonding Orbital
-0.3 -0.4
Bonding Orbital
-0.5
Asymptotic limit of EH as R
4 5
1
2
3
R/a0 Calculated Minimum Energy Emin(cal) = -0.565 au at Rmin(cal) = 2.49 a0 = 1.32
Chapter 10: Slide 31
( R Vplus Vminus Line )
Comparison with Experiment Emin(cal) = -0.565 au <V>+ <V>Antibonding Orbital
-0.4
-0.1 -0.2
Rmin(cal) = 1.32 Dissociation Energy De(cal) = EH Emin(cal) = -0.5 au (-0.565 au) = +0.065 au27.21 eV/au = 1.77 eV Rmin Cal. 1.32 Expt. 1.06 De 1.77 eV 2.79
E (au)
-0.3
Bonding Orbital
-0.5
1
2
3
4
5
R/a0 Calculated Minimum Energy
The calculated results aren't great, but it's a start. We'll discuss improvements after looking at the wavefunctions.
Chapter 10: Slide 32
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
Chapter 10: Slide 33
H2+ Wavefunctions (aka Molecular Orbitals)
The LCAO Wavefunction: = ca 1sa + 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 + ( H ab - E S ab ) cb = 0
(H
ab
- E S ab ) ca + ( H bb - E ) cb = 0
We then solved the Secular Determinant of the matrix coefficients to get two values for <E> H - H ab H aa + H ab E - = aa E += 1 - S ab 1 + S ab 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.
Chapter 10: Slide 34
Bonding Wavefunction
(H
aa - E ) ca + ( H ab - E S ab ) cb = 0
Plug in
E
+
=
H aa + H ab 1 + S ab
H + H ab H + H ab H aa - aa ca + H ab - aa S ab cb = 0 1 + S ab 1 + S ab
( H aa (1 + S ab ) - H aa - H ab ) ca + ( H ab (1 + S ab ) - H aa S ab - H ab S ab ) cb = 0 ( H aa + H aa S ab - H aa - H ab ) ca + ( H ab + H ab S ab - H aa S ab - H ab S ab ) cb = 0 ( H aa S ab - H ab ) ca + ( H ab - H aa S ab ) cb = 0
H aa S ab - H ab H - H aa S ab ca + ab H S -H H S - H cb = 0 ab ab aa ab aa ab ca - cb = 0 cb = ca
Note: Plugging into the second of the two linear equations gets you the same result.
Chapter 10: Slide 35
cb = ca + = ca 1sa + cb1sb
+ = ca (1sa + 1sb )
or + = N + (1sa + 1sb ) N+ (=ca) is determined by normalizing +
Normalization:
* 1 = + + d = + +
2 1 = N + (1sa + 1sb ) N + (1sa + 1sb ) = N + ( 1sa 1sa + 1sa 1sb + 1sb 1sa + 1sb 1sb 2 2 1 = N + (1 + S ab + S ab + 1) = N + ( 2 + 2S ab )
)
N+ =
1 2 + 2 S ab
+ = N + (1sa + 1sb ) 1 (1sa + 1sb ) = 2 + 2 S ab
Chapter 10: Slide 36
Antibonding Wavefunction
(H
aa - E ) ca + ( H ab - E S ab ) cb = 0
Plug in
E
-
=
H aa - H ab 1 - S ab
H - H ab H - H ab H aa - aa ca + H ab - aa S ab cb = 0 1 - S ab 1 - S ab
HW
ca + cb = 0
cb = -ca 1 (1sa - 1sb ) 2 - 2S ab
- = ca (1sa - 1sb ) = N - (1sa - 1sb ) =
HW Note: Plugging into the second of the two linear equations gets you the same result.
Chapter 10: Slide 37
Plotting the Wavefunctions
+ = N + (1sa + 1sb )
- = N - (1sa - 1sb )
2
0
1
2
3
0
1
2
3
Nuc a
Nuc b
Nuc a
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.
Chapter 10: Slide 38
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: 1 - ra 1 - rb 1sa = 1sa = e and 1sb = 1sb = e
(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 - Z 'ra e
and b =
Z '3 - Z 'rb e
The expectation value for the energy, <E>, is now a function of both Z' and R.
Chapter 10: Slide 39
= caa + cbb = ca
Z '3 - Z 'ra Z '3 - Z 'rb e + cb e
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 Cal.(Z=1) Cal.(Z'=1.24) Expt. 1.32 1.06 1.06 De 1.77 eV 2.35 2.79
Chapter 10: Slide 40
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 2 p za + c41sb + c5 2sb + c6 2 p zb
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.
Chapter 10: Slide 41
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 Cal.(Z=1) Cal.(Z'=1.24) Cal.(Big) Expt. 1.32 1.06 1.06 1.06 De 1.77 eV 2.35 2.78 2.79
Chapter 10: Slide 42
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
Chapter 10: Slide 43
MO Treatment of the H2 Molecule
The H2 Electronic Hamiltonian
1
r1a
r1b R r2a
2
a
b
r2b
1 2 1 1 1 1 1 1 H = - 1 - 2 - - - - + 2 2 2 r1a r1b r2 a r2b r12
KE e1 KE e2 PE e-N Attr PE e-N Attr PE e-N Attr PE PE e-N e-e Attr Repuls
c.f. H2+ Electronic Hamiltonian 1 2 1 1 - - - = Eelect = E 2 ra rb
Chapter 10: Slide 44
The LCAO Molecular Orbitals
E- = H aa - H ab 1 - S ab
- = N - (1sa - 1sb )
Antibonding Orbital
Energy
1sa
EH = H aa
EH = H aa
1sb
E+ =
H aa + H ab 1 + S ab
+ = N + (1sa + 1sb )
Bonding Orbital
H aa = 1sa H 1sa
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.
Chapter 10: Slide 45
Notation
Antibonding Orbital
* - = N - (1sa - 1sb ) = u 1s = u 1s
Text
Bonding Orbital
Mine
+ = N + (1sa + 1sb ) = g 1s = g 1s
g 1s
e- density max. on internuclear axis
Combin. of 1s orbitals
1s
* u
antibonding
symmetric w.r.t. inversion
antisymmetric w.r.t. inversion
Chapter 10: Slide 46
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 =
1 g 1s (1)1 g 1s (1) 1 2! g 1s (2) 2 g 1s (2) 2
MO =
1 g 1s(1)1 g 1s (2) 2 - g 1s (1) 1 g 1s (2) 2 2
[
]
MO = [ g 1s (1) g 1s (2)]
spat
1 [1 2 - 1 2 ] = spat spin 2 spin
Chapter 10: Slide 47
MO = [ g 1s (1) g 1s (2)]
spin =
1 [1 2 - 1 2 ] 2
1 [1 2 - 1 2 ] 2
The spin wavefunction is already normalized (see Chap. 8 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 ) ]
Chapter 10: Slide 48
The MO Energy of H2
spat = g 1s (1) g 1s (2) = [ N + (1sa + 1sb ) ][ N + (1sa + 1sb ) ]
1 2 1 1 1 1 1 1 H = - 1 - 2 - - - - + 2 2 2 r1a r1b r2 a 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
Chapter 10: Slide 49
Energy (au)
-1.0
2EH 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
De(cal)= 2EH Emin(cal) De(cal)= +0.099 au = 2.69 eV
Rmin Cal.(Z=1) Expt. 0.85 0.74
De 2.69 eV 4.73
Chapter 10: Slide 50
Improving the Results
As for H2+, one can add a variational parameter to the atomic orbitals used in g1s. Z '3 - Z 'ra Z '3 - Z 'rb a = e and b = e (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 Cal.(Z=1) Cal.(Var. Z') Expt. 0.85 0.73 0.74 De 2.70 eV 3.64 4.73
Chapter 10: Slide 51
An Even Better Improvement: More Atomic Orbitals
Z-Direction
a
b
As for H2+, one can make the orbital bonding a Linear Combination of more than two atomic orbitals; e.g.
g 1s = c11sa + c2 2sa + c3 2 p za + c41sb + c5 2sb + c6 2 p zb
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 Cal.(Z=1) Cal.(Var. Z') Cal.(HF-Big) Expt. 0.85 0.73 0.74 0.74 De 2.70 eV 3.49 3.62 4.73
Chapter 10: Slide 52
Question: Hey!! What went wrong??
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.
The reason is that HF does not include correlation (remember chapter 9)
Chapter 10: Slide 53
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 11. We used a form of "Configuration Interaction", called QCISD(T), to calculate the Correlation Energy and, thus, a new value for De Rmin Cal.(Z=1) Cal.(Var. Z') Cal.(HF-Big) Expt. 0.85 0.73 0.74 0.74 De 2.70 eV 3.49 3.62 4.69 4.73 That's Better!!
Cal.(QCISD(T)) 0.74
Chapter 10: Slide 54
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
Chapter 10: Slide 55
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 = (nB nA) nB = number of electrons in Bonding Orbitals nA = number of electrons in Antibonding Orbitals
Chapter 10: Slide 56
Bonding in He2+
He2+ has 3 electrons
Electron Configuration
* u 1s = - = N (1sa - 1sb )
Config = ( 1s ) ( 1s )
2 g * u
1
Antibonding Orbital Energy
1sa
1sb BO = (2-1) = 1/2
g 1s = + = N (1sa + 1sb )
Bonding Orbital
Chapter 10: Slide 57
Slater Determinant: He2+
* Config = ( g 1s ) ( u 1s ) 2 1
MO
* 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 * g 1s (1)1 g 1s (2) 2 u 1s(3) 3 3!
Shorthand Notation
Chapter 10: Slide 58
He2
He2 has 4 electrons
Electron Configuration
* u 1s = - = N (1sa - 1sb )
Config = ( 1s ) ( 1s )
2 g * u
2
Antibonding Orbital Energy
1sa
1sb BO = (2-2) =0
g 1s = + = N (1sa + 1sb )
Bonding Orbital
Actually, He2 forms an extremely weak "van der Waal's complex", with Rmin 3 and De 0.001 eV [it can be observed at T = 10-3 K.
Chapter 10: Slide 59
Second Row Homonuclear Diatomic Molecules
We need more Molecular Orbitals to describe diatomic molecules with more than 4 electrons. + 2sa and + 2sb and Sigma () MO's Max. e- density along internuclear axis Sigma () MO's Max. e- density along internuclear axis Pi () MO's
2pza
+
and
+
2pya
2pyb
2pxa and 2pxb also combine to give MO's
-
-
+
+
2pzb
Max. e- density above/below internuclear axis
Chapter 10: Slide 60
Sigma-2s Orbitals
* u 2s = N ( 2sa - 2sb )
Antibonding Orbital Energy
+
2sa
2sb
+
g 2 s = N ( 2sa + 2 sb )
Bonding Orbital
Chapter 10: Slide 61
Sigma-2p Orbitals
* u 2 p = N ( 2 p za + 2 p zb )
Antibonding Orbital Energy
2pza
g 2 p = N ( 2 p za - 2 p zb )
Bonding Orbital
Note sign reversal from 2s and 1s orbitals.
Chapter 10: Slide 62
-
-
+
+
2pzb
Pi-2p Orbitals
* g 2 p = N ( 2 p ya - 2 p yb )
Antibonding Orbital Energy
+
+
2pya
u 2 p = N ( 2 p ya + 2 p yb )
Bonding Orbital
2pyb
There is a degenerate u2p orbital and a degenerate g*2p orbital arising from analogous combinations of 2pxa and 2pxb
Chapter 10: Slide 63
Homonuclear Diatomic Orbital Energy Diagram
* u 2p
* * g2p g2p
2 p xa
2 p ya
2 p za
g2p
u 2p
u 2p
2 p xb
2 p yb
2 p zb
* u 2s
2 sa
g 2s
* u 1s
2 sb
1sa
g 1s
1sb
Chapter 10: Slide 64
Consider Li2
* u 2p
* g2p
(a) What is the electron configuration? (b) What is the Bond Order? (c) What is the spin multiplicity? (Singlet, Doublet or Triplet)
g2p
u 2p
6 Electrons
* u 2s
* ( g 1s) 2 ( u 1s ) 2 ( g 2 s ) 2
g 2s
* u 1s
BO = (4-2) = 1 S = 0 : Singlet
g 1s
Chapter 10: Slide 65
Consider F2
* u 2p
* g2p
(a) What is the electron configuration? (b) What is the Bond Order? (c) What is the spin multiplicity? (Singlet, Doublet or Triplet)
g2p
u 2p
18 Electrons
* u 2s
* * * ( g 1s) 2 ( u 1s ) 2 ( g 2 s ) 2 ( u 2 s ) 2 ( u 2 p ) 4 ( g 2 p ) 2 ( g 2 p ) 4
g 2s
* u 1s
BO = (10-8) = 1 S = 0 : Singlet
g 1s
Chapter 10: Slide 66
Consider O2
* u 2p
* g2p
(a) What is the electron configuration? (b) What is the Bond Order? (c) What is the spin multiplicity? (Singlet, Doublet or Triplet)
g2p
u 2p
16 Electrons
* u 2s
* * * ( g 1s) 2 ( u 1s ) 2 ( g 2 s ) 2 ( u 2 s ) 2 ( u 2 p ) 4 ( g 2 p ) 2 ( g 2 p ) 2
g 2s
* u 1s
BO = (10-6) = 2 S = 1 : Triplet
g 1s
Chapter 10: Slide 67
Consider O2 , O2+ , O2* u 2p
* g2p
(a) Which has the longest bond? (b) Which has the highest vibrational frequency? (c) Which has the highest Dissociation Energy? O2: 16 Electrons BO = 2
g2p
u 2p
O2+: 15 Electrons BO = 2.5 O2-: 17 Electrons BO = 1.5
* u 2s
g 2s
* u 1s
O2- has the longest bond. O2+ has the highest vibrational frequency. O2+ has the highest Dissociation Energy. O2
Chapter 10: Slide 68
g 1s
A More General Picture of Sigma Orbital Combinations
2pza
+
2sa
+
1sa
MO = c11sa + c2 2sa + c3 2 p za + c41sb + c5 2 sb + c6 2 p zb
-
-
+
+
2pzb
+
2sb
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.
+
1sb
Chapter 10: Slide 69
The MO's of C2 (STO-3G)
+28 eV +7 eV
* * 3 u u 2 p = ( - 0.12 1sa + 1.09 2 sa + 1.16 p za ) + ( 0.17 1sb - 1.09 2 sb + 1.16 p zb )
* * 1 g g 2s = ( 0.82 2 p xa ) + ( - 0.82 p xb )
* * 1 g g 2s = ( 0.82 2 p ya ) + ( - 0.82 p yb )
g2p and u*2p contain significant 2s orbital contributions. Energy
-14 eV -15 eV 3 g g 2 p = ( - 0.07 1sa + 0.40 2 sa + 0.60 p za ) + ( - 0.07 1sb + 0.40 2 sb - 0.60 p zb ) 1 u u 2 s = ( 0.63 2 p xa ) + ( 0.63 p xb ) 1 u u 2 s = ( 0.63 2 p ya ) + ( 0.63 p yb )
* * 2 u u 2 s = ( - 0.17 1sa + 0.75 2 sa + 0.25 p za ) + ( + 0.17 1sb - 0.75 2 sb + 0.25 p zb )
-20 eV -38 eV
2 g g 2 s = ( - 0.17 1sa + 0.50 2 sa - 0.23 p za ) + ( - 0.17 1sb + 0.50 2 sb + 0.23 p zb )
g2s and u*2s contain significant 2pz orbital contributions.
-422 eV -422 eV
* * 1 u u 1s = ( 0.70 1sa + 0.03 2 sa ) + ( - 0.70 1sb - 0.03 2 sb )
1 g g 1s = ( 0.70 1sa + 0.01 2 sa ) + ( 0.70 1sb + 0.01 2sb )
g1s and u*1s are degenerate. There is no bonding.
Chapter 10: Slide 70
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
Chapter 10: Slide 71
Heteronuclear Diatomic Molecules
Whenever a Molecular Orbital can be taken as a Linear Combination of 2 Atomic Orbitals, one obtains a 2x2 Secular Determinant, which can be solved for the energies, and then the coefficients of the wavefunctions.
MO = caa + cbb
H aa - E H ab - E S ab
H ab - E S ab =0 H bb - E
For Homonuclear diatomic molecules, Hbb = Haa, and the MO energy levels are almost symmetrically displaced.
EA =
Antibonding (A)
H aa - H ab 1 - S ab
A = N ( a - b )
a
E = H aa
H + H ab EB = aa 1 + S ab
E = H aa
b
Note that the magnitudes of the orbital coefficients, ca and cb are the same.
cb = ca
Chapter 10: Slide 72
Bonding (B)
B = N ( a + b )
Heteronuclear Diatomic Molecules
H aa - E H ab - E S ab H ab - E S ab =0 H bb - E
H bb H aa
Therefore, the energies are not symmetrically displaced, and the magnitudes of the coefficients are no longer equal.
cb ca
Antibonding (A)
' ' A = caa + cbb
a
E = H aa
E = H bb
Bonding (B)
b
B = caa + cbb
Chapter 10: Slide 73
Interpretation of Secular Determinant Parameters
H aa - E H ab - E S ab H ab - E S ab =0 H bb - E
* S ab = a b = ab d
Overlap Integral
+
+
Large Sab
+
+
Small Sab
Generally, Sab 0.1 0.2 Commonly, to simplify the calculations, it is approximated that Sab 0
Chapter 10: Slide 74
H aa - E H ab - E S ab
* H aa = a H a = a Ha d * H bb = b H b = b Hb d
H ab - E S ab =0 H bb - E
Energy of an electron in atomic orbital, a, in an unbonded atom. 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 IE(2s)=20.8 eV 2s Carbon
Chapter 10: Slide 75
H 2 s , 2 s (C ) -20.8 eV H 2 p , 2 p (C ) -11.3 eV
H aa - E H ab - E S ab
* H ab = a H b = a Hb d
H ab - E S ab =0 H bb - 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 + H bb H ab K aa S ab 2
K 1.75
Wolfsberg-Helmholtz Formula (used in Extended Hckel Model) Hab < 0
Chapter 10: Slide 76
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 )
2 2 1 = N 2 ca a a + cb b b + 2ca cb a b 2 2 1 = N 2 ca + cb + 2ca cb S ab
[
]
Overlap
[
]
S ab = a b Orbital
N=
1
2 2 ca + cb + 2ca cb S ab
Chapter 10: Slide 77
MO =
If Sab 0:
1 c + c + 2ca cb S ab
2 a 2 b
( caa + cbb )
b
MO =
2 ca fa = 2 2 ca + cb 2 cb fb = 2 2 ca + cb
ca c +c
2 a 2 b
a +
cb c +c
2 a 2 b
Fraction of electron density in orbital a
Fraction of electron density in orbital b
General:
MO = N cii
ci2 fi = ci2
Fraction of electron density in orbital i
Chapter 10: Slide 78
2 ca fa = 2 2 ca + cb
2 cb fb = 2 2 ca + cb
H aa - E H ab - E S ab
H ab - E S ab =0 H bb - E
Homonuclear Diatomic Molecules:
H bb = H aa cb = ca f a = f b = 0.50
Heteronuclear Diatomic Molecules:
H bb H aa cb ca fb f a
Chapter 10: Slide 79
Antibonding (A)
' ' A = caa + cbb
a
E = H aa
E = H bb
Bonding (B)
b
B = caa + cbb
H aa - E H ab - E S ab
H ab - E S ab =0 H bb - E
(H
aa
- E )( H bb - E ) - ( H ab - E S ab )( H ab - E S ab ) = 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.
Chapter 10: Slide 80
A Numerical Example: Hydrogen Fluoride (HF)
+
1sa(H)
2pzb(F)
H aa - E H ab - E S ab
MO = N ( caa + cbb ) = N ( ca 1sa ( H ) + cb 2 p zb ( F ) )
Matrix Elements
H aa = a H a = 1sa ( H ) H 1sa ( H ) = -13.6 eV H bb = b H b = 2 p zb ( F ) H 2 p zb ( F ) = -17.4 eV H ab = a H b = 1sa ( H ) H 2 p zb ( F ) = -2.0 eV S ab = a b = 1sa ( H ) 2 p zb ( F ) 0
Chapter 10: Slide 81
+
H ab - E S ab =0 H bb - E
H aa - E H ab - E S ab
H ab - E S ab =0 H bb - E
H aa = -13.6 eV H bb = -17.4 eV H ab = -2.0 eV S ab 0
- 13.6 - E -2
-2 =0 - 17.4 - E
( - 13.6 -
E )( - 17.4 - E ) - ( - 2 )( - 2 ) = 0
2
E + 31.0 E + 232.64 = 0 - 31.0 (31.0) 2 - 4(1)(232.64) E = 2 E = - 31.0 - 30.44 = -18.26 eV 2 E = - 31.0 + 30.44 = -12.74 eV 2
Chapter 10: Slide 82
B
A
- 13.6 - E -2
-2 =0 - 17.4 - E
( - 13.6 - E )c - 2c + ( - 17.4 -
a
a
E ) cb = 0
- 2cb = 0
Bonding MO
Antibonding MO
[ - 13.6 - (-18.26)]ca - 2cb = 0
cb = 2.33 ca cb = 2.33ca
[ - 13.6 - (-12.74)]ca - 2cb = 0
cb = -0.430 ca cb = -4.30ca
B = ( caa + cbb ) = ca ( a + 2.33b ) = N ( a + 2.33b )
B =
1 1 + (2.33)
2
A = ( caa + cbb ) = ca ( a - 0.430b ) = N ( a - 0.430b )
(a + 2.33b ) 2
A =
1 1 + (-0.430)
2 2
(a - 0.430b )
B = 0.394a + 0.919b = 0.394 1sa ( H ) + 0.919 2 p zb ( F )
MO = 0.919a - 0.394b = 0.919 1sa ( H ) - 0.394 2 p zb ( F )
Chapter 10: Slide 83
Electron Densities in Hydrogen Fluoride
Bonding Orbital
B = 0.394a + 0.919b = 0.394 1sa ( H ) + 0.919 2 p zb ( F )
2 ca f a = f H = 2 2 = (0.394) 2 = 0.16 ca + cb 2 cb f b = f F = 2 2 = (0.919) 2 = 0.84 ca + cb
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 p zb ( F )
2 ca f a = f H = 2 2 = (0.919) 2 = 0.84 ca + cb 2 cb f b = f F = 2 2 = (0.394) 2 = 0.16 ca + cb
The situation is reversed in the Antibonding MO. However, remember that there are no electrons in this orbital.
Chapter 10: Slide 84
( x yH yL )
f a = 0.84 f b = 0.16
A = 0.919a - 0.394b
E
A
= -12.74 eV
-13
HH
aa
= -13.6 eV
-15
-17
H bb = -17.4 eV
F
E
-19
B
= -18.26 eV f b = 0.84
f a = 0.16
B = 0.394a + 0.919b
Chapter 10: Slide 85
Statistical Thermodynamics: Electronic Contributions to Thermodynamic Properties of Gases
Here they are again.
ln Q U = kT 2 T V , N ln Q ln Q H = kT 2 + kT T V , N ln V T , N S = k ln Q + U T
U CV = T V , N
H CP = T P , N
A = U - TS = - kT ln Q ln Q G = H - TS = - kT ln Q + kT ln V T , N
Chapter 10: Slide 86
The Electronic Partition Function
q elect = g i e
i =0 - -
i kT
= g 0e
-
-
0 kT
+ g1e
-
-
1 kT
+ g 2e
-
2 kT
+
or q
elect
=e
0 kT
g e
i =0 i i
i - 0 kT
=e
0 kT
- - - 1 0 - 2 0 kT kT + g 2e + g 0 + g1e
- 1 - 2 kT kT + g 2e + g 0 + g1e
q
elect
=e
-
0 kT
g e
i =0 i
-
i kT
=e
-
0 kT
q
elect
=e
-
0 kT
g e
i =0
-
ei T
=e
-
0 kT
- e1 - e2 g 0 + g1e T + g 2 e T +
ei =
i hc ei = k k ~ is the absorption frequency (in cm-1) to this state.
Chapter 10: Slide 87
I is the energy of the i'th electronic level above the ground state, 0
q elect = e
-
0 kT
gi e
i =0
-
ei T
=e
-
0 kT
- e1 - e2 T T + g 2e + g 0 + g1e
ei =
i hc i = 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
e
hc i 6.63 x10 -34 3.00 x1010 20,000 ei = = = 2.88 x10 4 K k 1.38 x10 - 23
- ei T
=e
2.88 x10 4 - 3, 000
= e -9.60 = 0.00007
- ei T
At room temperature (298 K):
e
1x10 - 42
Thus, one can ignore all but very low-lying electronic states, except at elevated temperatures.
Chapter 10: Slide 88
For convenience, we will consider only systems with a maximum of one "accessible" excited state:
q
elect
=e
-
0 kT
- e1 T g 0 + g1e
Internal Energy and Enthalpy
U
elect
ln Q elect = kT T V , N
2
Qelect independent of V
H
elect
ln Q elect ln Q elect + kT = U elect = kT T ln V V , N T , N
2
H
elect
=U
elect
elect elect 2 ln q 2 ln Q = kT T = kT T V , N
[
]
N
elect 2 ln q = nRT T V , N V , N
Chapter 10: Slide 89
q elect = e
-
0 kT
- e1 T g 0 + g1e
ln q elect
- e1 0 =- + ln g 0 + g1e T kT
H
elect
=U
elect
d d ln q elect 2 = nRT - 0 + = nRT T V , N dT kT dT
2
- e1 ln g 0 + g1e T
= nRT 2 0 2 + kT
nRT
2 - e1 T
g 0 + g1e
g1 e1 2 e T
- e1 T
nRg1 e1e = nN A 0 + - e1 g +ge T
R NA = k
0 1
-
e1 T
H0elect
Hthermelect
H elect = U elect = nE0 +
nRg1 e1e g 0 + g1e
-
e1 T
-
e1 T
E0 = N A 0 ei =
GS Electronic Energy per mole
elect elect U elect = H elect = H 0 + H therm
1 hc i = k k
Chapter 10: Slide 90
H elect = nE0 +
nRg1 e1e g 0 + g1e
-
e1 T
-
e1 T
ei =
1 hc i = 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 T T g 0 + g1e = g 0 + g1e
which would have led to:
H elect = nRg1 e1e g 0 + g1e
- e1 T - e1 T
Chapter 10: Slide 91
H elect = U elect =
nRg1 e1e g 0 + g1e
-
e1 T
-
e1 T
ei =
1 hc i = 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
=U
elect
=
nRg1 e1e g 0 + g1e
-
e1 T
-
e1 T
nRg1 e1e - = =0 - g 0 + g1e
By the same reasoning, Helect = Uelect = 0 for molecules with ground states of any multiplicity if the excited states are inaccessible.
Chapter 10: Slide 92
H elect = U elect =
nRg1 e1e g 0 + g1e
-
e1 T
-
e1 T
ei =
1 hc i = 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.
hc i (6.63x10 -34 J s )(3.00 x1010 cm / s )(7882 cm -1 ) = = 11,360 K ei = - 23 1.38 x10 J / K ) k H elect = nRg1 e1e g 0 + g1e
- e1 T
-
e1 T
=
1mol (8.31 J / mol K )( 2)(11360 K ) e 3 + 2e
11360 - 3000
-
11360 3000
= 1400 J = 1.40 kJ
At 298 K: H elect = 2 x10 -15 kJ 0
Chapter 10: Slide 93
Entropy
S
elect
= k ln Q
elect
U elect + = k ln q elect T
(
)
N
U elect U elect elect + = Nk ln q + T T
S
elect
= nR ln q
elect
U elect + T
because Nk = nNAk = nR
q
elect
=e
-
0 kT
- e1 - e1 T g 0 + g1e = g 0 + g1e T
because we're assuming that 0 = 0.
Chapter 10: Slide 94
S elect
U elect elect = nR ln q + T
q
elect
- e1 = g 0 + g1e 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 .
hc i (6.63x10 -34 J s )(3.00 x1010 cm / s )(7882 cm -1 ) = = 11,360 K ei = - 23 1.38 x10 J / K ) k
Let's calculate Select for one mole at 3000 K. Uelect = Helect = 1400 J (from earlier calculation.
q elect
11, 360 - - e1 3000 T = g 0 + g1e = 3 + 2e = 3 + 0.045 = 3.045
Chapter 10: Slide 95
S elect
U elect elect = nR ln q + T
q
elect
- e1 = g 0 + g1e T = 3.045
Uelect = 1400 J
S
elect
= nR ln q
elect
U elect J 1400 J + = 1 mol 8.31 ln(3.045) + T mol K 3000 K J K
S elect = 9.25 + 0.47 = 9.72
At 25 oC = 298 K:
Uelect = 0.00 J qelect = g0 = 3 Select = Rln(3) + 0/298 = 9.13 J/K
Chapter 10: Slide 96
S elect
U elect elect = nR ln q + T
q elect
- e1 = g 0 + g1e T
At 25 oC = 298 K:
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.
Chapter 10: Slide 97
O2 Entropy: Comparison with experiment O2: Smol(exp) = 205.1 J/mol-K at 298.15 K
S tran + S rot + S vib = 43.8 + 151.9 + 0.035 = 195.7 J / mol K
Chap. 3 Chap. 4 Chap. 5
We were about 5% too low. Let's add in the electronic contribution.
S tran + S rot + S vib + S elect = 43.8 + 151.9 + 0.035 + 9.1 = 204.8 J / mol K
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.
Chapter 10: Slide 98
Helmholtz and Gibbs Energy
Aelect = U elect - TS elect = - kT ln Q elect
Qelect independent of V
G
elect
=H
elect
- TS
elect
= - kT ln Q
elect
ln Q elect elect + kT ln V = A 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 .
hc i (6.63x10 -34 J s )(3.00 x1010 cm / s )(7882 cm -1 ) = = 11,360 K ei = 1.38 x10 - 23 J / K ) k
Let's calculate Gelect ( = Aelect) for one mole of O2 at 3000 K.
Chapter 10: Slide 99
G
elect
=A
elect
= -kT ln Q
elect
q elect
- e1 = g 0 + g1e T
e1 = 11,360 K g0 = 2 g1 = 3
q elect
- - e1 3, 000 K T = g 0 + g1e = 3.045 = 3 + 2e
11, 360 K
G elect = -kT ln Q elect = -kT ln ( q elect )
N
= - NkT ln q elect = -nRT ln q elect
Nk = nNAk = nR
G elect = -nRT ln q elect = -1 mol
8.31 J 3000 K ln(3.045) mol K
G elect = Aelect = -27,760 J = -27.8 kJ
Chapter 10: Slide 100
O2: Entropy
300
250
S [J/mol-K]
200 Expt T TR TRV TRVE 0 1000 2000 3000 4000 5000
150
Temperature [K]
The agreement of the calculated entropy including electronic contributions with experiment is close to perfect.
Chapter 10: Slide 101
O2: Enthalpy
200
150
H [kJ/mol]
100
50
0 0 1000 2000 3000 4000
Expt T TR TRV TRVE 5000
Temperature [K]
The agreement of the calculated enthalpy including electronic contributions with experiment is close to perfect.
Chapter 10: Slide 102
O2: Heat Capacity
40 35
CP [J/mol-K]
30 25 20 15 Expt T TR TRV TRVE 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
Chapter 10: Slide 103