Slide 1 Chapter 8 Diatomic Molecules. Slide 2 Outline Math Prelim.: Systems of Linear Equations –...

Chapter 8

Diatomic Molecules

Outline

• Math Prelim.: Systems of Linear Equations – Cramer’s Rule

• MO Treatment of the H2 Molecule

• LCAO Treatment of H2+

• Homonuclear Diatomic Molecules

• H2+ Energies

• H2+ Wavefunctions

• Heteronuclear Diatomic Molecules

• Hydrogen Molecular Ion: Born-Oppenheimer Approximation

Hydrogen Molecular Ion: Born-Oppenheimer Approximation

The simplest molecule is not H2. Rather, it is H2+, which has two

hydrogen nuclei and one electron.

a b

e

Rab

ra rb

The H2+ Hamiltonian (in au)

KENuc a

KENuc b

KEElect

PEe-NAttr

PEe-NAttr

PEN-N

Repuls

2 2 21 1 1 1 1 1

2 2 2a b ea b a b ab

HM M r r R

Electrons are thousands of times lighter than nuclei.Therefore, they move many times faster

Born-Oppenheimer Approximation

The Born-Oppenheimer Approximation states that since nucleimove so slowly, as the nuclei move, the electrons rearrange almostinstantaneously.

With this approximation, it can be shown that one can separatenuclear coordinates (R) and electronic coordinates (r), and get separate Schrödinger Equations for each type of motion.

Nuclear Equation

Eel is the effective potential energy exerted by the electron(s) onthe nuclei as they whirl around (virtually instantaneously on thetime scale of nuclear motion)

2 21 1 1

2 2a b el ab nuc aba b ab

E R E RM M R

Electronic Equation

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

21 1 1

2 electa b

E Er r

Outline





• H2+ Energies



• Hydrogen Molecular Ion: Born-Oppenheimer Approximation.

Systems of Linear Equations: Cramer’s Rule

In these equations, the aij and ci are constants.

We want to solve these two equations for the valuesof the variables, x1 and x2

Cramer’s Rule

11 1 12 2 1

21 1 22 2 2

a x a x c

a x a x c

1 12

2 221

11 12

21 22

c a

c ax

a a

a a

" "Det of Augmented matrix

Det of original matrix of coefficients

11 1

12 22

11 12

21 22

a c

a cx

a a

a a

" "Det of Augmented matrix

Det of original matrix of coefficients

A Numerical Example

11 1 12 2 1

21 1 22 2 2

a x a x c

a x a x c

1 2

1 2

2 5 1

3 2 11

x x

x x

1 12

2 221

11 12

21 22

c a

c ax

a a

a a

1 5

11 2

2 5

3 2

1( 2) 5(11) 573

2( 2) 5(3) 19

11 1

12 22

11 12

21 22

a c

a cx

a a

a a

2 1

3 11

2 5

3 2

2(11) 1(3) 191

2( 2) 5(3) 19

Homogeneous Linear Equations

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!!

11 1 12 2

21 1 22 2

0

0

a x a x

a x a x

Let’s try Cramer’s rule.

11 1 12 2

21 1 22 2

0

0

a x a x

a x a x

12

221

11 12

21 22

0

0

a

ax

a a

a a

11

122

11 12

21 22

0

0

a

ax

a a

a a

11 12

11 12 21 22

21 22

00 0

a aunless

a a a a

a a

11 12

11 12 21 22

21 22

00 0

a aunless

a a a a

a a

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”.

11 1 12 2

21 1 22 2

0

0

a x a x

a x a x

11 121 2

21 22

0 0a a

x x unlessa a

Outline





• H2+ Energies




LCAO Treatment of H2+

Molecular Orbitals

When we dealt with multielectron atoms, we assumed that the totalwavefunction is the product of 1 electron wavefunctions (1 for eachelectron), and that one could put two electrons into each orbital, onewith 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 of1 electron wavefunctions (“Molecular Orbitals”), and that we can puttwo electrons into each orbital.

Actually, that’s not completely correct. We really use aSlater Determinant of product functions to get an Antisymmetrizedtotal wavefunctions (just like with atoms).

1 1 1 2 2 3 2 41,2,3,... (1) (2) (3) (4) ...MO MO MO MON

Linear Combination of Atomic Orbitals (LCAO)

Usually, we take each Molecular Orbital (MO) to be aLinear Combination of Atomic Orbitals (LCAO), where each atomicorbital 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 ofhydrogen atom 1s orbitals on each atom:

a b

1sa 1sb

Assume that 1sa and 1sb

are each normalized.

1 1a sa b sbc c shorthand

1 1a a b bc s c s

Expectation Value of the Energy

Our goal is to first develop an expression relating the expectation valueof the energy to ca and cb.

Then we will use the Variational Principle to find the best set ofcoefficients; i.e. the values of ca and cb that minimize the energy.

1 1a a b bc s c s

*

*

H dE

d

H

Num

Denom

1 1a a b bc s c s

1 1 1 1a a b b a a b bDenom c s c s c s c s

2 21 1 1 1 1 1 1 1a a a a b a b b a b a b b bDenom c s s c c s s c c s s c s s

Remember that because 1sa and 1sb are normalized.1 1 1 1 1a a b bs s s s

Define: , where Sab is the overlap integral.1 1 1 1ab a b b aS s s s s

2 22a a b ab bDenom c c c S c

Note: For this particular problem, Hbb=Haa by symmetry.

However, this is not true in general.

1 1a a b bc s c s

1 1 1 1a a b b a a b bNum H c s c s H c s c s

2 21 1 1 1 1 1 1 1a a a a b a b b a b a b b bNum c s H s c c s H s c c s H s c s H s

1 1

1 1

1 1

1 1

aa a a

bb b b

ab a b

ba b a

H s H s

H s H s

H s H s

H s H s

abH because H is Hermitian

2 22a aa a b ab b bbNum c H c c H c H

Minimizing <E>: The Secular Determinant

It would seem relatively straightforward to take the derivativesof 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 onlyrelatively 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”

In order to find the values of ca and cb which minimize <E>, we

want: 0 0a b

E Eand

c c

2 22a aa a b ab b bbNum c H c c H c H 2 22a a b ab bDenom c c c S c

H NumE

Denom

2 2 2

2 2

2

2a aa a b ab b bb

a a b ab b

c H c c H c H

c c c S c

Differentiate both sides w.r.t. ca: Use product rule on left side

Set and group coefficients of ca and cb0

a

E

c

2 2

2 2

2

2a aa a b ab b bb

a a b ab b

c H c c H c HE

c c c S c

2 2 2 22 2a a b ab b a aa a b ab b bbc c c S c E c H c c H c H

2 2 2 22 2a a b ab b a aa a b ab b bb

a a

c c c S c E c H c c H c H

c c

2 22 2 2 0 2 2 0a a b ab b a b ab a aa b aba

Ec c c S c E c c S c H c H

c

This is one equation relating the two coefficients, ca and cb.

We get a second equation if we repeat the procedure, exceptdifferentiate w.r.t. cb and set the derivative =0.

2 2 2 2a b ab a aa b abE c c S c H c H

0 aa a ab ab bH E c H E S c

or 0aa a ab ab bH E c H E S c

The secondequation is: 0ab ab a bb bH E S c H E c

a b ab a aa b abE c E c S c H c H After dividing bothsides by 2

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. Theonly 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.

The Secular Determinant

0aa a ab ab bH E c H E S c

0ab ab a bb bH E S c H E c

0aa ab ab

ab ab bb

H E H E S

H E S H E

Extension to Larger Systems

The 2x2 Secular Determinant resulted from using a wavefunctionconsisting of a linear combination of atomic orbitals.

If, instead, you use a linear combination of N orbitals, then you getan NxN Secular Determinant

A simple way to remember how to build a Secular Determinant isto use the “generic” formula:

After you have made the Secular Determinant, set the diagonaloverlaps, Sii = 1.

0ij ijH E S

Then the Secular Determinant is:

0ij ijH E S

For example, if a a b b c cc c c

0aa aa aB ab ac ac

ab ab bb bb bc bc

ac ac bc bc cc cc

H E S H E S H E S

H E S H E S H E S

H E S H E S H E S

0aa aB ab ac ac

ab ab bb bc bc

ac ac bc bc cc

H E H E S H E S

H E S H E H E S

H E S H E S H E

Setting diagonal Sii = 1

Outline





• H2+ Energies




H2+ Energies

Linear Equations

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.



Secular Determinant

0aa ab ab

ab ab bb

H E H E S

H E S H E

Expansion of the Secular Determinant

This expression can be expanded, yielding a quadratic equationin <E>. This equation can be solved easily using the quadraticformula.

Therefore, Hbb=Haa (by symmetry)

The equation then simplifies to:

0aa ab ab

ab ab bb

H E H E S

H E S H E

20aa bb ab abH E H E H E S

or 20aa bb ab abE H E H H E S

However, let’s remember that for this problem: 1 1

1 1

aa a a

bb b b

H s H s

H s H s

2 2

aa ab abE H H E S

Solving for the Energies

2 2

aa ab abE H H E S

aa ab ab ab abE H H E S H E S

ab aa abE E S H H 1 ab aa abE S H H

Therefore:1aa ab

ab

H HE

S

or and1aa ab

ab

H HE

S

1aa ab

ab

H HE

S

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 involvetwo-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 aspecial “confocal elliptic” coordinate system. We will just presentthe results. They are functions of the internuclear distance, R.

and1aa ab

ab

H HE

S

1aa ab

ab

H HE

S

21 1 11

2R

aaH eR R

11

2R

ab abH S R e

3

13

Rab

RS e R

<E>+ and <E>- represent the electronic energy of the H2+ ion.

1aa ab

ab

H HE

S

1aa ab

ab

H HE

S

1 1

1aa ab

ab

H HV E

R S R

1 1

1aa ab

ab

H HV E

R S R

The total energy, <V>+ and <V>- , also includes the internuclear repulsion, 1/R

Asymptotic limitof EH as RV() = -0.50 au

Antibonding Orbital

<V>+

1 2 3 4 5

-0.5

-0.4

-0.3

-0.2

-0.1

R Vplus Vminus Line( )

R/a0

E (

au

)

<V>-

Bonding Orbital

Calculated Minimum Energy

Emin(cal) = -0.565 au at Rmin(cal) = 2.49 a0 = 1.32 Å

1 1

1aa ab

ab

H HV E

R S R

1 1

1aa ab

ab

H HV E

R S R

Antibonding Orbital

<V>+

1 2 3 4 5

-0.5

-0.4

-0.3

-0.2

-0.1

R Vplus Vminus Line( )

R/a0

E (

au

)

<V>-

Bonding Orbital

Calculated Minimum Energy

Emin(cal) = -0.565 au

Rmin(cal) = 1.32 Å

Comparison with Experiment

Dissociation Energy

De(cal) = EH – Emin(cal)

= -0.5 au – (-0.565 au)

= +0.065 au•27.21 eV/au

= 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.

Outline





• H2+ Energies




H2+ Wavefunctions (aka Molecular Orbitals)

Remember that by using the Variational Principle on the expressionfor <E>, we developed two homogeneous linear equations relating ca and cb.

We then solved the Secular Determinant of the matrix coefficientsto get two values for <E>

We can now plug one of the energies (either <E>+ or <E>-)into either of the linear equations to get a relationship betweenca and cb for that value of the energy.

The LCAO Wavefunction: 1 1a a b bc s c s



1aa ab

ab

H HE

S

1aa ab

ab

H HE

S

Bonding Wavefunction

Note: Plugging into the second of the two linear equations gets you the same result.

Plug in 1aa ab

ab

H HE

S


01 1aa ab aa ab

aa a ab ab bab ab

H H H HH c H S c

S S

1 1 0aa ab aa ab a ab ab aa ab ab ab bH S H H c H S H S H S c

0aa aa ab aa ab a ab ab ab aa ab ab ab bH H S H H c H H S H S H S c

0aa ab ab a ab aa ab bH S H c H H S c

0aa ab ab ab aa aba b

aa ab ab aa ab ab

H S H H H Sc c

H S H H S H

0a bc c b ac c

N+ (=ca) is determined by normalizing +

b ac c + 1 1a a b bc s c s 1 1a a bc s s

or 1 1a bN s s

Normalization: *1 d

1 1 1 1 1a b a bN s s N s s 2 1 1 1 1 1 1 1 1a a a b b a b bN s s s s s s s s

2 21 1 1 2 2ab ab abN S S N S

1

2 2 ab

NS

1 1

11 1

2 2

a b

a b

ab

N s s

s sS

Antibonding Wavefunction

Note: Plugging into the second of the two linear equations gets you the same result.

HW

0aa a ab ab bH E c H E S c Plug in 1

aa ab

ab

H HE

S

01 1aa ab aa ab

aa a ab ab bab ab

H H H HH c H S c

S S

HW

0a bc c b ac c

11 1 1 1 1 1

2 2a a b a b a b

ab

c s s N s s s sS

Plotting the Wavefunctions

ba ssN 11 ba ssN 11

0 1 2 3

Nuca

Nucb

0 1 2 3

Nuca

Nucb

2

Note that the bonding MO, +, has significant electron density inthe region between the two nuclei.

Note that the antibonding MO, - , has a node (zero electron density inthe region between the two nuclei.

Improving the Results

One way to improve the results is to add more versatility to theatomic orbitals used to define the wavefunction.

We used hydrogen atom 1s orbitals:

Instead of assuming that each nucleus has a charge, Z=1, we can usean effective nuclear charge, Z’, as a variational parameter.

The expectation value for the energy, <E>, is now a functionof both Z’ and R.

1 1

1 11 1a br r

sa a sb bs e and s e

(in atomic units)

3 3' '' 'a bZ r Z r

a b

Z Ze and e

Rmin De

Cal.(Z=1) 1.32 Å 1.77 eV

Cal.(Z’=1.24) 1.06 2.35

Expt. 1.06 2.79

This expression for the wavefunction can be plugged into theequation for <E>. The values of Z’ and R which minimize <E>can then be calculated. The best Z’ is 1.24.

3 3' '' 'a bZ r Z r

a a b b a b

Z Zc c c e c e

An Even Better Improvement: More Atomic Orbitals

a b

Z-Direction

Instead of expanding the wavefunction as a linear combination of justone orbital on each atom, put in more atomic orbitals. e.g.

Note: A completely general rule is that if you assume that a MolecularOrbital is an LCAO of N Atomic Orbitals, then you will get anNxN Secular Determinant and N Molecular Orbitals.

1 2 3 4 5 61 2 2 1 2 2a a za b b zac s c s c p c s c s c p

a b

Z-Direction

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

Outline





• H2+ Energies




MO Treatment of the H2 Molecule

The H2 Electronic Hamiltonian

a b

1

2

R

r1a r1b

r2a r2b

KEe1

KEe2

PEe-NAttr

PEe-e

Repuls

PEe-NAttr

PEe-NAttr

PEe-NAttr

2 21 2

1 1 2 2 12

1 1 1 1 1 1 1

2 2 a b a b

Hr r r r r

The LCAO Molecular Orbitals

1sb1sa

Antibonding Orbital

En

erg

y

Bonding Orbital

We can put both electrons in H2 into the bonding orbital,+, one with spin and one with spin.

1aa ab

ab

H HE

S

1aa ab

ab

H HE

S

(1 1 )a bN s s

(1 1 )a bN s s

H aaE HH aaE H

is the energy of an electron in a hydrogen1s orbital.

1 1aa a aH s H s

Notation

e- density max. oninternuclear axis symmetric w.r.t.

inversion

Combin. of1s orbitals

antisymmetric w.r.t.inversion

antibonding

Antibonding Orbital

1 1a bN s s *1u s

1g s

Bonding Orbital

1 1a bN s s

1g s *1u s

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.

spat spin

1 1

2 2

1 (1) 1 (1)11 (2) 1 (2)2!

g gMO

g g

s s

s s

1 2 1 2

11 (1) 1 (2) 1 (1) 1 (2)

2MO g g g gs s s s

1 2 1 2

11 (1) 1 (2)

2MO g gs s spat spin

Because the Hamiltonian doesn’t operate on the spin, the spinwavefunction has no effect on the energy of H2.

This independence is only because we were able to write the totalwavefunction as a product of spatial and spin functions.This cannot be done for most molecules.

1 2 1 2

11 (1) 1 (2)

2MO g gs s

The spin wavefunction is already normalizeD(see Chap. 8 PowerPoint for He).

1 2 1 2

1

2spin

1 (1) 1 (2) 1 1 1 1spat g g a b a bs s N s s N s s

The MO Energy of H2

The (multicenter) integrals are very messy to integrate, but canbe integrated analytically using confocal elliptic coordinates, toget E as a function of R (the internuclear distance)

1 (1) 1 (2) 1 1 1 1spat g g a b a bs s N s s N s s

2 21 2

1 1 2 2 12

1 1 1 1 1 1 1

2 2 a b a b

Hr r r r r

The expectation value for the ground state H2 electronic energyis given by:

using the wavefunction and Hamiltonian above.

spat spatE H

The total energy is then:1

( ) ( )V R E RR

De: Dissociation Energy

0.0 0.5 1.0 1.5 2.0 2.5

-1.0

Ene

rgy

(au)

R (Angstroms)

2•EH

Rmin De

Cal.(Z=1) 0.85 Å 2.69 eV

Expt. 0.74 4.73

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

Improving the Results

As for H2+, one can add a variational parameter to the atomic orbitals

used in g1s.

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

(in atomic units)3 3

' '' 'a bZ r Z r

a b

Z Ze and e

1 (1) 1 (2)spat g g a b a bs s N N

An Even Better Improvement: More Atomic Orbitals

a b

Z-Direction

As for H2+, one can make the bonding orbital a Linear Combination

of more than two atomic orbitals; e.g.

We performed a Hartree-Fock calculation on H2 using an LCAOthat 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??

1 2 3 4 5 61 1 2 2 1 2 2g a a za b b zas c s c s c p c s c s c p

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 electroncorrelation energy.

Don’t you remember anything from Chapter 7??

Question: Does your watch also say 3:00 ? Tee Time – Gotta go!!!

Inclusion of the Correlation Energy

There are methods to calculate the Correlation Energy correctionto the Hartree-Fock results. We’ll discuss these methods inChapter 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. 0.74 4.73

That’s Better!!

Outline





• H2+ Energies




Homonuclear Diatomic Molecules

We showed that the Linear Combination of 1s orbitals on two hydrogen atoms form 2 Molecular Orbitals, which we used to describethe bonding in H2

+ and H2.

These same orbitals may be used to describe the bonding inHe2

+ 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 describethe 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

Bonding in He2+

En

erg

y

1sa 1sb

He2+ has 3 electrons

BO = ½(2-1) = 1/2

Bonding Orbital

1 1 1g a bs N s s

Antibonding Orbital

*1 1 1u a bs N s s

Electron Configuration

12 *1 1g uConfig s s

Slater Determinant: He2+

*1 1 1

*2 2 2

*3 3 3

1 (1) 1 (1) 1 (1)1

1 (2) 1 (2) 1 (2)3!

1 (3) 1 (3) 1 (3)

g g u

MO g g u

g g u

s s s

s s s

s s s


Shorthand Notation

*1 2 3

11 (1) 1 (2) 1 (3)

3!MO g g us s s

He2E

ner

gy

1sa 1sb

He2 has 4 electrons

BO = ½(2-2) = 0

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.

Bonding Orbital

1 1 1g a bs N s s

Antibonding Orbital

*1 1 1u a bs N s s

Electron Configuration


Second Row Homonuclear Diatomic Molecules

We need more Molecular Orbitals to describe diatomic moleculeswith more than 4 electrons.

+ +and Sigma () MO’s

Max. e- density alonginternuclear axis2sa 2sb

+- +-2pza 2pzb

and Sigma () MO’s

Max. e- density alonginternuclear axis

+

-

2pya

and Pi () MO’s

Max. e- density above/belowinternuclear axis

+

-

2pyb

2pxa and 2pxb also combine to give MO’s

Sigma-2s Orbitals

* 2 2 2u a bs N s s Antibonding Orbital

2 2 2g a bs N s s Bonding Orbital

En

erg

y

2sa 2sb+ +

Sigma-2p OrbitalsE

ner

gy

+-2pza

+-2pzb

Note sign reversal of 2p from 2s and 1s orbitals.

Bonding Orbital

2 2 2g za zbp N p p

Antibonding Orbital

* 2 2 2u za zbp N p p

Pi-2p OrbitalsE

ner

gy +

-

2pya

+

-

2pyb

There is a degenerate u2p orbital and a degenerate g*2p

orbital arising from analogous combinations of 2pxa and 2pxb

Bonding Orbital 2 2 2u ya ybp N p p

Antibonding Orbital

* 2 2 2g ya ybp N p p

Homonuclear Diatomic Orbital Energy Diagram

1 bs1 as

*1u s

1g s

2 bs2 as

* 2u s

2g s

2 xbp 2 ybp 2 zbp2 zap2 xap 2 yap

2u p 2u p

2g p

* 2g p * 2g p

* 2u p

Consider Li2(a) What is the electron configuration?

(b) What is the Bond Order?

(c) What is the spin multiplicity? (Singlet, Doublet or Triplet)

6 Electrons

BO = ½(4-2) = 1

S = 0 : Singlet

1g s

*1u s

2g s

* 2u s

2u p

2g p

* 2g p

* 2u p

22 2*1 1 2g u gs s s

Consider F2

(a) What is the electron configuration?



18 Electrons

BO = ½(10-8) = 1

S = 0 : Singlet

1g s

*1u s

2g s

* 2u s

2u p

2g p

* 2g p

* 2u p

2 2 42 2 24* * *1 1 2 2 2 2 2g u g u u g gs s s s p p p

Consider O2

(a) What is the electron configuration?



16 Electrons

BO = ½(10-6) = 2

S = 1 : Triplet

1g s

*1u s

2g s

* 2u s

2u p

2g p

* 2g p

* 2u p

2 2 22 2 24* * *1 1 2 2 2 2 2g u g u u g gs s s s p p p

Consider O2 , O2+ , O2

-

(a) Which has the longest bond?

(b) Which has the highest vibrational frequency?

(c) Which has the highest Dissociation Energy?

O2

O2- has the longest bond.

O2+ has the highest vibrational frequency.

O2+ has the highest Dissociation Energy.1g s

*1u s

2g s

* 2u s

2u p

2g p

* 2g p

* 2u p

O2: 16 Electrons – BO = 2

O2+: 15 Electrons – BO = 2.5

O2-: 17 Electrons – BO = 1.5

A More General Picture of Sigma Orbital Combinations

+-

2pza

+-

2pzb

+

2sa

+

2sb

+

1sa

+

1sb

The assumption in the past section thatonly identical orbitals on the two atomscombine to form MO’s is actually a bitsimplistic.

In actuality, each of the 6 MO’s is reallya combination of all 6 AO’s.

1 2 3 4 5 61 2 2 1 2 2MO a a za b b zbc s c s c p c s c s c p

Approximate vs. Accurate MO’s in C2

1 2 3 4 5 61 2 2 1 2 2MO a a za b b zbc s c s c p c s c s c p

0.70 1 0.01 2 0.70 1 0.01 2Accur a a b bs s s s

0.70 1 0.70 1Approx a bs s

0.17 1 0.50 2 0.23 2 0.17 1 0.50 2 0.23 2Accur a a za b b zbs s p s s p

1s MO (E -420 eV)

2s MO (E -40 eV)

2p MO (E -15 eV)

0.70 2 0.70 2Approx za zbp p

0.70 2 0.70 2Approx a bs s

0.07 1 0.40 2 0.60 2 0.07 1 0.40 2 0.60 2Accur a a za b b zbs s p s s p

Outline





• H2+ Energies




Heteronuclear Diatomic Molecules

Therefore, the energies are not symmetrically displaced,and the magnitudes of the coefficients are no longer equal.

a

b

Antibonding (A)

Bonding (B)

aaE H

bbE H

B a a b bc c

' 'A a a b bc c

0aa ab ab

ab ab bb

H E H E S

H E S H E

bb aaH H

b ac c

Slide 71

Interpretation of Secular Determinant Parameters

+ + Large Sab

+ + Small Sab

Generally, Sab 0.1 – 0.2

Commonly, to simplify the calculations,it is approximated that Sab 0

0aa ab ab

ab ab bb

H E H E S

H E S H E

Overlap Integral

*ab a b a bS d

Haa , Hbb < 0

Commonly, Haa is estimated as –IE, where IE is the Ionization Energyof an electron in the atomic orbital, a.

Carbon

IE(2s)=20.8 eV

IE(2p)=11.3 eV

2s

2p

C+

Traditionally, Haa and Hbb are called “Coulomb Integrals”

0aa ab ab

ab ab bb

H E H E S

H E S H E

Energy of an electron in atomic orbital,a, in an unbonded atom.

*aa a a a aH H H d

Energy of an electron in atomic orbital,b, in an unbonded atom.

*bb b b b bH H H d

2 ,2

2 ,2

( ) 20.8

( ) 11.3s s

p p

H C eV

H C eV

Hab is approximately proportional to: (1) the orbital overlap (2) the average of Haa and Hab

Hab < 0

Traditionally, Hab is called the “Resonance Integral”.

0aa ab ab

ab ab bb

H E H E S

H E S H E

Interaction energy between atomic orbitals,a and b .

*ab a b a bH H H d

Wolfsberg-Helmholtz Formula(used in Extended Hückel Model)

1.752

aa bbab ab

H HH K S K

Interpretation of Orbital Coefficients

Let’s assume that an MO is a linear combination of 2 normalized AO’s:

Normalization:

MO a a b bN c c

21 MO MO MOd

1 a a b b a a b bN c c N c c

2 2 21 2a a a b b b a b a bN c c c c

2 2 21 2a b a b abN c c c c S

OrbitalOverlap

ab a bS where

2 2

1

2a b a b ab

Nc c c c S

Fraction of electron density in orbital b

Fraction of electron density in orbital a

Fraction of electron density in orbital i

2 2

1

2MO a a b b

a b a b ab

c cc c c c S

If Sab 0: 2 2 2 2

a bMO a b

a b a b

c c

c c c c

2

2 2a

aa b

cf

c c

2

2 2b

ba b

cf

c c

General: MO i iN c 2

2i

ii

cf

c

Homonuclear Diatomic Molecules:

Heteronuclear Diatomic Molecules:

2

2 2a

aa b

cf

c c

2

2 2b

ba b

cf

c c

0aa ab ab

ab ab bb

H E H E S

H E S H E

bb aaH H

b ac c

0.50a bf f

bb aaH H

b ac c

b af f

One has a quadratic equation, which can be solved to yieldtwo values for the energy, <E>.

One can then determine cb/ca for both the bondingand antibonding orbitals.

Antibonding (A)

Bonding (B)

a

b

aaE H

bbE H

B a a b bc c

' 'A a a b bc c

0aa ab ab

ab ab bb

H E H E S

H E S H E

0aa bb ab ab ab abH E H E H E S H E S

A Numerical Example: Hydrogen Fluoride (HF)

+ -2pzb(F)

+

1sa(H)

Matrix Elements

0aa ab ab

ab ab bb

H E H E S

H E S H E

1 ( ) 2 ( )MO a a b b a a b zbN c c N c s H c p F

1 ( ) 1 ( ) 13.6aa a a a aH H s H H s H eV

2 ( ) 2 ( ) 17.4bb b b zb zbH H p F H p F eV

1 ( ) 2 ( ) 2.0ab a b a zbH H s F H p F eV

1 ( ) 2 ( ) 0ab a b a zbS s H p F

0aa ab ab

ab ab bb

H E H E S

H E S H E

13.6

17.4

2.0

0

aa

bb

ab

ab

H eV

H eV

H eV

S

13.6 20

2 17.4

E

E

13.6 17.4 2 2 0E E

231.0 232.64 0E E

231.0 (31.0) 4(1)(232.64)

2E

31.0 30.4418.26

2BE eV

31.0 30.4412.74

2AE eV

13.6 2

02 17.4

E

E

13.6 2 0

2 17.4 0

a b

a b

E c c

c E c

Bonding MO Antibonding MO

13.6 18.25 2 0a bc c

2.33 2.33bb a

a

cc c

c

2.33

2.33

B a a b b

a a b

a b

c c

c

N

2

12.33

1 2.33B a b

0.394 0.919

0.394 1 ( ) 0.919 2 ( )B a b

a zbs H p F

0.919 0.394

0.919 1 ( ) 0.394 2 ( )A a b

a zbs H p F

0.430 0.430bb a

a

cc c

c

0.430

0.430

A a a b b

a a b

a b

c c

c

N

2

10.430

1 0.430A a b

13.6 12.74 2 0a bc c

Electron Densities in Hydrogen Fluoride

Bonding Orbital

Over 80% of the electron density of the two electrons in the bonding MO resides on the Fluorine atom in HF.

Antibonding Orbital

The situation is reversed in the Antibonding MO.However, remember that there are no electrons in this orbital.

0.394 0.919 0.394 1 ( ) 0.919 2 ( )B a b a zbs H p F

22

2 20.394 0.16a

a Ha b

cf f

c c

22

2 20.919 0.84b

b Fa b

cf f

c c

0.919 0.394 0.919 1 ( ) 0.394 2 ( )A a b a zbs H p F

22

2 20.919 0.84a

a Ha b

cf f

c c

22

2 20.394 0.16b

b Fa b

cf f

c c

Statistical Thermodynamics: Electronic Contributions to Thermodynamic Properties of Gases

Here they are again.

2

,

ln

V N

QU kT

T

,

VV N

UC

T

lnU

S k QT

lnA U TS kT Q

,

lnln

ln T N

QG H TS kT Q kT

V

,P

P N

HC

T

2

, ,

ln ln

lnV N T N

Q QH kT kT

T V

The Electronic Partition Function

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.~

0 1 2

0 1 20

...i

elect kT kT kT kTi

i

q g e g e g e g e

or0 0 0 1 0 2 0

0 1 20

...i

elect kT kT kT kT kTi

i

q e g e e g g e g e

0 0 1 2

0 1 20

...i

elect kT kT kT kT kTi

i

q e g e e g g e g e

0 0 1 2

0 1 20

...ei e e

elect kT T kT T Ti

i

q e g e e g g e g e

eiiei

hc

k k

The ground state energy, 0, is arbitrary (depending upon the point ofreference. It is sometimes taken to be zero.

Excited state energies above ~5,000-10,000 cm-1 don't contributesignificantly to qelect (or to thermodynamic properties). For example,

Thus, one can ignore all but very low-lying electronic states, exceptat elevated temperatures.

eiiei

hc

k k

120,000

3,000ei cm

T K

34 104

23

(6.63 10 )(3.00 10 )(20,000)2.88 10

1.38 10ei

ei

hc x xx K

k x

42.88 10

9.60 53,000 7 10ei x

Te e e x

At room temperature (298 K): 421 10ei

ke x

0 0 1 2

0 1 20

...ei e e

elect kT T kT T Ti

i

q e g e e g g e g e

For convenience, we will consider only systems with a maximum of one "accessible" excited state:

Internal Energy and Enthalpy

Qelect independent of V

0 1

0 1

eelect kT Tq e g g e

2

, ,

ln ln

ln

elect electelect elect

V N T N

Q QH kT kT U

T V

2

,

ln electelect

V N

QU kT

T

2

,

ln electelect elect

V N

QH U kT

T

2

,

ln elect

V N

qnRT

T

2

,

lnNelect

V N

qkT

T

H0elect Htherm

elect

A

RN

k

0 1

0 1

eelect kT Tq e g g e

1

00 1ln ln

eelect Tq g g e

kT

1

1

1 10

0 1

e

e

Telect elect e

T

nRg eH U nE

g g e

12 0

0 1lne

Td d

nRT g g edT kT dT

1

1

22 0 1 1

2 2

0 1

e

e

e T

T

gnRTnRT e

kT Tg g e

1

1

1 10

0 1

e

e

Te

A

T

nRg enN

g g e

2

,

ln electelect elect

V N

qH U nRT

T

0elect elect elect elect

thermH U H H

GS Electronic Energyper mole

0 0AE N

111

ee

hc

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 electronicground state.

In this case, we could have started with:

which would have led to:

1

1

1 10

0 1

e

e

Telect e

T

nRg eH nE

g g e

111

ee

hc

k k

0 1 1

0 1 0 1

e eelect kT T Tq e g g e g g e

1

1

1 1

0 1

e

e

Telect e

T

nRg eH

g g e

Numerical Example #1

The electronic ground state of CO is a singlet, with no accessibleexcited 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:

By the same reasoning, Helect = Uelect = 0 for molecules with ground states of any multiplicity if the excited states are inaccessible.

1

1

1 1

0 1

e

e

Telect elect e

T

nRg eH U

g g e

111

ee

hc

k k

1

1

1 1

0 1

e

e

Telect e

T

nRg eH

g g e

01 1

0 1

enRg e

g g e

Numerical Example #2

The electronic ground state of O2 is a triplet (refer back to orbital diagramfor 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.

1

1

1 1

0 1

e

e

Telect elect e

T

nRg eH U

g g e

111

ee

hc

k k

34 10 1

11 23

6.63 10 3.00 10 / 788211,360

1.38 10 /e

e

x J s x cm s cmhcK

k x J K

1

1

11360

30001 1

11360

30000 1

1 8.31 / (2) 11360

3 2

e

e

Telect e

T

mol J mol K K enRg eH

eg g e

1400 1.40J kJ

At 298 K: 152 10 0electH x kJ

Entropy

because Nk = nNAk = nR

because we're assumingthat 0 = 0.

lnelect

elect elect US k Q

T ln

electNelect U

k qT

lnelect

elect UNk q

T

lnelect

elect elect US nR q

T

0 1 1

0 1 0 1

e eelect kT T Tq e g g e g g e

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 .

Let's calculate Select for one mole at 3000 K.

Uelect = Helect = 1400 J (from earlier calculation)

lnelect

elect elect US nR q

T

1

0 1

eelect Tq g g e

34 10 1

11 23

6.63 10 3.00 10 / 788211,360

1.38 10 /e

e

x J s x cm s cmhcK

k x J K

1 11360

30000 1 3 2

eelect Tq g g e e

3 0.045 3.045

At 25 oC = 298 K: Uelect = 0.00 J

qelect = g0 = 3

Select = Rln(3) + 0/298 = 9.13 J/K at 298 K

Uelect = 1400 J

lnelect

elect elect US nR q

T

1

0 1 3.045e

elect Tq g g e

lnelect

elect elect US nR q

T

14001 8.31 / ln(3.045)

3000

Jmol J mol K

K

9.25 0.47 9.72 / 3000electS J atK K

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 amultiplicity, g0 > 1, then Select >0, even if there are noaccessible excited states.

lnelect

elect elect US nR q

T

1

0 1

eelect Tq g g e

O2 Entropy: Comparison with experiment

O2: Smol(exp) = 205.1 J/mol-K at 298.15 K

Chap. 3 Chap. 4 Chap. 5

We were about 5% too low. Let's add in the electronic contribution.

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.

151.9 43.8 0.035 195.7tran rot vibS S S

151.9 43.8 0.035 9.1 204.8tran rot vib electS S S S

Helmholtz and Gibbs Energy

Qelect independent of V

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 .

Let’s calculate Gelect ( = Aelect) for one mole of O2 at 3000 K.

lnelect elect elect electA U TS kT Q

,

lnln

ln

electelect elect elect elect

T N

QG H TS kT Q kT

V

electA

34 10 1

11 23

6.63 10 3.00 10 / 788211,360

1.38 10 /e

e

x J s x cm s cmhcK

k x J K

Nk = nNAk = nR

e1 = 11,360 Kg0 = 2g1 = 3

lnelect elect electG A kT Q 1

0 1

eelect Tq g g e

1 11360

30000 1 3 2 3.045

eelect Tq g g e e

ln lnNelect elect electG kT Q kT q ln electNkT q ln electnRT q

ln 1 8.31 / (3000 )ln(3.045)elect electG nRT q mol J mol K K

27,760 27.8elect electG A J kJ

0 1000 2000 3000 4000 5000

150

200

250

300

Expt T TR TRV TRVE

S

[J/m

ol-K

]

Temperature [K]

O2: Entropy

The agreement of the calculated entropy including electroniccontributions with experiment is close to perfect.

0 1000 2000 3000 4000 5000

0

50

100

150

200

Expt T TR TRV TRVE

H

[kJ/

mol

]

Temperature [K]

O2: Enthalpy

The agreement of the calculated enthalpy including electroniccontributions with experiment is close to perfect.

0 1000 2000 3000 4000 500015

20

25

30

35

40

Expt T TR TRV TRVE

CP

[J/m

ol-K

]

Temperature [K]

O2: Heat Capacity

The calculated heat capacity is the most sensitive function of anyneglect in the calculations.

We have neglected several subtle factors, including vibrationalanharmonicity, centrifugal distortion and vibration-rotation coupling

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