Chemistry 431

Lecture 11

Chemistry 431

Lecture 11

The Born-Oppenheimer ApproximationThe Born Oppenheimer ApproximationThe electronic structure of molecules

NC State University

Valence-bond (VB) orbitals for diatomics( )We consider the orbital on atoms A and B. Following thetext we shall represent electron 1 on atom A as A(1).A(1) i t i f ti Lik i l t 2 iA(1) is an atomic wave function. Likewise, electron 2 islocated on atom B and the wave function is B(2).

A(1)B(2) represents the electrons localized on each atomand is not a bonding picture. If the electrons trade placeswe have A(2)B(1) which is also not a bonding situationwe have A(2)B(1), which is also not a bonding situation.

However, the combination A(1)B(2) + B(2)A(1) is a bondingi t ti B th l i A(1)B(2) B(2)A(1) iinteraction. By the same logic A(1)B(2) - B(2)A(1) is ananti-bonding interaction.

We must consider the anti-symmetry issue. In this descriptionOnly the anti-bonding interaction is anti-symmetric.

The role of spin in the VB pictureIf we include electron spin then we have two possible spin wave functions for the electrons:

iα = spin upβ = spin downTherefore, the spin wave function for the two electrons can be either symmetricα(1)α(2)β(1)β(2)β(1)β(2)α(1)β(2) + α(2)β(1)or anti-symmetric

(1)β(2) (2)β(1)α(1)β(2) - α(2)β(1)Note that there are three symmetric combinations, and onlyone anti-symmetric combination. If we combine the spin andspatial parts of the wave functions we can satisfy the anti-symmetry requirement.

Total VB wave function for diatomicsThe combined wave functions explain the fact that spinpairing results in a singlet for the bonding combination and

t i l t f th ti b di bi tia triplet for the anti-bonding combination.

Anti-bonding combination (triplet)[A(1)B(2) - B(2)A(1)][α(1)α(2)][A(1)B(2) - B(2)A(1)][β(1)β(2)][A(1)B(2) - B(2)A(1)][α(1)β(2) + α(2)β(1)]

(2)[A(1)B(2) B(2)A(1)][α(1)β(2) + α(2)β(1)]

Bonding combination (singlet)[A(1)B(2) + B(2)A(1)][ (1)β(2) (2)β(1)][A(1)B(2) + B(2)A(1)][α(1)β(2) - α(2)β(1)]

Extension of VB to polyatomicsIf we consider more than two nuclei the same principles canbe applied. However, this approach does not explain theb d l l t i f l t i l lobserved molecule geometries of polyatomic molecules.

For example, H2O does not have a 90o bond angle as 2predicted by simple VB theory.

This led to the idea of a hybrid orbitalThis led to the idea of a hybrid orbital.sp3

h 1 = s + px + py + pz

sp2

h 1 = s + 12

pysph 1 = s + px

h 2 = s – px – py + pz

h 3 = s – px + py – pz

h 4 = s + p – p – p

2h 2 = s + 3

2 py – 12

py

h = s 3 p 1 p

1 px

h 2 = s – px

Note that these are all orthogonal.

h 4 s + px py pz h 3 = s – 2 py –2

py

A note on orthogonalityThe hydrogen atom wave functions are orthogonal to oneanother. This means that they have no overlap. We can

i ifi l Th l i t l fexamine some specific examples. The overlap integral ofpx and pz can be written as follows:

We look back to the solution of the hydrogen atom to find

pxpzdτall space

= 0

We look back to the solution of the hydrogen atom to findThe mathematical forms of these p orbitals (and the volumeelement dτ).

p = 3 cosθpz 4π cosθ

px = 34π sinθcosφ

dτ = sinθdθdφ

A note on orthogonalityWe can write all of this out:

3 3π2π

pz*pxdτ

all space= 3

4π cosθ 34π sinθcosφsinθdθdφ

00

3 d i 2 dπ2π

= 34π cosφdφ cosθsin2θdθ

00

3 cosφdφ z2dz02π

0= 34π cosφdφ z2dz

00

= 0

let z = sin θ , dz = – cos θ

Of course the orbitals are normalized so

* d 1pz*pzdτ

all space= 1

A note on normalization of sp hybridsWe apply this logic to the two sp hybrid orbitals. Are they normalized? Are they orthogonal?

h 2* h 2 dτ = (s – pz) *(s – pz)dτ

= s *s d s *p d p *s d + p *p d= s *s dτ – s *pz dτ – pz*s dτ + pz*pzdτ

= 1 + 1 = 2

OK. The text book gave us non-normlized sp orbitals. However, if we multiply each orbital by a normalization

t t th th ill b li dconstant , then they will be normalized.

1 (s – p ) *(s – p )dτ = 1 ∴h 2 = 1 (s – p )2 (s pz) (s pz)dτ 1 ∴h 2 2(s pz)

A note on orthogonality of sp hybridsWe continue to address the question: Are they orthogonal?

1 h * h d 1 ( + ) *( )d12 h 1* h 2 dτ = 1

2 (s + pz) *(s – pz)dτ

=12 s *s dτ – 1

2 s *pz dτ + 12 pz*s dτ – 1

2 pz*pzdτ

They have zero overlap They are orthogonal This concept

= 12 – 1

2 = 0

They have zero overlap. They are orthogonal. This conceptapplies not only to hybrid orbitals in VB theory but to the linear combinations of orbitals in molecular orbital theory.

Heteronuclear diatomicsIn the molecular orbital picture we will consider variouslinear combinations of atomic orbitals. The simplest case i di t i hi h h t bit lis a diatomic, which has two orbitals:

ψ = cAA + cBB

This wave function will be normalized if cA

2 + cB2 = 1

In fact, for a homonuclear diatomic these coefficients must be the same by symmetry so there is not much work to do.H if id HF th l l th ill b tHowever, if we consider HF, then clearly there will be a netdisplacement of electrons towards fluorine. We can thinkabout sharing of electrons (covalency), but that sharingis not necessarily equal (i.e. cA is not equal to cB).

Heteronuclear diatomicsTo make the HF example more precise we can considerthe valence orbitals that interact.

H F

ψ = c1sH + c2pzF

We already know that 1s orbital of hydrogen is 13.6 eVbelow the ionization limit If we are given that the pz orbital

ψ c1sH c2pzF

below the ionization limit. If we are given that the pz orbitalof fluorine is 18.6 eV below the ionization limit then wecan see that these two will not contribute equally to the b di i t tibonding interaction.

1s H2pz F

ElectronegativityThe electronegativity of an element is a measure of its abilityto attract electrons to itself. The Pauling scale of electro-

ti it i b d b d di i ti i Ifnegativity is based on bond dissociation energies. IfWe call the electronegativity of an atom A, χA then

1

The greater the electronegativity the greater the polar

|χA – χB| = 0.102 D(A – B) – 12 D(A – A) + D(B – B)

The greater the electronegativity, the greater the polarcharacter of the chemical bond. The consequence of thischarge asymmetry is that the molecule has a dipole moment.

Dipole momentThe dipole moment of a molecule is a measure of its tendencyto align in an externally applied electric field.

No field E = 0 Applied field E = V/d

Thermal fluctuations tend to cause the dipoles to be random.

Random orientation Dipoles tend to align

Thus at high temperature the dipoles are harder align.

Dipole momentA dipole on a molecule is caused by displacement of chargewithin the molecule. The molecular charge is not affected,b t th di t ib ti i t t i lbut the distribution is not symmetrical.

Negative Positive

We can define the dipole moment as a charge displaced through a distance. μ = qdwhere q is the charge and d is the distance. So for example,if one electron is transferred over a distance of 1 Å the dipole

μ qd

moment is (1 electron)(1 Å) = 1 eÅ. We can also write thisin units of Coulomb-meters. 1 eÅ = 1.62 x 10-29 Cm.

Dipole moments due to a charge cloudWe have seen that we cannot consider the electron to be adiscrete unit that can be placed somewhere in space, butR th it i h l d ( t k) It i i bit lRather it is a charge cloud (so to speak). It is in an orbital.We return to the idea of using wave functions to calculateaverage properties. In the case of the dipole moment we need a dipole operator. This is an operator that representsthe charge displacement.

We have assumed that the charge moves in the z direction.U i thi d fi iti th t h i l di l t

μ = ez

Using this definition the quantum mechanical dipole momentis:

μ = Ψ*ezΨdτμ Ψ ezΨdτ

Typical dipole momentsThe commonly-used unit for dipole moments is the Debye

1 D b 3 33 10 30 C1 Debye = 3.33 x 10-30 Cm4.8 Debye = 1 eÅ

For the series HF, HCl, HBr, HI we can see a trend in thedipole moment that follows the electronegativity. However,the bond length also plays a role since the dipole momentthe bond length also plays a role since the dipole momentdepends on both charge and distance.

Application of the variation principle to a h t l di t i l lheteronuclear diatomic molecule

The name of game is to find the coefficients c and c sinceThe name of game is to find the coefficients cA and cB sincethese will determine the nature of the chemical bond andproperties such as the dipole moment. We will use the variation principle to do this. Starting with our trial wave function.

ψ = c A + c B

The energy is

ψ = cAA + cBB

E =ψ *Hψdτ

ψ *ψdτ

Minimize the energy (it will always be t th th t )greater than the true energy)

We can use calculus to find the minimum of the functionWe can use calculus to find the minimum of the function.Here we will minimize with respect to the coefficients cA andcB. We can call these the variational parameters. SinceThere are two coefficients, there are two equations,

∂E = 0 ∂E = 0

Let’s write out each of the integrals in terms of the coefficients,∂cA

= 0 ,∂cB

= 0

E =ψ *Hψdτ

ψ *ψdτ

Plug in the wave function and work outg

ψ *Hψdτ = cAA + cBB H cAA + cBB dτψ Hψdτ cAA cBB H cAA cBB dτ

= cA2 AHAdτ + cAcB AHBdτ+ cAcB BHAdτ + cB

2 BHBdτ2 2 β 2= cA2αA + 2cAcBβ + cB

2αB

cAA + cBB cAA + cBB dτ = cA2 A2dτ + 2cAcB ABdτ + cB

2 B2dτ

We have made the following definitions

= cA2 + 2cAcBS + cB

2

αA = AHAdτ , αB = BHBdτ Energy

β = AHBdτ = BHAdτ Resonance

Take the derivatives and set up the l tisecular equation

The energy is gy

Now take the derivative with respect to each coefficient

E = cA2αA + 2cAcBβ + cB

2αB

cA2 + 2cAcBS + cB

2

Now take the derivative with respect to each coefficient.

∂E =2 cAαA – cAE + cBβ – cBSE

= 0∂cA cA2 + 2cAcBS + cB

2 0

∂E =2 cBαB – cBE + cAβ – cASE

= 0∂cB=

cA2 + 2cAcBS + cB

2 = 0

cA αA – E + cB β – SE = 0cA αA E cB β SE 0cB αB – E + cA β – SE = 0

Take the derivatives and set up the l tisecular equation

In matrix form the secular equation is:q

αA – E β – SEβ – SE αB – E

cA

cB= 0

The equations have a solution if the determinant vanishes:

α E β SE

This is relatively easy to solve exactly for a homonuclear

αA – E β – SEβ – SE αB – E = 0

y y ydiatomic, where αA = αΒ.

For the heteronuclear case (and all other problems we willFor the heteronuclear case (and all other problems we willwork here) we can make the assumption that S = 0.

Solve the secular equation with lzero overlap

Expand the determinant as shown and solve for the energy:p gy

αA – E ββ αB – E = αA – E αB – E – β2 = 0

E2 – αA + αB E + αAαB – β2 = 02 2 2

E =αA + αB ± αA

2 + αB2 + 2αAαB – 4αAαB + 4β2

2α + α ± α – α

2 + 4β2

This leads to a geometric solution where

E =αA + αB ± αA – αB + 4β

2

1 2βThis leads to a geometric solution where ζ = 1

2arctan 2βαA – αB

Wave functions and energies for th h t l di t ithe heteronuclear diatomic

Expand the determinant as shown and solve for the energy:

E– = αB – β tan ζ , ψ = – A sinζ + B cosζE+ = αA + β tan ζ , ψ = A cosζ + B sinζ

As the energy different between αA and αB increases the Magnitude of ζ decreases. In other words when αA – αB >> βThe orbitals essentially just the unmixed atomic orbitals.The orbitals essentially just the unmixed atomic orbitals. Therefore, bonding will be strongest when αA ~ αB. The approximate values of the energy are

β2

E– = αB + βαA – αB

E = α β2

where we have used the fact that β < 0.

E+ = αA –αA – αB

Application to polyatomicspp p yThe same principles apply to polyatomic molecules.The general form of the molecular orbitals is:g

ψ j = cijχ iΣi = 1

N

A simple and useful kind of system to treat is are linear and cyclic π systems.

Polyenes Aromatic hydrocarbonsPolyenes Aromatic hydrocarbons

Huckel theoryHuckel theoryBasic assumptionsBasic assumptions- Overlap S = 0.- Carbon Coulomb integrals are α for π e-.Carbon Coulomb integrals are α for π e .- Resonance integrals β for neighbor atoms.- β = 0 for non-neighbor atoms.β 0 o o e g bo ato sForm of secular determinant- diagonal elements α - E.d ago a e e e s α- off-diagonal elements β for adjacent atoms.- off-diagonal elements β = 0 for others. g β

EtheneEthene

For ethene there are two carbon atoms and twoπ electrons. Huckel theoryyignores the σ electrons.

C CHH

2π electrons

The secular determinant isC C

HH

α – E ββ α – E = α – E 2 – β2 = 0 , E = α ± β

Bonding energy and electronic g gyspectra from a simple theory

The energy solutions give an MO diagramE- = α - β LUΜΟ

Lowest unoccupiedmolecular orbital

E2p = α

E+ = α + β ΗΟΜΟ

Highest occupiedmolecular orbital

The highest occupied gmolecular orbital of ethene

π

The lowest unoccupied molecular orbital of ethene

π∗π∗

1 d1 π node

Huckel theory for butadieneHuckel theory for butadieneEach carbon contributes α.The resonance integral for each is β.

E β 0 0α – E β 0 0β α – E β 00 β E β0 β α – E β0 0 β α – E

We know how to expand a 4x4 determinant. However,We cannot solve the resulting 4th order polynomial. Oneg yapproach that makes this tractable is to use symmetry.A second approach is to use the free electron model (FEM).

Application of the free electron ppmodel to polyenes

Application of the free electron model to a polyene involves the assumption that the "box" contains the n atoms of the polyene. This is shown for butadiene below. The wave-function coefficients are derived from the amplitude of the sine function obtained from the solution of the particle in asine function obtained from the solution of the particle in a box. The particle in a box solutions are:

ψ = 2 sin jπx

where a is the length of the box and j is the quantum number

ψ j = a sin ja

for a given state.

Application of the free electron ppmodel to polyenes

If we imagine that an electron is placed in a box that contains N atoms at positions na/N along the box then the free electron model (FEM) states that the wavefunction coefficients for a polyene will be given by:

NΨ j = cnjφnΣ

n = 1

N

cnj = 2N sin jnπ

N + 1This is illustrated for butadiene below. The four orbitals shown correspond to the four π orbitals of butadiene.

N N + 1

FEM MOs for butadieneFEM MOs for butadiene

This is the lowest π orbital. It has no nodes.

FEM MOs for butadieneFEM MOs for butadiene

This is the highest occupied π orbital. It has one node.

FEM MOs for butadieneFEM MOs for butadiene

This is the lowest unoccupied π orbital. It has two nodes.

FEM MOs for butadieneFEM MOs for butadiene

This is the highest π orbital. It has three nodes.

Wave functions for butadieneWave functions for butadieneHere we graphically represent thevarying coefficients and sign of theWave function the results for butadiene.

Ψ4 = 0.41φ1 – 0.67φ2 + 0.67φ3 – 0.41φ4

Ψ3 = 0.67φ1 – 0.41φ2 – 0.41φ3 + 0.67φ4Ψ3 0.67φ1 0.41φ2 0.41φ3 0.67φ4

Ψ2 = 0.67φ1 + 0.41φ2 – 0.41φ3 – 0.67φ4

Ψ1 = 0.41φ1 + 0.67φ2 + 0.67φ3 + 0.41φ4

2 i jnπcn = 2N sin j

N + 1φ1 φ2 φ3 φ4

An electronic wavefunction d t h l lcorresponds to each energy level

NODES3

2

1

0

We can construct molecular orbitals of benzene using the six electrons in π orbitalssix electrons in π orbitals

H

C

C CHH

C

C

CHH

C

H Electrons arespin-paired

Benzene Structure Electronic Energy Levels

The Perimeter Model

The benzene ring has D6h symmetry.The aromatic ring has 6 electronsThe aromatic ring has 6 electrons.The π system approximates circular electron path. 5-5

44

-5

3

4

-3

-4

1

2

3

1-2

3

Φ 1 imφ Δm=1Δm=3

m = 01-1Φ = 1

2πeimφ

The Perimeter Model

The porphine ring has D4h symmetry.The aromatic ring has 18 electronsThe aromatic ring has 18 electrons.The p system approximates circular electron path. 5-5

N N44

-5Δm=1Δm=9

NN3

4

-3

-4

1

2

3

1-2

3

Φ 1 imφm = 0

1-1Φ = 12π

eimφ