Chemistry 431 Lecture 11 The Born-Oppenheimer Born Oppenheimer Approximation The electronic structure of molecules

NC State University

Valence-bond ((VB)) orbitals for diatomics We consider the orbital on atoms A and B. Following the text we shall represent electron 1 on atom A as A(1). A(1) iis an atomic t i wave ffunction. ti Lik Likewise, i electron l t 2 iis located on atom B and the wave function is B(2). A(1)B(2) represents the electrons localized on each atom and is not a bonding picture. If the electrons trade places we have A(2)B(1) A(2)B(1), which is also not a bonding situation situation. However, the combination A(1)B(2) + B(2)A(1) is a bonding i t interaction. ti B By th the same llogic i A(1)B(2) - B(2)A(1) iis an anti-bonding interaction. We must consider the anti-symmetry issue. In this description Only the anti-bonding interaction is anti-symmetric.

The role of spin in the VB picture If we include electron spin then we have two possible spin wave functions for the electrons: α = spin i up β = spin down Therefore, the spin wave function for the two electrons can be either symmetric α(1)α(2) β(1)β(2) α(1)β(2) + α(2)β(1) or anti-symmetric α(1)β(2) (1)β(2) - α(2)β(1) (2)β(1) Note that there are three symmetric combinations, and only one anti-symmetric combination. If we combine the spin and spatial parts of the wave functions we can satisfy the antisymmetry requirement.

Total VB wave function for diatomics The combined wave functions explain the fact that spin pairing results in a singlet for the bonding combination and a triplet t i l t for f the th anti-bonding ti b di combination. bi ti 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)] (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)][ B(2)A(1)][α(1)β(2) (1)β(2) - α(2)β(1)] (2)β(1)]

Extension of VB to polyatomics If we consider more than two nuclei the same principles can be applied. However, this approach does not explain the observed b d molecule l l geometries t i off polyatomic l t i molecules. l l For example, H2O does not have a 90o bond angle as predicted by simple VB theory. This led to the idea of a hybrid orbital orbital. sp 3 h 1 = s + px + py + pz h 2 = s – px – py + pz h 3 = s – px + py – pz h 4 = s + px – py – pz

sp 2 h 1 = s + 1 py 2 3p – 1 p h2 = s + y 2 y 2 3p – 1 p h3 = s – y 2 y 2

Note that these are all orthogonal.

sp h 1 = s + px h 2 = s – px

A note on orthogonality The hydrogen atom wave functions are orthogonal to one another. This means that they have no overlap. We can examine i some specific ifi examples. l Th The overlap l iintegral t l off px and pz can be written as follows: p xp zdτ = 0 all space

We look back to the solution of the hydrogen atom to find The mathematical forms of these p orbitals (and the volume element dτ). 3 pz =

px =

cosθ 4π 3 sinθcosφ 4π

dτ = sinθdθdφ

A note on orthogonality We can write all of this out: 2π

π

3 cosθ 4π

p z* p xdτ = all space

0

0

= 3 4π

π

2π

cosφdφ d

cosθsin i θdθ d 2

0

0 2π

= 3 4π

3 sinθcosφsinθdθdφ 4π

0

z2dz = 0

cosφdφ 0

0

let z = sin θ , dz = – cos θ

Of course the orbitals are normalized so p z* p zdτ d =1 all space

A note on normalization of sp hybrids We apply this logic to the two sp hybrid orbitals. Are they normalized? Are they orthogonal? h 2* h 2 dτ = =

s *s dτ d –

(s – p z) *(s – p z)dτ s *p z dτ d –

p z*s dτ d +

p z*p zdτ d

=1+1=2

OK. The text book gave us non-normlized sp orbitals. However, if we multiply each orbital by a normalization constant t t , then th they th will ill b be normalized. li d 1 2

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

A note on orthogonality of sp hybrids We continue to address the question: Are they orthogonal? 1 2

h 1* h 2 dτ d = 1 ((s + p z) *(s *( – p z)dτ )d 2 = 1 s *s dτ – 1 s *p z dτ + 1 p z*s dτ – 1 2 2 2 2 = 1 – 1 =0 2 2

p z*p zdτ

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

Heteronuclear diatomics In the molecular orbital picture we will consider various linear combinations of atomic orbitals. The simplest case i a di is diatomic, t i which hi h h has ttwo orbitals: bit l ψ = c AA + c BB

This wave function will be normalized if c 2A + c B2 = 1

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

Heteronuclear diatomics To make the HF example more precise we can consider the valence orbitals that interact. H

F

ψ = c 1s H + c 2p zF We already know that 1s orbital of hydrogen is 13.6 eV below the ionization limit limit. If we are given that the pz orbital of fluorine is 18.6 eV below the ionization limit then we can see that these two will not contribute equally to the b di iinteraction. bonding t ti 1s H 2pz F

Electronegativity The electronegativity of an element is a measure of its ability to attract electrons to itself. The Pauling scale of electronegativity ti it iis b based d on b bond d di dissociation i ti energies. i If We call the electronegativity of an atom A, χA then

|χ A – χ B| = 0.102

D(A – B) – 1 D(A – A) + D(B – B) 2

The greater the electronegativity, electronegativity the greater the polar character of the chemical bond. The consequence of this charge asymmetry is that the molecule has a dipole moment.

Dipole moment The dipole moment of a molecule is a measure of its tendency to align in an externally applied electric field.

No field E = 0 Random orientation

Applied field E = V/d Dipoles tend to align

Thermal fluctuations tend to cause the dipoles to be random. Thus at high temperature the dipoles are harder align.

Dipole moment A dipole on a molecule is caused by displacement of charge within the molecule. The molecular charge is not affected, b t th but the di distribution t ib ti iis nott symmetrical. ti l

Negative

Positive

We can define the dipole moment as a charge displaced through a distance.

μ = qd

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

Dipole moments due to a charge cloud We have seen that we cannot consider the electron to be a discrete unit that can be placed somewhere in space, but R th it iis a charge Rather h cloud l d ((so tto speak). k) It iis iin an orbital. bit l We return to the idea of using wave functions to calculate average properties. In the case of the dipole moment we need a dipole operator. This is an operator that represents the charge displacement.

μ = ez We have assumed that the charge moves in the z direction. U i thi Using this definition d fi iti th the quantum t mechanical h i l di dipole l momentt is:

μ=

Ψ*ezΨdτ Ψ ezΨdτ

Typical dipole moments The commonly-used unit for dipole moments is the Debye 30 Cm 1D Debye b =3 3.33 33 x 10-30 C 4.8 Debye = 1 eÅ

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

Application of the variation principle to a h t heteronuclear l di t i molecule diatomic l l The name of game is to find the coefficients cA and cB since these will determine the nature of the chemical bond and properties such as the dipole moment. We will use the variation principle to do this. Starting with our trial wave function. ψ = c AA + c BB

The energy is ψ *Hψdτ E=

ψ *ψdτ

Minimize the energy (it will always be greater t than th the th true t energy)) We can use calculus to find the minimum of the function function. Here we will minimize with respect to the coefficients cA and cB. We can call these the variational parameters. Since There are two coefficients, there are two equations, ∂E = 0 , ∂E = 0 ∂c A ∂c B

Let’s write out each of the integrals in terms of the coefficients, ψ *Hψdτ E=

ψ *ψdτ

Plug g in the wave function and work out ψ *Hψdτ =

c AA + c BB H c AA + c BB dτ

= c 2A AHAdτ + c Ac B AHBdτ+ c Ac B BHAdτ + c B2 BHBdτ = c 2Aα A + 2c 2 Ac Bβ + c B2 α B c AA + c BB c AA + c BB dτ = c 2A A 2dτ + 2c Ac B ABdτ + c B2 B 2dτ = c 2A + 2c Ac BS + c B2

We have made the following definitions αA =

AHAdτ , α B =

β=

AHBdτ =

BHBdτ

BHAdτ

Energy Resonance

Take the derivatives and set up the secular l equation ti The energy gy is

c 2Aα A + 2c Ac Bβ + c B2 α B E= c 2A + 2c Ac BS + c B2

Now take the derivative with respect to each coefficient coefficient. ∂E = ∂c A ∂E = ∂c B

2 c Aα A – c AE + c Bβ – c BSE 2 A

c + 2c Ac BS + c

2 B

=0

2 c Bα B – c BE + c Aβ – c ASE 2 A

c + 2c Ac BS + c

2 B

c A α A – E + c B β – SE = 0 c B α B – E + c A β – SE = 0

=0

Take the derivatives and set up the secular l equation ti In matrix form the secular equation q is: α A – E β – SE c A =0 β – SE α B – E c B

The equations have a solution if the determinant vanishes: α A – E β – SE =0 β – SE α B – E

This is relativelyy easyy to solve exactlyy for a homonuclear diatomic, where αA = αΒ. For the heteronuclear case (and all other problems we will work here) we can make the assumption that S = 0.

Solve the secular equation with zero overlap l Expand p the determinant as shown and solve for the energy: gy αA – E β = α A – E αB – E – β 2 = 0 β αB – E E 2 – α A + α B E + α Aα B – β 2 = 0 E= E=

α A + αB ± α A + αB ±

α 2A + α 2B + 2α Aα B – 4α Aα B + 4β 2 2 α A – αB 2

2

+ 4β 2

This leads to a geometric solution where

2β ζ = 1 arctan α – α A B 2

Wave functions and energies for th heteronuclear the h t l di t i diatomic Expand the determinant as shown and solve for the energy: E– = α B – β tan ζ , E+ = α A + β tan ζ ,

ψ = – A sinζ + B cosζ 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 >> β 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– = α B + α A – αB β2 E+ = α A – α A – αB

where we have used the fact that β < 0.

Application pp to p polyatomics y The same principles apply to polyatomic molecules. general form of the molecular orbitals is: The g

ψj =

Σcχ N

i=1

ij

i

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

Polyenes

Aromatic hydrocarbons

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

Ethene For ethene there are two carbon atoms and two π electrons. Huckel theory y 2π electrons ignores the σ electrons. H H The secular determinant is

C H

C H

2 α–E β 2 = α–E –β =0,E=α±β β α–E

Bonding g energy gy and electronic spectra from a simple theory The energy solutions give an MO diagram E- = α - β LUΜΟ

Lowest unoccupied molecular orbital

E2p = α

E+ = α + β ΗΟΜΟ

Highest occupied molecular orbital

The highest g occupied molecular orbital of ethene π

The lowest unoccupied molecular orbital of ethene π∗ π∗

1 π node d

Huckel theory for butadiene Each carbon contributes α. The resonance integral for each is β.

α–E β 0 0 β α–E β 0 0 β α–E β 0 0 β α–E We know how to expand a 4x4 determinant. However, g 4th order polynomial. y One We cannot solve the resulting approach that makes this tractable is to use symmetry. A second approach is to use the free electron model (FEM).

Application pp of the free electron model 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 wavefunction coefficients are derived from the amplitude of the sine function obtained from the solution of the particle in a box. The particle in a box solutions are:

ψj =

j 2 sin jπx a a

where a is the length of the box and j is the quantum number for a given state.

Application pp of the free electron model 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: Ψj = c nj =

Σ N

n=1

c njφn 2 sin jnπ N N+1

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

FEM MOs for butadiene

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

FEM MOs for butadiene

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

FEM MOs for butadiene

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

FEM MOs for butadiene

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

Wave functions for butadiene Here we graphically represent the varying coefficients and sign of the Wave function the results for butadiene. Ψ4 = 0.41φ1 – 0.67φ2 + 0.67φ3 – 0.41φ4 Ψ3 = 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 φ1

φ2

φ3

φ4

cn =

jjnπ 2 sin i N N+1

An electronic wavefunction corresponds d tto each h energy llevell NODES 3

2

1

0

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

H

H C

C

C

C H C H

Benzene Structure

Electrons are spin-paired

Electronic Energy Levels

The Perimeter Model The benzene ring has D6h symmetry. The aromatic ring has 6 electrons. electrons The π system approximates circular electron path.

1 imφ Φ= e 2π

-5

5

-44

4

-33

3

-2 Δm=3 -11

2 Δm=1 1 m=0

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

N

5

Δm=9

Δm=1

-44

4

-33

3

-2 -11

2

N

N

1 imφ Φ= e 2π

1 m=0