Chemistry 431Chemistry 431Lecture 10

Diatomic moleculesDiatomic moleculesBorn-Oppenheimer approximation

LCAO MO li ti t H +LCAO-MO application to H2+

The potential energy surfaceMOs for diatomic molecules

NC State University

Born-OppenheimerBorn Oppenheimer approximation

• Molecular motion includes the motion of both nuclei and electrons.

• The time scale of the motion of the nuclei is orders of magnitude slowernuclei is orders of magnitude slower than electron motion due to the difference in mass:difference in mass:

me = 9.109 x 10– 31 kg27mp = 1.672 x 10– 27 kg

The potential energy surface

• The idea that nuclear and electronic motions can be separated implies that p pthere is a nuclear potential energy surface U(R) where R is the internuclear ( )distance.

• The potential surface has an equilibriumThe potential surface has an equilibrium geometry Re and well depth De.

Hydrogen molecule ion wave functions

The spatial wavefunction on each of two H atoms formsThe spatial wavefunction on each of two H atoms formslinear combinations: e-

rΨH1sA or 1sA ΨH1sB or 1sB

rA rB

A BR

The atomic wave functions form linear combinations to make molecular orbital wave functions.

Y = 1s ± 1sY± = 1sA ± 1sB

The hamiltonian for H +The hamiltonian for H2The potential involves three particles, one electrons and two protons. In atomic units it is given by

V = 1 + 1 + 1V = – 1r A

+ 1rB

+ 1R

The hamiltonian includes the kinetic energy terms for theThe hamiltonian includes the kinetic energy terms for theelectron only since the Born-Oppenheimer approximation allows separation of nuclear and electronic motion. The i t l di t R i fi d d th l ki tiinternuclear distance R is fixed and the nuclear kinetic energy is zero.

H = 1∇ 2 + VH = – 2∇ + V

Setting up the energy l l ti f H +calculation for H2

+

The energy is obtained by evaluating the gy y gexpectation value:The denominator E ′ =

Ψ *HΨdτ

gives the requirednormalization.

E =Ψ *Ψdτ

Ψ *Ψ dτ = 1sA* +1sB

* 1sA +1sB dτ

= 1sA* 1sAdτ + 1sA

* 1sBdτ+ 1sB* 1sBdτ+ 1sB

* 1sAdτ

= 1 + S + 1 + S, where 1sA* 1sAdτ = 1 and 1sA

* 1sBdτ = S

Significance of the overlap integral

The wave functions 1sA and 1sB are not orthogonalsince they are centered on different nuclei. The overlap integral S is a function of the internuclear

A B

overlap integral S is a function of the internuclear distance.

A B

Overlap regionOverlap region

Normalized LCAO wave ffunctions

The LCAO wave functions for the H + molecule ionThe LCAO wave functions for the H2 molecule ionare

Ψ = 1 1s + 1sΨ + = 12(1 + S)

1sA + 1sB

and

Ψ – = 12(1 – S)

1sA – 1sB2(1 S)These wave functions are orthogonal as well as

li dnormalized.

Energy levels in H2+

Explicit substitution of the hamiltonian gives1 1 1 1E = Ψ *HΨdτ = Ψ * – 12∇

2 – 1rA

– 1rB

+ 1R Ψdτ

= 1sA* – 1

2∇2 – 1

rA– 1

rB+ 1

R 1sAdτ

+ 1sA* – 1

2∇2 – 1

rA– 1

rB+ 1

R 1sBdτ

+ 1sB* – 1

2∇2 – 1

rA– 1

rB+ 1

R 1sBdτ

+ 1sB* – 1

2∇2 – 1

rA– 1

rB+ 1

R 1sAdτ

Energy levels in H2+

2

= 1sA* E1s – 1

rB+ 1

R 1sAdτ+ 1sA* E1s – 1

rB+ 1

R 1sBdτB B

+ 1sB* E1s–

1rA

+ 1R 1sBdτ+ 1sB

* E1s – 1rB

+ 1R 1sAdτB 1s r A R B B 1s rB R A

SinceSince– 1

2∇2 – 1

rA1sA =E1s1sA

– 12∇

2 – 1rB

1sB =E1s1sB

Energy levels in H2+gy 2

= E1s 1+S + 1sA* – 1

rB+ 1

R 1sAdτ + 1sA* – 1

rB+ 1

R 1sBdτ

+ E1s 1+S + 1sB* – 1

rA+ 1

R 1sBdτ+ 1sB* – 1

rB+ 1

R 1sAdτ

To further evaluate these integrals we define theCoulomb integralCoulomb integral

J = 1sA* – 1

rB+ 1

R 1sAdτ = –1sA

* 1sArB

dτ + 1R

And the exchange integral1s* 1s SK = 1sB

* – 1rB

+ 1R 1sAdτ = –

1sB1sArB

dτ + SR

Energy levels in H2+

E ′ =Ψ +

*HΨ +dτ=

E1s2 1 + S + 2 J + KE+ =

Ψ +*Ψ +dτ

=2 1 + S

′ J + KE+′ =E1s + J + K

1 + S

E–′ =

Ψ –*HΨ –dτ

=E1s2 1 – S + 2 J – K

2 1 SΨ –*Ψ –dτ 2 1 – S

E ′ E J – KE– =E1s + J K1 – S

Diagram of H2+ energy levelsDiagram of H2 energy levels

J – K

anti-bonding1 – S

E1sE1s

bonding

J + KJ1 + S

Note that the anti bonding level is moreNote that the anti-bonding level is more destabilizing than the bonding level is stabilizing.

Potential energy surface for H2+gy 2

Using elliptical integrals that S, J, and K integralscan be solved analytically to yieldy y y

S(R) = e–R 1 + R + R2S(R) = e R 1 + R + 3

1J(R) = e–2R 1 + 1R

K(R) = SR – e–R 1+R

MO treatment of H2

E1E1s

anti-bonding

E1s1s

bonding

Th t l t t h it i d bThe two electrons must have opposite spins,a and b.The wave function must be anti-symmetricwith respect to electron exchangewith respect to electron exchange.

Ψ MO =Ψ b(1)Ψ b(2) 12α(1)β(2) – α(2)β(1)

2Spatial part Spin part (anti-symmetric)

Application to diatomic moleculesLi2 B2Be2 C2 N2 O2 F2

2π∗4σ∗

2p

1

3σ2π2pz

2px,y

1

2σ∗1π

2s1σ

• Considering only valence electrons we can fill the molecular orbitals of diatomicscan fill the molecular orbitals of diatomics.• Linear combinations of 2s, 2pz give s orbitals.• Linear combinations of 2px,y give p orbitals.

Th l ti d i d d th• The relative energy ordering depends on thenumber of electrons in occupied orbitals.

Molecular Orbital TheoryMolecular Orbital Theory

• In MO theory electrons are treated asIn MO theory electrons are treated as including the entire molecule.– Each MO is built up from a linear combination of

i bi l (LCAO)atomic orbitals (LCAO).

ΨMO = ciφiΣI

where φi are atomic orbitals

– The coefficients are optimized by the self-i t t fi ld (SCF) th d

MO iφii = 1φi

consistent field (SCF) method.– The variational principle justifies minimization of

the energy by adjustment of the coefficients ci.gy y j i

Nitrogen Molecular OrbitalsNitrogen Molecular OrbitalsExample of a Homonuclear Diatomicp

• Isosurfaces represent Ψ*Ψ of orbitalIsosurfaces represent Ψ Ψ of orbital showing 90% of total probability.

• The spatial wavefunction is an LCAO• The spatial wavefunction is an LCAO.• Core electrons are not included.

• There are five electrons for each N atom.

Dinitrogen 1σ MO

N2 2σ* MO

N2 1π MO

Thi i d bl d t bit lThis is a doubly degenerate orbital. Only one of the two is shown.

N2 3σ MO

N2 2π* MO

Thi i d bl d t bit lThis is a doubly degenerate orbital. Only one of the two is shown.

N2 4σ* MO

Energy level diagram for NEnergy level diagram for N2

• Negative energies represent bonding 1π∗

4σ∗

interactions (< 0 eV)• For N2 all there are

t l t llten electrons so all orbitals are filled through 3σ 2

1π3σ

through 3σ• Only valence

orbitals are shown1σ1σ

2σ∗

orbitals are shown

Molecular Properties

• Bond length (structure).• Vibrational frequency:q y

– Calculated at stationary point.– Depends on accuracy of second p y

derivative matrix with respect to nuclear displacement.

Di l t ( l l f N )• Dipole moment (clearly zero for N2).• Absorption spectrum.