A tutorial on Molecular Orbitals

Truman State University

Description

Molecular Orbitals of BenzeneM. Samiullah,Nov 18, 2009Suchi’s OLED groupUMC-PhysicsOverview• Absorption spectrum of benzene film• Molecular Orbitals– Quantum Problem– Approximations• Linear Combination of Atomic Orbitals (LCAO)– Assignment of absorption bandsTakashi Inagaki J. Chem. Phys. 1972UV Absorption Spectrum (thin film of Benzene between quartz plates) -3Y-scale = x 10 Extremely 1900 Å weakStrongestY-scale = x 12100 Å 2550 Å4SWhat are the origins of these bands? (known since 1880s).3SThree transitions from ground state to low lying states.2S1SElectronic problem of benzene• Full electronic Hamiltonian for 54 particles ( 42 e and 12 nuclei) 1. H = T + T + V + V + Ve n nn en ee2T = KE =1⁄2 m v ; V = PE=(1/4πε ) q q / r ; 0 1 2 12(e = electrons; n = nuclei).2. Solve for stationary states : HΨ(r,R)=EΨ(r,R) Assumptions1. Divide the electrons into valence (e’) and core electrons (e ).02. Born-Oppenheimer adiabatic approximation.– Expand around equilibrium positions of the nuclei.→ H ≈ T + V + V (no vibrations) (6+24 = 30 e problem!) +e’ e’n0 e’e’3. Chemical properties linked to pπ electrons (e”) of C only. -→ H = T + V + V (6 e + 12 nuclei).e’’ e’’n0 e’’e’’4. Mean Field Approximation . Assume each of the six electrons moves in some effective potential called mean field. → (1 e + 12 nuclei)5. Ignore H-nuclei. → (1 e moving in potential of 6 nuclei)Electronic problem of benzene (Summary)1. Solve H Ψ =E Ψ with ...

Informations

• Publié par: Uffe
• Nombre de lectures: 45
• Langue: English

Extrait

Molecular Orbitals of Benzene

M. Samiullah, Nov 18, 2009 Suchi’ sOLED group UMC-Physics

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Overview

Absorption spectrum of benzene film

Molecular Orbitals –Quantum Problem

–Approximations •Linear Combination of Atomic Orbitals (LCAO)

–Assignment of absorption bands

2 S1 4 S3 S

What are the origins of these bands? (known since 1880s). Three transitions from ground state to low lying states.

Takashi Inagaki J. Chem. Phys. 1972 UV Absorption Spectrum (thin film of Benzene between quartz plates)

S

Y-scale = x 10 -3 1900 Å Strongest

Extremely weak Y-scale = x 1

2100 Å

2550 Å

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Electronic problem of benzene

Full electronic Hamiltonian for 54 particles ( 42 e and 12 nuclei) 1. H = T e + T n + V nn + V en + V ee T = KE = 1⁄2 m v 2 ; V = PE=(1/4πε 0 ) q 1 q 2 / r 12 ; (e electrons; n = nuclei). = 2. Solve for stationary states : HΨ(r,R)=EΨ(r,R)

”e(  snortceleπpo  tednklis iert  V ’e + ’   T  ’e H =     →  nly. C o) oflap orep  hCmeci3. +    n12leuc.i)+ 0n’’e V  ’’e’ e 6(

1. Divide the electrons intovalence (e’ ) and core electrons (e 0 ).

Assumptions

+ -

2. Born-Oppenheimer adiabatic approximation. –Expand around equilibrium positions of the nuclei. → H ≈ T e’ + V e’ n0 + V e’ e’  = 30 e problem!) (6+24(no vibrations)

5.Ignore H-nuclei. moving in potential of 6 nuclei)→ (1 e

4.Mean Field Approximationeach of the six electrons moves in. Assume some effective potential called mean field. → (1 e + 12 nuclei)

Electronic problem of benzene (Summary)

1. Solve H 0 Ψ n =E 0n Ψ n with H 0 = T e’ ’ + V e’ ’ n0 (1 e in the field of 6 nuclei). 2. Solutions are 1-e states, called Molecular OrbitalsΨ n . 3.Fill MOs one e at a timeto obtain the effective ground state of the molecule |0 . > Achtung!! Achtung!!Molecular orbitals are not quantum states of the molecule! 4. Treat H 1 (= V e’ ’ e” ) perturbatively: ΔE 0n =<Ψ n |H 1 |Ψ n >.

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Use symmetry to solve H 0 Ψ n = E 0 Ψ n

Point Symmetry Group D 6h H 0 is the Hamiltonian with e-e interactions ignored. → *H 0 , R i ] = 0 for all operations R i in D 6h. → Simultaneouseigenstates of H 0 and R i . → Stationary states form basis for irreducible representations of D 6h .

There are 12 irreducible reps of D 6h . Which ones will be useful for benzene?

Character Table of D 6h

D6h

A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u

E

1 1 1 1 2 2 1 1 1 1 2 2

Irreps of point group D 6h 2C 6 2C 3 C 2 3C 2 3C 2 ”i 2S 3 2S 6 σ h (z) (z) (z) (c) (b) (z) (z) (c)

1 1 -1 -1 1 -1

1 1 1 1 -1 -1

same

1 1 -1 -1 -2 2

1 -1 1 -1 0 0

’

1 -1 -1 1 0 0

1 1 1 1 2 2 -1 -1 -1 -1 -2 -2

1 1 -1 -1 1 -1

1 1 1 1 -1 -1

1 1 -1 -1 -2 2

3σ d   (b)

1 -1 1 -1 0 0

times -1

3σ v (c)

1 -1 -1 1 0 0

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Steps in solution (Once again) Pick atomic orbitals (AO)of interest –In benzene, the low lying excited states “ must” involve only pπatomic orbitals. We denote them here by {α 1 ,α 2 ,α 3 ,α 4 ,α 5 ,α 6 }. Construct Molecular Orbitals –forming linear combinations (LCAO) that transform according toby “ relevant” irreducible representationsof the symmetry group. –LCAO is one among many other ways of obtaining MOs, which are loosely defined as wavefunctions that extend over a molecule. [Block wave would be an example of a MO extending over a very large molecule]. Energy of orbitals –Find theexpectation value of Hamiltonianin these molecular orbitals. Afbau principle –Fill each orbital with one up spin and one down spinelectron till all electrons get used up. Construct the ground state –From the electron configurationconstruct the ground stateof the molecule by properly symmetrizing the product of orbitals. Generate first excited state – highest occupied MO (HOMO) to the lowest unoccupiedPromote one electron from MO (LUMO) togenerate first excited state properly and find energy of. Symmetrize excited state(s).

Construct MO : Full representationГ π α1 α2 α3 α4 α5 α6

•Arrange orbitals on C atoms in a vector. •this vector change when you actHow does on the molecule by operations of D 6h ? •C 6 α1 = α6  C 6 represented by: α2α3 0 0 0 0 0 1 α3α 0001 4 00 α4α 0 5 00010 α5α6 001000 α6α 100000 1 1 0 0 0 0 0

Same procedure for other operations. The 6x6 matrices “ represent” the operations in the space of the atomic orbitals α i .