A tutorial on Molecular Orbitals
28 pages
English

A tutorial on Molecular Orbitals

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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 ...

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Molecular Orbitals of Benzene
M. Samiullah, Nov 18, 2009 Suchi’ sOLED group UMC-Physics
Overview
Absorption spectrum of benzene film
Molecular Orbitals Quantum Problem
Approximations Linear Combination of Atomic Orbitals (LCAO)
Assignment of absorption bands
2S14S3S
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
Electronic problem of benzene
Full electronic Hamiltonian for 54 particles ( 42 e and 12 nuclei) 1. H = Te+ Tn+ Vnn+ Ven+ Vee T = KE =1⁄2m v2; V = PE=(1/4πε0) q1q2/ r12; (e electrons; n = nuclei). = 2. Solve for stationary states : HΨ(r,R)=EΨ(r,R)
e(  snortceleπpo  tednklis iert Ve+  T eH =    nly. C o) oflap orep  hCmeci3.+    n12leuc.i)+0neV ee 6(
1. Divide the electrons intovalence (e’ ) and core electrons (e0).
Assumptions
+ -
2. Born-Oppenheimer adiabatic approximation. Expand around equilibrium positions of the nuclei. → H ≈ Te’+ Ve’ n0+ Ve’ 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 H0Ψn=E0nΨnwith H0= Te’ ’+ Ve’ ’ n0 (1 e in the field of 6 nuclei). 2. Solutions are 1-e states, called Molecular OrbitalsΨn. 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 H1(= Ve’ ’ e”) perturbatively: ΔE0n=<Ψn|H1|Ψn>.
Use symmetry to solve H0Ψn= E0Ψn
Point Symmetry Group D6h H0is the Hamiltonian with e-e interactions ignored. → *H0, Ri] = 0 for all operations Riin D6h. → Simultaneouseigenstates of H0and Ri. → Stationary states form basis for irreducible representations of D6h.
There are 12 irreducible reps of D6h. Which ones will be useful for benzene?
Character Table of D6h
D6h
A1g A2g B1g B2g E1g E2g A1u A2u B1u B2u E1u E2u
E
1 1 1 1 2 2 1 1 1 1 2 2
Irreps of point group D6h 2C62C3C23C23C2i 2S32S6σh (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
3σd  (b)
1 -1 1 -1 0 0
times -1
3σv (c)
1 -1 -1 1 0 0
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 {α1,α2,α3,α4,α5,α6}. 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 representationГπ 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 D6h? C6α1=α6 C6represented by: α2α30 0 0 0 0 1 α3α0001400 α4α0500010 α5α6001000 α6α1000001 1 0 0 0 0 0
Same procedure for other operations. The 6x6 matrices “ represent” the operations in the space of the atomic orbitals αi.
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