Expect to see:



C. Molecular Orbital Theory

Now what we’ve seen are transitions to electronic states of certain symmetry

Problem – how do we determine symmetries of the excited states (or ground state)

Idea – follow same procedure as other quantum mechanical problems of multi dimensions

Try to separate H into sum over hi(qi) -- where hi(qi) is a single particle Hamiltonian’s

Problem - electron repulsion: [pic] – never really separable,

Solution: effectively treat as a correction in a perturbation or variation sense

For sake of discussion -- let [pic]

where

hi(ri) = [pic] + V(ri) i.e. assume V(ri) separable

How? Let V(ri) = [pic] + (average repulsion( of all other electrons

then H0ψ0 = E0ψ0

ψ0 = [pic] – multi electron state

hiφi0 = εiφi0(ri) – one electron - Hamiltonian

φi0 → orbital – one electron w/f solution to molecular potential

so must correspond to representation of group

if φi’s are functional representation of group

then ψ = [pic] will be a direct product representation called configuration

when specify occupied φi -- completely analogous to AOs and atomic state/configuration

(For high symmetry( Due to large number electrons

This would be a real mess if did not have the Pauli principle

Lowest energy wave functions will have electrons paired up

φi = φj but ms = ±1/2

These give rise to only A1g – i.e. only one way to put 2e- in one orbital or so - no degeneracy

→ Γi ( Γ i = A1g + …

( To determine symmetry of ψ the multi-electron state:

Only need to consider unpaired electrons

a) if all electrons paired, typically ground (A) state

ψ symmetry – total symmetry representation “A1g”

b) if only one electron unpaired (radical)

ψ symmetry = φi symmetry for unpaired electron i--i.e. orbital symmetry=state sym.

c) if more electrons unpaired (e.g. excited state)

ψ symmetry = Γφi ( Γj …

i) if all representations non-degenerate → multiply out reps, will have unique ψ & sym

ii) if representations degenerate and 1e-/orbital then Γψ = direct product

iii) if multi electron in degenerate orbital - must count possiblities, since elect. identical

example--all orbitals non-degenerate:

H2O ground state – all paired: (1a1)2 (2a1)2 (2b2)2 (3a1)2 (b1)2 -- 1A1

H2O+ ion state – one unpaired: (1a1)2 (2a1)2 (2b2)2 (3a1)2 (b1)1 -- 2B1

H2O* excited state – two unpaired: (1a1)2 (2a1)2 (2b2)2 (3a1)2 (b1)1 (b2)1 -- 3A2, 1A2

[c,iii.--last case is like -- for overtones: e x e = a1 + e (no a2) t x t = a1 + e + t2 (no t1)]

perfect analogy to p2 → 1S, 3P, 1D, -- only choices due to Pauli Principle

NH3 ground state – all paired: (1a1)2 (2a1)2 (1e)4 (3a1)2 -- 1A1

NH3* excited state – two un-paired: (1a1)2 (2a1)2 (1e)4 (3a1)1 (2e)1 -- 1,3E

NH3* doubly excited state – two un-paired: (1a1)2 (2a1)2 (1e)4 (3a1)0 (2e)2 -- 1E, 1A1, 3A1

In this case consider possibilities

__ __ , __ __ , __ __ , __ __ --> 1A1, 3A1 ____ ____ + ____ ____ --> 1E,

Problem → now need to focus on symmetry of φi’s

i.e. How to get a1 b1 etc. orbital symmetries above

can solve equations and see what symmetries develop for 1e- w/f (like in computer demo)

But – know that H → A1g blocks mixing of representations

so no energy terms between states different symmetry

Choose AO’s to make MO’s → make linear combination of AO’s to form representation

Symmetry adapted AOs

Advantage → choose a basis set of functions that have symmetry of molecule

→ will block diagonal H, simplify calculation

Energy Matrix

Diagonalization → maintains symmetry; just makes “unitary transform” of one basis →

To a new one (diagonal) best solution to one electron h – i.e. still big approximation

Last time looked at H2O

AO’s can be symmetry adapted

“reduce” O(1s) → a1

O(2s) → a1

O(2p) → a1z + b1y + b2x

2H(1s) → a1 + b2

Number of MO = number of AO this is just a coordinate transformation

representations repeat so convention is to just number them

again total number of AO = number of MO for each symmetry

fill them with 2e- each ( Pauli Principle

bonding orbital → decrease in energy → compact atom

non bonding orbital → different symmetry (b1) → no mix

anti bonding → increase in energy

# * = # b on sense of large excursion from AO

This is all group theory → reduce representations or projection

Energies for using this with Hamiltonian

ψ = [pic] state is product of filled orbital

[pic] Energy is sum of filled orbital energies

State representation is product of orbital representation

Optimize → diagonalize [pic] {symmetry represents block diagonal

Can choose any basis, but traditional LCAO-MO idea, use as a basis linear combination AO whose energies are similar and symmetry LCAO → symmetry molecule

Now choose more AO’s → get more MO’s

just a coordinate (basis) transform

these are 1-e- solutions

the lowest E ones will have e-

Identical orbital move into each other by symmetry operator → Project

Minimal Basis: φ1 = (1S0) – a1 φ4 = (2px0) – b1

φ2 = (2S0) – a1 φ5 = (2py0) – b2

φ3 = (2pz0) – a1

φ6 (1S1H + 1S2H) – a1 φ6 (1S1H + 1S2H) – b2

Block diagonal due to symmetry basis

To get energies (eigen values) solve

[pic]

Now solution of this would demand evaluating the actual integrals. Group Theory tells which bases mix.

10e- problem fill lowest to highest – ground state

[pic]

(

ground state above to be anti symmetry

note each solution 2-φi’s express as determinant w/f

ex. φ1a1( + φ1a1( … ( determinant w/f

excited states – any other filling mechanism

ex: (1a1)2 (2a1)2 (1b2)2 (3a1)2 (1b1)1 (2b2)1

b1 ( b2 excitation

Eex – Egrd = εεex – εεgrd = ε2b2 – ε1b1

These are called configurations, since potential is angular of all other states, simple addition not very exact – configuration interact ([pic])

Actual spectroscopic (i.e. excited state) calculation must then include configuration interaction to be accurate – leads to shifts

1) Note only configuration of same symmetry interact due to [pic] ~ A1g

2) Note use only orbitals as a basis –

h ( orbitals with MO properties as bonding anti bonding

E lower (same symmetry) bond / E up – anti/no charge

Analogous to result from H2+ ion but have not biased by selection

3) If include more orbitals ( get more MO’s

state depends on which ones filled

[pic] 4 H’s – identical, consider 1s

[pic]

ΓH1s = 4 0 0 0 0 4 0 0

= Ag + B1g + B2u + B3u

ΓC2s = 2 0 0 2 0 2 2 0

= Ag + B3u

ΓC2px = 2 0 0 2 0 2 2 0

= Ag + B3u

Γ2pz = 2 0 0 -2 0 2 -2 0

= B1g + B2u

Γ2pz = 2 0 0 -2 0 -2 2 0

= B2g + B1u ( note do not interact with any other

CH bonds ( φC + φH same symmetry

come from Ag – 2s, 2px

B1g – 2py

B2u – 2py

B3u – 2s, 2px

C–C bonds ( φC - φC all as listed

basis – 12 orbitals (neglect C1S) 12 x 12 determinant

|[pic] | | | | | | | | | |

| |[pic] | | | | | | | | |

| | |[pic] | |[pic] | | | | | |

| | | |[pic] | | | | | | |

| | | | |[pic] | |[pic] | | | |

| | | | | |[pic] | | | | |

| | | | | | |[pic] | |[pic] | |

| | | | | | | |HH | | |

| | | | | | | | |H2s | |

| | | | | | | | | |H2px |

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