5.4. Electronic structure of water
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1 5.4. Electronic structure of water Water belongs to C 2v point group, we have discussed the corresponding character table. Here it is again: C 2v E C 2 σ v (yz) σ v (xz) A A B B Remember the discussion on symmetry: the orbitals must belong to one of the four irreps. The orbitals will be labeled according to the irreps. Orbitals can be obtained from Independent Particle Approximation, using the following atomic functions as basis: H: 1s a, 1s b O: 1s, 2s, 2p x, 2p y, 2p z One obtains the following orbitals (called canonical orbitals) and orbital energies: 1a 1 orbital ( hartree) Edge = 4.35 Space = Psi = 1 1a 1 : 1s (core) 54
2 2a 1 orbital ( hartree) Edge = 4.35 Space = Psi = 2 2a 1 : 2s ( 2p z )+ 1s a + 1s b bonding 1b 1 orbital ( hartree) Edge = 4.35 Space = Psi = 3 1b 1 : 2p y + 1s a 1s b bonding 3a 1 orbital ( hartree) Edge = 4.35 Space = Psi = 4 3a 1 : 2p z (+ 2s) non-bonding 55
3 1b 2 orbital ( hartree) Edge = 4.35 Space = Psi = 5 1b 2 : 2p x non-bonding 4a 1 orbital ( 0.33 hartree) Edge = 4.35 Space = Psi = 6 4a 1 : 2s + 2p z 1s a 1s b anti-bonding 2b 1 orbital ( 0.49 hartree) Edge = 4.35 Space = Psi = 7 2b 1 : 2p y 1s a +1s b anti-bonding Configuration: (1a 1 ) 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 2 (1b 2 ) 2 56
4 State: 1 A 1 (orbitals are all occupied total symmetric singlet) Excited states: Configuration: (1a 1 ) 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 2 (1b 2 ) 1 (4a 1 ) 1 state: 3 B 2 or 1 B 2 Configuration: (1a 1 ) 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 2 (1b 2 ) 1 (2b 1 ) 1 State: 3 A 2 or 1 A 2 Localized orbitals The bonding orbitals extend to all three atoms: these are not the types of orbitals we really like in chemistry. Chemical intuition requires different orbitals!!!! Remember: in case of anti-symmetric wave function (Slater determinant) the orbitals have no physical meaning : in a determinant we can form any linear combination of the occupied orbitals, the total wave function does not change. Let us form linear combination of the occupied orbitals! 2a 1 1b 1 Edge = 4.35 Space = Psi = 1 Edge = 4.35 Space = Psi = 2 Edge = 4.35 Space = Psi = 3 Edge = 4.35 Space = Psi = 4 2a 1 1b 1 2a 1 +1b 1 We obtain to bonding orbitals which correspond to chemical intuition! 57
5 Now consider the two non-bonding orbitals: 3a 1 1b 2 Edge = 4.35 Space = Psi = 1 Edge = 4.35 Space = Psi = 2 Edge = 4.35 Space = Psi = 3 Edge = 4.35 Space = Psi = 4 3a 1 + 1b 2 3a 1 1b 2 We obtained two non-bonding orbitals which correspond to chemical intuition (lone pairs). Comparison of localized and canonic orbitals: cononic localized orbital energy yes no symmetry yes no bond of an atom pair no yes lone pair of an atom no yes 58
6 5.5. The Hückel approximation For larger molecules the description by the MO theory becomes complicated. In same cases it is enough to consider only a subset of orbitals. For example, in case of conjugated molecules, we only consider the 2p orbitals perpendicular to the molecular plane. In the Hückel theory, in addition, also some further approximations are introduced, what we do not detail here. We mention only the that the MOs of the π system are obtained by the linear combination of the 2p z orbitals of the carbon atoms (one each), and the total energy is the sum of the energy of the occupied orbitals Ethylene Orbitals: Orbital energies: ε 1 = α + β ε 2 = α β Energy diagram: Total energy: E = 2α + 2β (49) 59
7 Butadiene Point group: C 2h Four orbitals from the four 2p z orbitals: 2b g 2a u 1b g 1a u 60
8 Orbital energies 2b g : ε 1 = α β 2a u : ε 2 = α β 1b g : ε 3 = α 0.62β 1a u : ε 4 = α 1.62β Energy diagram: Configuration: (1a u ) 2 (1b g ) 2 State: 1 A g (total symmetric) Total energy: E butadien = 4α β Delocalization energy: E butadien 2E etilen = (4α β) (4α + 4β) = 0.48β 61
9 Benzene Point group: D 6h Orbitals (six from six 2p z orbitals): 1b 2g 1e 2u (degenerate) 1e 2u (degenerate) 62
10 1a 2u Orbital energies: a 2u : e 1g : e 2u : b 2g : ε 1 = α + 2β ε 2 = α + β ε 3 = α β ε 4 = α 2β Orbitals e 1g and e 2u orbitals are degenerate due to their symmetry! Energy diagram: Configuration: (a 2u ) 2 (e 1g ) 4 State: 1 A 1g Energy: E = 6α + 8β Delocalization energy: E benzol 3E etilen = (6α + 8β) 6(α + β) = 2β 63
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