Review: Molecules Start with the full amiltonian Ze e = + + ZZe A A B i A i me A ma ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB Use the Born-Oppenheimer approximation elec Ze e = + + A A B i i me ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB ZZe Neglect the electron-electron interactions. elec is then a sum of MO. MO Ze A = 1 m 4πε r r e A 0 1 A The molecular orbital amiltonian can be solved numerically or by the Linear Combinations of Atomic Orbitals (LCAO)
LCAO example: O Guess that the solution to mo can be written as a linear combination of atomic orbitals. For O: ψ = cφ + c φ + cφ + c φ + cφ + O O O O mo 1 1s s 3 px 4 py 5 pz ψ mo = Eψ mo Construct the amiltonian matrix. O O O O φ1s mo φ1s φ1s mo φs φ1s mo φ pz c φ1s φ1s φ1s φs φ1s φ pz 1 c1 O O O O O O φs mo φ1s φs mo φs c φs φ1s φs φs c = E O O O O O c 5 O O O φpz mo φ1s φpz mo φs φpz mo φpz O O φpz φ1s φpz φs φ c5 pz φ pz amiltonian matrix Overlap matrix S
Linear combination of atomic orbitals The overlap matrix S 1. O O φ1s mo φ1s φ1s mo φs φ1s mo φ pz c1 c1 O O O φs mo φ1s φs mo φs c c = E O O O O O φ 5 5 pz mo φ1s φpz c c mo φs φpz mo φpz This is an eigenvalue problem. Solve to find the eigenenergies and the coefficients
Molecular orbitals Construct the many-electron wave function from the molecular orbitals. Ψ ( r, r ) = 1 N 1 N! ψmo 1 ( r1) ψmo 1 ( r1 ) ψmo, N ( r1) ψ ( r ) ψ MO1 ( r ) ψ ( r ) MO1 N MO, N N elec Ze = + A i i me ia, 4πε 0riA i< j4πε 0rij Evaluate the energy of the many-electron wave function. e E = Ψ elec ΨΨ Ψ
Benzene Assume the valence molecular orbital is Ψ = cφ + c φ + cφ + c φ + cφ + cφ C C C C C C MO 1 p 1 p 3 p 3 4 p 4 5 p 5 6 p 6 z z z z z z
ückel model benzene = C C 11 pn MO pn = φ z C C 1 pn MO pn 1 z φ φ φ + z z 0 0 0 1 11 1 0 0 0 0 1 11 1 0 0 0 0 1 11 1 0 0 0 0 0 0 0 11 1 1 1 11 1 1 1 11 ückel model: matrix elements of next nearest neighbors =0. Overlap matrix S=1.
Translation operator Tu 0 1 0 0 0 0 u1 u 0 0 1 0 0 0 u u 3 0 0 0 1 0 0 u u = = 3 4 0 0 0 0 1 0 u4 u5 0 0 0 0 0 1 u5 u6 1 0 0 0 0 0 u6 u1 T u 0 0 1 0 0 0 u1 u3 0 0 0 1 0 0 u u 4 0 0 0 0 1 0 u u = = 3 5 0 0 0 0 0 1 u4 u6 1 0 0 0 0 0 u5 u1 0 1 0 0 0 0 u6 u T and T have the same eigenvectors
Translation operator T N = I T N u 1 0 0 0 0 0 u1 u1 0 1 0 0 0 0 u u 0 0 1 0 0 0 u u = Iu = = 3 3 0 0 0 1 0 0 u4 u4 0 0 0 0 1 0 u5 u5 0 0 0 0 0 1 u 6 u 6 0 0 0 0 0 1 u1 u6 1 0 0 0 0 0 u u 1 0 1 0 0 0 0 u u 1 N 1 3 T u = T u = =. 0 0 1 0 0 0 u4 u3 0 0 0 1 0 0 u5 u4 0 0 0 0 1 0 u 6 u 5
Eigen values of the translation operator T N Tu = λu N u = λ u = u λ N = 1 T-λ I u = 0 For each eigenvalue, solve ( ) to determine the eigenvectors.
Eigen vectors of the translation operator 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 T = 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 e e e e e iπ j/3 i π j/3 i π j/3 iπ j/3 j = 1,, 6 1 1 1 1 1 1 /6 4 /6 8 /6 1 i π i i e π e π 1 e i10 π /6 e i4 π/6 i8 π/6 i16 π/6 i0 π/6 1 i π/6 e i4 π/6 e 1 i8 π/6 e i10 /6 e π 1, ; e, ; e, ; 1, ; e, i6 π/6 i1 π/6 ; e, i4 π/6 i30 π/6 1 e e 1 e e i8 π/6 i16 π/6 1 e e 1 i3 π/6 i40 /6 e π e i10 π/6 i0 π/6 i40 π/6 i50 π/6 1 e e 1 e e iπ j
ückel model benzene = C C 11 pn MO pn = φ z C C 1 pn MO pn 1 z φ φ φ + z z 0 0 0 1 11 1 0 0 0 0 1 11 1 0 0 0 0 1 11 1 0 0 0 0 0 0 0 11 1 1 1 11 1 1 1 11 ückel model: matrix elements of next nearest neighbors =0. Overlap matrix S=1.
ückel model benzene 11 1 0 0 0 1 1 11 1 0 0 0 0 0 0 = T + T + T 0 0 0 0 0 0 1 11 1 1 0 0 0 1 11 ( ) 1 11 1 N 1 11 1 1 11 1 The amiltonian matrix and the translation operator commute and have the same eigenvectors.
ückel model benzene 0 0 0 1 1 11 1 1 i j/3 i j/3 1 11 1 0 0 0 π e π e i π j/3 i π j/3 0 1 11 1 0 0 e ( iπ j/3 iπ j/3 e ) i j 11 1 e e π = + + iπ j 0 0 1 11 1 0 e e i π j/3 i π j/3 0 0 0 1 11 1 e e iπ j/3 iπ j/3 1 0 0 0 1 11 e e e + e = cos 3 j i π j /3 i π j /3 π j = 1,, N
ückel model benzene ψ = ϕ + e ϕ + e ϕ + e ϕ + e ϕ + e ϕ C iπ j/3 C i π j/3 C iπ j C i π j/3 C iπ j/3 C j p 1 p p 3 p 4 p 5 p 6 z z z z z z c 1 1 i j/3 c π e i π j/3 c 3 e = iπ j c4 e i π j/3 c 5 e iπ j/3 c6 e from: Blinder, Introduction to Quantum Mechanics π j Ej = 11 + 1 cos j = 1,, 6 3
1 6 7 8 9 19 0 1=homo =lumo 3 4 http://www.stolaf.edu/people/hansonr/jmol/mo/
Molecular orbitals benzene http://www.chemcomp.com/journal/molorbs.htm -0.57 ev 0.151 ev -0.340 ev -0.491 ev http://www.stolaf.edu/people/hansonr/jmol/mo/ π j Ej = 11 + 1 cos j = 1,, 6 3-0.508 ev Some e-e effects included
http://lampx.tugraz.at/~hadley/ss1/skriptum/outline.php
Gaussian http://www.gaussian.com/g_prod/gv5b.htm
ückel model rings Ψ = cφ + c φ + c φ C C C MO 1 p 1 p N p N z z z Assume S = S = 1 S = 0 11 1 0 0 0 c c 11 1 1 1 1 1 11 1 0 0 0 c c 0 1 11 1 0 c3 c3 = E 0 0 1 11 c4 c4 0 0 1 1 0 0 1 11 cn cn
ückel model, rings The general formula for N atoms in a ring is: π j EMO, j = 11 + 1 cos j = 1,, N N ψ 1 iπ nj = exp j = 1,, N N C MO, j φp zn N n= 1 N
Particles confined to a ring ψθ ( ) ψθ ( ) = = Eψθ ( ) m mr θ ψ n inθ e = n = 0, ± 1, ±, π E n n = mr Aromatic molecules obey ückel's 4n + rule Molecules that don't obey the 4n+ rule are radicals
Particles confined to a ring cyclobutane benzene 4n +
Radicals Molecules are most stable with a closed shell configuration. Ar O 10 electrons N 3 C 4 Radicals are electrically neutral but chemically reactive. O radical C methylene radical
Particles confined to a tube R L ψ ψ ψ = = m mr θ m x Eψ n inθ ikx e e ψ n = π RL π 4π = 0, ± 1, ±, k = 0, ±, ±, L L E n k = + mr m nk, 4n + rule
Particles confined to a line ψ = m x Eψ ψ n nπ x = sin L L n = 1,, 3, L
Linear chains ethene butadiene C C
ückel model - linear chains Ψ = cφ + c φ + c φ C C C MO 1 p 1 p N p N z z z Assume S = S = 1 S = 0 11 1 0 0 0 0 c c 11 1 1 1 1 11 1 0 0 0 c c 0 1 11 1 0 c3 c3 = E 0 0 1 11 c4 c4 0 0 1 0 0 0 1 11 cn cn
ückel model - linear chains Ψ = cφ + c φ + c φ C C C MO 1 p 1 p N p N z z z Eigen values: π Ej = 11 + 1 cos j j = 1,,3, N N + 1 c jn, π jn = sin N + 1 N + 1 Eigen vectors: π jn Ψ = N C MO, j sin φpz N + 1 n= 1 N + 1 L E. eilbronner und. Bock, Das MO-Modell und seine Anwendung
Polyacetylene ideki Shirakawa, Alan J. eeger, and Alan G MacDiarmid Nobel Prize in Chemistry in 000