Review: Molecules = + + = + + Start with the full Hamiltonian. Use the Born-Oppenheimer approximation

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "Review: Molecules = + + = + + Start with the full Hamiltonian. Use the Born-Oppenheimer approximation"

Αγάπη Βονόρτας
5 χρόνια πριν
Προβολές:

1 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)

2 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

3 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

4 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

5 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

6 ückel model benzene = C C 11 pn MO pn = φ z C C 1 pn MO pn 1 z φ φ φ + z z ückel model: matrix elements of next nearest neighbors =0. Overlap matrix S=1.

7 Translation operator Tu u1 u u u u u = = u4 u u5 u u6 u1 T u u1 u u u u u = = u4 u u5 u u6 u T and T have the same eigenvectors

8 Translation operator T N = I T N u u1 u u u u u = Iu = = u4 u u5 u u 6 u u1 u u u u u 1 N 1 3 T u = T u = = u4 u u5 u u 6 u 5

9 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.

10 Eigen vectors of the translation operator T = e e e e e iπ j/3 i π j/3 i π j/3 iπ j/3 j = 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

11 ückel model benzene = C C 11 pn MO pn = φ z C C 1 pn MO pn 1 z φ φ φ + z z ückel model: matrix elements of next nearest neighbors =0. Overlap matrix S=1.

12 ückel model benzene = T + T + T ( ) N The amiltonian matrix and the translation operator commute and have the same eigenvectors.

13 ückel model benzene i j/3 i j/ π e π e i π j/3 i π j/ e ( iπ j/3 iπ j/3 e ) i j 11 1 e e π = + + iπ j e e i π j/3 i π j/ e e iπ j/3 iπ j/ e e e + e = cos 3 j i π j /3 i π j /3 π j = 1,, N

14 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 = cos j = 1,, 6 3

15 =homo =lumo 3 4

16 Molecular orbitals benzene ev ev ev ev π j Ej = cos j = 1,, ev Some e-e effects included

19 Gaussian

20 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 = c c c c c3 c3 = E c4 c cn cn

21 ückel model, rings The general formula for N atoms in a ring is: π j EMO, j = cos j = 1,, N N ψ 1 iπ nj = exp j = 1,, N N C MO, j φp zn N n= 1 N

22 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

23 Particles confined to a ring cyclobutane benzene 4n +

24 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

25 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

26 Particles confined to a line ψ = m x Eψ ψ n nπ x = sin L L n = 1,, 3, L

27 Linear chains ethene butadiene C C

28 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 = c c c c c3 c3 = E c4 c cn cn

29 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 = 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

30 Polyacetylene ideki Shirakawa, Alan J. eeger, and Alan G MacDiarmid Nobel Prize in Chemistry in 000

Σχετικά έγγραφα

Solutions to the Schrodinger equation atomic orbitals. Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz

Solutions to the Schrodinger equation atomic orbitals Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz ybridization Valence Bond Approach to bonding sp 3 (Ψ 2 s + Ψ 2 px + Ψ 2 py + Ψ 2 pz) sp 2 (Ψ 2 s + Ψ 2 px + Ψ 2 py)