advertisement

MIT OpenCourseWare http://ocw.mit.edu 5.04 Principles of Inorganic Chemistry II �� Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8: N-Dimensional Cyclic Systems This lecture will provide a derivation of the LCAO eigenfunctions and eigenvalues of N total number of orbitals in a cyclic arrangement. The problem is illustrated below: There are two derivations to this problem. Polynomial Derivation The Hückel determinant is given by, x 1 1 x 1 1 x O 1 O O O O O DN(x) = = 0 O O O where x= α −E β O O 1 O x 1 1 x From a Laplace expansion one finds, DN(x) = xDn–1(x) – DN–2(x) where D1(x) = x D2 (x) = x 1 1 x = x2 − 1 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8 Page 1 of 6 With these parameters defined, the polynomial form of DN(x) for any value of N can be obtained, D3(x) = xD2(x) – D1(x) = x(x2–1) – x = x(x2–2) D4(x) = xD3(x) – D2(x) = x2(x2–2) – (x2–1) and so on The expansion of DN(x) has as its solution, x = −2 cos 2π j N (j = 0, 1, 2, 3 ... N − 1) and substituting for x, E = α + 2β cos 2π j N (j = 0, 1, 2, 3 ... N − 1) Standing Wave Derivation An alternative approach to solving this problem is to express the wavefunction directly in an angular coordinate, θ For a standing wave of λ about the perimeter of a circle of circumference c, c ψ j = sin θ λ The solution to the wave function must be single valued ∴ a single solution must be obtained for ψ at every 2nπ or in analytical terms, 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8 Page 2 of 6 c c ψ = sin (θ + 2π ) = sin θ λ λ = sin c λ θ • cos c λ 2π + sin must go to 1 λ 2π • c λ 2π m j N c c θ = sin θ λ λ 2π = 2π j (j = 0, 1, 2 ... N − 1) condition for an integral number of λ’s about the circumference of a circle c =j λ Thus the amplitude of ψj at atom m is, (where ψ j(m) = sin cos must go to 0 iff ∴ c c λ = j and θ = 2π m) N (j = 0, 1, 2 ... N − 1) Within the context of the LCAO method, ψj may be rewritten as a linear combination in φm with coefficients cjm. Thus the amplitude of ψj at m is equivalent to the coefficient of φm in the LCAO expansion, N ψ j = ∑ c jmφm m=1 where c jm = sin 2π m j N (j = 0, 1, 2 ... N − 1) The energy of each MO, ψj, may be determined from a solution of Schrödinger’s equation, Hψ j = E jψ j H − Ej ψ j = 0 N H − E j ∑ c jmφm = 0 m The energy of the φm orbital is obtained by left–multiplying by φm, N φm H − E j ∑ c jmφm = 0 m 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8 Page 3 of 6 but the Hückel condition is imposed; the only terms that are retained are those involving φm, φm+1, and φm-1. Expanding, α 1 β 0 ⎡c φ H φ − c E φ φ ⎤ + ⎡c ⎤ m jm j m m ⎥ j(m +1) φm H φ m +1 − c j(m +1)E j φm φm +1 ⎥ ⎢⎣ jm m ⎢ ⎣ ⎦ ⎦ + ⎡c j(m–1) φm H φm–1 − c j(m–1)E j φm φm–1 ⎤ = 0 ⎢⎣ ⎥⎦ β 0 Evaluating the integrals, [ ] α c jm − c jm E j + β c j(m+1) + c j(m−1) = 0 [ ] α c jm + β c j(m+1) + c j(m−1) = c jmE j Substituting for cjm, α sin ⎛ 2π m 2π (m + 1) 2π (m − 1) ⎞⎟ 2π m j + β ⎜⎜ sin j + sin j⎟ = E j sin j N N N N ⎝ ⎠ Dividing by sin 2πm j, N ⎛ α + β ⎜⎜ sin ⎝ 2π (m + 1) 2π (m − 1) ⎞⎟ j + sin j ⎟ N N ⎠ = E j 2π m sin j N Making the simplifying substitution, κ = Ej = α + ( 2π j N ) β sin κ (m + 1) + sin κ (m − 1) sin κm ⎛ sin κm • cos κ + sin κ • cos κm + sin κm • cos κ − sin κ • cos κm ⎞ ⎟ E j = α + β ⎜⎜ ⎟ sin κm ⎝ ⎠ Ej = α + 2βcosκ E j = α + 2β cos 2π j N (j = 0, 1, 2 ... N − 1) 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8 Page 4 of 6 Let’s look at the simplest cyclic system, N = 3 N = 3, so E j = α + 2β cos 2π j where j = 0, 1, 2 N E0 = α + 2 β 2π = α −β 3 4π = α + 2β cos = α −β 3 E1 = α + 2β cos E2 Continuing with our approach (LCAO) and using Ej to solve for the eigenfunction, we find… ψ j = ∑ eijθ φm m ⎧ N ⎪± for N even ⎪ 2 for j = 0, ± 1, ± 2 ... ⎨ ⎪± (N − 1) for N odd ⎪⎩ 2 Using the general expression for ψj, the eigenfunctions are: i(0) i(0)0 ψ0 = e φ1 + e i(1) ψ +1 = ei(1)0φ1 + e 2π 3 2π 3 i( −1) ψ −1 = ei(−1)0φ1 + e i(0) φ2 + e i(1) φ2 + e 2π 3 4π 3 φ3 4π 3 φ3 i( −1) φ2 + e 4π 3 φ3 Obtaining real components of the wavefunctions and normalizing, ψ 0 = φ1 + φ2 + φ3 → ψ0 = ψ +1 + ψ −1 = 2φ1 − φ2 − φ3 → ψ 1 = ψ +1 − ψ −1 = φ2 − φ3 → 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera ψ2 = 1 3 1 6 1 2 (φ + φ2 + φ3 1 (2φ 1 (φ 2 ) − φ2 − φ3 − φ3 ) ) Lecture 8 Page 5 of 6 Summarizing on a MO diagram where α is set equal to 0, 1 2 1 –2 2 2 –1 E/ 6 0 1 1 2 6 1 3 5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 8 Page 6 of 6