Diagonalization and Hückel Molecular Orbital Theory

Transcription

1 Diagonalization and Hükel Moleular Orbital Theory Solving the HMO seular equation for omplex moleules an beome very diffiult by hand. However, we may enlist the help of the omputer. The solutions to the seular equation for butadiene: α-e 0 0 α-e 0 0 α-e 2 α-e 0 0 = are E 1, E 2,. The energies may also be obtained as the eigenvalues of the equation: α 0 0 α 0 0 α 0 0 α 2 2 = E where we have just rearranged Equation 1 to put the energy term on the right-hand side. The mathematial proedure for solving Equation 2 is alled diagonalization. There are several effiient ways to diagonalize matries; our omputer program uses Jaobi s method. Sine we don t know the values of α and, we must rewrite Equation 1 in a slightly different way. Dividing both sides by gives: We now define x = x x x x = 0.. Substituting x into Equation gives: = One again, rearranging Equation 4 to put the x terms on the right-hand side gives: = -x where the eigenvalue is x. Define the eigenvalues of Equation 5 as e i.

2 = e i Then the eigenvalues of Equation 6 are e i = -x i, and e i = - (α-e i) Notie that the eigenvalues are the negative of the normal x values used in HMO theory, and E i = α + e i 8. The omputer program to diagonalize a matrix is available on the Web at All you need to do is to input the Hükel matrix with 0 s on the diagonal (for arbon; see below for heteroatoms). The HMO matrix is always symmetri, so we only need to input the lower triangular part. 7. Example: Let s use butadiene as an example: A = B - = D. The input for butadiene is: The output is: Eigenvetor 1: E= Eigenvetor 2: E= Eigenvetor : E= Eigenvetor 4: E=

3 Giving the following wavefuntions and energies, from highest energy to lowest: Eigenvetor 4: E= α Ψ 4 = - p za + p zb - p z + p zd Eigenvetor : E= α Ψ = p za - p zb - p z + p zd Eigenvetor 2: E= α Ψ 2 = - p za - p zb + p z + p zd Eigenvetor 1: E= α Ψ 1 = p za + p zb + p z + p zd Heteroatoms Heteroatoms are introdued into the HMO matrix using two parameters, h and k. The diagonal element is the oulomb integral, whih for arbon is α. The off-diagonal elements are the resonane integrals, whih for diretly bonded arbon atoms are. The diagonal element for a heteroatom is hanged to α + h and the off-diagonal element for diretly bonded atoms is hanged to k. A table of h and k is given below. Atom Bond Type π eletrons for atom h k -= N -=N- (pyridine) N =-N< (pyrrole) N -N=N- (azo) O -=O (arbonyl) O =-O- (furan) F =-F l =-l Br =-Br S =-S- (thiophene) For example, arolein, A = B - =O D, has four total pi eletrons and the input matrix is:

4 4 The lowest energy orbital is Eigenvetor 1: E= The oeffiient for oxygen of shows that extra eletron density resides on the oxygen, beause of its larger eletronegativity. Another example is vinyl flouride, A = B -F, with four total pi eletrons and the input matrix: The two bonding MO s are: Eigenvetor 1: E= Eigenvetor 2: E= with the lowest energy orbital being primarily a lone-pair on fluorine and the next higher bonding orbital being primarily the double bond on the two arbons.

5 5 Homework You will be assigned one of the following moleules. Ions and free radials an also be studied: H + H 2 H H 2 + H 2 H 2 H 2. Please answer the questions that follow. 1. Give the Hükel determinant in terms of x s and 1 s: 2. Determine the energies and the orbital oeffiients using a matrix diagonalization program. List those below.

6 6. Sketh the orbitals below with the appropriate phase for eah p orbital. 4. Give the number of nodes in eah wavefuntion. lassify eah orbital as bonding or antibonding. 5. Draw the energy level diagram, below. Give the eletron filling.

7 7 6. alulate the π-bond order for eah unique bond in the moleule. P ij = n k ki kj k 7. alulate the π-bond deloalization energy. (Remember for eah isolated double bond with two eletrons, E = 2(α+). ) 8. alulate the π-eletron density on any two atoms of your hoosing. 2 d i = n k ki k 9. On the energy level in part 5, indiate the lowest energy eletroni transition with a vertial arrow. Label the HOMO and LUMO (or SOMO if you have a radial).