Hückel Molecular Orbital (HMO) Theory

Transcription

1 Hückel Molecular Orbital (HMO) Theory A simple quantum mechanical concept that gives important insight into the properties of large molecules Why HMO theory The first MO theory that could be applied to large molecules. A theory that can be implemented without the aid of a computer. Has been the starting point for very successful theoretical interpretation of electronic (UV-vis) spectra of conjugated and aromatic hydrocarbons. Has been the starting point for more advanced theories. 1

2 Important conjugated and aromatic H compounds 3 Benzene arotene H 3 N N H 3 Mg N N H 3 H 3 O 2 H 3 O O O H 3 H 3 hlorophyl H 3 3 Photosynthetic apparatus Molecular electronics OH 3 S H 3 O S H 3 F F OH 3 F F F F H 3 Graphene H 3 O Molecular memory Image of graphene taken with TEAM 0.5 NANO LETTERS, 2008, 8,

3 The basis of the Hückel approach 1 2p orbital Ethene σ-skeleton and unhybredized 2p-orbitals σ-skeleton The geometries and special properties of conjugated and aromatic compounds may be interpreted in terms of the sp 2 - hybridization. The unhybridized 2p AOs perpendicular to the sp 2 hybride orbitals are available for the formation of the π-electron structure. The σ-electrons are disregarded, except for their role in establishing the molecular geometry. S. M. Lindsay Introduction to Nanoscience pp B.H. Bransden and.j. Joachain Physics of Atoms and Molecules 2 nd Edn. pp The basis of the Hückel approach 2 H H H H H H Benzene σ-skeleton There is no bonding interaction between σ- and π-orbitals in planar conjugated molecules. The σ-bonds between the sp 2 -hybridized atoms and their neighbours (the σ-orbitals) lie in the molecular plane. The σ-orbitals are symmetrical with respect to reflection in the plane of the molecule. The unhybridized 2p AOs are perpendicular to the molecular skeleton made up by the sp 2 hybride atoms. Any MO (π-orbital) formed from them will therefore be antisymmetric with respect to reflection in the plane of the molecule. The two types of electrons move in different regions of space: the σ- electrons occupy the space in the plane of the conjugated molecule, the π- electrons are located above and below the that plane. (The energies of the σ-electrons are on the whole lower than those of the π-electrons. The σ-electrons are involved in strong bonds and the π- electrons in weaker ones.) 3

4 Hückel theory i j 2p z (φ) orbital 1 2 x a b y n The total π-electron energy E π is: = = Where r I is the number of electrons in molecular orbital I. =1 Hückel theory i j 2p z (φ) orbital 1 2 x a b y n The problem: find the set of π-electron MOs (Ψ I ) formed as linear combinations of the 2p z (φ) orbitals at the atoms in the conjugated system. = = The variation theorem gives that the best wave function is the one that gives the lowest energy E I. S. M. Lindsay Introduction to Nanoscience pp B.H. Bransden and.j. Joachain Physics of Atoms and Molecules 2 nd Edn. pp

5 Expectation value of the MO energy ε I The expectation value of the energy ε I is: = H = H = H =1 =1 =1 =1 = =1 H =1 =1 =1 =1 =1 =1 =1 Where = H is the resonans or interaction integral and = is the overlap integral. Rearrange and drop the index I since the equation applies to any of the π molecular orbitals = =1 =1 =1 =1 Variational approach To find the minimum energy, differentiate both sides of the expression for ε I with respect to the coefficients c a, where a runs from 1 to n. Rearrange and differentiate =1 =0 = =1 = 0, =0, thus + = = The derivative will be non-zero only for those terms where j = a. S. M. Lindsay Introduction to Nanoscience pp 63 and

6 = Take the complex conjugate of each side. But =, = = = Rearrange. = ( ) =0 =1 There are n simultaneous equations, one for each coefficient a set of homogeneous linear equations the secular equations. Secular equations There are n simultaneous equations, one for each coefficient a set of homogeneous linear equations the secular equations. 1 ( ) + 2 ( ) + + ( 1 1 ) =0 1 ( ) + 2 ( ) + + ( 2 2 ) =0 1 ( 1 1 ) + 2 ( 2 2 ) + + ( ) =0 Rewritten in matrix notation or ( ) ( ) ( ) ( ) ( 1 1 ) ( 2 2 ) where is a column vector. ( ) = 0 1 ( 1 1 ) ( 2 2 ) 2 =0 ( ) 6

7 Hückel s assumptions The overlap integral S ij is a measure of the extent of overlap between the 2p z AOs on carbon atoms i and j. i = j: S ji = 1, because the overlap of an AO with itself is unity for a normalized wave function. i j: S ij = 0. A drastic assumption* The resonance integral H ij represents the energy of a single electron moving in the average field of all the nuclei, all the σ-electrons and all other π-electrons. i = j: H ii = α. The integral represents the energy of an electron in a free carbon 2p z AOs in a molecule. arbon i adjacent to j = i ± 1: H i(i ± 1) = β.** H ij negligable (= 0) if the two carbons i and j are separated by two or more bonds.** * The overlap integral for two 2p z orbitals, one each on two adjacent carbon atoms in a conjugated compound, is approximately It falls of rapidly with increasing distance. ** The molecular structure (connectivity) has to be known on beforehand. Apply Hückel s approximations The secular equations. Simplifies to: Equals: ( ) ( ) ( ) ( ) ( 1 1 ) ( 2 2 ) ( ) ( ) (/0 0) ( ) /0 ( ) ( ) (/0 0) ( ) /0 Divide by and define = ( ). ( 1 1 ) ( 2 2 ) 2 =0 ( ) (/0 0) (/0 0) 2 =0 ( 1) 1 1 /0 /0 2 =0 ( ) 1 7

8 Matrix algebra 1 1 1//0 1 1/0 1/0 2 = //0 1/0 0 1/ = //0 1/ 0 1/ =0 1 Extrapolate to solve for all orbital coefficients c Ij and orbital energies ε I [all x(n)] for all n π-orbitals simultaneously //0 Topology matrix 1/ 0 1/ =0 1 Extrapolate to solve for all orbital coefficients c Ij and orbital energies ε I [all x(n)] for all n π-orbitals simultaniously //0 Topology matrix T or 1/ 0 1/ ( ) = = I 0 (1) 0 0 (1) 1 (2) 1 () (2) 0 (1) 2 (2) 2 () 2 = () (1) (2) () Eigenvalue matrix X Solved according to Eigenvector matrix = 8

9 Butadiene as an example H H 1 2 H H 3 4 H H no S 11 = 1 H 11 = α 2 S 21 = 0 H 21 = β 3 S 31 = 0 H 31 = 0 4 S 41 = 0 H 41 = 0 S 12 = 0 H 12 = β S 22 = 1 H 22 = α S 32 = 0 H 32 = β S 42 = 0 H 42 = 0 The topology matrix: S 13 = 0 H 13 = 0 S 23 = 0 H 23 = β S 33 = 1 H 33 = α S 43 = 0 H 43 = β S 14 = 0 H 14 = 0 S 24 = 0 H 24 = 0 S 34 = 0 H 34 = β S 44 = 1 H 44 = α = The π-orbital energies = (α ) = ± and ± = ± and ± ε 4 = α β ε 3 = α β Antibonding orbitals ε 2 = α β ε 1 = α β Bonding orbitals Next, determine the shape of the π-orbitals 9

10 Orbital coefficients Ψ I oefficient Ii for carbon number I Energy α β α β α β α β β is a negative quantity representing the (π) bonding energy between two adjacent carbon atoms. The order of the energy levels follows the number of nodes. The term α is common to all energies. It is the energy of a non-bonded 2p z electron IN the molecule. MOs of energy of lower energy than α are bonding and those of energy greater than α are antibonding. E Orbital shapes 10

11 Add in the electrons Total energy is equal to the sum of orbital energies times the occupation = E = =1 Add in electrons according to: 1. Aufbau principle 2. Pauli principle 3. Hund s rule Each carbon contributes with one electron. The stabilization due to the π-electrons is: 4α 2(α β) 2(α β) = β S. M. Lindsay Introduction to Nanoscience pp and 58. Linear polyenes General solutions i j =+2 ( +1) yclic polyenes 1 2 x a b y n, = 2 1/2 +1 ( +1) j= atom number in linear π-system. I = 1, 2, 3, n (MO number). n= total number of atoms in π-system 2( 1) =+2, = 1 1/2 1)( 1) 2( Nb (i) 2 = 1 I = 1, 2, 3, n (MO number) 11

12 Energy General solution yclic polyenes α 2β α β αβ α + β α + 2β π 4π/3 5π/3 Phase π 2π/3 π/3 0 π/3 2π/3 π angle =+2 2 n even: =0,±1,±2,, 2 n odd: = 0, ±1, ±2,, ( 1) 2, = 1 1/2 Nb (i) 2 = 1 1) 2( E Orbital shapes 12

13 yclic polyenes Benzene 13

14 Aromaticity For a molecule to experience the stabilization of aromaticity it must fulfill the following requirements: It must contain a cyclic sequence of atoms. All the atoms of the cyclic sequence must lie in the same plane, or nearly so. Each atom in the cyclic sequence must contribute with a p-orbital aligned perpendicular to the plane of the cyclic structure. The cyclic, conjugated system must contain (4n + 2) electrons. Parameterized method The parameter β is composite parameter representing the electron-nucleus attraction, the interelectronic repulsion and the kinetic energy. The atomic 2p energy level for carbon: E(2p Z )=α (~ ev) The resonance integral for adjacent 2p orbitals H ij (=) = β (~ -3 ev) 14