CHE-30042 Inorganic, Physical & Solid State Chemistry Advanced Quantum Chemistry: lecture 3 Rob Jackson LJ1.16, 01782 733042 r.a.jackson@keele.ac.uk www.facebook.com/robjteaching.

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CHE-30042 Inorganic, Physical & Solid State Chemistry Advanced Quantum Chemistry: lecture 3 Rob Jackson LJ1.16, 01782 733042 r.a.jackson@keele.ac.uk www.facebook.com/robjteaching @robajackson

che-30042 Advanced QC lecture 3 Lecture 3 contents Secular determinants revised Overlap, resonance and Coulomb integrals Introducing the Hückel approximation Applying the Hückel approximation to conjugated hydrocarbons Delocalisation energy Aromatic systems che-30042 Advanced QC lecture 3

Secular determinants revised From lecture 2 we wrote the secular determinant for a diatomic molecule as: H11 -  S11 H12 -  S12 = 0 H21 -  S21 H22 -  S22 This can be expanded to give a quadratic equation that can be solved to give two values of , one for each orbital, provided we know H and S. che-30042 Advanced QC lecture 3

Overlap, resonance and Coulomb integrals A reminder of the integral expressions Slk =  l k d (Overlap integral) Their value depends on the degree of overlap of the orbitals. (Remember they are = 1 if l = k) Hll =  l H l d (Coulomb integral) The energy an electron would have in a particular orbital. Hlk =  l H k d (Resonance integral) = 0 if there is no overlap. che-30042 Advanced QC lecture 3

The Hückel approximation The main problem with molecular orbital calculations is associated with calculating the integrals introduced in the last lecture, and revised on slide 4.. Later we will see how to try to calculate them in full. The Hückel approach neatly sidesteps the problem by making assumptions about the values of the integrals, which are listed on the next slide: che-30042 Advanced QC lecture 3

Approximating the integrals In the Hückel method it is assumed that: All overlap integrals, Slk = 0 unless l = k, when they = 1 All Coulomb integrals, H11, H22 etc. are set to  All resonance integrals between neighbouring atoms (e.g. H12, H23) are given the value  (normally negative). All resonance integrals between non-neighbouring atoms (e.g. H13, H14) are set to 0 Enables secular determinants to be solved more easily! che-30042 Advanced QC lecture 3

Background: molecular orbitals for conjugated hydrocarbons Conjugated hydrocarbons are characterised by having a mix of  and  bonding; the  bonds are responsible for holding the molecule together, while the  bonding is delocalised, with the  electrons free to move around the molecule. The Hückel approach (Hückel, 1930) provides a more accurate description of  bonding. Hückel, E. (1930) Zeitschrift für Physik 60 (7–8): 423–456. doi:10.1007/BF01341254. che-30042 Advanced QC lecture 3

che-30042 Advanced QC lecture 3 Applying the Hückel approximation to simple hydrocarbons containing  electrons Although the Hückel approximation is most useful for conjugated and aromatic hydrocarbons, we will apply it to a simple example first to illustrate how it works for localised  orbitals. We will apply it to ethene, remembering that we are only looking at the  molecular orbitals, which result from overlap of the pz orbitals on each C atom. che-30042 Advanced QC lecture 3

Wavefunctions and energies of Hückel orbitals of ethene – (i) We start with a wavefunction in terms of 1 atomic orbital on each C atom:  = c1 1 + c2 2 Following the procedure explained in lecture 2, we obtain the energy by solving the secular determinant: H11 - E S11 H12 - E S12 = 0 H21 - E S21 H22 - E S22 But the determinant is easier to solve making the Hückel approximation. -E  = 0  -E che-30042 Advanced QC lecture 3

Wavefunctions and energies of Hückel orbitals of ethene – (ii) Expanding the determinant gives a quadratic equation: ( - E)2 = 2 The solutions to this (obtained by expanding the equation and applying the quadratic formula) are: E=    These are the energies of the orbitals. What about the wavefunctions? The secular equations are (slide 15 lecture 2): c1 ( - E) + c2  = 0, c1  + c2 ( - E) = 0 che-30042 Advanced QC lecture 3

Wavefunctions and energies of Hückel orbitals of ethene – (iii) If we substitute the two possible values for E in either of these equations, we find the following: When E =  + , c1 = c2, so b = c1 1 + c1 2 When E =  - , c1 = -c2, so a = c11 - c22 Here ‘b’ and ‘a’ denote bonding and antibonding orbitals. They will be sketched in the lecture, or you can find a better diagram in Hayward p 172. However, the Hückel method was really devised for larger molecules, particularly with alternating double and single bonds. che-30042 Advanced QC lecture 3

Application of the Hückel approach to butadiene First we note that the framework of the molecule is held together by  bonds formed by overlap of sp2 hybrid orbitals on the C atoms, and that further  bonds are formed by overlap of C sp2 orbitals and H 1s orbitals. The 2pz orbitals on the C atoms are oriented perpendicular to the plane of the molecule, and overlap to form  orbitals. The Hückel approach is used for these orbitals. che-30042 Advanced QC lecture 3

The Hückel approach to -bonding in butadiene – (i) We form a molecular orbital from the 2pz orbitals on the 4 C atoms:  = c1 1 + c2 2 + c3 3 + c4 4 This leads to 4 secular equations: c1(H11 - ES11) + c2(H12 - ES12) + c3(H13 - ES13) + c4(H14 - ES14) = 0 c1(H21 - ES21) + c2(H22 - ES22) + c3(H23 - ES23) + c4(H24 - ES24) = 0 c1(H31 - ES31) + c2(H32 - ES32) + c3(H33 - ES33) + c4(H34 - ES34) = 0 c1(H41 - ES41) + c2(H42 - ES42) + c3(H43 - ES43) + c4(H44 - ES44) = 0 Now form the secular determinant! che-30042 Advanced QC lecture 3

The Hückel approach to -bonding in butadiene – (ii) This leads to a secular determinant: H11 - ES11 H12 - ES12 H13 - ES13 H14 - ES14 = 0 H21 - ES21 H22 - ES22 H23 - ES23 H24 - ES24 H31 - ES31 H32 - ES32 H33 - ES33 H34 - ES34 H41 - ES41 H42 - ES42 H43 - ES43 H44 - ES44 Simplify the determinant, making the approximations on slide 6. Pay particular attention to resonance integrals. Hayward (p 168) then shows how the determinant is solved, and energies and weighting coefficients determined. Do the simplification step on the board. che-30042 Advanced QC lecture 3

Delocalisation energy in butadiene The total  electron energy in butadiene (Hayward p169 fig. 8.26) is: 2 x ( + 1.62) + 2 x ( + 0.62) = 4 + 4.48 Compare this with the  electron energy of 2 ethene molecules (slide 8): 2 x 2 ( + ) = 4 + 4 Because  is negative (slide 6), this means that the  electron energy in butadiene is lower than the 2 localised  bonds in ethene by 0.48. This is called the delocalisation energy. Explain meaning of delocalisation energy here! che-30042 Advanced QC lecture 3

Hückel theory applied to aromatic molecules The stability of aromatic molecules can be related to the delocalisation of the  electrons. Taking benzene as an example, hybridisation of the C orbitals gives sp2 orbitals which overlap to give  bonding within the plane of the molecule (Hayward figure 8.30). The C 2pz orbitals are at 90 and form a separate molecular orbital: Show how the theory is applied to benzene http://en.wikipedia.org/wiki/Aromaticity che-30042 Advanced QC lecture 3

Delocalisation in aromatic molecules Following the procedure on slide 9, we write the Hückel molecular orbitals as:  = c1 1 + c2 2 + c3 3 + c4 4 + c5 5 + c6 6 The secular determinant is written, using the notation on slide 4: -E  0 0 0  = 0  -E  0 0 0 0  -E  0 0 0 0  -E  0 0 0 0  -E   0 0 0  -E Note the difference with butadiene in that there are now 2 extra  terms to account for H16 and H61 because in the ring they are connected. che-30042 Advanced QC lecture 3

Delocalisation energy in benzene – (i) Expansion of this secular determinant gives 6 roots which give the energies of the 6 Hückel orbitals in benzene. They are shown on p 174, fig. 8.32 of Hayward The energies are:   2,   ,    We can compare the energy of the delocalised orbitals with localised ones, as we did for butadiene: che-30042 Advanced QC lecture 3

Delocalisation energy in benzene – (ii) In benzene, the 6  electrons are in the  + 2, and doubly degenerate  +  levels, so the energy is 2 x ( + 2) + 4 x ( + ) = 6 + 8 Compare with 3 delocalised  bonds whose energy is 3 x (2 + 2) = 6 + 6 This gives a delocalisation energy of -2. Again, the energy is lowered because the  electrons are dissociated over the whole molecule. che-30042 Advanced QC lecture 3

che-30042 Advanced QC lecture 3 Lecture summary The Hückel approximation has been introduced and applied to: Simple hydrocarbons Conjugated hydrocarbons Aromatic hydrocarbons Delocalisation energy has been defined and calculated for conjugated and aromatic hydrocarbons. che-30042 Advanced QC lecture 3