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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.

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Presentation on theme: "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."— Presentation transcript:

1 CHE-30042 Inorganic, Physical & Solid State Chemistry Advanced Quantum Chemistry: lecture 3
Rob Jackson LJ1.16, @robajackson

2 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 Advanced QC lecture 3

3 Secular determinants revised
From lecture 2 we wrote the secular determinant for a diatomic molecule as: H11 -  S11 H12 -  S = 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 Advanced QC lecture 3

4 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 Advanced QC lecture 3

5 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 Advanced QC lecture 3

6 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 Advanced QC lecture 3

7 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: /BF che Advanced QC lecture 3

8 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 Advanced QC lecture 3

9 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 Advanced QC lecture 3

10 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 Advanced QC lecture 3

11 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 Advanced QC lecture 3

12 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 Advanced QC lecture 3

13 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 Advanced QC lecture 3

14 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 Advanced QC lecture 3

15 Delocalisation energy in butadiene
The total  electron energy in butadiene (Hayward p169 fig. 8.26) is: 2 x ( ) + 2 x ( ) = 4  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 Advanced QC lecture 3

16 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 che Advanced QC lecture 3

17 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  -E  0  -E  0 0 0 0  -E  0  -E    -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 Advanced QC lecture 3

18 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 of Hayward The energies are:   2,   ,    We can compare the energy of the delocalised orbitals with localised ones, as we did for butadiene: che Advanced QC lecture 3

19 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 Advanced QC lecture 3

20 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 Advanced QC lecture 3


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