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I am trying to calculate the energy levels of biphenyl (see picture below) using Hückel method. Since this is my first time using Hückel I am a little confused.

enter image description here

Biphenyl has 12 carbon atoms, therefore I should get a 12x12 matrix like the one below.

\begin{array}{cccccccccccc} \alpha-E & \beta & 0 & 0 & 0 & \beta & \beta & 0 & 0 & 0 & 0 & 0 \\ \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 \\ \beta & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 \\ \beta & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & \beta \\ 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta \\ 0 & 0 & 0 & 0 & 0 & 0 & \beta & 0 & 0 & 0 & \beta & \alpha-E \end{array}

Is this matrix correct for biphenyl, if I neglect dihedral angles etc.?

If I take det(Matrix)=0 and solve this for the Energy term I should get several solutions. How can I assign these solutions to each energy level correctly?

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    $\begingroup$ The eigenvalue of the solution is the energy of the orbital and the eigenvector itself tells you the components in that solution. There's no work to do for assignment. $\endgroup$ – Zhe Aug 10 '18 at 18:53
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    $\begingroup$ For a planar molecule this looks ok. If the molecule is twisted then remove the $\beta$ connecting atom 6 and 7 as now conjugation will be removed. The parameter $\beta$ is a negative number so the lowest energy is will be of the form $\alpha + C\beta$ where coefficient $C$ has the largest positive value. Note that some levels may be degenerate. The wavefunction with the lowest eigenvalue has the fewest nodes, the highest energy one has the most nodes between atoms. $\endgroup$ – porphyrin Aug 11 '18 at 8:28
  • $\begingroup$ @porphyrin Thanks for your comment! I have another question: If I want to calculate the excitation energies for 1->2 then I have to fill all energy levels with the 12 electrons and then I build the difference between HOMO and LUMO right? $\endgroup$ – p_punkt Aug 11 '18 at 20:26
  • $\begingroup$ Yes add all the electrons them move one from homo to lumo for the lowest possible excitation. $\endgroup$ – porphyrin Aug 12 '18 at 7:37
  • $\begingroup$ @porphyrin If I want to calculate the second lowest transition it would be HOMO->LUMO+1, right? $\endgroup$ – p_punkt Aug 12 '18 at 14:30

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