On writing down the Hamiltonian for the particle-in-a-ring problem, and solving the Schrodinger to get the quantized energy in terms of the radius (R) of the ring, for the two examples of Benzene & Coronene.
Why is it acceptable during calculating the excitation energy, like in Benzene, to deal with the model as the the energy is driving the ring from quantum state (n=1) to (n=2)? I mean why are the pairs distributed as: 1) One pair in n=0 2) Two pairs in n=1 ? And similar for Coronene, up to n=6.
Maybe i lead to similar question but now comparatively for Benzene and Coronene & β-Carotene: Why do the distribution of the pairs begin from state (n=1) to (n=11) sequentially each level is filled only by one pair, in the case of β-Carotene, and for the other two examples from (n=0) with one pair, and up to any states by two pairs.
Hopefully, it is clear enough.
Unfortunately the question is not so clear
However, the presence of what you call the "pair" in the electronic structure for benzene is well known: in a huckel model, there are 6 pi-electrons on benzene and when you diagonalise the hueckel matrix for benzene, you get 3 doubly occupied electronic states: there is the lowest energy state and then two HOMOs that are doubly degenerate due to the symmetry in benzene (D6h group). This is well known for aromatic ring systems.
Coronene is a super benzene and from its symmetric structure will also show these degenerate states (https://www.google.co.uk/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0CAMQjxxqFQoTCN3ns46F28gCFQKFGgodBIoPjg&url=http%3A%2F%2Fpubs.rsc.org%2Fen%2Fcontent%2Farticlehtml%2F2009%2Fcp%2Fb904614c&bvm=bv.105841590,d.d2s&psig=AFQjCNFLQzs4xSy1g8W7PxMJvKvj8NV61A&ust=1445773743269494)
beta-carotene is not an aromatic ring so will not show these degenerate states
Thanks so much, Michael for your reply!
Sorry for the lack of clearness. But you got my point right, and yes i meant the degeneracy of the states, which exist in the cases of benzene and coronene. In other words, in terms of the latter molecules, what is the physical reason for why only one LUMO state and a doubly degenerate state, what is going to happen if all the electrons in the degenerate quantum state are photo-excited by E2-E1, will they all occupy the only one LUMO state?
For β-Carotene, why aren't the two conjugations at the two ends of the molecular structure exhibiting sort of degeneracy as well?
Zyad,
hope it helps.
Benzene HOMO and LUMO are both made up of two degenerate orbitals which is a consequence of the symmetry of benzene and for coronene. If you break the symmetry, that is lower the symmetry, you remove the degeneracy.
In terms of excitation, I guess you excite an electron from a HOMO into a LUMO, without identifying which of the degenerate orbitals you excited from and into. Possibly a TDDFT calculation of this would give some quantitative answers
The beta-carotene is conjugated but not in the way benzene is (not aromatic) - it has low symmetry and the terminal rings are not aromatic so this question does not arises. Conjugation and aromaticity are not always found together.
M
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