It's my understanding that for a particle in a 1-D box, the change in energy from the HOMO to LUMO can be approximated with the following equation: $$\Delta E=\frac{h^2 ( N + 1 )}{8 m_\mathrm{e} L^2}$$ I am told that $N$ represents the number of π electrons in the molecule.

If you're studying a molecule like the one shown below and you want to predict the absorption wavelength from UVVIS, would you count all of the π electrons, or just the conjugated ones? It doesn't make much sense to me why you'd want to include the π electrons from the phenyl group next to the neutrally charged Nitrogen because it's not conjugated with the rest of the molecule to the same extent the other phenyl group is, and thus seems like it would be grossly inaccurate to assume that those π electrons contribute the same as the others.

image of dye molecule with extended pi system

  • 2
    $\begingroup$ I disagree with your assertion that this molecule is not fully conjugated. I believe both phenyl rings are coplanar with each other and conjugated together. Don't believe me? Draw the resonance structure where the uncharged donates electrons into the 2-position, shifting the methylene pi-bond to the 2' position, dumping a lone pair back on to the originally charged nitrogen. More importantly, it can't be resolved without a crystal structure. This is one of those times where you'd have to calculate several options and let experiment tell you which is correct. $\endgroup$
    – Lighthart
    Feb 3, 2016 at 22:26
  • $\begingroup$ Yes I just thought of that. I was thinking that the sulfur and nitrogen are sp3 hybridized without an open p orbital, but I forgot about rehybridization. Thank you makes a lot of sense. So I'm counting 16 pi electrons, using the equation I get a theoretical absorption wavelength of 272 nm. Experimental is 425. Seems too far off $\endgroup$
    – Plex ASM
    Feb 3, 2016 at 22:31
  • 1
    $\begingroup$ Molecule is symmetric and only ethyl groups' carbons don't participate in conjugation (nitrogens and sufurs do) . BTW I cant see how 1-D box would be of much use here. $\endgroup$
    – Mithoron
    Feb 3, 2016 at 22:36
  • $\begingroup$ And then Mithoron's comment is great. You'd have n-pi* transitions for this molecule $\endgroup$
    – Lighthart
    Feb 3, 2016 at 23:53

1 Answer 1


In a very basic first order approximation you can treat dyes with an extended conjugated π-system with the one-dimensional particle in the box approach. I simplified your system and calculated it at the DF-BP86/def2-SVP level of theory. The length of the box is about $L=1.20~\mathrm{nm}$, the following graphic shows it in Ångström.

box of dye

With that there are a number of problems associated. Counting the number of electrons in the π-system should not be among them. Just assume that all non-hydrogen atoms are sp² hybridised. Then count the electrons that occupy π/p-type orbitals. Here is an image of the lowest occupied π-type orbital of that molecule; you can see that the box stretches over the whole molecule. The thumbnail next to it gives you all occupied π-type orbitals and the lowest unoccupied molecular orbital.

lowest occupied pi-type molecular orbital of the dye  all occupied pi-type molecular orbitals of the dye

In this example each phenyl ring contributes six π electrons (12), two sulfurs contribute two each (4), the nitrogen double bond (2), the carbon-carbon double bond (2) and the last nitrogen (2). That makes a total of 22 π electrons.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.