# How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach?

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.

• 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. – Lighthart Feb 3 '16 at 22:26
• 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 – Plex ASM Feb 3 '16 at 22:31
• 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. – Mithoron Feb 3 '16 at 22:36
• And then Mithoron's comment is great. You'd have n-pi* transitions for this molecule – Lighthart Feb 3 '16 at 23:53

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.