The experimental observed wavelength for this molecule is 425 nm.

We are supposed to estimate this wavelength theoretically using this equation (energy change in a 1D box), where $N$ is the number of pi electrons. $$\Delta E=\frac{h^2(N+1)}{8m_\mathrm{e}L^2}$$

$L$ is given as 11.84 angstroms.

I'm getting a wavelength of 200 nm. It seems to be too far off.

Molecule: enter image description here

I'm counting 22 pi electrons and using that as N

This question is different from 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? in that this question is specifically addressing the discrepancy between the theoretical and experimental wavelengths, whereas the previous question was specific to the number to use for $N$.

Also how does this even make sense? As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as $N$ increases as does the energy gap. To me it makes no sense, how can it be explained?

  • 2
    $\begingroup$ Possible duplicate of What pi electrons to include when calculating Energy change $\endgroup$
    – Mithoron
    Feb 3, 2016 at 22:54
  • 1
    $\begingroup$ No the previous question was specifically addressing what to use for N. $\endgroup$
    – Plex ASM
    Feb 3, 2016 at 22:54
  • $\begingroup$ Why did you post it again?! You can edit and any additional info $\endgroup$
    – Mithoron
    Feb 3, 2016 at 22:55
  • $\begingroup$ This question is specifically asking why there is such a large discrepancy between the theoretical and experimental wavelengths. The previous question was asking only about how many pi electrons to include for N $\endgroup$
    – Plex ASM
    Feb 3, 2016 at 22:55
  • 2
    $\begingroup$ Well you have wrong number for starters and you should simply edit old stuff to include it. This is still about the same task. $\endgroup$
    – Mithoron
    Feb 3, 2016 at 23:00

3 Answers 3


The energy level of the particle in a one-dimensional box are given by $$E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}.\tag1$$ We see that the energy is quantised and dependent on the quantum number $n$. We also see that the energy difference between those levels gets larger with $n$. The only variable in this equation is the length of the box $L$. The larger the box the denser the states.

For an extended pi system like the ones in the dyes we make various simplification. First and foremost we reduce a multidimensional problem to one dimension. We also treat the barriers of the box as infinitely large potentials. Then we can only estimate the length of the pox. We are completely ignoring the fact that the molecule will have vibrations. And there are a couple of more assumption we make. All of them will influence the accuracy of the calculation. You should discuss these issues thoroughly and come to your own conclusions.

pH13 outlined a procedure that has been established in the literature, which neatly correlates the experimental findings with the calculations.[1] I personally find this method a little bit sketchy and not very intuitive at all. My first issue is the choice of the box itself; it seems rather arbitrary to me, as they decide to ignore the phenyl rings, add half a length of a bond, choose the average bond distance to be the one of benzene.

In this case, the box extends from nitrogen to nitrogen plus one-half of a bond length beyond each nitrogen. That is, $L = (p + 3)l$, where $p =$ the number of carbon atoms in the chain and $l$ is the average bond distance in the chain. The number of electrons in the chain, $N$, is $p + 3$.
[...] we let $l = 1.39 \times 10^{-8}~\mathrm{cm}$ (the bond distance in benzene) [...]

For a slightly simpler model I have estimated the length of the box to be $L=1.20~\mathrm{nm}=1.2\times10^{-9}$ and the number of π electrons to be $N= 22$.
With the reasonable assumption that orbitals are doubly occupied, we can hence find the the quantum number of the energy level of the HOMO $E_\mathrm{HOMO}$ to be $n_\mathrm{HOMO}=N/2$ and in analogy to that for $E_\mathrm{LUMO}$ as $n_\mathrm{LUMO}=N/2+1$. The energy of the photon to excite one electron from the HOMO to the LUMO is in first approximation it's difference $\Delta E_\mathrm{gap}$. Therefore we will find the formula you used after some transformations. \begin{align} \Delta E &= E_\mathrm{LUMO} - E_\mathrm{HOMO}\\ &= \frac{n_\mathrm{LUMO}^2 h^2}{8m_\mathrm{e}L^2} - \frac{n_\mathrm{HOMO}^2 h^2}{8m_\mathrm{e}L^2}\\ &= \frac{h^2}{8m_\mathrm{e}L^2}\left[n_\mathrm{LUMO}^2 -n_\mathrm{HOMO}^2\right]\\ &= \frac{h^2}{8m_\mathrm{e}L^2}\left[\left(\frac{N}{2}\right)^2 -\left(\frac{N}{2}+1\right)^2 \right]\\ &= \frac{h^2}{8m_\mathrm{e}L^2}\left[N+1\right]\tag2\\ \Delta E &= \frac{hc}{\lambda}\\ \lambda &= \frac{8m_\mathrm{e}cL^2}{h(N+1)}\tag3 \end{align}

Plugging in the data, we find $\Delta E = 5.965~\mathrm{eV}$ and $\lambda=207~\mathrm{nm}$. Well, that's almost the same as you got, and that is very disappointing. Could it be, that there are a couple of approximations, that do not work well?

Farrell also admits that his approach is flawed:

Agreement between predicted transition energies and experimental energies, Table 1, is not good (although the agreement is rather surprising in view of the approximations made).

Therefore a parameter $\gamma$ which is adjusted to give a good fit to the experimental results is introduced. The length of his box becomes $L=p+3+\gamma$.
Elongating the conjugated bridge between the nitrogen atoms, lengthening the box, will result in a better fit, even without that adjustment. In other words, the less important the end groups (phenyl moieties) become, the better the simple model performs. That is not surprising at all. The following table is reproduced from that paper (with slight modification). The number of carbons in the chain is $p$.

\begin{array}{lcccc} p & \lambda_\mathrm{exp}/\mathrm{nm} & \Delta E_{\mathrm{exp}}/\mathrm{eV} & \Delta E_{\mathrm{th.}\, \gamma=0}/\mathrm{eV} & \Delta E_{\mathrm{th.}\, \gamma=0.68}/\mathrm{eV}\\\hline 3 & 422 & 2.94 & 3.78 & 3.05\\ 5 & 555 & 2.23 & 2.74 & 2.32\\ 7 & 650 & 1.91 & 2.14 & 1.88\\ 9 & 760 & 1.63 & 1.76 & 1.57\\ \end{array}


The failure of the model for short chains is not surprising. It will get better the longer the box is and the less influential the end moieties are. This system especially is well suited for discussions of such approximations and gain a basic understanding of quantum chemistry. Stay sceptical.


  1. John J. Farrel J Ed Chem 1985, 62, 351-352 [mirror]


As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as N increases as does the energy gap. Makes no sense

It actually does. The more particles you fill in such a well, the higher the energy of the same ensemble gets. What makes the energy gap smaller and the states more dense is the extension of the actual box. The larger $L$ the closer are the states together. If the box size grows larger faster than the electrons you fill in it, then the gap shrinks. The gap is linear dependent on $N$, but inverse quadratic in $L$.


Luckily I found a source containing your molecule as an example.

You need your equation for the energy difference between the HOMO and the LUMO, which is (nearly) defined as you wrote: $$\Delta E = \frac{(N+1)h^2}{8 m_e L^2}$$

As you want a wavelength, you need to convert the energy difference, which yields then: $$\lambda = \frac{8m_ecL^2}{h(N+1)}$$

Now they state, that the length of the box can be guessed by the amount of carbons (p) between both nitrogens by using this formula: $$L = (p+3+\alpha)l$$ ... with $l$ being the length of typical benzene-C-C-bonds, thus being 0.139 nm and $\alpha$ is an empirical correction factor for polarizable end groups.

The amount of pi-electrons (N) that you need to inlude can be calculated by a similar equation: $$N = p+3$$ ... where the additional "+3" come from both nitrogens (would be actually +4, but it's +3 because one nitrogen atom had to form an ion).

This leads to their equation 4: $$\lambda = 63.7\frac{(p+3+\alpha)^2}{p+4}$$ ... where all the physical constants have been combined to give 63.7.

As you have three carbon atoms between both nitrogens, you come up with 328 nm and when you include their "correction factor for polarizable end groups" of $\alpha = 0.675$, you come up with 405 nm, which is more than close to the experimental result.

When I use your box length and use 12 electrons (3 from the "left" benzene ring, 4 from the two double bonds to the "right" nitrogen" and 1 from the lone pair of this nitrogen), I come up with about 356 nm, which is not that far away as it sounds. In eV this is an error of 20% (the 405 nm are wrong by about 5%), whereas your 200 nm would have been an error of 120%!


When one "arbitrarily" insists that the box is of a certain length, that is where any method fails. The molecular geometry provides the boundary conditions on the standing waves for the electrons, so people fix things by adding an extra length (fudge factor) to account for the discrepancy in the model. So, the Kekule structure(s) (Lewis structure(s)) people start with to model the box length are very approximate.

So, can we make a better approximation without having to account for every electron? Yes, but we need to include those electrons in the molecule that are leading to the measured "characteristic length". What is that thing, but the "conjugation length". You are gonna say, "the whole thing is conjugated", but look closer.

We all learn that conjugation is alternating double and single bonds right, but what is that here and how does it help? Although the conjugation method is limited too, it does offer a better approximation to a "length" since it starts with a good fragment (stable "vinylic like CH2=CH-" chunks).

Use what you know about MO's from organic chemistry and try assigning phases to the Kekule structures and see what you need to do to make the longest length of alternating (++)_(--) units. That is, try alternating phase pairs of the pi orbitals and see which configuration(s) gives the longest length. You may need to change the number of alpha and beta phases that appear on one side of the molecule to get the longest length (flip a pi-bond over to get its phase out of the way of conjugation path). How many bonds long is that, and is that an integer multiple of the measured length (or a reasonably close approximation to an integer multiple of a bond length)? Compare that best guess and shape to what the HOMO and LUMO look like. What do you think conjugation is now and how does the end group affect conjugation? Cheers.

  • $\begingroup$ I appreciate that you are trying to get the OP to think through what the answer would be, but it would be preferable to add a definitive answer at the end so that others could benefit from this in the future. Thanks! $\endgroup$
    – jonsca
    Sep 29, 2019 at 18:41
  • $\begingroup$ Better than presenting the wrong answer such as that with pure fudge factors that are erroneously repeated as some kind of understanding or knowledge; L=(p+3+α)l...This equation is totally arbitrary and without any physical significance as it is completely contrived. (tinkered) The fact that it approximates the solution is what the student should be thinking about. Eventually the student should consider Huckel MO theory and phases of p orbitals. Thank you. $\endgroup$
    – LDF
    Jan 16, 2020 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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