@riboswitch

I apologize for using terminology in my previous post that you had no reason to know. Also, there were one or two errors in my post that arose because I was working off of memory rather than actually taking the time to work through the math from the beginning. So here I'll set the record straight.

First, your answer:

Using the information provided, you did the problem correctly.

Commentary:

The free electron model of the porphyrin (particle-on-a-ring) was first solved by William Simpson in 1949 (J Chem Phys 17, 1218-21). Actually, porphyrins play a very important historical role in the development of quantum chemical theory. Simpson defined a porphyrin as an 18 member ring with 18 electrons.

(You will see that the ring of the porphyrin actually has 20 carbons, and 4 nitrogens. When considering the actual conjugation pathway, it is common to exclude four of the carbons and include 2 of the nitrogens, which is a shorter conjugation path. There are two mutually perpendicular ways to do this. I do not remember the reason that this is done, but it's probably because of the problem encountered in the free electron model when using 20 electrons.)

Back when I was writing my PhD dissertation, I pretty much reproduced every theoretical model there was on porphyrin electronic structure. My dissertation includes a 150 page introduction to this topic. When I reproduced the free electron method using 18 electrons, I calculated an energy separation for the lowest energy transition of 3.43 x 10

^{-19} J, which is very close to what you calculated. The agreement is coincidental. You are using a radius of 440 pm. I used a radius of 400 pm. These values derive from an estimate of the C-C bond length multiplied by the number of bonds (circumference) divided by 2 pi to get the radius. For instance, an 18 center porphyrin has 18 bonds. Taking benzene as a model with an average bond length of 139 pm, you arrive at a radius of 398 pm. Likewise, a 20 center ring gives a radius of about 442 pm. Back when Simpson wrote his paper, I presume the bond lengths of various organic molecules were not know with much precision. To estimate a radius to use in the free electron energy formula, Simpson "calibrated" his value with the absorption wavelength of benzene. You can read his paper if you want to know more about how he did this, but the end result is that he calculated an energy separation of 615 nm. The actual experimental absorption energy of porphyrin depends quite a bit on what substituents are on the macrocycle, and "plain" (unsubstituted) porphine isn't soluble in anything, so it's hard to know what to compare these values to. Simpson used bacteriochlorophyll, which has a transition "center of gravity" (more on that in a minute) of 549 nm, and he was pretty happy with the level of agreement.

Regardless of the precision of the calculated transition wavelength, using 20 carbons/electrons is very wrong for a more important reason, and that is that it predicts the HOMOs contain two unpaired electrons. If this were the case, porphryin would be magnetic, and it isn't. 18 electrons gives rise to a fully filled m = 4 level. All electrons paired: no magnetism. Before I used the formalism "triplet" and "singlet". This is called the multiplicity, which is a formalism of spectroscopy. The multiplicity (M) is just equal to 2S + 1, where S is the total electron spin. If you have two unpaired electrons, each with a spin of 1/2, the multiplicity is 3 (2*0.5 + 1), which we call a triplet. No unpaired electrons is a multiplicity of 1 (2*0 + 1), which we call a singlet. These designations are very important for fluorescence and phosphorescence, as you've indicated. The ground state of a porphyrin (with a closed shell metal center) is a singlet, meaning all electrons are paired, so the 20-electron system cannot be right, at least in the free electron model. In fact, virtually all organic molecules are ground state singlets. If they weren't, they'd very readily react with oxygen (a ground-state triplet) and life probably wouldn't exist. So if a model predicts a common organic molecule is a triplet in its ground state, the model probably isn't right.

Here's something else interesting about the free-electron model of porphyrin. You'll notice that there are two degenerate HOMOs, one with a m = 4 and one with m = -4. Both of these have the same energy, and they are both fully filled with two electrons. The m = 5 (and -5) level is empty. This means that the lowest energy transition between levels 4 and 5 is actually four fold degenerate. You can have a transition from m = 4 to m = 5, m = 4 to m = -5, m = -4 to m = 5 and m = -4 to m = -5. Each of these transitions has the same energy equal to what was calculated above.

(Quick note: using 20 electron system means in your ground state you have 1 electron in the n = 5 orbital and one electron in the m = -5 orbital. Here it becomes unclear what the lowest lying transition is. Is it a transition from the filled m = 4 orbitals to the partially filled m = 5 orbital? Or is it from the partially filled m = 5 orbital to the fully unfilled m = 6 orbitals?)

Let's define the electron configurations as follows: {w,x;y,z}, where the four electrons are in orbitals w, x, y and z. So the ground state configuration is { +4,+4;-4,-4} because we have two electrons in the m = 4 orbital and two electrons in the m = -4 orbital. There are, as I've said, four possible excited state configurations: {+4,+5;-4,-4}, {+4,-5;-4,-4}, {+4,+4;-4,-5}, and {+4,+4;-4,+5}. (I have a nice figure that illustrates this graphically, but I can't access image sharing sites from my ISP here; I'll upload later if I get a chance.) In the Zero Order approximation, each of these excited states have the same energy, because the energy only depends on the orbital energy, which only depends on the n value. Heretofore what we've basically been doing is calculating orbital energies using one electron, and then filling in the other electrons after the fact, assuming that putting all these extra electrons in the orbitals doesn't change anything. That's completely wrong! In reality, when you have multiple electrons in a system, the electrons interact with each other, and certain configurations will have more favorable interactions than other.

You may have heard by now of Hund's Rules, possibly even in general chemistry. According to these rules, states with the highest spin multiplicity are the lowest energy (all things being equal, spins like to be unpaired), followed by those with higher orbital angular momentum. Thus we can conclude that the four excited-states we predict in the free electron model don't in reality all have the same energy. Can we say more about this?

In the free electron model, all angular momentum is in the z-direction (perpendicular to the ring) and all wavefunctions are eigenfunctions of the angular momentum operators.

[tex]\hat{L}^2 = -\hbar^2 \frac {d^2}{d \phi^2}[/tex]

[tex]\hat{L_z} = i\hbar \frac {d}{d \phi}[/tex]

The wavefunctions from which the energy formula derives are

[tex]\psi(\phi) = (2 \pi)^{-1/2} e^{im\phi}[/tex]

If we take the electrons a functionally independent, the ground state configuration(s) then are found to have a total orbital angular momentum of 0, and the excited-state configurations have angular momentum of ± 1 and ± 9 (in factors of ħ; basically, just adding up all the possible m values).

Although it's difficult at this stage to quantify the effects of electron-electron interaction in the excited-states of the free-electron model, but we can order them qualitatively using Hunds rules. Accordingly, the states with high orbital momentum (L

_{z} = ± 9ħ) are predicted to be lower in energy (to some degree) than the states with low orbital momentum (L

_{z} = ± ħ). Therefore instead of having a single 4-fold degenerate transition, we actually will have two 2-fold degenerate transitions: a lower energy transition with an orbital momentum change of (L

_{z} = ± 9ħ) and a higher energy transition with an orbtial momentum change of L

_{z} = ± ħ.

Now, take a look at an experimental porphyrin spectrum. You can find one here:

http://www.thno.org/ms/getimage.php?name=thnov02p0916g03.pngYou will see that the spectrum includes a very intense peak at around 400 nm (5 x 10

^{-19 } J) and a series of weak transitions around 600 nm (3.3 x 10

^{-19 } J). The strong band is called the Soret or B-band and the weak transitions are collectively called the Q-bands. "Q" and "B" are historical designations; "Soret" is named after a person. Let's ignore the fact that there are multiple Q-band transitions (these are vibronic in origin and also due to the fact that two nitrogens in the free base form of porphyrin have hydrogens, which lowers the symmetry). The important thing is that there are TWO transitions in porphyrin spectrum and the free base model, along with Hunds' rules, predicts that there will be two transitions with different energies. We can conveniently define the Q-band as the transition to orbitals with high orbital angular momentum and the B-band to the transition to orbtials with low orbital angular momentum. The different orbital momentum of the states has been verified experimentally (Barth et al, Ann. New York Acad. Sci, 1973, 206, 223-246). Importantly, the transition energy you calculated is the "center of mass" or average of these two transitions, and this seems to be right as well. We estimated ~570 nm, and we do have one transition on the low energy side of that and one on the high energy side of that. The real center of mass is actually somewhere in the 450-480 nm range, suggesting that the free electron model actually doesn't do a great quantitative job of predicting where the center of mass is, despite William Simpson's claims to the contrary, but given its simplicity, it's not too bad.

Ok, one more thing, and this is where I have to correct myself. The free-electron model actually can account for the intensity disparity between the Q- and B-bands, although it takes a few steps to get there. I won't take you through the math, but you can show in simple approximation that for an electron transition to be allowed in the free electron model, the orbital angular momentum has to change by ± ħ. Any transition with a different change in momentum would be strictly forbidden. Well, as it happens the B-band, which is very intense, involves a transition from a state with momentum = 0 to a state with momentum of ± ħ. And the Q-band, which is very weak, involves a transition from a state with monentum = 0 to a state with moentum of ± 9ħ. The Q-band transition should be strictly forbidden. (In reality, it's a little bit allowed, because this determination is based on a first-order approximation.)

I know that was very long-winded but maybe it helps you see some of the value in these models. The free electron model is very simple; the fact that it can predict, at least qualitatively, so many features of such a complex molecule is pretty amazing. Of course, there are better models out there that can be used to quantify these effects. We have also neglected here that the excited state has higher order electron spin multiplicities, which also complicates things further.

Also, answering this question of yours:

You said that 20 electrons is unacceptable if we're using the "particle on a ring" energy equation because only levels up to n=4 is allowed in reality. So I'm guessing that a ring with 16 electrons is still acceptable, or a ring with 12 electrons.

There is nothing special about 18 electrons specifically. It's just that for porphyrin, 18 electrons gives the best fit to empirical observation. For benzene, you'd use a ring of 6 electrons.