[dcrosen]

The energy carried by molecules is generally divided into electronic, translational, vibrational and rotational.  I would like to clarify what is meant by electronic energy.  Some sources seem to focus specifically on the energy associated with electrons.  Other sources consider nucleus-electron, electron-electron and nucleus-nucleus interactions.  I'd like to sort this out more clearly. 

What makes sense to me is that the electronic energy is the minimum potential energy without vibrations and rotations taken into account.  For example, this is the energy that would exist in the bond between two hydrogens in an H2 molecule if the two hydrogen atoms were stationary at the bond distance (not possible of course).  Is this correct? 

If the previous paragraph is in fact correct, wouldn't it be clearer to substitute the term "base potential energy" or something similar?  The term "electronic energy" seems overly broad since the energies associated with vibration, rotation and translation also involve electronic energies (energies associated with motion and position with respect the electromagnetic force).  Any comments would be of interest.

Thanks!

[Irlanur]

The electronic Energy is given by the Eigenvalues of  the electronic Hamiltonian. this includes electron kinetic energy, electron-electron, electron-nucleus and often also nucleus-nucleus interactions. why? because the Schrödinger equation is typically solved for a given nuclei-konfiguration, the electronic energy thus forms a potential energy (hyper-)surface. vibrational energies are usually added with harmonic potential approximations. rotational energy can be calculated with the moment of inertia.

[Enthalpy]

Since the other mentioned forms are translational, vibrational and rotational, I interpret this "electronic energy" as a degree of freedom, that is, electrons getting a higher energy when the temperature rises. At moderate temperatures it won't happen where the energy levels are well separated, say in N₂, but at heat it does happen with CO₂ and with molecules that have broadly delocalized electrons.

[Corribus]

@Enthalpy, while this is based on a generally true principle - in that lower lying states have a greater chance of being thermally populated; and increasing the temperature raises the probability of thermal population of electronic states - your conclusion is misleading. Practically speaking, almost no molecule would fare well at the temperatures that would be required to observe any reasonable probability of thermal population of an excited electronic state. The energies involved are just too large. Using some real numbers will help show why,

Let's start with nitrogen and carbon dioxide.

As far as I could tell from a quick Google search, the lowest lying electronic levels of nitrogen molecule (A ³Sigma_u⁺) and carbon dioxide (A ³Sigma_u⁺) are ~7.63 and ~7.35 eV above the respective ground electronic states. 

Refs:

http://www.wag.caltech.edu/publications/sup/pdf/52.pdf
Gillan, J. Phys. B: At. Mol. Opt. Phys. 29 (1996) 1531–1547

This would correspond to electronic transitions in the 163-169 nm region (UV). As far as molecules go, these aren't very low lying electronic states, but since you brought them up, let's go with them. It is true that the electronic state of CO₂ is lower than that of N₂, and as a result would have a higher likelihood of being thermally populated. But by how much?

We can use a very basic statistical mechanical treatment of partition functions to estimate the probability of thermal population of these states. Assuming a two state model, with a ground state and the excited electronic state, the probability of finding a molecule in the excited state at any time is given by

[tex] \frac{1}{Z} e^{-\beta E_s} [/tex]

Where

[tex] \beta = \frac{1}{k_B T} [/tex]

And Z is the partition coefficient, E_s is the state's energy (relative to the ground state, which we'll assign as 0) and k_b is Boltzmann's constant.

In the event that the upper state and lower state are far apart in energy, Z is essentially equal to 1. It's very simple to calculate probabilities in these circumstances.

At 298 K, the probability of the lowest state of carbon dioxide being thermally populated is about 4.7 x 10^-125. Ok, what about 1000 K? The probability is still staggeringly low: 9.07 x 10^-38. To put that in a little perspective, if earth's entire atmosphere were flash heated to 1000 K, of the 2 x 10¹⁵ kg of carbon dioxide currently present, only about 2500 of those molecules would be thermally excited to the lowest lying electronic state. If you're willing to heat the atmosphere to 10,000 K, the probability of thermal population of the lowest state would be ~2 x 10^-4, or about 1 in 5000 molecules. Of course, at that temperature, you probably don't have much carbon dioxide left, to say nothing of the people who exhale it!

In most cases those probabilities are so ridiculously low that comparisons become meaningless, but in the 10,000 K Armageddon scenario, the probability of nitrogen molecules being thermally excited are about 1.5 x 10^-4, or 1 in 7000. Yeah, lower than the likelihood of carbon dioxide being thermally excited, but not really the either/or scenario you painted, where it never occurs with nitrogen but frequently does with CO₂.

Ok, so I think we can agree that, for all intents and purposes, electrons in carbon dioxide and nitrogen are never going to be thermally excited. What about other molecules?

The lowest-lying electronic transitions are usually in the near-infrared. There are some organic molecules that have very low lying electronic states, but a good example most people may be familiar with are the states of dioxygen. Dioxygen has a low lying singlet state, with an optical transition at about 1270 nm. Using our same means of estimation, we can show that the probability of thermal excitation of this state at 1000 K is about 1.24 x 10^-5. That's certainly not negligible, but it's not huge either. At combustion temperature (roughly 2000 K), this probability becomes 0.003. Now we're talking! In fact, it is easily concluded that one of the reasons oxygen is such a vigorous combustion agent is that at suitable temperature, singlet oxygen can be thermally prepared in reasonable portion, and singlet oxygen is a beast at reacting with organic fuels.

Even so, at 1000 - 2000 K, few molecules, and certainly few organic molecules, are going to hold together. And the example of singlet oxygen is rather extreme, it being among the lowest lying molecular electronic states you'll ever encounter.

Is there, then, ever any reasonable expectation of thermal population of electron states? You bet! Atoms and monoatomic ions have electronic states as well, and they can't fall apart so easily. In fact, a very commonly used analytical technique called ICP-AES uses thermal population of electronic states to detect trace quantities of elements in complex samples.  This is because those thermally excited atoms and ions quickly relax to their ground states, and in the process emit photons of characteristic wavelengths.

For example, one analyte I've frequently looked for in my own research is aluminum, which has a strong optical transition at about 396 nm, corresponding to about 5 x 10^-19 J. ICP-AES works by introducing the sample into an extremely hot argon plasma, with temperatures that can easily range up to 6000 K or even hotter. At this temperature, the probability of thermal population of aluminum's 396 nm equivalent state (again, assuming only a two-state system) is about 0.0024, or one in approximately 400 atoms. That's apparently quite enough to be able to detect at concentrations as low as 5 ppb.

So, yes it's quite possible to thermally excite electrons, but it requires temperatures that are for the most part too high to be practically observed for most molecules.

[Enthalpy]

My mistake, I thought CO₂ had much lower excited electronic states. At 7eV it's clear they are essentially empty. Thanks, +.

No different result from bigger molecules. Pentacene has its first excited electronic state at 1.82eV
http://arxiv.org/pdf/cond-mat/0211420.pdf
far too much for a significant population. And since the much longer and conjugated Lycopene absorbs in the visible region, I should have guessed that CO2 can't be excited:
http://infohost.nmt.edu/~jaltig/LycopeneSpectrum.pdf
OK, I won't check up to coal. It's a clear case.

[mjc123]

NO has an electronic transition at 121 cm^{-1  2}Π_3/2  ²Π_1/2

[Irlanur]

There will be another problem if you have low-lying electronic states. the whole assumption that you can express the wavefunction as a product of electronic, vibrational, rotational ... wavefunctions won't be correct anymore. you will get couplings between electronic and vibrational states, the BO approx might fail completely and the term "electronic energy" won't be very feasible anymore...