# Why is orbital energy not the mean between ionization energy and electron affinity when the orbital has two electrons?

In this question it is asked why ferrocene is colored when the HOMO-LUMO gap seems to be beyond the visible light range. I tried to answer that orbital energies change with electronic transitions and therefore the transition energy need not match the difference in orbital energies. However, my answer crumbled under unexpected pressure and I wound up deleting it.

I tried to use a hydrogen atom as an example of changing orbital energy levels. When an electron is added to make a hydride ion, the combined energy of both electrons in the 1s orbital should be -13.6 eV (from the ionization energy of the neutral atom) plus -0.8 eV (from the affinity for the additional electron). With this total energy distributed between two electrons in the orbital the 1s orbital energy seemingly should br -7.2 eV (per electron) in the anion, versus -13.6 eV for the neutral atom when there is only one electron in the orbital. But I was barraged with comments that this analysis is wrong; the orbital energy in the hydride ion is only -0.8 eV from just the electron affinity. I do not understand why, and am baffled because this rendering seems inconsistent with conservation of energy.

• Why does only the electron affinity figure into the orbital energy?

• How does such a rendering square with conservation of energy?

Please don't tell me I am wrong, direct me to getting right.

I think the confusion stems from a misunderstanding of the meaning of the orbital energies.

Consider the hydrogen atom and its orbitals. You can obtain the energy of the electron in its groudn state using the Schrödinger equation $$\hat{H} \Psi = E \Psi$$ For this atom you can write the hamitltonian (in atomic units) as $$\hat{H} = -\frac{1}{2}\nabla^2 -\frac{1}{R}$$ where $$-\frac{1}{2}\nabla^2$$ represents the electron kinetic energy (operator) and $$\frac{1}{R}$$ represents the Coulomb atraction between the electron and the hydrogen nucleus. Thus, the energy of the electron calculated corresponds to the binding energy of the electron to the nucleus or, if you prefer, the energy needed to ionize the atom.

Consider now the H$$^-$$ anion, the atomic system is different, you have an hydrogen nucleus and two electrons that interact amongst themselves. Thus, the hamiltonian is different too, and, accordingly, the orbitals and its energies are different from those of hydrogen.

The hamiltonian for the anion has the form:
$$\hat{H} = -\frac{1}{2}\nabla^2(1) -\frac{1}{2}\nabla^2(1) -\frac{1}{R_1} -\frac{1}{R_2} + \frac{1}{r_{12}}$$ where there appears a new term in the hamiltonian $$+\frac{1}{r_{12}}$$ that represents the repulsion of the electrons.

As @Infinite has mentioned in his answer, you can describe the properties of the anion ignoring the electron-electron interaction.

• With this method we would get the energy of the two electrons in the orbital. Would the orbital energy then become half that, per electron? Jul 6 at 1:33
• When you solve the Schröedinger equation, besides the total energy $E$, you also obtein both the orbitals $\phi_i$ and the energy of electrons in the orbitals ($\epsilon_i$). However the total energy, $E$, is not simply the sum of the energy of the electrons, $E \neq \epsilon_1 + \epsilon_2$. Why? Because the energy of the electrons in the orbital includes the repulsion energy with the other(s) electron(s) and simply adding the energies of the electrons will overstimate the interlectronic repulsion energy. Hope this help.
– PAEP
Jul 6 at 21:08
• I accepted this answer with one caveat. If you spell Schrödinger with the umlaut over the o, you do not also use the following e. Jul 7 at 0:45
• Thanks pointing the errata.
– PAEP
Jul 7 at 10:28

The orbital energy simply refers to the ionisation energy. So the orbital energy of the hydride ion should be numerically equal to the electron affinity value of the hydrogen atom.

And there is no violation of law of conversation of energy. For removing the first electron we should supply 0.8 eV and for removing second electron we should supply 13.6 eV for an atom. And I see that there is no loss of energy ( $$\because$$ the total energy of hydride ion is (13.6+0.8) eV in magnitude).

An explanation to the increased orbital energy from -13.6 eV to -0.8 eV is due to the electron-electron repulsions.

Please correct me if I did anything wrong because I don't have a good background in quantum mechanics.

• Thank you, but what I needed to know is why the electronic energies do not add up as I expected they would. Jul 5 at 14:11
• If energy is distributed in such manner then 7.2 eV of energy is required for the removal electron but actually we need just 0.8 eV of energy for removing that electron. This energy is lesser than the energy expected by you, it is because electron-electron repulsions makes our task easy. Jul 5 at 14:22
• my interpretation would be 7.2 eV is required to remove either electron but 6.4 eV out of that is transferred from the remaining electron, this transfer being due to the disappearance of interelectronic repulsion. Jul 5 at 14:26
• Let us say we have supplied 1 eV of energy. If we don't observe any removal of electron then you interpretation could be correct. But I don't have enough sources for supporting this. May be someone can look into this with experimental results. Jul 5 at 14:36
• Orbital energy but for H cannot be equal to ionization energy, as total energy of the other electrons before and after ionization is not the same. Jul 5 at 14:52