3
$\begingroup$

This question already has an answer here:

If two electrons that get paired occupy the same orbital, then wouldn't there be a heavy amount of repulsion between the two? As is, since electrostatic force ${\displaystyle \propto }$ $\ce{ 1/ (distance)^2}$ so since the distance between two electrons is way smaller than between the electrons and nucleus, I don't really assume it has any role here.

Does this have anything to do with the strong nuclear force? As far as I know its range is too short and exists only between protons and neutrons.

Summing up, what causes two paired electrons to stay in the same orbital rather than get thrown away due to "the force"?

$\endgroup$

marked as duplicate by airhuff, Jon Custer, Todd Minehardt, Klaus-Dieter Warzecha, M.A.R. ಠ_ಠ Mar 23 '17 at 19:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ No, it has nothing to do with the strong nuclear force, which operates, well, in the nucleus. The pairing of electrons in atomic wave functions is a natural consequence of quantum mechanics. $\endgroup$ – Jon Custer Mar 22 '17 at 23:16
4
$\begingroup$

No strong nuclear force is involved. Instead, tt's the same force as always: electromagnetic interaction. By staying in the same low orbital as the other electron, the electron in question will continue to be subject to a stronger attractive force from the nucleus. Note that the MO diagrams (see e.g. wikipedia) typically already account for the intra-orbital electron-electron repulsion, and you still have HOMO-LUMO gap.

It is also helpful to remember that electrons are "spread out" over the orbital, so once you do the math, you realize that the average distance between electrons does not have to be small compared to the average electron-nucleus distance.

Let's get semi-quantitative: $\ce{H_2}$, 1.4 bohr, HF/STO-6G($\zeta = 1.20$). After the SCF is completed, I transformed the integrals from the AO basis to the MO basis, which yields the following, relevant integrals in Mulliken notation:

(1|h|1) -1.24964
(2|h|2) -0.49918
(11|11)  0.65720
(22|22)  0.68091
(11|22)  0.64686
(12|12)  0.17440

Nuclear repulsion is 0.71429, all values in hartree. We can then calculate the low-spin, ground-state energy as: $$ E_\text{gs}^\text{RHF} = E_{nn} + 2\cdot(1|h|1) + (11|11) = -1.12778, $$ wherein we add the nuclear repulsion energy, the one-electron energy of the lower MO (once for each electron) (one-electron meaning that this is the sum of the electron's kinetic energy and its nuclear attraction) and the relevant two-electron integral for the Coulomb repulsion. Since the electrons are of opposite spin, no exchange term (as is responsible for the spin-pairing energy, cf. Hund's rule) is needed.

The energy of the high-spin, "excited" state can be calculated according to: $$ E_\text{triplet}^\text{RHF} = E_{nn} + (1|h|1) + (2|h|2) + (11|22) - (12|12) = -0.56207, $$ where an additional exchange term enters the equation to account for partly-correlated electrons due to Pauli-repulsion.

This calculation is semi-quantitative because RHF in such a small basis set does not yield quantitative results. One would have to go to natural orbitals obtained through full CI in a larger basis set for fully quantitative results. However, for illustrating the relative size of the quantities involved, this level of calculation is sufficient.

$\endgroup$

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