# When a diatomic ion dissociates, how does the electron density on each atom change?

Let's say, we have a fluorine molecule (F2), and we take 3 electrons from it, so now its bond starts to stretch, and at last the bond breaks and the two atoms are farther and farther away from one another. When they're far enough from one another, one of them must have one more electron than the other. But what does this electron do during the dissociation process - does it choose one atom and stay there, or does it go back and forth between the two atoms? Do the two atoms share the electrons equally during the early stage and then they start to compete for one electron, making each atom's electron density fluctuate? Or does one of the atom's electron density keeps going down while the other's electron density keeps going up during the whole process?

• Perhaps better suited for Chemistry.SE? – Apoorv Khurasia Apr 19 at 23:33
• Isn't this a physics problem? – XYZ Apr 19 at 23:34
• Maybe it's more of a 'chemical physics' problem, but there isn't a tag called chemical physics or molecular physics here. – XYZ Apr 20 at 15:14

The above is based on assuming an adiabatically slow separation of the dimer. If time-dependence comes into play in an ultrafast separation, experiments on the dissociation of $${{\rm H}_2}^+$$ show that after only $$15\cdot 10^{-15}$$ seconds, the electron has "decided" which of the two protons to pick at a distance of 8 Bohr and cannot be driven force and back with a laser between the two anymore [Xu et al. Observing electron localization in a dissociating $${{\rm H}_2}^+$$ molecule in real time. Nat. Commun. 8, 15849 (2017) (open access)].