Why don't gases of elements with negative electron affinities exist as ions in nature?

Question

Fluorine has an electron affinity of about -300 kJ/mol. Lithium has an electron affinity of about 60 kJ/mol.

As I understand it when $\ce{Li^-}$ loses an electron about 60 kJ/mol is released but when $\ce{F}$ gains an electron about 300 kJ/mol is released.

Furthermore, when $\ce{Li^+}$ gains an electron about 520 kJ/mol is released because that is the first ionization energy of lithium and when $\ce{F^+}$ gains an electron about 1700 kJ/mol of energy is released because that is the first ionization energy of fluorine.

Shouldn't the cases that release the most energy be $\ce{Li}$ and $\ce{F^-}$ and so shouldn't those be the states that gases of those elements occupy?

Is it the case that $\ce{2F^-}$ is less stable than $\ce{F2}$ or something? Or is it the case that $\ce{F^-}$ could only exist in a highly artificial situation such as a magnetic contraption that trapped ions?

Why don't gases of elements with negative electron affinities exist as ions in nature?

Answer 1

It totally does not work like that. You can't say that $\ce{2F^-}$ is less stable (or more stable) than $\ce{F2}$; you can't compare them at all. People would say "B is more stable than A" when there is a spontaneous reaction $\ce{A\to B}$, but this is not the case here. A reaction must be balanced in charge and in all elements, otherwise it is just a bunch of letters that makes no sense. There is no reaction $\ce{F2<->2F^-}$ (either way), much like there is no reaction $\ce{F2->Au}$.

There is indeed a reaction $\ce{F2 + 2e- -> 2F^-}$, and it is quite favorable energetically (as you already know), but that's another story. You need the electrons for it, and they are just not there. Fluorine can't wrestle those electrons from another fluorine, as explained by IT Tsoi. And if you have lithium around, why, of course they would react to form $\ce{LiF}$, but that's a chemical compound, not just $\ce F$ by itself; also, at room temperature it is solid.

There is more to it. A gas (as well as a liquid or solid, for that matter) formed of negative ions alone is very much impossible. The ions will repel with such force that no material in the world will hold them even for a second. (I mean macroscopic amount, not a thin trickle of ions inside a mass spectrometer.) So you can't have fluorine in this state even if you have a stash of electrons which you are willing to share. $\ce{F-}$ ions do exist, of course, and there is nothing exotic about it (arguably, almost all fluorine in the world is found in this form), but they exist as a part of various ionic compounds, and not by themselves.

Answer 2

So consider a vessel wherein which we have Fluorine atoms (F) in gaseous state.

Now, they have seven outer shell electrons and would really love to get the 8th one for an octet. What would have to happen is:

$$\ce{2F -> F+ +F-}$$

It's not possible for both to become fluoride ions because there would be no charge balance.

The total change in energy is then given by:

$$\Delta E = -300 + 1700 = + 1400\ \mathrm{kJ/mol}$$

So it is not energetically favorable for one Fluorine atom to completely give up its electron in favor of the other. Hence, they prefer to share one electron each in a covalent bond. And this results in the release of $156.9\ \mathrm{kJ/mol}$ of energy.