Chemistry - Why exactly is a fluoride anion a stronger base than iodine anion?

Solution 1:

First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).

That is correct.

Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.

I don’t agree with that explains; but the observation is correct, iodide ($\ce{I-}$ is the weakest base and $\ce{HI}$ is the strongest acid.

But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.

No. The fluoride anion is not an oxidising agent. If it were, it would take up electrons to become a fluoride dianion ($\ce{F^2-}$), inserting an electron into the 3s level—not happening.

What you probably meant is fluorine (the element; the diatomic molecule) being the strongest oxidising agent out there. That might be correct or it might not (I am not up to date with extremely strong oxidising agents) but it is not a statement I could dismiss at first glance. Certainly, fluorine is a stronger oxidising agent than iodine.

So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$?

Thermodynamic stability is sometimes a tricky beast, but in general a fluoride ion is more stable than an iodide ion which is prone to oxidation to give iodine or iodate. This is connected to the oxidising ability of fluorine (strong) versus iodine (weak), not to the strengths of the acids, though.

And hence does not give away its electrons easily and therefore is a weaker base?

The conclusion is invalid. Not wanting to give away one’s electrons means not wanting to get oxidised. As mentioned, fluorine is a strong oxidising agent so to reduce fluorine to fluoride is a favourable process. But acid-base chemistry has nothing to do with gaining or losing electrons: it is an entirely electrostatic process and builds on different principles.

The high oxidising ability of fluorine can be thought of as a side effect of its high electronegativity: it has a tendency to strongly attract electrons since it has a highly positively charged nucleus that is shielded by only two core and seven valence electrons. Adding an eighth electron is comparatively easy. For iodine, there are four core shells and a valence shell between the nucleus and the incoming electron; this shielding effect can be thought of as reducing electronegativity and oxidising ability.

The high acidity of $\ce{HI}$ and low basicity of iodide is a direct consequence of its size: the negative charge is distributed across a much larger volume meaning it is more stabilised. In fluoride, the negative charge is confined to a much smaller volume meaning that positive charges such as a proton are attracted stronger. This is the underlying reason for the observed acidities/basicities.

Solution 2:

Second period nonmetals form much stronger bonds with hydrogen than their heavier congeners and so tend to bind more strongly with protons than we might otherwise expect. Not only is fluoride ion a stronger Bronsted-Lowry base than chloride ion (and heavier halides), so are hydroxide ion versus hydrosulfide ion and ammonia versus phosphine. The last pair has the property that phosphine is much more highly flammable, which can be traced to the phosphorous-hydrogen bonds being easy to break while nitrogen-hydrogen bonds are more robust.