1
$\begingroup$

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

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.

So far so good.

But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there. So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$? And hence does not give away its electrons easily and therefore is a weaker base?

Where am I going wrong here?

I have a little bit of gut feeling that this is because of the large size of iodide ion which can hold the negative charge better than the tiny fluoride ion. But I have found nothing to back this up.

$\endgroup$
  • 3
    $\begingroup$ I would rather say F- is the weakest oxidizing agent, as there are similar complementary rules for redox pairs as for acid-base pairs. $\endgroup$ – Poutnik Aug 29 at 16:38
  • 3
    $\begingroup$ There is no connection whatsoever between the "strength" of the acid base pairs and the redox pairs. Your gut feeling is basically right, but dont drag the redox chemistry in to the argument. $\endgroup$ – Karl Aug 29 at 17:13
  • 1
    $\begingroup$ If the fluoride anion, $\ce{F^-}$, were an oxidizer, than because it would removes at least one electron from an other atom, ion, or molecule. Yet, as fluoride anion, $\ce{F^-}$ already possess completely filled s- and p-orbitals. And as member of the second period (counting the one with H and He as the first one), the octet rule typically is observed strictly. There use of d orbitals (like for sulfur, one period later) isn't typical either. So, while while the fluorine molecule ($\ce{F-F}$) is capable to oxidize all up and including oxygen, the same by fluoride ($\ce{F^-}$) is unclear. $\endgroup$ – Buttonwood Aug 29 at 20:11
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/34818/… $\endgroup$ – Karsten Theis Aug 29 at 22:01
  • 2
    $\begingroup$ You are mixing several concepts in your question: being an acid/base is very different to being an oxidizer/reducing agent $\endgroup$ – SteffX Aug 30 at 14:33
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yes I did mean Fluorine. Thanks a lot! $\endgroup$ – The Jade Reaper Aug 31 at 1:52
2
$\begingroup$

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.

$\endgroup$

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