[OrganicSynthesis]

Many people will already know the answer, but when I found the answer, I found it so interesting that I decided that I wanted to pose the question for people who never thought about it.

Despite the trend that hydrohalic acids tend to be strong, HF is surprisingly weak. Explain its unexpected (or expected depending on the individual) weakness in acidity.

[Schrödinger]

Vf vg orpnhfr bs gur Ulqebtra obaqvat?
V qba'g xabj vs guvf vf jung lbh'er ybbxvat sbe, ohg V guvax gung rkcynvaf gur geraq jvgu nyy gur bgure unyb npvqf...gurl qba'g unir Ulqebtra obaqvat

(rot13)

[stewie griffin]

I think the usual argument goes something like this: F is so electronegative that we'd expect HF to be a very strong acid, since HF should be really polarized. Yet HF is weak compared to HCl, HBr, and HI. The Cl-, Br-, and I- are simply larger anions and thus the charge density is decreased. Therefore the conjugate base of HCl, HBr, and HI are much more stable than the conjugate base of HF.
Other reasons I've read as to why acidity increases going down the halogen column is that the bond strength of HX gets weaker as we move down the column due to the large differences in atomic size between H and X. For example, I is much much larger than F, and the resultant orbital overlap between I and H is not nearly as great as the orbital overlap in F and H.
I think orgopete may have some interesting comments on this... 

[Schrödinger]

I think the usual argument goes something like this: F is so electronegative that we'd expect HF to be a very strong acid, since HF should be really polarized.

I think the fact that F is highly electronegative is a factor that can account for its weak acidity (due to Hydrogen bonding)

Other reasons I've read as to why acidity increases going down the halogen column is that the bond strength of HX gets weaker as we move down the column due to the large differences in atomic size between H and X. For example, I is much much larger than F, and the resultant orbital overlap between I and H is not nearly as great as the orbital overlap in F and H.

Isn't that true with all groups?

I think the OP was referring to an 'anomaly' in the halogen group...or atleast that's what I could make of the question.

[stewie griffin]

Schrodinger,  the anomaly is the fact that despite the fact that F is so electronegative, HF is still a weak acid. I tried list possible explanations in my previous post.

[Schrödinger]

Schrodinger, the anomaly is the fact that despite the fact that F is so electronegative, HF is still a weak acid.

It is true that when an element is electronegative, the H-E (E=Element) bond is quite polar, and hence a proton can be easily lost. But when the element is as electronegative as F, you should also look for factors like Hydrogen bonding, that tend to decrease its acidity.

That's what I've been thinking. Am I wrong?

[MissPhosgene]

Schrodinger, I think you are right H-F---H.

[MrTeo]

Schrodinger, I think you are right H-F---H.

Well, I think the HF₂^- bonds are even more interesting: especially for their extremely high bond energy (≈230 kJ/mol) caused by a three-center two-electron bond (like in diborane). I don't really know if this applies to the answer but it could be helpful.

Another reason for their behaviour could be the high electronegativity coupled with the small radius (if compared with other halogens) which could cause them to form more stable ionic clouds around the hydronium ions.

[jeffrey.struss]

I was always told it was because while fluorine is highly electronegative, it is too small to handle the charge density and as such does not like being fully F-.