Is this the correct relative Brønsted acidity in these four acids?

Question

We are asked to sort these four acids by increasing Brønsted acidity:

The problem is, I am confused why $\ce{H2F+}$ is the strongest acid of them all. I understand that stronger acids will have weaker bonds, and the $\ce{F-H}$ bonds should be the most electronegative of them all, leading me to believe that they are the strongest bonds. So then why are these bonds the most easily broken? What have I forgotten here? Is the correct order really from weakest to strongest acid as follows? $\ce{CH3OH}$, $\ce{(CH3)2OH+}$, $\ce{CH3SH2+}$, $\ce{H2F+}$

Answer 1

Fluorine being the most electronegative element in the periodic table, it doesn’t like bearing a positive charge. Thus $\ce{H2F+}$, which has the positive charge on the fluorine nucleus, is not very stable and willing to expel the extra proton. So, it is indeed a strong acid.

$\ce{H2F+}$ is formed, e.g., by autoionization of liquid $\ce{HF}$, but only because that reaction also involves creating a $\ce{F-}$ ion (associated with another molecule of $\ce{HF}$ to form $\ce{HF2-}$), which is very favorable:

$$\ce{3 HF <=> H2F+ + HF2-}$$