If the central atom is more polar, it pulls electron density away from the H+, making it more detachable. But why, then, are ionic compounds (which are even MORE polar than acidic covalent compounds) held tightly together?

  • $\begingroup$ The acidity trend goes to the right and down. There is the electronegativity effect and the size effect, but the size effect is more important! $\endgroup$ – gannex Apr 15 '16 at 3:07

Ionic compounds make a network of electrostatic attraction. Molecules have covalent intramolecular bonds which are very strong, but each molecular unit is weakly attracted to one another (relatively) due to lack of ionic attraction. So, essentially, in the solid form, each $\ce{NaCl}$ has more attraction to another $\ce{NaCl}$ than does each molecule to another in a molecular system. That does not, however, translate to each $\ce{Na+}$ ion of an $\ce{NaCl}$ unit being more tightly bound to a $\ce{Cl-}$ ion than two atoms of a covalent bond are to each other.

The idea that ionic bonds are "stronger" than covalent bonds is confusing to early chemistry student (and maybe more so to chemists, but it's tough to convey concisely what is actually occuring) but, for some reason, seems to be nearly ubiquitous in general chemistry courses. It is much easier for compounds like water to stabilize $\ce{Na+}$ and $\ce{Cl-}$ than to dissociate something like paraffin wax into ions in order to solvate the molecules as ions.

So I guess what I'm trying to say is: the ions of ionic compounds aren't usually held more tightly together than two atoms of a covalent bond.

As for the idea that it is a higher electronegativity difference in $\ce{H-A}$ bonds that allow for higher acidity, that can be disproven by looking at $\ce{HF}$ vs $\ce{HI}$. $\ce{HF}$ has a higher electronegativity difference but is a weaker acid. This is because $\ce{I-}$ is larger, so it can spread the negative charge out, essentially stabilizing the negative charge on its own. $\ce{HF}$ produces $\ce{F-}$, and because fluorine is more electronegative than iodine, that may be expected to be a stronger acid than $\ce{HI}$. However, because $\ce{F-}$ is much more of a point-charge rather than a diffuse charge like $\ce{I-}$, it has to be stabilized by solvation effects to a larger extent.

So what causes an acid to be more acidic is the thermodynamic stability of the resultant conjugate base. Because the charge in $\ce{I-}$ is more stabilized by the ability to spread out its own charge, it more readily donates $\ce{H+}$ than does $\ce{HF}$.


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