In HOI, the oxygen would be more negatively charged than the oxygen in HOCl because iodine is less electronegative than chlorine. Thus the O-H bond in HOI would be more polarized, which would mean it should be a stronger acid. The O-H bond in HOI being stronger is nonrelevant because according to my textbook (Zumdahl, General Chemistry), the more polarized a X-H bond is, the more acidic it is regardless of the strength, giving a example of C-H vs Cl-H where the latter bond is slightly stronger but much more acidic because of its polarity.

So why is HOI in reality a weaker acid? I felt like I have made some trivial mistake.

  • $\begingroup$ HOI is even unstable and rapidly disproportionates to HIO3 and iodine. $\endgroup$ Commented Apr 30 at 11:26

1 Answer 1


$\ce{HOCl}$ is a stronger acid then $\ce{HOI}$ for similar reasons as why $\ce{HOI}$ is a stronger acid than $\ce{HOH}$.

Because the electronegativity decreases $\ce{Cl > I > H}$.

For the general schema of oxoacids $\ce{H-O-R}$, the more polarized bond $\ce{O-R}$ means the less polarized bond $\ce{H-O}$ and the weaker acid.

The weaker pulling of electrons by the oxygen atoms from the atom attached to oxygen, the stronger the acid. If the oxygen atom already has a partial negative charge from $\ce{R}$, it is less favorable to increase this charge even further by polarization of $\ce{H-O}$ bond.

Polarization of $\ce{O-H}$ bond helps its heterolytic cleavage. The anion is then stabiliaze by weaking of the charge on the oxygen atom with relatively electronegative $\ce{R}$, often boosted by the charge delocalization like in $\ce{ClO4-}$.

What I mean is, $\ce{O}$ takes the most electron density from $\ce{H}$, than from $\ce{I}$, than from $\ce{Cl}$. Therefore it takes the least electron density from the other $\ce{H}$ of $\ce{H2O}$, more from $\ce{H}$ of $\ce{HOI}$ and the most from $\ce{H}$ in $\ce{HOCl}$. Therefore $\ce{H-O}$ bond is the most polarized and acidic in $\ce{HOCl}$, than $\ce{HOI}$ than $\ce{HOH}$.

And for the anions, the charge on $\ce{O}$ is most concentrated in $\ce{OH-}$, less in $\ce{IO-}$ and the least of $\ce{ClO-}$.

So there are 2 parallel reasoning for relative strength for all 3 pairs of conjugate acid/base.

Just illustratively, not reflecting real absolute nor relative values of partial charges, it could be like this:

$$\ce{H^{(+0.25)}-O^{(-0.50)}-H^{(+0.25)}}$$ $$\ce{H^{(+0.30)}-O^{(-0.45)}-I^{(+0.15)}}$$ $$\ce{H^{(+0.35)}-O^{(-0.40)}-Cl^{(+0.05)}}$$

$$\ce{O^{(-0.95)}-H^{(-0.05)}}$$ $$\ce{O^{(-0.90)}-I^{(-0.10)}}$$ $$\ce{O^{(-0.85)}-Cl^{(-0.15)}}$$

  • $\begingroup$ I'm confused by your third and fourth paragraphs. Are you saying that Cl takes electron density from O and O then takes it from H and so HOCl is more acidic than HOI since I is not so electronegative? $\endgroup$
    – porphyrin
    Commented Apr 30 at 10:36
  • $\begingroup$ @porphyrin The answer has been updated. $\endgroup$
    – Poutnik
    Commented Apr 30 at 11:08

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