Why is $\ce{HClO4}$ the strongest oxyacid?

My effort: The structure of $\ce{HClO4}$ should be tetrahedral. So when there is a negative charge on oxygen, it should not be in resonance with other oxygen atoms as a condition for resonance is that the atoms taking part in resonance should lie in a plane.

  • $\begingroup$ What about inductive effects? $\endgroup$ – Lighthart Mar 18 '16 at 22:12
  • $\begingroup$ Actually $\ce{ClO4^{-}}$ is tetrahedral. I forget, to which oxygen does the hydrogen attach?!? $\endgroup$ – MaxW Mar 18 '16 at 23:00
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    $\begingroup$ Geez, stop thinking with primitive mesomeric models. All oxygens in anion are single bonded and identical. That's it. $\endgroup$ – Mithoron Mar 19 '16 at 1:21
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    $\begingroup$ @DGS possibly pi back-donation from oxygen to P-H antibonding orbitals is working on hypohosphorous acid (in a sense passing some positive formal charge from P to O). P-H bonds are weak so the antibonding orbitals are at low enough energy for pi acceptance. $\endgroup$ – Oscar Lanzi Mar 19 '16 at 23:47

Assume an acid $\ce{HA}$. An acid is strong if

  1. protons are donated easily, i.e., the $\ce{H-A}$ bond is readily broken;
  2. the conjugate base $\ce{A-}$ is stable, i.e., $\ce{A-}$ is reluctant to combine back to $\ce{HA}$.

When are protons easily donated?

First parameter to consider is the polarity of the $\ce{H-A}$ bond. More polar bonds mean less electron density on the bond. Thus the dissociation equilibrium

$$\ce{{{HA_{(aq)}}+H2O_{(l)}}<=>H3O_{(aq)}+ + A_{(aq)}-}$$

is tilted strongly towards the right. For this reason, $\ce{HF}$ is a stronger acid than $\ce{NH3}$.

In addition, the size of $\ce{A}$ is important. If $\ce{A}$ is very big, essentially it is harder to hold on to the hydrogen. Thus the $\ce{H-A}$ bond is weaker. This is why $\ce{HI}$ is a better acid than $\ce{HF}$.

Stability of the conjugate base

Charge delocalisation is key here. If all atoms bonded to a central atom are equal, then the hydrogen ion does not have a preferred location at which to bond. A large charge in a small volume would also translate into an unstable ion.


  1. Look at the number of oxygens bonded to the central atoms. More oxygens -> stronger acid, e.g., $\ce{HNO3}$ is stronger than $\ce{HNO2}$.
  2. If the number of oxygens is equal, consider the electronegativity of the central atom. Higher electronegativity -> stronger acid. For instance, $\ce{HBrO3}$ is stronger than $\ce{HIO3}$.
  3. When electronegativity is similar as well ($|\Delta EN|\le 0.1$), it sometimes helps to look at the size of the central atom. For example, $\ce{H2SeO4}$ is stronger than $\ce{H2SO4}$.

The case at hand

Polarity of the $\ce{O-H}$ bond in $\ce{HClO4}$ is guaranteed via

  • the electronegativity of the $\ce{O}$ atom itself;
  • the electronegativity of the central atom $\ce{Cl}$;
  • the electronegativity of the three additional $\ce{O}$ atoms.

Thus, the electron density is pulled from the $\ce{O-H}$ bond and the proton can be easily donated.

Here is a picture where you can compare the charge delocalisation of various chlorine oxoanions.

Chlorine oxoanions

Clearly, the perchlorate ion $\ce{ClO_4-}$ is the most stable. Every valence orbital and valence electron of chlorine takes part in the bond formation.


This reply discussed relative acid strength with respect to Brønsted–Lowry acid–base theory (in water). Do not forget that these are pointers; there will be other parameters to consider and exceptions to take into account.

$\ce{HOF}$ is the single known fluorine oxoaxid. The analog $\ce{HFO4}$ is simply not possible. It would violate the octet rule.

EDIT: Answer to comment

You asked: "When you say charge delocalisation in ClO4- anion is it due to resonance? Is resonance possible?"

From IUPAC Gold Book: "[Resonance] refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. Resonance among contributing structures means that the wavefunction is represented by 'mixing' the wavefunctions of the contributing structures."

Therefore, the answer is yes. The main resonance structures are given in the picture below.

Main resonance structures of the perchlorate anion

The resulting ion is a resonance hybrid. In classical valence bond theory, charge delocalisation is not due to resonance; rather, resonance is delocalisation.

Furthermore, coplanarity is not a requirement. It is fair to say that there is (almost) always some delocalisation (again, see the definition given). Still, having atoms that lie in the same plane can be advantageous. Qualitative models of aromaticity and conjugation may even give coplanarity as a necessary condition. This is not always the case, however. Quantitative models are used to more rigorously predict aromaticity and antiaromaticity (see NICS, QTAIM models).

P.S. Resonance energy is important. High, positive resonance energy indicates increased stability; slightly positive or even negative values point to unstable particles. For instance, antiaromatic molecules generally have low resonance energies. And even though planarity is a prerequisite of qualitative antiaromaticity, too, it is not actually a rule, instead a rough indicator. Example: 2,3,4,5-tetraphenylcyclopenta-2,4-dienone here and there.


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