# Relation between oxoanion acidity and number of oxygen double bonds

https://en.m.wikipedia.org/wiki/Oxyanion

Was reading the above Wiki page when I came across the 'heuristic for acidity' section which mentions that the oxoanion's acidity can be predicted by the number of oxygen double bonds, with more double bonds making the oxoanion more acidic. My question is, why is this so? Can someone explain it to me by the definition of an acid, and is this prediction basis only limited to oxoanions or can it be extended to other molecules?

• This rule is doomed to fail. The only oxyacids with element-oxygen double bonds are carbonic acid and nitric acid, both counting 1. Unless you want to include boronic acid’s resonance structure containing a double bond.
– Jan
Oct 27 '17 at 5:35
• Oct 27 '17 at 13:03

$$\ce{HA} \rightarrow \ce{H^+} + \ce{A^-}$$
proceeds. One of the products will always be the same for all acids ($\ce{H+}$), so no influence there. The other product is the anion ($\ce{A-}$) and the basic question is: "How stable is that anion?", or equivalently, "how well can that anion stabilize its negative charge?"
Many factors go into answering that question, the one under consideration in the rule alluded to in the question is: Are there atoms surrounding the central atom that can stabilize the charge effectively? Oxygen atoms bound only to the central atom (in the protonated form) are a prime example, because oxygen is a very electronegative element. (Another example not involving oxyanions/oxyacids is trifluoroacetic acid, which is more acidic than acetic acid.) The more such oxygens are in your acid, the stronger it will be - $\ce{H2SO3}$ is weaker than $\ce{H2SO4}$.
Jan commented on the question that double bonds between the central atom and oxygens only exist in the first row of main group elements (i.e. $\ce{C},\ce{N}$). This is true, however, the rather basic rule cited in the question does not go into this level of detail and simply assumes that the outdated picture of double bonds in say, $\ce{ClO4-}$, is correct.