# Is there an error in a Wikipedia article explaining the influence of oxidation states?

Referring to the series of oxoacids of chlorine: $$\ce{HClO, HClO2, HClO3},$$ and $$\ce{HClO4}$$, tabulated in the Wikipedia article on electronegativity, the article states, in the section near the end titled "Variation of electronegativity with oxidation number":

As the oxidation state of the central chlorine atom increases, more electron density is drawn from the oxygen atoms onto the chlorine, reducing the partial negative charge on the oxygen atoms …” and illustrates it with a table. (NB There is an ambiguity in the use of the word "reducing" which is the opposite of "oxidising", but the intended meaning in terms of increasing or decreasing electron density is clear in the Wiki statement.)

But as I understand it (and as also given by the rules for assigning oxidation numbers), insofar as oxidation state represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element, the table is correct, but shouldn’t the textual statement be the converse:

“As the oxidation state of the central chlorine atom increases, more electron density is drawn from the central chlorine atom onto the oxygen atoms, increasing the partial negative charge on the oxygen atoms ...”

I reason it thus And just to be clear, let's go back one step to start with the even simpler case of $$\ce{HCl}$$, where clearly the oxidation states must be ($$\ce{H} = +1, \ce{Cl} = -1$$, respectively; total 0):

When we move on to $$\ce{HClO}$$, then it becomes (+1, +1, -2 respectively) as the single oxygen with stronger electronegativity than chlorine is assigned the oxidation state -2, so that balancing the total means the electronegativity of the central chlorine must change from -1 to +1 (in agreement with the table on the Wiki page, but not with the article’s statement about the direction of change of electron density.

And as we move along the series to $$\ce{HClO4}$$, the central chlorine is having more and more electron density sucked off it by the four more electronegative oxygen atoms, resulting in the final oxidation states (+1, +7, -8), in agreement with the table in the above-cited Wiki article, but not with the article's textual explanation.

So I conclude that the text statement in the Wiki article is currently wrong (and if so I will amend the Wikipedia article) - or else I'm wrong, and would appreciate someone pointing out my own error…?

• This is a bit awkward wording there, but nevertheless it's not wrong, somewhat ambiguous, perhaps. – Mithoron Jul 17 '20 at 17:30
• "drawn from the central chlorine atom onto the oxygen atoms, increasing the partial positive charge on the chlorine atom" would be better then your version, as partial charge on each one oxygen does indeed decrease, not their sum though. – Mithoron Jul 17 '20 at 17:34
• @Mithoron - Excellent point. So you agree that basically I'm right and the Wikipedia article should be amended, but using your refinement. Many thanks. – iSeeker Jul 17 '20 at 19:35
• Well, no it shouldn't! Such statement has nothing to do with the point the article makes - that Cl +7 doesn't release its electrons as eagerly as Cl +1. – Mithoron Jul 17 '20 at 20:57
• OK - gottit. Thanks – iSeeker Jul 17 '20 at 21:21

Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.

What the article is trying to say is that the electron density on each oxygen is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with $$\ce{HClO}$$. Here, the Cl has oxidation state +1 and the O is -2. But of course the oxygen doesn't have a full negative 2 charge. Some of that charge is shared with the chlorine and the hydrogen, neither of which has a full charge of +1.

To get to $$\ce{HClO2}$$, we can imagine reacting $$\ce{HClO}$$ with a neutral oxygen atom (oxidation state 0). After the reaction, the oxygen will have oxidation state -2, and the Cl will now be +3. Again, these are not the formal charges, but give us a hint about where charge will be located. The oxygen will have a partial negative charge, which means the chlorine will have to lose some of its electron density, increasing its actual charge. The original oxygen from $$\ce{HClO}$$ is still there and feels a stronger "pull" on its electron density, so it gives more to the chlorine, and the oxygen's partial negative charge becomes less negative.

Completely making numbers up, we could imagine that in $$\ce{HClO}$$, the charge on oxygen is -1, on chlorine is +0.25 and on H is +0.75. After adding the second oxygen to make $$\ce{HClO2}$$, the charge on the new oxygen might be -0.5, on the already attached O might be -0.75, +0.75 on H and +0.5 on Cl. Do you see how the negative charge on each O gets smaller in magnitude, but the total negative charge on oxygen(s) gets larger in order to offset the increased positive charge of the more oxidized chlorine? The Wikipedia statement refers to the individual O's not the sum.

• Many thanks for the careful and detailed explanation, resulting in the well-clarified Wiki entry inserted by KarstenTheis . – iSeeker Jul 18 '20 at 10:20

There is a nice illustration of the partial charges of the protonated and deprotonated species on the Libretexts site. The $$\mathrm{p}K_\mathrm{a}$$ values quoted at Wikipedia and the textbook referenced above don't match up, especially those below zero, so there is still room for consolidation and improvement.