How useful are oxidation states?

Introduction

Paradigm shifts often occur when an established theory has to be ever more distorted to account for observable facts and to make accurate predictions. I wonder if the time is coming soon when the concept of oxidation states will join phlogiston, equivalent weights, normality and relative density in the dustbin of chemical history (see Dinosaur chemistry)?

Oxidation states are man-made and provide a model used by most, if not all, chemists to determine whether a reaction is a redox reaction. Oxidation can be defined as an increase in oxidation state and conversely reduction as a decrease in oxidation state. The model can also be used to determine the number of electrons involved when balancing half-equations – but does it always work? It is worth exploring some Nature of Science by considering how valid and useful the concept of oxidation states actually is.

It is confusing enough that in the new programme the IB has switched the way in which oxidation numbers and oxidation states should be represented. This may be what IUPAC recommends but most textbooks and websites still use Roman numerals, e.g. II, IV etc. for oxidation states and Arabic numbers, e.g. +2, −7 etc. for oxidation numbers. Since this is a website for IB teachers lets go with the new definition and represent oxidation states by using Arabic numerals.

Have oxidation states passed their sell-by date?

The rules for determining the oxidation state of an element in a compound are well known. If the element or species is an ion then the oxidation state is simply equal to the charge on the ion. This is straightforward and uncontroversial when there is just one element containing the charge under consideration as, for example, in ionic compounds such as sodium chloride, Na+Cl−. It could be asked though, why use the man-made concept of “oxidation state” in this situation when we could just use the change in “charge” when determining which species is oxidized or reduced etc.?

The problem comes when covalent bonding is involved, either in a complex ion or in a covalent compound. The assumption is made that covalent bonds act as if they are ionic with the electrons moving to the more electronegative atom. This assumption is clearly false and only works in rather limited circumstances. It is almost completely meaningless when it comes to carbon compounds. Consider ethanoic acid, CH3COOH. If we apply the rules, then the four hydrogen atoms give a total of +4, the two oxygen atoms give a total of −4 so the oxidation states of the two carbon atoms is zero. Go down the homologous series of carboxylic acids and the oxidation state of carbon in propanoic acid, C3H6O2, works out to be −2/3. For carbon in butanoic acid, C4H8O2, and pentanoic acid, C5H10O2, it is −1 and −1.2 respectively, which is clearly not very helpful. Of course you could isolate the carboxyl functional group by replacing the alkyl group with a hydrogen atom to give a fixed value of +2 for the oxidation state of the carboxyl carbon atom in all carboxylic acids (or, by simply ignoring the alkyl group completely, to give a value of +3) but there is nothing in the rules to tell you to do this and it is again making a huge assumption.

Carbon compounds are generally mainly or completely covalent so one might expect a problem with organic compounds, but even some well-known inorganic ions can cause difficulty. For example, compare the sulfate ion, SO42−, and the thiosulfate ion, S2O32−.

The central sulfur atom in the sulfate ion has an oxidation state of +6 and the surrounding oxygen atoms each have an oxidation state of –2. If one of the oxygen atoms is simply substituted by a sulfur atom (an element in the same group as oxygen) to give the thiosulfate ion, S2O32–, then according to the rules sulfur now has an oxidation state of +2. In reality the central sulfur atom has not changed so should still be +6 and the other sulfur atom should be –2 (although note that Wikipedia states that one is +5 and the other is –1 due to the S−S bond). The reason for the difficulty is because two or more atoms of the same element are bonded together. Even the rules themselves have to take this into account with oxygen (the oxidation state of oxygen is minus one in peroxides) but they do not mention it for other elements. Another ion that also contains S−S bonds is formed when the thiosulfate ion is oxidised to the tetrathionate ion, S4O62–. According to the rules sulfur now has the non-integer oxidation state of +2.5 in this complex ion. Even so, the change in the oxidation state of sulfur can still be used to balance the half-equation for the oxidation reaction.

2S2O32– → S4O62– + 2e–

2 x (2 x +2) → (4 x +2.5) + (–2)

Perhaps where the rules really do cause difficulties is with the triiodide ion, I3−. Pure iodine, which is purple, is not soluble in water. It does dissolve in potassium iodide solution to form the triiodide ion, which is responsible for the yellow/brown colour of the solution.

The triiodide ion is a complex ion.

The Lewis structure shows a central iodine atom surrounded by two iodine atoms and three non-bonding pairs of electrons, i.e. it has expanded its octet and has five electron domains. The ‘parent’ shape of the molecular ion is trigonal bipyramidal with the actual shape being linear as the three non-bonding pairs of electrons lie in the same plane.

What actually happens when the triiodide ion is formed?

I2(aq) + I−(aq) → I3− (aq)

According to the rules the oxidation number of iodine in the triiodide ion is –1/3. If this logic is followed then the formation of the triiodide ion is a disproportionation reaction as iodine has been simultaneously reduced from zero to –1/3 and oxidised from –1 to –1/3. However, some chemists consider the triiodide ion to behave simply as if it is a mixture of molecular iodine and the iodide ion. In which case no redox reaction occurs when iodine dissolves in potassium iodide. So is it a redox reaction or not?

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