# How to find the oxidation numbers of individual ions of a transition metal with an average fractional oxidation state

In the classical example compound of $$\ce{Fe3O4}$$, we know that $$\frac 23$$ of Fe ions display +3 oxidation state and $$\frac 13$$ of Fe ions display +2 oxidation state fulfilling charge neutrality condition. However, this is the only solution considering the most common oxidation states of Fe are +2 and +3.

Now, consider the compound $$\ce{La(Ni_{0.1}Mn_{0.9})O3} \rightarrow \ce{La10NiMn9O30}$$,

Assume, the oxidation numbers of La, Ni and O are +3, +2 and -2. Therefore, under charge neutrality condition, the charges of 9 Mn ions should add up to 28. Knowing that the most common oxidation states of Mn are +2, +4, +7, we will have two solutions where charges add up to 28.

1. 4 Mn ions showing oxidation number of +2 and 5 Mn ions showing oxidation number of +4
2. 7 Mn ions showing oxidation number of +2 and 2 Mn ions showing oxidation number of +7

How could I decide what is actually happening?

If it helps, I need to know this to calculate the average ionic radius of a B-site cation of a perovskite structure.

EDIT

Edited the typo by changing ox. state of La to +3.

Furthermore, I am looking for an answer that makes sense theoretically (real world formation may be different). It would be great if somebody could guide me for a relevant publication, if there is any.

• I don't think you can do this without a practical analysis of the compound in question. – Aniruddha Deb May 19 '20 at 12:14
• For clarity to quote from Wikipedia on Fe3O4: "sometimes formulated as FeO ∙ Fe2O3". So, look for related mixed salt compounds. – AJKOER May 19 '20 at 17:35
• @AJKOER Please don't suggest OP the outdated "mixed salt" concept. First, it's called a formal addition compound or a double oxide at best, not "mixed salt". – andselisk May 19 '20 at 18:37
• @AJKOER Second, writing "FeO ∙ Fe2O3" was acceptable in the past when there were no determined crystal structures available. These days the structure of $\ce{Fe3O4}$ is well-defined (cubic; monoclinic) and using "FeO ∙ Fe2O3" makes little to no sense as there are very little structural similarities and the stoichiometry as well as oxidation numbers can be well expressed with $\ce{Fe3O4}$ formula unit. – andselisk May 19 '20 at 18:37

## 1 Answer

To accurately determine oxidation states in compounds, you need to have a good idea of what the actual underlying bonding model is like. Only in rare cases you can resort to shortcuts.

One such shortcut which is often very helpful for binary compounds is to take one element’s ‘obvious’ oxidation state and use maths to derive the other one. That, for example, works well with ethane (hydrogen must have $$\mathrm{+1}$$, so carbon has $$\mathrm{-III}$$), iron(III) chloride ($$\ce{Fe^{+III}/Cl^{-I}}$$) or manganese dioxide ($$\ce{Me^{+IV}/O^{-II}}$$).

In some cases, such a shortcut will lead to a fractional oxidation state such as in $$\ce{Fe3O4}\ (\ce{Fe^{+\frac83}/O^{-II})},$$ $$\ce{HN3}\ (\ce{H^{+I}/N^{-\frac13})},$$ or $$\ce{S4O6^2-}\ (\ce{S^{+\frac52}/O^{-II})}.$$ One might be tempted to take the closest common or known oxidation states of the element in question and use maths to derive a potential distribution. That works by accident in $$\ce{Fe3O4}$$ ($$\frac83$$ is $$2\frac23$$ which suggests $$\ce{Fe1^{+II}Fe2^{+III}}$$) and $$\ce{HN3}$$ ($$-\frac13$$ suggests $$\ce{N2^{\pm0}N1^{-I}}$$) but it already fails with $$\ce{S4O6}$$ because although $$\ce{S3^{+II}S1^{+IV}}$$ gives an overall oxidation state of $$+\frac52$$ so does the correct solution of $$\ce{S2^{\pm0}S2^{+V}}$$.

Since the simple model already broke down for binary compounds, there is little hope it will be of much use in ternary or even quarternary compounds—especially if there are less leads to go on. So while it might be possible to use the simple shortcut to derive the oxidation states in aluminium potassium sulphate or potassium alum ($$\ce{KAl(SO4)2}$$) and by extension that of chromium in chromium potassium sulphate or chrom alum ($$\ce{KCr(SO4)2}$$), it is a lot harder if you have two redox-promiscuous elements (nickel and manganese) to consider.

In short, there is no simple way to derive the oxidation states just by looking at the formula and I myself would have to research what publications say and how they arrived at this conclusion.

• (+1) Nice answer. By S4O6 did you mean S4O5? I do understand the problem complexity increases with number of elements in the compound, especially when the compound includes transition metals. But please do come back here if you find a theoretical investigation on such compounds. I will too. – Achintha Ihalage May 21 '20 at 9:26
• He probably means the tetrathionan anion $\ce{S4O6^2- = O^- - SO2 - S - S - SO2 - O^-}$, formed by iodine oxidation of thiosulphate $\ce{S2O3^2-}$ – Poutnik May 21 '20 at 9:44
• @AchinthaIhalage As Poutnik guessed, I meant tetrathionate ($\ce{^-O-SO2-S-S-SO2-O-}$). Sadly for you, I don’t really intend on researching that oxide and coming back to it; it’s really out of my field of interest. – Jan May 21 '20 at 13:15