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.


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.

  • 3
    $\begingroup$ I don't think you can do this without a practical analysis of the compound in question. $\endgroup$ Commented May 19, 2020 at 12:14
  • $\begingroup$ For clarity to quote from Wikipedia on Fe3O4: "sometimes formulated as FeO ∙ Fe2O3". So, look for related mixed salt compounds. $\endgroup$
    – AJKOER
    Commented May 19, 2020 at 17:35
  • 2
    $\begingroup$ @AJKOER Please don't suggest OP the outdated "mixed salt" concept.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". $\endgroup$
    – andselisk
    Commented May 19, 2020 at 18:37
  • 3
    $\begingroup$ @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. $\endgroup$
    – andselisk
    Commented May 19, 2020 at 18:37
  • $\begingroup$ Mn(VII) is unlikely in this context... $\endgroup$ Commented Oct 25, 2022 at 3:56

1 Answer 1


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 ($\ce{Mn^{+IV}/O^{-II}}$).

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 chrome 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.

  • $\begingroup$ (+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. $\endgroup$ Commented May 21, 2020 at 9:26
  • 3
    $\begingroup$ He probably means the tetrathionan anion $\ce{S4O6^2- = O^- - SO2 - S - S - SO2 - O^-}$, formed by iodine oxidation of thiosulphate $\ce{S2O3^2-}$ $\endgroup$
    – Poutnik
    Commented May 21, 2020 at 9:44
  • 2
    $\begingroup$ @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. $\endgroup$
    – Jan
    Commented May 21, 2020 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.