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.


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