How do you explain the formula Fe3O4 using the ionic compound theory? I don't understand at all how the ionic bonding in $\text{Fe}$$3$$\text{O}$$4$ works. The oxygens all together have $4(–2e)=–8e$ net charge. But we cannot give the three irons equal positive charges to balance this: $8/3$ isn't an integer!
And anyways, what kind of iron is in $\text{Fe}$$3$$\text{O}$$4$? It can only be either Fe(II) or Fe(III), and none of them fits in here.

A diagram demonstrating the chemical structure of $\text{Fe}$$3$$\text{O}$$4$ with supporting explanation would be preferable. I'm only an 8th grader, so advanced chemical concepts won't be comprehensible to me. The lattice structure would be too huge for me to handle, so pointing out a single molecule/piece of that compound will be more helpful.
Iron forms both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions. The first combine with oxygen ions to $\ce{FeO}$; the second $\ce{Fe2O3}$.

$\ce{Fe3O4}$ can be thought of as a mixture of both compounds in equal amounts. So $\ce{Fe3O4}$ contains both Fe(II) and Fe(III) ions; twice as much Fe(III) as Fe(II).

In the picture below, showing the crystal structure, green atoms are Fe(II) and brown atoms Fe(III). All Fe(III) atoms are linked with six O atoms, sharing 'half an electron' with each oxygen; all Fe(II) atoms are linked with four O atoms, and all O atoms are linked with four Fe atoms.



(source: Manoj B. Gawande, Paula S. Branco, and Rajender S. Varma, Chem. Soc. Rev. 2013, 42, 3371-3393. DOI: 10.1039/C3CS35480F See also: ResearchGate
)