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.


1 Answer 1


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.

enter image description here

(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 )

  • $\begingroup$ The lattice structure would be too huge for me to handle, so pointing out a single piece of that compound will be more helpful. $\endgroup$ Aug 2, 2017 at 11:11
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    $\begingroup$ But it's a crystal, not a single molecule. If you insist on one, here you go. But those molecules don't exist in the wild. $\endgroup$
    – Glorfindel
    Aug 2, 2017 at 11:12
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    $\begingroup$ There is no single piece. Just think of it this way: some Fe are +2, and some are +3. $\endgroup$ Aug 2, 2017 at 11:12
  • $\begingroup$ @Glorfindel I do understand it's not reality. Diamond is a giant structure, and at the first glance, one will think "All I see is just carbons arranged randomly". But if you point out that each carbon bonds to 4 other, he can understand better. C-C-C-C. $\endgroup$ Aug 2, 2017 at 17:30
  • 1
    $\begingroup$ But diamond is C, not C4. $\endgroup$
    – Glorfindel
    Aug 2, 2017 at 17:33

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