Why is Fe3O4 a non-stoichiometric compound? [closed]








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Fe combines with O in a whole number ratio 3:2. Even the ions Fe2+ and Fe3+ are in the ratio 1:2. Then, why Fe3O4 is a non-stoichiometric compound?




Stoichiometric compounds have composition with fixed and exact atomic ratio in small integer numbers.


Nonstoichiometric compounds have their real composition expressed by atomic ratios in rational numbers.


There is also a gray zone, where the compound is not exactly stoichiometric, but the deviation is small in considered context and we use for it a stoichiometric formula that fits it "good enough". Many metal oxides fall in this category.


The fact a compound has mixed oxidation numbers like $\ce{Fe3O4}=\ce{Fe^{II}Fe^{III}2O4}$ or $\ce{Pb3O4}=\ce{Pb2^{II}[Pb^{IV}O4]}$, so it cannot be expressed by a single integer oxidation number, does not *alone* make the compound non-stoichiometric. But deviations in composition, leading to rational atom ratios, do make such a compound non-stoichiometric.


Iron oxides are famous for that, even if $\ce{Fe3O4}$ is generally closer to stoichiometry than $\ce{FeO}$ or $\ce{Fe2O3}$. That is general phenomena for transition metals with multiple close oxidation states and oxide properties. There are vacancies or extra atoms and changes in oxidation states in frequent lattice defects.


E.g. the formal [FeO(s)](https://en.wikipedia.org/wiki/Iron(II)_oxide) is usually a non-stoichiometric oxide with real composition $\ce{Fe\_{0.84−0.95}O(s)}$. There to produce its stoichiometric variant too:


[Wikipedia:](https://en.wikipedia.org/wiki/Iron(II)_oxide)



> 
> Stoichiometric $\ce{FeO}$ can be prepared by heating $\ce{Fe\_{0.95}O}$ with metallic iron at $\pu{770^{\circ}C}$ and $\pu{36 kbar}$.
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Other typical cases of non-stoichiometric compounds are


* manganese oxides
* mixed oxide ceramics, e.g. high temperature superconductors
* Li-ion cell cathode materials
* "hydroxides"/"oxidohydroxides"/"hydrated oxides" of transition metals
* hydrides of transition metals


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The exact formulation of $\ce{Fe3O4}$ is indeed stoichiometric and requires a *fixed* amount of $\ce{Fe}$ and $\ce{O}$ to react as stated in the question statement. However, usually, this won't be the case; if you try to synthesize stoichiometric $\ce{Fe3O4}$, chances are that you will find some deviation from this formula due to defects and varying amounts of $\ce{Fe(II)}$ and $\ce{Fe(III)}$. Therefore, iron oxides are usually characterized as nonstoichiometric. As pointed out in the comments, stoichiometric iron oxides have also been synthesized.


[Nonstoichiometric compounds](https://en.wikipedia.org/wiki/Non-stoichiometric_compound) are known to exist in various, sometimes simple and sometimes not, such ratios and the composition may be affected by various factors such as synthesis conditions, reactivity etc. Common examples of such compounds are:


1. Iron oxides
2. Iron sulfides
3. Palladium hydrides
4. Tungsten oxides


Do note that the nonstoichiometric behavior does have applications such as catalysis, hydrogen/sulfur storage, ion conductance, and more.


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