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?
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$\begingroup$ Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, misunderstanding, objections, question down-voting or even question closure. $\endgroup$– PoutnikCommented May 2, 2023 at 10:59
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2 Answers
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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) is usually a non-stoichiometric oxide with real composition $\ce{Fe_{0.84−0.95}O(s)}$. There to produce its stoichiometric variant too:
Stoichiometric $\ce{FeO}$ can be prepared by heating $\ce{Fe_{0.95}O}$ with metallic iron at $\pu{770^{\circ}C}$ and $\pu{36 kbar}$.
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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$\begingroup$ Then is Fe2O3 non stoichiometric? $\endgroup$ Commented May 1, 2023 at 16:59
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$\begingroup$ I have thought it was clear. Its ideal composition is not. Its real composition is. Same as for FeO, Fe2O3, MnO, Mn3O4, MnO2. They at special cases and careful preparation may approach stoichiometry, but will not achieve it exactly. $\endgroup$– PoutnikCommented May 1, 2023 at 17:12
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$\begingroup$ Although I agree with the notion that mixed valence compound can be non stoichiometric at times (due to its deviation from ideal integer ratio of atoms and it is very much a case when you think those compounds are adduct of two small compounds, also non stoichiometric at times and so in practical cases, the atoms doesn't always reach integer values, e.g ($\ce{Fe3O4}$ = $\ce{FeO.Fe2O3}$ but for simplicity we give them a stoichiometric formula), it's better to classify them separately. Not all non stoichiometric compounds are ionic (so don't have mixed valence). E.g. Palladium hydride. $\endgroup$ Commented May 2, 2023 at 4:06
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$\begingroup$ @NilayGhosh I have not said nor assumed all nonstoichiometric compounds are ionic. See also related closed question. The answer was intended to address the specific iron oxides case and not as general classification, which you correctly commented in the follow up question the user could find it himself. I would added there metal hydrides too, if remembered in the time of writing. $\endgroup$– PoutnikCommented May 2, 2023 at 5:12
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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 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:
- Iron oxides
- Iron sulfides
- Palladium hydrides
- 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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1$\begingroup$ @ananta I suspect your answer is downvoted because you first sentence assumes that the iron oxide is always exactly stoichiometric which avoids the question rather than answering it. it might be better to say, if the formulation is exact, then it isn't non-stoichiometric but the substance often isn't depending on the exact conditions of its creation. $\endgroup$ Commented May 2, 2023 at 10:00
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1$\begingroup$ @matt_black It could be also said the formula Fe3O4 itself is stoichiometric, but real composition may not be, if it statistically significantly deviates from the given atomic ratio. $\endgroup$– PoutnikCommented May 2, 2023 at 10:20
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1$\begingroup$ @ananta The point by matt_black is also good. You are assuming that iron(II,III) oxide is always stoichiometric. If you were to take the formula literally then yes, iron(II,III) oxide is stoichiometric but it oxidizes in air leaving behind $\ce{\gamma-Fe2O3}$ and you are left with a $\ce{Fe2O3-Fe3O4}$ mixture which is not at all stoichiometric. Although you may call it partially oxidized magnetite, it is nonetheless magnetite and better to formulate it $\ce{Fe_{3−δ}O4}$ with $\ce{\delta}$ values reaching upto 0.02. $\endgroup$ Commented May 2, 2023 at 12:22
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$\begingroup$ @matt_black thank you for the suggestion. I made some changes to the answer. Hope this clarifies the discrepancy. $\endgroup$– anantaCommented May 2, 2023 at 13:38