Can we determine the lowest and highest possible oxidation numbers of an element? Since if we know this, then it is easy to determine whether the element will undergo disproportionation.

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    $\begingroup$ Well, if you want to determine if an compound will undergo disproportionation, looking at a Latimer Diagram is a really quick way to find out. I don't think you can really look at a compound and say it will undergo disproportionation without any knowledge of its thermodynamic properties. This link provides a good introduction to Latimer Diagrams: chemwiki.ucdavis.edu/Textbook_Maps/… $\endgroup$
    – Nanoputian
    Apr 30, 2016 at 9:06
  • $\begingroup$ @Nanoputian You really should expand this into an answer. $\endgroup$ Apr 30, 2016 at 9:16

3 Answers 3


Expanding on my original comment, I disagree with the OP's statement that knowing the possible oxidation states of an element will easily allow you determine if a disproportionate reaction will occur.

To predict such a reaction, thermodynamic data such as the $E_\mathrm{cell}^\circ$ of the two reactions must be known. Latimer diagrams are a really simple and compact way of showing the $E_\mathrm{cell}^\circ$ values for a variety of reduction and oxidation half-equations between different oxidation states of the same element. Below is an example of a Latimer diagram for manganese in acidic medium:

enter image description here

Just from this I can tell the $E_\mathrm{cell}^\circ$ potential for any oxidation or reduction reaction between any pair of manganese species (in fact there is extra information that isn't even required). For example by looking at this diagram I can tell what $E_\mathrm{cell}^\circ$ will be for the reduction of $\ce{Mn^7+}$ to $\ce{Mn^3+}$ (I won't go into how you would do that in this post, I suggest that if you want to know how to, have a look at the website that I have linked).

But wait, there is more! Now the neat thing about Latimer diagrams is that I can easily tell if a particular oxidation state of manganese is stable or not (in other words, will it undergo disproportionate reaction?). All you have to do is follow one very simple rule:

'if the number on the left is lower than the number in the right, the species is unstable to disproportionation'

For example, let's consider $\ce{Mn^3+}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO2}$ to $\ce{Mn^3+}$ is $0.95~\mathrm{V}$ (that is the number on the 'left') while the $E_\mathrm{cell}^\circ$ for the reduction of $\ce{Mn^3+}$ to $\ce{Mn^2+}$ is $1.51~\mathrm{V}$ (this is the number on the 'right'). Since the number on the left is smaller than the number on the right, $\ce{Mn^3+}$ will spontaneously disproportionate to form $\ce{MnO2}$ and $\ce{Mn^2+}$. The reaction is below:

$$\begin{align} \ce{2Mn^3+ + 2H2O} &\ce{-> MnO2 + Mn^2+ + 4H+} & E_\mathrm{cell}^\circ = +0.56~\mathrm{V} \end{align}$$

Note how for this reaction $E_\mathrm{cell}^\circ > 0$, hence it is spontaneous.

However these 3 species that are involved don't have to be right next to each other. For example, let's consider $\ce{MnO3-}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO4-}$ to $\ce{MnO3-}$ is $0.419~\mathrm{V}$ while the E cell for the reduction of $\ce{MnO3-}$ to $\ce{Mn^3+}$ is $2.61~\mathrm{V}$. Again, the number on the left is smaller than the number on the right, hence $\ce{MnO3-}$ will disproportionate spontaneously.

This may not be the answer that you were expecting originally, but hopefully this will help you with your original questions.

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    $\begingroup$ just a tip: mhchem on chem.SE is a little odd, so $\ce{Mn^{3+}}$ gives a weird space between the 3 and the +: $\ce{Mn^{3+}}$. If you omit the curly braces it looks better: $\ce{Mn^3+}$ renders as $\ce{Mn^3+}$. $\endgroup$ Apr 30, 2016 at 11:46
  • $\begingroup$ @orthocresol thanks for the tip, I always thought you needed the curly brackets. It definitely looks much better now. $\endgroup$
    – Nanoputian
    Apr 30, 2016 at 23:54

The primary and secondary question are in fact two separate issues since we can artificially bombard elements or ions with photons or electrons to try to get other oxidation states. So in principle I can force into existence Na2+ but whether or not it survives in nature is a separate issue. Hence the 'highest' and 'lowest' oxidation states are rather arbitrary, depending on how much energy you want to put into the endeavor.

However, we have experimentally determined the typical oxidation states of elements in the periodic table and Compound Interest has made a version of the Periodic Table which shows such common oxidation states availablePeriodic Table from Compound Interest

Whether or not the said element undergoes disproportionation then becomes a question of thermodynamics and the specific reaction in question.


The concept may sound strange, but it does happen. Alkali metals such as sodium can be combined with an appropriate chelating ligand which forms a stable complex with $\ce{Na+}$. Then where is the negative charge? On sodium anions!


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