Is there a way to Determine if an Element can undergo Disproportionation?
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: 



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:

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

Note how for this reaction $E_\mathrm{cell}^\circ &gt; 0$, 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.