Transition metals can form stable ions with different oxidation states. But I am confused why doesn't only the most stable state exist. Let me clarify my question more:

$$\ce{Ti^{2+} -> Ti^{3+} + e^-} $$ With $E^0 = +0.5~\mathrm{V}$.

This just shows that $\ce{Ti^{3+}}$ is more stable. So my question is that why doesn't $\ce{Ti^{3+}}$ only exist in its most stable form?

  • $\begingroup$ Understanding Latimer, Pourbaix and Frost diagrams might help you understand the relative interconversion of different oxidation states. here and here is a good link. I will post an answer if you still want any clarifications. $\endgroup$ Feb 9, 2014 at 10:59
  • 1
    $\begingroup$ if you look at the electron configurations of the Titanium cations you can see what is stable (titanium 4+ is most stable because it has the same electron configuration as Argon) and Ti(3+) has a half filled 4s-orbital and is relatively stable... $\endgroup$
    – user2117
    Feb 9, 2014 at 13:52

1 Answer 1


Thermodynamics proposes, kinetics disposes. The most stable form can have a high activation energy barrier surrounding it. The medium matters. Hard Lewis bases (water) stabilize high oxidation states (Cu(II), tungstate). Soft Lewis bases (acetonitrile) stabilize low oxidation states (Cu(II); $\ce{W(CO)6}$ and http://www.google.com/patents/EP1995347A1?cl=en).

$\ce{Ti^{+4}}$ is the basement, especially with oxygen. Now rationlalize titanium suboxide Magnéli phases. They are spectacular as inert electrodes under extreme conditions, re Ebonex.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.