# Determine stable oxidation state from Ionization energy

Question:

Given the first, second and third ionization energies of an element (7 eV, 12.5 eV and 42.5 eV respectively), how would one find the most stable oxidation state (or charge on the resultant ion) of that element?

I think that the element has to ionize with oxygen to reach a certain oxidation state, since oxygen has a 6 valence electrons, the element has to lose 2 electrons to attain stability. So the second ionization energy will the give the most stable oxidation state. So the answer is 12.5 eV. Now that happens to be the correct answer but can someone verify my reasoning?

• You should what you think about the problem - your try in solving, otherwise it probably will be closed. Also it's duplicate - i'll search earlier q. – Mithoron Jul 17 '15 at 15:58
• As far as I know there is no direct relation between ionization energy and ionization state stability. Could you perhaps give me a hint to get started? – Abhishek Mhatre Jul 17 '15 at 16:38
• Abhishek this raises some alarms of being homework. Would you mind adding the info in the comment and any additional thoughts to the problem in your question? – M.A.R. Jul 17 '15 at 16:39
• you don't have lone ions but compounds with counterions and have to ionise with oxidant, does it help? – Mithoron Jul 17 '15 at 16:45

You don't necessarily need to have your element binding with oxygen to be oxidized, per se. Unless it is, of course, specified that we're talking about a reaction with oxygen. SO, don't misguide yourself on focusing on the oxygen here :) After all, if we, for instance, think about burning of potassium, it is being oxidized with oxygen, however, it still only makes $\ce{K+}$ and not $\ce{K^2+}$!