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Question: Given the first, second and third ionization energies of an element (7 eV, 12.5 eV and 42.5 eV respectively), 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 my answer is +2.

Now that happens to be the correct answer but can someone verify my reasoning?

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  • $\begingroup$ 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? $\endgroup$ Jul 17, 2015 at 16:38
  • $\begingroup$ 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? $\endgroup$
    – M.A.R.
    Jul 17, 2015 at 16:39
  • $\begingroup$ you don't have lone ions but compounds with counterions and have to ionise with oxidant, does it help? $\endgroup$
    – Mithoron
    Jul 17, 2015 at 16:45

3 Answers 3

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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 $\ce{K^2+}$!

The trick in this question is to look for the values, or a magnitude, of those ionization energies in question. To remove the first electron, you need to provide $\pu{7 eV}$, to remove the second one, you need to provide $\pu{12.5 eV}$. So the difference between 7 and 12.5 is not that big if we compare it with the difference between $\pu{12.5 eV}$ and $\pu{42.5 eV}$ that is required to remove the third electron. If explaining it in simple terms, the first to energy "bumps" are easy to "climb", while the last one is way high and you really need to provide a sufficient amount of energy to get that high. Thus, removing the first two electrons is relatively easy, while removing the third one is not.

This gives you the most stable oxidation state for this element to be +2. If we perform a reaction of this element (which looks like $\ce{Sr}$) with any other electronegative element, say chlorine, it would still have a charge of +2.

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Have a look at your data booklet to see some of these patterns on the Ionisation energies table eg its quite striking with Aluminium - you will see that it can make an Al3+ ion.

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Always remember that the difference in ionization enthalpy of two successive states is approx or equal to $10$ to $\pu{15 eV}$ or less. The lower oxidation state is not stable; conversely if the difference is more than $\pu{15 eV}$, the lower oxidation state is more stable. Here the $3^\mathrm{rd}$ and $2\mathrm{nd}$ ionization enthalpy difference is more than $\pu{15 eV}$; that's why second ionization enthalpy is more stable

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