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The following question arises from a question I found in my book.

Experimentally it has been determined that the value of the first energy of ionization of oxygen is lesser than the first energy of ionization of nitrogen. Select the statement which explains this:

  1. The higher electronegativity of oxygen
  2. The lesser electron affinity of nitrogen
  3. The lesser radius of oxygen
  4. The higher stability of nitrogen
  5. The higher electron affinity of nitrogen

According to my book the answer is the fourth option, but I couldn't find a way to justify this. What sort of explanation can do this?.

My guess is that nitrogen forms a triple bond with itself which is higher than the oxygen and this makes the molecule more stable but I don't know if this can be translated into a higher energy of ionization nitrogen rather than the oxygen which does have a sigma and pi bonds. Am I right?

Can someone please illustrate this or bring some sort of diagram with some values to justify this?. It would help a lot if an answer could help me why should I discard the second option?.

Doesn't nitrogen has smaller electron affinity than the oxygen? Wouldn't it mean that having a smaller electron affinity cause it to be easier to strip out an electron from its outer shell rather than the opposite as the question indicates?

Martin - マーチン
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Chris Steinbeck Bell
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  • But wait it's ionization of atomic element by default. If it was about molecules, it would have to be clearly stated! – Mithoron Feb 21 '20 at 21:40
  • @Mithoron Actually it does in part but it doesn't explain more details than stability due half filled orbitals. Could stability be explained using orbital molecular theory or $\pi$ bonding as one of the answerers suggested?. – Chris Steinbeck Bell Feb 22 '20 at 07:45
  • It's for atoms, so just a matter of half filled orbitals. – Mithoron Feb 22 '20 at 14:42
  • Please cite the textbook from which this exercise is from. – Martin - マーチン Feb 22 '20 at 16:38
  • Regardless of whether it’s atoms or molecules (I assume it’s atoms), the fourth option in the question is a meaningless phrase. It... just doesn’t say anything. What kind of “stability” is this? How can the “stability” of lone atoms even be compared? What does it even have to do with ionisation energies? – orthocresol Feb 22 '20 at 16:49

1 Answers1

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Two electrons of the oxygen molecule are located on a $\pi $* level. They are relatively easy to ionize. It is not the case in $N_2$ : all electrons are located in binding orbitals.

Maurice
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    That was also my thought, but that's barely an idea for an answer. – Mithoron Feb 21 '20 at 21:38
  • @Maurice In my comment I guessed that it had to do with the electrons in the $\pi^{*}$ bonds. In your answer you mentioned that this is not the case. Is it because the overlap occurs side to side and this makes the electrons easy to remove?. But doesn't this also occurs in nitrogen?. – Chris Steinbeck Bell Feb 22 '20 at 07:41
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    If you tried to actually write answers that backup your claims, then you might had realised that what you write is incorrect. – Martin - マーチン Feb 22 '20 at 16:32