Oxygen Exception Ionization Energy

[Ian_Lee_1E]

Hello, I am confused to why Oxygen is the only ionization energy exception.

I get that since the p orbital is already half full, it creates repulsion and leads the 4th electron to be easily accessible, leading to lower ionization potential. However, why doesn't this apply to flourine as well? with that logic, flourine would have more repulsion, wouldn't it?

[Benjamin_Hugh_3F]

Atoms have a tendency to either form a full octet or a half-filled octet. Fluorine has 7 electrons in the outer shell, nearly achieving a full octet, and it has a higher nuclear charge, making it hard to remove the electrons. On the other hand, Oxygen has 1 more electron in the 2p orbital than the half-filled orbital, so it will readily give away the electron to reach a stable state.

[David Chibukhchian 2G]

I believe that a big part of this is because fluorine is a halogen. Because its electron affinity is so high, this discrepancy that we see in oxygen when it comes to ionization energy doesn't really happen in group 17 elements. The way I think about this exception is that we only pay attention to it when there is some kind of break in the trend, but it's not something that we have to actively assume (if that makes sense). For example, when we see that oxygen's ionization energy is lower, we just have to remember that the reason is because of that paired orbital. I hope that helps!

[Anna Yakura 2F]

When/where did we learn this exception? Just wondering bc this is the first time I heard of it oops

[Samantha Pedersen 2K]

When/where did we learn this exception? Just wondering bc this is the first time I heard of it oops

The first place I saw it was Problem 1.13 from Outline 2 of the textbook, and the solutions manual does a good job of explaining it. I hope this helps!

[Emily Ding 1J]

Also, flourine has a higher effective nuclear charge as it's to the right of oxygen, so it's harder to remove the 2p electrons of F than O, resulting in a higher ionization energy.

[David Chibukhchian 2G]

I think we learned about this exception when covering trends in the periodic table. It was mentioned when the textbook talked about ionization energy, electron affinity, atomic radius, etc. In lecture, I think Dr. Lavelle talked about this rule for oxygen in lecture 12 (since that was the one that covered periodic table trends).

[EnricoArambulo3H]

The atoms favor a half-filled or a full subshell. Since the 4th 2p electron in oxygen disrupts the favorable electron states of the p subshell, it will cause more repulsion. This also means that it is more ready to give up its electron so that the orbital may be only half filled. I hope this helped!

[AnnaNovoselov1G]

Oxygen has the electron configuration 1s^2 2s^2 2p^3, so the p shell is half. Nitrogen, on the other hand, has the electron configuration 1s^2 2s^2 2p^4. In oxygen, each electron in the p shell occupies its own shell while in nitrogen, there are two electrons in one of the orbitals, which results in electron repulsion. Thus, removing an electron from oxygen makes it more stable, so it's easier.
(First ionization energy of oxygen is 1313.9 kJ⋅mol−1 and that of nitrogen is 1402.3 kJ⋅mol−1. However, the second ionization energy of oxygen is 3388.67 kJ⋅mol−1 while the second ionization energy is 2856 kJ⋅mol−1 for nitrogen).

This exception applies to other group 16 elements like S and Se (it's easier to remove an e- from them than from group 17 elements) as well as to group 2 elements (like B) vs group 3 elements (like Be).

[Veronica Macias 3K]

Oxygen has the electron configuration 1s^2 2s^2 2p^3, so the p shell is half. Nitrogen, on the other hand, has the electron configuration 1s^2 2s^2 2p^4. In oxygen, each electron in the p shell occupies its own shell while in nitrogen, there are two electrons in one of the orbitals, which results in electron repulsion. Thus, removing an electron from oxygen makes it more stable, so it's easier.
(First ionization energy of oxygen is 1313.9 kJ⋅mol−1 and that of nitrogen is 1402.3 kJ⋅mol−1. However, the second ionization energy of oxygen is 3388.67 kJ⋅mol−1 while the second ionization energy is 2856 kJ⋅mol−1 for nitrogen).

This exception applies to other group 16 elements like S and Se (it's easier to remove an e- from them than from group 17 elements) as well as to group 2 elements (like B) vs group 3 elements (like Be).

Hi, I believe, the electron configuration of oxygen is 1s^2 2s^2 2p^4 not 1s^2 2s^2 2p^3 and for nitrogen it's 1s^2 2s^2 2p^3 not 1s^2 2s^2 2p^4. I think you may have accidentally switched them around.

[Veronica Macias 3K]

When/where did we learn this exception? Just wondering bc this is the first time I heard of it oops

The first place I saw it was Problem 1.13 from Outline 2 of the textbook, and the solutions manual does a good job of explaining it. I hope this helps!

Yup. This explanation helped me understand better. The textbook said that as we move across period 2, oxygen is the first element where the p-electrons have to be paired and this added electron-electron repulsion energy causes the IE to be lower than of Nitrogen and Fluorine.

[Jessica Hu 3L]

What about the elements below Oxygen also in Group 16, such as Sulfur? Wouldn't they also encounter a similar situation?

[George_Yin_3I]

Oxygen has a decrease in ionization energy despite the fact that the trend is all increasing generally across the period. That is because atoms have a tendency to either form a full octet or a half-filled octet.