Sapling Hw 2, 3, 4 Question 20

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Jessie Hsu 1C
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Sapling Hw 2, 3, 4 Question 20

Postby Jessie Hsu 1C » Wed Oct 28, 2020 11:50 pm

I don't really understand how the first reason allows us to conclude that oxygen has a lower first ionization energy than both N and F. Can someone go through their logic when solving this problem? Thanks!

Identify the reasons why oxygen has a lower first ionization energy than both nitrogen and fluorine.

- Upon ionization, oxygen is relieved of electron‑electron repulsion.
- The 2p electrons in fluorine are more highly shielded from the nuclear charge than the 2p electrons in oxygen.
- The 2p electrons in fluorine experience a higher effective nuclear charge than the 2p electrons in oxygen.
- Oxygen has a smaller atomic radius than fluorine.

Stuti Pradhan 2J
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Re: Sapling Hw 2, 3, 4 Question 20

Postby Stuti Pradhan 2J » Thu Oct 29, 2020 12:05 am

In the lecture, Dr. Lavelle explains how one of the anomalies in the ionization energy, why oxygen has a lower ionization energy than nitrogen, is because oxygen has 4 electrons in the 2p orbital. This means that the 2px orbital has two electrons, unlike nitrogen, which only has one electron for each orientation (2px, 2py, 2pz). Sincethe two paired electrons in oxygen result in more electron-electron repulsions, it takes less energy to remove an electron from oxygen (a lower ionization energy). Therefore, the electron-electron repulsions are a major reason that oxygen has a lower ionization energy than nitrogen.

Additionally, as the number of protons increase for any element (as the atomic number of the element increases), there is a greater nuclear charge, which results in a higher ionization energy, as there is stronger attraction between the nucleus and the electrons. This can be seen when comparing oxygen and fluorine, as fluorine has more protons and, therefore, a greater nuclear charge.

When electrons are in the same subshell (the 2p orbital for example), they all receive the same amount of shielding from the core electrons, so that does not result in any difference in the ionization energy.

Finally, atomic radius decreases as you go up and to the right of the periodic table, so fluorine has a smaller atomic radius than oxygen.

Hope this helps!

Benjamin_Hugh_3F
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Re: Sapling Hw 2, 3, 4 Question 20

Postby Benjamin_Hugh_3F » Thu Oct 29, 2020 12:16 am

We know that N and F are more stable than O because N has a half-filled stable orbital, and F has a high nuclear charge, which decreases the radius of the atom, increasing the ionization energy. The electron-electron repulsion in O explains why it has lower ionization energy than N because an additional electron is being added to a half-filled stable orbital, making it easier to remove the electron.

Michael Sun Dis 3G
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Re: Sapling Hw 2, 3, 4 Question 20

Postby Michael Sun Dis 3G » Thu Oct 29, 2020 8:22 pm

Oxygen has 4 electrons in the p-subshell, meaning one of the p-orbitals contains two electrons. Because both electrons are negatively charged, they will naturally repel each other, making it easier to remove an electron from that orbital. Nitrogen, on the other hand, has only one electron in each p-orbital, and so it experiences weaker repulsions. Fluorine has more protons than oxygen, and so its electrons experience stronger electrostatic attractions.

Xinying Wang_3C
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Re: Sapling Hw 2, 3, 4 Question 20

Postby Xinying Wang_3C » Fri Oct 30, 2020 3:46 am

I also had the same question, and this is really helpful! Thank you.


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