Focus 1 Exercise 1.13

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Fiona Latifi 1A
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Joined: Sat Sep 14, 2019 12:16 am

Focus 1 Exercise 1.13

Postby Fiona Latifi 1A » Mon Oct 21, 2019 2:12 pm

This question asks us to explain why the ionization energy for oxygen is lower than that of either nitrogen and fluorine. I do not understand the explanation that the solution manual provides. Can someone explain this anomaly in a different way? Thanks!

asannajust_1J
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Re: Focus 1 Exercise 1.13

Postby asannajust_1J » Mon Oct 21, 2019 2:16 pm

IE for Oxygen is lower than N and F because having a half full or completely full orbital is more stable and favorable. Thus, since oxygen has 4 valence electrons, it is 1 electron too many to have a half full orbital (3 e-out of 6). Therefore, the Ionization Energy (energy to remove an electron) is lower since it "wants" to lose an electron to accomplish this more stable state of a half-full p orbital.

Ananta3G
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Joined: Wed Sep 18, 2019 12:19 am

Re: Focus 1 Exercise 1.13

Postby Ananta3G » Mon Oct 21, 2019 2:21 pm

In other words, adding another electron to an already half full orbital of electrons, it causes electron-electron repulsion which makes it easier for an electron to be removed or for the ionization energy to be much lower.

RoshniVarmaDis1K
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Re: Focus 1 Exercise 1.13

Postby RoshniVarmaDis1K » Mon Oct 21, 2019 4:20 pm

It makes sense that O would have a lower first ionization energy than F because we know that ionization energy increases from left to right across a period.

O has a lower ionization energy than N because O is the first element to have a paired electron in the 2p orbital. This creates extra electron-electron repulsion that lowers the ionization energy.


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