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Postby ChristianM3F » Mon Oct 28, 2019 10:32 pm

Ok, so the solutions manual explained why oxygen has a lower ionization energy than fluorine and nitrogen, and basically said that since oxygen is the first element encountered in which the p-electrons must be paired.
So if it's based on p-electrons being paired creating repulsion energy, then why wouldn't fluorine have a lower ionization energy..?
Also, if this is just a special case for oxygen not following the ionization energy pattern, is there any other special cases I should remember?

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Re: 1.13

Postby 905416023 » Mon Oct 28, 2019 11:41 pm

Honestly I don't understand this either. It seems like there are so many exceptions to rules in Chemistry and it just makes no sense.

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Re: 1.13

Postby Mashkinadze_1D » Tue Oct 29, 2019 2:27 am

The way you can see this is because oxygen is going to be the fourth electron in the p orbital, it will become less geometric. The two most favorable states for the orbitals to be occupied as is half full or full. Nitrogen is half full and therefore it will be harder to take away an electron because it is in a relatively stable configuration, whereas oxygen has one electron that is making it less geometrically favorable and therefore it will be easier to take away. Florine is significantly higher because it is one electron away from having a full shell and being similar to its noble gas in configuration and therefore would not want to give up one of its electrons, while in addition the increase in effective nuclear charge also makes fluorine be greater than oxygen as is seen in almost all elements when we go from left to right and the ionization energy increases. Hope this helps!

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