ionization energy

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Monica Soliman 3F
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Joined: Wed Sep 30, 2020 9:57 pm

ionization energy

Postby Monica Soliman 3F » Sun Nov 01, 2020 7:30 pm

can someone explain this question?
Screen Shot 2020-11-01 at 7.30.16 PM.png

Melody Haratian 1B
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Re: ionization energy

Postby Melody Haratian 1B » Sun Nov 01, 2020 9:26 pm

Hello! Oxygen will also have a lower first ionization energy than fluorine because of fluorine’s higher atomic number. Because fluorine has a higher atomic number, it has more protons, thus a higher effective nuclear charge. This positive nuclear charge will attract electrons more, and result in a higher ionization energy ( energy needed to remove the electrons). Electrons experiencing a stronger effective nuclear charge are harder to remove, and require more energy to be removed.

Oxygen also has electron-electron repulsions in its 2p orbitals, so that contributes to its lower ionization energy.

Marc Lubman 3B
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Re: ionization energy

Postby Marc Lubman 3B » Sun Nov 01, 2020 9:39 pm

This question asks you to consider what is going on in the Oxygen's orbitals, as well as its other properties in order to explain why it takes less energy to remove one electron from it than to do the same to fluorine or nitrogen.
The first option, mentioning electron-electron repulsion, makes sense because oxygen has four electrons in p orbitals. There are three p suborbitals, and when electrons are filling orbitals they prefer to be alone in their own orbitals because two negatively charged particles repel each other. Since oxygen has 4 p-electrons, one p suborbital is occupied by two electrons and thus it's eager to lose one of those to get to the more stable arrangement of one electron in each suborbital.
The second option doesn't make sense because not only does oxygen have a greater atomic radius than fluorine, if it did have a smaller radius than fluorine then it would have a higher first ionization energy since the electrons would be more tightly bound to the nucleus.
The third option actually describes the scenario I just mentioned in regards to the second option. Effective nuclear charge is how much the positive charge of the nucleus attracts the negatively charged electrons, and since fluorine has more protons in the nucleus and a smaller atomic radius, its outer electrons are more attracted to the nucleus and thus it takes more energy to remove one of those tightly bound electrons in fluorine than it does in oxygen.
The fourth option doesn't make sense because if it were true, then those shielded electrons in fluorine would experience less attraction to the nucleus and would be easier to remove, so fluorine would have a lower first ionization energy than oxygen. Also, fluorine's valence electrons are in the same orbital as oxygen's, so there's no reason that they would be more shielded from nuclear charge.

Sam Wentzel 1F 14B
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Re: ionization energy

Postby Sam Wentzel 1F 14B » Sun Nov 01, 2020 10:16 pm

First Ionization Energy means the energy required to remove the outermost electron of the atom.

Oxygen's electron structure is 1s2 2s2 2p4.

This means that the outermost electron of that 2p sub-shell is subject to to more electron-electron repulsion than is the outermost electron in nitrogen.

Flourine has more protons than oxygen, so its force of nuclear attraction will be higher, "drawing" its electrons in with more strength than does oxygen's nucleus.

Choose the options that show why Flourine's last electron is harder to remove than Oxygen's.

Hope this helped!

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