1.13 Ionization energies usually increase on going from left to right across the periodic table. The ionization energy for oxygen, however, is lower than that of either nitrogen or fluorine. Explain this anomaly.
I've read explanations online but I don't really understand what they mean. Can someone explain in a really simple(?), easier way?
Textbook Question Focus 1.13
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Re: Textbook Question Focus 1.13
Oxygen is one of the exceptions due to the electron repulsion that it experiences in the 2p orbitals.
Nitrogen has each of its orbitals in the 2p subshell half-filled.
Oxygen, on the other hand, has 4 e- in the 2p subshell and will have a paired e- in one of its orbitals.
This additional electron added to the orbital results in more e- repulsion and makes it so Nitrogen has higher first ionization energy.
Nitrogen is more stable with all half-filled orbitals and will require more energy to remove an e-
Nitrogen has each of its orbitals in the 2p subshell half-filled.
Oxygen, on the other hand, has 4 e- in the 2p subshell and will have a paired e- in one of its orbitals.
This additional electron added to the orbital results in more e- repulsion and makes it so Nitrogen has higher first ionization energy.
Nitrogen is more stable with all half-filled orbitals and will require more energy to remove an e-
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Re: Textbook Question Focus 1.13
From my understanding, O has lower ionization enthalpy than N because N has a half-filled electron configuration. The lower ionization enthalpy of O is less than F because of the higher electronegativity and high effective nuclear charge of F. Hope this helps!
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