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The ionization energy of Oxygen is lower than that of Fluorine because, as we go across the period, the atomic radii decreases as more protons are added and therefore the nucleus exerts a tighter pull on the electrons bringing them closer to the nucleus. It then requires more energy to remove an electron from the atom (this is one of the trends we learned about). However, there are some exceptions to this rule. One of them then being that Nitrogen has a higher ionization energy then Oxygen. This is because Nitrogen has a half filled 2p state orbital and is more stable since the electrons repel each other equally (they are all in a separate orbitals). Oxygen would readily lose an electron to reach this state, while Nitrogen has a greater resistance to losing stable configuration (therefore higher ionization energy).
So if we were to draw a nucleus on the periodic table to show which elements are closer to it in this case, would it go in the upper right corner of the periodic table or the bottom left? (I think I'm thinking too hard about this)
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