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A question on oxygen electron binding energies

  1. Feb 18, 2009 #1
    Hello, I'd like to ask a question about the electron binding energies of oxygen atom. As is known, when an oxygen atom gains one additional electron to form an anion with charge -1 it gives some 1.4ev, while it gives nearly 7.8ev if two additional electrons are bound to it. Considering that the second electron should feel a strong repulsion as it gets closer to the negatively charged anion, this seems remarkable. The only possibility to resolve this seems to take into account additonal energies. But what are they? And what are their magnitudes?
  2. jcsd
  3. Feb 24, 2009 #2


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    Well, that's the Octet Rule from high-school chemistry. O2- has eight electrons and is more stable than O- because it has a filled shell. Searching, this has come up before, but I don't seem to find a satisfactory answer.

    Classically, your observation would be correct. However, it was the complete failure of classical physics in explaining atomic phenomena which lead to the development of quantum physics in the first place. Bound electrons can't have just any value for their energy, and this has some pretty big consequences.

    The states (orbitals) that they're 'allowed' to occupy are characterized by four 'quantum numbers',
    n, l, m and s. Which are the principal, azimuthal, magnetic and spin quantum numbers.
    The following rules apply: n is an integer, l is an integer <= n, m is an integer from -l to +l. and spin is -1/2 or +1/2. Each n corresponds to an old-fashioned electron 'shell', n=0 is the K shell, n=1 is the L shell and so on.

    If n=0, then l=0 and m = 0 and we have only the two possible spin states. The K shell can contain two electrons. (where l=0 these are known as 's' orbitals) This is the valence shell for hydrogen and helium.

    When n=1, you have another pair of s obitals for l=0, m=0, and you also have n=1, l=1, m=-1,0,+1 - which (together with spin) holds six electrons (l=1, 'p orbitals').
    So the L shell (the second row of the periodic table) holds 8 electrons in total and O2- has a filled L shell.

    Okay, now to answer your question: Why does a filled shell have low energy?
    The answer to this is in what the quantum numbers represent. The magnetic ('m') quantum number represents the orientation of the electrons angular momentum. In an oxygen atom with 7 electrons, m=0 is occupied, together with either m=1 or -1 (it doesn't matter). In other words the electrons of the atom have what's called orbital angular momentum.

    But if the orbital is filled, the +m and -m electrons angular momentum will cancel out. So an atom with a filled shell has no (total) orbital angular momentum. And that's the main contributing factor to the lower energy.

    There's a second factor though - spin. (which is also a kind of angular momentum, sort of). An atom with an odd number of electrons (a radical in chemistry terminology) also has electronic spin. Whereas if it has an even number of electrons, the electrons can form a pair with -1/2 and +1/2 and cancel out their spin as well. Which also lowers the energy. (In chemistry that's the 'spin-pairing energy') This is also why radicals tend to be chemically reactive.

    I can't give any exact quantities offhand, but the spin pairing energy is generally the smaller of the two. (they're not entirely independent quantities) Note though, that your observation also plays a role: Charge matters, which is why elements on the edges of the periodic table (closer to a closed-shell configuration) are easier to oxidize or reduce, and why oxygen naturally 'prefers' to gain two electrons rather than lose six of them.
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