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Principal Quantum Number of the Excited Oxygen atom.

  1. Apr 13, 2013 #1
    Normally, Oxygen has 8 electrons in its neutral form that is 1s2 2s2 2p4. In this case, its principal quantum number (n) is two.

    But what happens if it got excited and its electronic configuration becomes 1s2 2s2 2p3 3s1? In this case, Is the principal quantum number (n) of oxygen two or three?

    Thanks.
     
    Last edited: Apr 13, 2013
  2. jcsd
  3. Apr 14, 2013 #2
    Quantum numbers are used to describe the state(s) of electrons not whole atoms AFAIK. You may be referring to the fact that many Chemists will refer to things such as 2p elements etc, this is really just a reference to their place in the periodic table which may be useful in certain conversations.

    It really doesn't make sense to give a quantum number to an entire atom/molecule as these numbers are used to describe the state of the electron(s) in an atom/molecule. For instance a 1s electron has n=1 (principle quantum number), l=0, ml=0 and ms= +/- 1/2. A 2s electron will have the same numbers except n=2. A 2p electron is n=2; l=0,1; ml=-1, 0, 1 and ms=+/- 1/2.

    One more thing to keep in mind is that Oxygen is diatomic, meaning its elemental form is O2 so a description using atomic orbitals is severely lacking. A more thorough description of Oxygen and its excited states should make use of molecular orbital theory.
     
  4. Apr 14, 2013 #3
    Oh, you are right. I should have asked like this: the electron of oxygen atom that is denoted by n=2, l=1 transferred to n=3, l=0 after the excitation process. At this moment, what is the electron's principal quantum number, two or three? But I think I got it after your message. It will be three.
     
  5. Apr 14, 2013 #4
    I also misspoke actually. A 2p electron would have n=2, l=1, ml=-1,0,1 and ms=+/- 1/2. Mistake being that the letter "p" denotes the azimuthal quantum number (l) of 1.
     
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