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Quantum Numbers and Valence Electrons

  1. Oct 21, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    1) Apparently it's a true statement that the quantum numbers 2, 0, 0, 1/2 can apply to any of Cl's electrons. But chlorine's electron configuration is [Ne]3s^2 3p^5. What happened to the n = 3 electrons?

    2) How many valence electrons can a ground state oxygen atom have with the quantum numbers m_l = 0 and m_s = +1/2? Those are the m-sub-l (magnetic) and spin quantum numbers, respectively.

    2. Relevant equations

    For l = 0, m_l can only = 0.

    For l = 1, m_l can equal -1, 0, or 1.

    l = 0 corresponds with subshell s
    l = 1 corresponds with p.

    3. The attempt at a solution

    For the first question, I'm thinking the question is only considering "core" electrons without explicitly stating it as such.

    Second question: oxygen has 6 valence electrons. 2 are in its 2s subshell. Four are in its 2p subshell. Exactly half the electrons of each subshell have a spin of +1/2. The other two electrons have opposing spins.

    The question only specifies that m_l = 0. That means l = 0 or 1. Both the s and p subshells must be considered. In that case there are 4 electrons that fulfill the criteria of having m_l = 0 and a spin of +1/2. One electron from the s subshell, and 3 from the p subshell. The key says 2, however.
     
  2. jcsd
  3. Oct 21, 2013 #2

    Qube

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    Okay scratch that about question 2. I see what I did wrong. Key is correct. Magnetic quantum numbers correspond to the orientation of an orbital; for the s orbital it's spherical so there's only one orientation. P orbitals have three orientations, Px, Py, and Pz. m_l = 0 specifies exactly one of these orbitals.
     
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