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Example · Example 9

Q.Write the set of four quantum numbers for the last (differentiating) electron of a chlorine atom (Z=17Z = 17).

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Chlorine has Z=17Z = 17 electrons, giving the configuration 1s2 2s2 2p6 3s2 3p51s^2\,2s^2\,2p^6\,3s^2\,3p^5. The last (differentiating) electron is the fifth electron entering the 3p3p subshell.

By Hund's rule, the three 3p3p orbitals (ml=−1,0,+1m_l = -1, 0, +1) are first filled singly, one electron each with parallel spin (electrons 1-3, all ms=+12m_s = +\tfrac{1}{2} by convention). The fourth and fifth electrons then begin pairing, starting again from ml=−1m_l = -1: the fourth electron pairs into the ml=−1m_l = -1 orbital and the fifth into the ml=0m_l = 0 orbital, each with the opposite spin, ms=−12m_s = -\tfrac{1}{2}. So the fifth (last) electron occupies ml=−1m_l = -1 (paired with the first electron placed there) with ms=−12m_s = -\tfrac{1}{2}.

For this last electron: …

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