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Now we resort to hunds rules rule 1 maximize spin so

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Unformatted text preview: l configuraBon, then add one electron with l=1. That gives L=1, S=½. But is J=L+S=3/2, or J=L ­S=½? Now we need the magic of Hund’s Rules…. Z 1 2 3 4 5 6 Element H He Li Be B C configura0on (1s) (1s)2 (1s)2(2s) (1s)2(2s)2 (He)(2s)2(2p) (He)(2s)2(2p)2 2S+1L J 2S 1/2 0 1S 2S For carbon this means L=2, 1, or 0. (D, P, or S) For the two spins we have either S=1 or S=0. Thus the allowed values of J are J=3, 2, 1, or 0. Now we resort to Hund’s Rules. Rule 1: maximize spin, so S=1 (as long as this doesn’t force two electrons to be in the same state). Rule 2: Maximize L. However with S=1 and L=2, both 2p electrons would have the same quantum numbers: nlmlms=(2,1...
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