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Molecular orbital configurations (bond order and electron config.)
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Apr21-12, 01:42 PM
I'm a little confused about calculating the bond order and choosing the correct electron configuration. Compare these two calculations (from my text book):
http://oi44.tinypic.com/2vt8krp.jpg and http://oi39.tinypic.com/sv19hc.jpg
As you can see, the electron configuration for O2 sums up to 16 electrons, the total number of electrons. Looking at the electron configuration the bond order then is just (8-2)/2 = 2.
But for NO they're not summing the total electrons, but rather the valence electrons through an orbital diagram, which then gives bond order (8-3)/2 = 5/2. Do you use different methods because O2 is a homonuclear diatomic molecule and NO is heteronuclear and is this a general rule?
Additional question: I thought p orbitals came as three and three (px, py, pz), which would suggest 3*2 = 6 electrons. Why do we limit ourselves to 2*2 in this case?
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