Molecular orbital configurations (bond order and electron config.)

[carle]

Hey guys.

I'm a little confused about calculating the bond order and choosing the correct electron configuration. Compare these two calculations (from my textbook):

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?

Thank you!

 

[jbsteele]

Yes, when it comes to calculating the bond order of homonuclear diatomic molecules, you're looking at the total number of electrons. For heteronuclear molecules, you're usually looking at the number of valence electrons (the outer shell electrons). This is because the electrons in the core shells are not involved in the formation of a bond. When it comes to the p orbitals, the number of electrons in p orbitals depend on the type of molecule. In NO, the two nitrogen atoms and one oxygen atom contribute three electrons each to the p orbitals. In O2, each oxygen atom contributes two electrons to the p orbital, giving a total of four electrons in the p orbitals.