Molecular Orbital diagram for CO ```Name: susan m buta Status: N/A Age: N/A Location: N/A Country: N/A Date: 1999 ``` Question: At a recent AP Chemistry conference the Molecular Orbital diagram for carbon monoxide was shown. The Atomic orbital diagram for oxygen was lower in energy then the carbon atom. Someone suggested that O has lower energy because of the increased nuclear attraction and the inverse square law predicts the increase of energy from like-charged electron repulsion. I disagreed with the logic because oxygen, with more electrons, would then be of higher energy then carbon. The instructor agreed but could not answer my question with an alternative answer. Can anyone help me with a better explanation? Replies: In the "Molecular Orbital" approximation, one guesses at the relative energies of the orbitals of the valence electrons in order to guess how the molecular orbitals will add together. The way that is usually successful is adopt the following rule; electronegativity is a measure of how strongly at atom within a molecule tends to attract electrons to itself. Thus, the atom with the lower electronegativity should have its atomic orbitals lower in energy than the atom with higher electronegativity. This tends to work pretty well overall. In the case of NO, we are talking about a very small difference in electronegativity...0.5 Pauling units. CO is a little more "extreme" with a difference of 1.0. M Robert Mullekin suggested that electronegativities can be predicted from atomic properties. He suggested that the electronegativity of an atom be given by half its ionization energy (IE) minus electron affinity (EA): X(Mulliken) = (IE - EA) / 2 So atoms that tend to pick up electrons easily (large, negative EA) and hold onto them strongly (large positive IE) have a large electronegativity. On the next "slide" let's look at some numbers for C and O. ```Atom Pauling EN IE EA Mulliken EN (dimensionless) (kJ/mol) (kJ/mol) (kJ/mol) C 2.5 1086 -122 604 N 3.0 1402 0 701 O 3.5 1314 -141 728 ``` As you can see, the trends are the same in the two scales. So Mulliken's formula really does predict how trends in electronegativity go. Now, notice that O > N > C. Second, notice that the trends are not uniform in either IE or EA! So, it is not sufficient to simply view the matter as due to an increase in Z. In fact, this little "blip" occurs in EVERY PERIOD in the periodic table...one way to interpret the data is to note that the N atom has a half-filled shell, an extremely stable situation due to a special property (probably not discussed in your training thus far) called Fermion exchange. In any event, the usual way to get the ordering of valence shell orbitals "right" for adding them up to form MOs is to use electronegativities. Since X_O > X_C, oxygen's levels should be placed lower in energy than carbon's. The most extreme example of this effect might be HF, in which there is such a big difference in Pauling electronegativity (about 2) that the 1s orbital of H is similar in orbital energy to the 2px/2py/2pz orbital energies, and thus bonding involves the combination/overlap of H(1s) with F(2pz). I hope this helps! This was an excellent question. -topper Click here to return to the Chemistry Archives

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