Lecture Notes from CHM 1341
3 July 1996

Molecular Orbitals

When atoms come together for either bonding or bouncing, their electron orbitals distort. This is easy to do because the electrons are thousands of times less massive than the nuclei, which seem to electrons to be lumbering behemoths. So the nimble electronic orbitals adjust almost as if the atoms weren't moving at all.

What they adjust to is what's interesting. That diagram above shows two hydrogen atoms apart then together; in the latter configuration, they've formed a bonding molecular orbital out of a combination of the (half-filled) 1s atomic orbital on each atom.

Clearly, this bonding has involved the valence electrons (is there any other kind in hydrogen?!?) being drawn toward both nuclei simultaneously. As long as the interatomic spacing isn't too close, their extra presence there helps the electrons shield those nuclei from one another. If the distance gets too cozy, not only do the nuclei repel but also confining the electrons raises their energy by constraining their wavelength. But at about 0.74 Å (that's 74 pm), the attractions and repulsions on the system are balanced and a stable hydrogen molecule is possible.

But there are other types of valence orbitals in other atoms. Take chlorine, for example; it can bond to another Cl via that half-filled 3p orbital. And that's what's happening in the diagram below.

Since the p orbitals are positive on one lobe and negative on the other (hey, they're waves, right?), we have to make sure that they interfere constructively when they make that molecular orbital bond! So we've organized 'em head-to-head, as it were.

Although we'll not pursue it in this course, it's instructive to think what would happen if we'd overlapped 'em head-to-tail...they'd have cancelled one another out in the very place where electron density is critical - between the nuclei. What that means is the electrons would be less likely to be found there, and not only would the molecule not be interested in bonding, it would be actively DISinterested in any nuclear approach. The chlorines would fly apart. Later courses will call this antibonding.

Notice that both molecular orbitals formed so far have been called "sigma" bonds. All that means is that looking up the tailpipe of either one (along the bond rather than above it as we are here), you'd see no nodes. Notice the contrast to the molecular orbitals made from one of the other p orbitals. (They're all perpendicular to one another, remember?)

This is how multiple bonds are really formed! (Tau bonds? Erase erase erase!) Instead this sideways overlap of p orbitals from neighboring atoms will produce an entirely new kind of bond called a pi bond. Since it isn't on a line between the nuclei (p's have a node there), it isn't quite as effective as a sigma bond. So double bonds, which are a sigma and a pi bond between the same two atoms, wouldn't be expected to be as strong as two single bonds...and it isn't.

Also the pi bonds are stuck out there vulnerable to chemical attack; indeed double and triple bonds are far more chemically reactive than single bonds.

They're called "pi" not just to distinguish 'em from "sigma" but to suggest that, just like the p orbitals from which they can be formed, a view up the tailpipe will show ONE node in the molecular orbital. And don't make the mistake of thinking that the two lobes of the pi are any more than a single orbital. After all atomic orbitals are whole only if count all their lobes.

And these molecular orbitals (MO) work just like atomic orbitals as far as Pauli's and Hund's rules are concerned: no more than two and preferably (with MOs of the same energy) only one electron occupancy.

Speaking of MOs with the same energy, that pi orbital above could also have been constructed from the p atomic orbitals lying not in the plane of the screen but rather perpendicular to it. That pi would be identical (in energy and everything else) to the one shown above, but it's "sausages" would lie above and below the screen. Indeed, this 2nd pi together with the 1st and a sigma bond between the same two atoms would constitute a triple bond!

Hybrid Orbitals

Molecular bond formation isn't all that happens to atomic orbitals as other atoms approach. The atomic orbitals on the same atom can rearrange themselves into more fortuitous shapes which obey the rules of VSEPR. Before combining across atoms, they can combine within an atom to produce hybrid (atomic) orbitals.

The best example is carbon. C has a ground state electronic configuration of

      1s²  2s²     2p²
     [ud] [ud] [u |u |  ]

where "u" is an "up" spin electron and "d" is a "down" spin one. But that gives C a valence of only 2. If carbon could scrounge from somewhere the small energy which separates the 2s from 2p levels, it could look excited as

      1s2  2s1     2p3
     [ud] [u ] [u |u |u ]

and it'd have a valence of 4 instead. That would mean twice as many bonds and the energy stabilization (lowering) which came with them. Say...carbon robs Peter to pay Paul, using the extra bonding energy to combine its one s and 3 p orbitals into hybrid orbitals like:

     {s + p + p + p}  =  4 new sp3 hybrid orbitals

And that quartet would be situated as far from one another as possible...at the corners of a tetrahedron as you might surmise from the image here. (The two colors are lobes of opposite sign.) So 4-valent carbon makes tetrahedrally bonded molecules if there are 4 atoms to be attached. But what if there are fewer bonding partners? In carbon dioxide, for example, there's only two bond partners, but being double bonded, they bring in more than two bonds worth of stability. How does carbon react then?

It need only hybridize its s orbital with one of its p orbitals to produce what's called (why are we not surprised?) an sp hybrid. Well...actually...2 sp hybrids since we have to make as many hybrid orbitals as we used of the garden variety orbitals. And lo, they direct bonding to opposite sides (linear bonding) of the carbon.

This makes one sigma bonding molecular orbital with each oxygen. But that exhausts the valence of neither atom! Carbon still has its two unhybridized p orbitals which overlap each with p orbitals from an oxygen to form the sigma+pi double bond with both.

And what do the oxygen atoms do all this time to prepare for this mating? They hybridize their atomic orbitals as well! Remember, oxygen has

      1s2  2s2     2p4
     [ud] [ud] [ud|u |u ]

So it hybridizes an s and two p orbitals to produce three new hybrids in the trigonal planar configuration which places them as far apart as possible. Into two of these new hybrids, it places its lone pair electrons, but into the 3rd, the one which will mate with carbon in a sigma molecular orbital, it places an unpaired electron. The remaining unhybridized p on oxygen forms the second half of the C=O double bond by forming a pi bond with one of carbon's free p orbitals (perpendicular to the screen from this perspective).

This new hybrid is as unimaginatively named as the other two:

     { s + p + p }  =  3 new sp²  hybrid atomic orbitals

And there was nothing unique about oxygen that it can do this. Had we chosen to bond carbon to 3 other atoms, as in formaldehyde:

     H
      \
       C = O:  then C (and O) would hybridize sp²
      /    ¨
     H

and carbon's hybridization would be manifest in the trigonal planar bonding of that molecule!

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

   Voice: (214) 883-2485
     Fax: (214) 883-2925
     BBS: (214) 883-2168 (HST) or -2932 (V.32bis)
Internet: parr@utdallas.edu  (Click on that address to send Chris e-mail.)