Readings for this section.
Petrucci: Section 87, 88
So far, we've seen that we can explain some experimentally observed properties using simple models like Lewis Dot structure and VSEPR. These models still do not explain many chemical properties. In this section, we will develop a new model that explains further some of the chemical reactivities we observe and also ties in our knowledge of atomic orbitals and expands it further.
So far, we've been saying that molecular bonding involves atomic orbitals in some way. We will now explore atomic orbitals and develop an understanding of their involvement in molecular orbitals (e.g. bonds).
Lets look at the VSEPR models identify the places where they are lacking in explanations. We'll then look at the model of localized orbitals and see how this new model is better.
To make a bond, we must use an atomic orbital from each atom involved in the bond. What do these atomic orbitals look like?
In the molecule BeCl_{2} , we saw that there are two electron domains on opposite sides of the beryllium atom involved in the bonds to the two chlorines. How do these domains arise? Our atom of beryllium has atomic orbitals for its valence shell (n = 2) of type s and p. If we try to make bonds using two of these orbitals, we find that we cannot come up with two orbitals that are identical and have lobes on opposite sides of the Be.
Image of probability domains of electrons in orbitals 2s and 2p. The color indicates the phase of the wavefunction (the wavelike properties of the electron). Inphase functions add positively (constructive interference) and out of phase functions add negatively (destructive interference). 
If we use one of these orbitals for one bond and the other for the other bond, we will obviously not have two identical bonds. We obviously need another model for the atomic orbitals in order to explain their participation in bonding. One thing many students forget when thinking of this process is that the orbitals them selves are not really entities and are not really the thing that matters. What matters is the overall energy of the system and its symmetry. An atom in free space is symmetrical (as much as its electrons will permit. The atomic orbitals s, p, d, f,... are merely the basis functions that we use to add up to the overall space occupied by the electrons. Note that any sub shell always adds up to be spherically symmetrical. We can easily choose a different set of basis functions (orbitals) to describe the space occupied by the electrons. Whatever basis set we use must exactly define the same space as our spdf model. We are free to divide up the space any way we want in order to more easily understand and calculate what we need.
What
we need to do is use different (linear) combinations of the spdf domains
to create new domains (orbitals). We divide up the space (sphere) in such a way that each domain (orbital) of the sphere is involved in only one bond. This makes it much easier for any calculations we might have to do. Let's first consider what happens in the BeCl_{2} case: First, we overlap the 2s and 2p in two different ways. (remember, we're just creating new functions, not moving electrons)
Here, we see that the two different ways of adding up the orbitals result in the phases interacting indifferent ways. In the 2s+2p case, the lefthand side will be smaller since the phases of the original s and p are opposite and vice versa for the 2s2p case. The two new orbitals we get will look like the following picture.


As an analogy, let's consider a point in 2 dimensional cartesian space.
If the point is represent in the normal set of cartesian coordinates (basis set), we need an x value and a y value. On the other hand we can represent the same point using the alternate (rotated set) coordinate system (alternate basis set) In the rotated system, notice that one of the coordinates is zero (It still exists though.) and we can more easily ignore it in any calculations involving the point. We do the same with the space occupied by the atom's electrons. Redefine the basis set we use so that we can simplify the mathematics.


Now, we can see that the two different orbitals (coloured blue and green) still have properties that look like the p orbitals (two lobes, of different phase) and some properties like an s orbital (large bulbous, surrounding the nucleus (a bit). Most importantly, the two will sum up to make the same shaped space as has using the spdf model so these two orbitals will exactly describe the same space as the original s and p orbitals did. Now, we can look at using these to make molecular bonds.
In the case of BF_{3}, we have three fluorenes bonded to the central boron with what are experimentally observed to be identical bonds. Again, we cannot find a way of using the spdf model to easily describe the orbitals used for this so we will reslice the sphere to make use of the space in a new way. First, we need only use three orbitals to create three bond so lets use s,p,p and leave one p orbital untouched on the atom.
An s and two p orbitals add up their shapes to give a fat disk (note the color coding is used here to indicate different orbitals, not different phases). The three orbitals are 120? apart. 
In the diagram here, we show the electron densities only resulting from each of three orbitals, s, p, p. colored differently If we add up the three sets of densities we get a picture of the space we need to divide up. It is a thick disk, sort of like a ball what is somewhat squashed in one direction. We now divide this up into three equal domains.
Three sp^{2} hybridized orbitals are made from an s and two p orbitals. Each hybrid has some p character and some s character. 
and see that these three also add up to regenerate the original fat disk the the spp did.
Here, we see the three orbital spaces added back together. Color coding helps remind us of the original sp^{2} hybridized orbitals. 
The next set of hybrid orbitals is not so easy to show using these electroncloud diagrams since the resulting set of hybrids will be evenly distributed in three dimensions rather than in just two. Hence, I will resort to stick diagrams to represent these orbitals.
In the molecule CH_{4} there are four equal electron domains on the carbon each involved in one bond with a hydrogen. How do we explain these using localized atomic orbitals.
First of all, recall that each sub shell will sum up to be a sphere. That means that if we add all the p orbitals with the s orbital for the n=2 level (or any n level), we still get a sphere. We divide the sphere into four equal parts and come up with domains that are 109.5? apart.
Here are the four sp^{3} hybrid orbitals after dividing up the space created by the s and the three p orbitals. The four new orbitals are all equal and have axes that are 109.5º apart. 
More complicated hybridization can be found for atoms in the n > 3 shell. In these shells, there are d orbitals that can participate in the hybridizing. Thus, if an atom forms 5 bonds with 5 other atoms, it will need 5 atomic orbitals to do this. We take s, p, p, p, d of our n = 3 available orbitals and combine them. The resulting set of sp^{3}d hybrids will have a trigonal bipyramidal arrangement exactly as describe in the VSEPR section. Finally, if there are six bond to six atoms, the hybridization will be sp^{3}d^{2} (6 atomic orbitals needed) and the geometric arrangement of the electron domains will be octahedral (see VSEPR section).
In the Valencebond approach, we assume that the molecules are made up by simply overlapping the atomic orbitals (hybridized if necessary) from the individual atoms. In this approach, the electrons from one atomic orbital AO don't interact with those of other AOs. We get great threedimensional pictures of what the molecules and bonds may look like but the information we can get from this approach is still not perfect. We'll see MO theory later that can be used in parallel to obtain other information that this approach does a poor job at calculating.
We've seen various ways of visualizing atomic orbitals. It is important to remember that these orbitals are only models of what happens on an atom in the gaseous state (unattached to anything). How we deal with bonding these atoms together to form molecules is a whole different story again.
To form a bond between two atoms, we must overlap one atomic orbital (hybridized or not) from each of the two atoms that are bonded.
Two H atoms approach each other. Their Atomic Orbitals begin to overlap.  
The overlap space is shaped like a rugby ball. This is a σ_{1s} bond. made by overlapping two 1s orbitals, one from each H atom. 
If we try to use VB theory to model the helium molecule, we run into problems. We see no place to put the four electrons as we only have one bond. Perhaps there is a lone pair, or something else. Experiment shows us that the He_{2} molecule does not exist in nature so that seems to correlate with the lack of places for all the electrons.
We will revisit this further. First, let's look at a different type of bond. The σ bond we just looked at is a bond that has it's major lobe on the axis joining the nuclei of the two atoms. A σ bond can be formed by the overlap of any type of orbitals contributed from the bonded atoms as long as the overlap occurrs along the axis joining the two nuclei. This is in contrast to a π bond where the overlap is above and below (or on opposing sides) of the axis.
Here, the nuclei are represented by the two black dots. In the bond, we see that the overlap occurs off axis. It may be noticeable that the overlap is not as great as it could be if the lobes were overlapping endtoend (σ bond)
This time, the overlap occurs along the bond axis so this is a σ bond. We now see that the definition of σ and π bonds does not depend on the type of orbitals used in their creation. They refer to the location of the overlap region with respect to the bond axis.
There are two different models that Chemists use when describing bonding in molecules. The one they choose depends on their purpose. For some purposes, VB theory is easiest; hybridized orbitals coming together to form σ and the leftover atomic orbitals (p and d) form π bonds. This is most easily seen in organic molecules. The simplest of these is, of course, CH_{4} in which we've already seen the carbon atom to be sp^{3} hybridized and the carbon sp^{3} orbitals form σ bonds with the hydrogen s orbital.
What about a molecule with double bonds? Let's look at the simplest of these; C_{2}H_{4}.
If we draw these orbitals, using solid lines to represent the shape of the orbitals rather than the electroncloud images, we get something like the following.
First, we hybridize an s and two p orbitals on each carbon to form trigonal planer sp^{2} orbitals which we use in making σ bonds.
The σ bonds on the ethene molecule.
Now we look at the p orbital from each carbon that we didn't yet touch and use it to form a π bond.
The π bonds on the ethene molecule
The complete picture looks like the following
The σ and π bonds on the ethene molecule
This scheme does a good job of helping us understand the properties of this molecule. We can see that the molecule is rigid and planar. It cannot rotate about the central C=C double bond because of the way the p orbitals overlap. We also see that the bond angle should be ca. the trigonal planar angle of 120?. the HCH angle is actually slightly smaller while the CCH angle is larger.
Triple bonds are even harder to draw. Consider the molecule ethyne C_{2}H_{2}. The carbons on this molecule are sp hybridized, leaving two p orbitals on each to form two π bonds. The final picture looks something like the following diagram.
The σ and π and π bonds on the ethene molecule.
Again, we see this model does a reasonably good job of modeling the shape of the real molecule. Ethyne is linear and has a C≡C triple bond.
In both the cases previously, the bonds used are localized orbitals.
Molecular Orbital (MO) theory is a more accurate way to use calculations to predict molecular geometries and energies than VB theory. In MO theory, there is no hybridization and all orbitals on the molecule (now called molecular orbitals) can be made by combining any and all of the atomic orbitals from the individual atoms, unlike VB theory where bonds are made from only one orbital from each atom. This allows more flexibility in building up the mathematical picture of the molecule and results in a better match between theory and experimental data than we can get via VB theory.
Let's revisit H_{2} and He_{2} using the MO approach. Using VB, we saw that we were unable to describe He_{2} and we concluded that the helium molecule must not exist. But that was somewhat unsatisfying. Using MO theory, we can build a model for He_{2} and we can use this model to explain why the molecule can not exist in nature.
To form a bond between two atoms, we must combine atomic orbitals from the two atoms in such a way that the energy level of the combination (molecular orbital, MO) is lower than the original atomic orbital (AO). This is most easily visualized using the s orbitals of two hydrogen atoms as they approach each other. Consider, for a moment the following diagrams.
Two H atoms approach each other. Their Atomic Orbitals begin to overlap.  
Overall space is shaped like a rugby ball. This looks the same as we saw in VB theory but now, we need to account for the proper space. We started with two orbitals so we must finish with 2 orbitals. We now divide this space into two new (Molecular) orbitals. 
The best way to do this is such that the energy of the resulting molecule is lowest. In the case of hydrogen, there are two electrons to place in the new orbitals. If we define them such that one has a much lower energy than the other then we will see the stable molecule form when both electrons are situated in the lowenergy orbital while the highenergy orbital is empty.
The division that best accomplishes this is one where the orbitals are added in phase (with normalization) and out of phase as is shown below.
inphase outofphase Molecular orbitals created by taking linear combinations of atomic orbitals (LCAO). On the left is the inphase combination (addition) of the two atomic s orbitals (σ orbital; Bonding orbital) and on the right is the outofphase combination (subtraction) of the same two orbitals (σ* orbital; antibonding). These two add up to give the same space as the two original atomic orbitals but now each is a single 'localized' orbital spanning the whole molecule. 
If we look at the energy of these two orbitals we come up with a correlation diagram like this
The Molecule H_{2} has an electron configuration (σ_{1s})^{2}. It has an overall bond order of 1.
The two electrons from the hydrogen have a lowenergy orbital and a highenergy orbital available to them. Provided they can lose energy (collision with the walls or a third atom or emission of light photon) the electrons will settle into the lower energy level and hence a net release of energy is observed. This released energy is called the bond energy.
The molecular electronic configuration can be named using molecular orbitals just as we saw in individual atoms. In this case, the only occupied orbital in the hydrogen atom is the sorbital that was created from the 1s orbitals of the individual atoms. This molecular orbital (MO) is thus called a σ_{1s} orbital. Thus, since both electrons are in this MO, we have a very simple molecular electron configuration (σ_{1s})^{2}.
If we try the same thing using Helium. we can quickly see from the correlation diagram below why the helium does not form a He_{2} diatomic molecule.
The He_{2} molecule will not remain stable since it's overall bond order is zero. The MO electronic configuration is (σ_{1s})^{2}(σ^{*}_{1s})^{2}, i.e., one bond and one antibond.
The two types of bond theory we're discussing in this section, VB and MO theories give virtually identical results for the H_{2} or He_{2} cases. Both predict the shapes and energies to be the same. The difference comes into play when there are more than only one orbital on each atom as in n ≥ 2. 
There are four electrons in the molecular orbitals which means that while two electrons can go to a lower energy level, the other two must go to the higher one. The total energy is unchanged. Hence, the two atoms don't form a bond at all since there is no net release of energy.
The previous model, for example is not useful at all in describing diatomic molecules like B_{2}, C_{2}, N_{2} and O_{2}. For this, we need a more complete Molecular Orbital approach where all atomic orbitals can interact simultaneously to form molecular orbitals. In this approach, we don't bother with the hybridization. All Atomic orbitals in the valence shell are involved in producing a molecular orbital scheme that mimics the experimentally observed properties. The four molecules mentioned above all use their n=2 valence shell to do the bonding. Let's look at the correlation diagram that result when the 2s and 2p orbitals from two such atoms interact. It is important to note that the MO energy levels (σ_{2p}) and (π_{2p}) may not remain in this sequence for all atomic pairs. For example, the (π_{2p}) levels are higher then the (σ_{2p}) for O_{2} and F_{2}. For our purposes, however, it will be sufficient to always use this one correlation diagram for any n=2 diatomic molecule, recognising that it is not necessarily an exact representation of the energy levels for certain molecules.
First, we'll consider the Boron molecule. It has three electrons from each atom for a total of 6 electrons that we must place in the molecular orbitals. So the final scheme will look like the diagram below.
In this diagram, we can easily see that the MO electronic configuration is (σ_{2s})^{2}(σ^{*}_{2s})^{2}(π_{2p})^{2} for an overall bond order of 1 ( (1 bond + 2 half bonds ? 1 antibond)) but where the (π_{2p}) orbitals contain unpaired electrons (Hund's Rule) because they are degenerate (Identical energies).
This model predicts that the boron diatomic molecule is paramagnetic (two unpaired electrons). This is exactly what is observed experimentally. Note also that the bond order is essentially 1. The σ bond and σ* antibond cancel out, leaving the two π bonds each half occupied. How does this correlate with the Lewis dot structure?
\[\cdot \dot{\mathrm{B}}: \dot{\mathrm{B}}\cdot\]
The molecule of C_{2} is different. Experimentally, it is diamagnetic (no unpaired electrons) and has a bond order of 2.
In this diagram, we can easily see that the MO electronic configuration is (σ_{2s})^{2}(σ^{*}_{2s})^{2}(π_{2p})^{4} for an overall bond order of 2 (3 bonds  1 antibond).
We see that there are no σ bonds in carbon dimer. It is held together with π bonds only. This contradicts the other model where hybridization always predicts the first bond in a multiple bond is a σ bond.
Nitrogen molecules are triple bonded as we can see in the following diagram.
In this diagram, we can easily see that the MO electronic configuration is (σ_{2s})^{2}(σ^{*}_{2s})^{2}(π_{2p})^{4}(σ_{2p})^{2} for an overall bond order of 3 (4 bonds  1 antibond).
We see that there is a (σ_{2p}) bond and two (π_{2p}) bonds holding the molecule together in this model. The (σ_{2s}) and (σ*_{2s}) cancel each other out and don't contribute to the bonding.
Finally, Oxygen is paramagnetic with a bondorder of two. Lewis dot and VB theory doesn't properly predict the paramagnetic properties. A good Lewis dot diagram of O_{2} shows a double bond with two lone pairs on each oxygen.
In this diagram, we can easily see that the MO electronic configuration is (σ_{2s})^{2}(σ^{*}_{2s})^{2}(π_{2p})^{4}(σ_{2p})^{2} (π_{2p})^{2} for an overall bond order of 3 (4 bonds  1 antibond). Again, the upper two electrons are unpaired in the π^{*}_{2p} degenerate orbitals. This molecule is paramagnetic because of Hund's Rule.
Hund's rule says we must put the last two electrons to into the π^{*} orbitals one at a time since they are degenerate orbitals. This gives us two unpaired electrons (Paramagnetic) and lowers the overall bond order to 2.
We can use these same diagrams for diatomic molecules or ions for any combination of n=2 elements. for elements of n=3 or n=4, there is no guarantee that the sequence of the MOs will remain the same. It will always remain as pictured above for n=2 valence levels.
Let's Try the NO molecule:
In this diagram, we can easily see that the MO electronic configuration is (σ_{2s})^{2}(σ^{*}_{2s})^{2}(π_{2p})^{4}(σ_{2p})^{2}(π^{*}_{2p})^{1}
for an overall bond order of 2.5 (4 bonds  1.5 antibonds). Here, the upper
electron is unpaired in one of the two π^{*}_{2p} orbitals.
They are no longer degenerate since one is empty. This molecule is paramagnetic,
containing an odd number of electrons.