Atomic orbitals come from treating electrons with the mathematics of waves, so orbitals have phase signs, like waves.

- A positive sign means a positive displacement of the wave ("up") and a negative sign means a negative displacement ("down").
- So that we do not confuse these signs with plus and minus charges, it is customary to represent signs with colors. Which colors doesn't matter, as long as we choose one color for positive and some other color for negative.

A p orbital, as shown, consists of two parts, usually called lobes. The two parts have opposite phase signs; which is positive and which negative is irrelevant. All that is significant is that they are opposite.

The place where the phase signs change is called a node.

- In the p orbital it lies in a plane perpendicular to the long axis of the orbital, and containing the nucleus.
- The plane is called the nodal plane.
- Nodes are places where the probability of finding the electron is
*zero*. The electron can't be there. - So, an electron in a p orbital can be either in the blue lobe or the magenta lobe, but cannot be in between.
- Its energy is the same no matter where it is in the orbital.

When atoms combine into molecules, their atomic orbitals (AOs) combine into molecular orbitals (MOs), which extend over two or more atoms.

- The number of MOs produced must equal the number of AOs combining, because we must have the same number of places to put electrons in the molecule that we had in the atoms.

In general, we can make two kinds of MOs from p orbitals. If we interact the AOs in an "end on" orientation, we get an MO that is symmetrical to rotation around a line connecting the nuclei.

Orbitals of this symmetry are called s MOs.

- As we shall encounter them, they typically involve only two AOs.
- Since two AOs are used, we must create two MOs; one will be lower in energy than the two AOs, the other higher.
- If each AO contains a single electron, as shown, the two incoming electrons can be put in the lower energy of the two MOs.
- The energy of the electrons will be lower than when we started, so the interaction is favorable.
- The two atoms are bonded; the orbital containing the electrons is a bonding MO.
- If additional electrons had to be accomodated, they would have to go into the higher energy orbital, which has a node between the nuclei.
- The increase in energy for these electrons would cancel the lowering of energy of the first two electrons, resulting in no bond formation. The higher energy orbital is an antibonding MO.

We also can make MOs from p orbitals by pushing them together sideways, as shown in the next picture. Here we introduce a convenient convention: instead of trying to draw the actual shape of the MO, we picture it by drawing the AOs from which it was made, in a way that illustrates the properties the MO will have.

- Again, from two AOs, we get two MOs, one bonding and one antibonding.
- As before, the antibonding orbital has a node between the nuclei.
- These orbitals are
*not*symmetrical about the internuclear line; we call this type p-orbitals.

Once again, if each atomic orbital contributes a single electron, as in ethylene (CH_{2}=CH_{2}), we can place both electrons in the bonding orbital, lowering their energy relative to what it was in the AOs.

It is hard to visualize how we might combine three AOs to make s-MOs, and indeed, three or more atom s-MOs are quite rare.

However, it is very easy to combine more than two p orbitals to make p-MOs; in fact, it is possible to imagine pushing an unlimited number of p orbitals together. Let's stay simple and use three, such as in allyl, CH_{2}=CHCH_{2}. Systems with three or more p orbitals on adjacent carbons are called *conjugated*.

Here we have viewed creating a set of three p-orbitals from three p AOs by taking the two p-MOs created above, and adding one more p.

- The addition can be done schematically in several ways; one such is to imagine inserting the additional p between the two AOs making up the two-atom p-MOs.
- Insert the p into the p in a bonding way, and we get the lowest energy orbital, bonding across all three atoms.
- Invert the p and insert it in an antibonding way, and we get the highest orbital, with nodes between each pair.
- Finally, take the p and insert it into the p*-MO; the interaction will be bonding on one side, antibonding on the other.
- This means no net contribution from the p on the middle carbon.
- Hence we draw that orbital with just the two end p's.

*nonbonding* - This means no net contribution from the p on the middle carbon.

If the system is the allyl cation, CH_{2}=CHCH_{2}^{+}, the only electrons are the ones from the double bond in the Lewis structure.

- They go in the lowest MO, the bonding one, and the interaction is stabilizing.

If we add one electron, we get the allyl free radical.

- The additional electron must go in the nonbonding orbital, and does not further stabilize the molecule.

A second additional electron produces the allyl anion. Again, the electron must go into the nonbonding orbital, and does not contribute to stabilizing the ion.

A little more nomenclature here: look at the allyl cation, with two electrons in the lowest energy orbital and the other two orbitals unoccupied.

- The orbital contining the electrons is the highest energy one that is occupied, the HOMO -
**h**ighest**o**ccupied**m**olecular**o**rbital. - The nonbonding orbital, conversely, is the LUMO - the
**l**owest**u**noccupied**m**olecular**o**rbital. - Together, these are the
*frontier*orbitals of the allyl cation, and they control the chemistry of the species. - Nucleophiles will interact with the LUMO, and electrophiles with the HOMO, creating new pairs of MOs.

Finally, here are the MOs for a system built from *four* p-orbitals: butadiene, CH_{2}=CH-CH=CH_{2}.

I am deliberately not showing the mode of formation of these orbitals. You should work this out for yourself, using either a scheme in which ethylene p and p* orbitals are added together together, or inserting one set between the p orbitals of the other. Note that I have marked the nodes with a yellow bar.

Butadiene has four p electrons. Two go in the lowest MO, and two in the next. All electrons thus occupy orbitals lower in energy than individual p orbitals, and thus the molecule is stabilized by the interaction of the two double bonds.

We shall only rarely need p-orbitals for more than a four-atom system. However, if we do, the easiest way in which to derive them is to look at the generalities found in our first three cases:

- The number of MOs is the same as the number of AOs combined to make them
- For an even number of MOs, divide them equally between bonding (lower in energy) and antibonding (higher)
- When the number of MOs is odd, a nonbonding orbital will always be found, with equal numbers of antibonding ones above it and bonding ones below it.
- Orbital energy follows the number of nodes; the lowest always has no nodes and is fullly bonding, the next one up has one node, the next two, and so on, until the highest has nodes between each pair of atoms and is thus fully antibonding
- The nodes will be symmetrically placed: one node goes exactly through the center, two nodes are placed halfway between the center and the ends, and so on. Nodes cancel the contribution of any atom through which they pass

*This page last modified*3:11 PM on Thursday September 30th, 2010.

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