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In my book, Organic Stereochemistry, the following picture is shown:

Overlap of atomic orbitals forming σ and π bonds

In the lowest diagram, in Figure 1.3, which shows a π* bond, the positive lobe of the p orbital of atom A is nearest to the negative lobe of the p orbital of atom B. By contrast, in the diagram above it, which shows a π bond, the positive lobe of the p orbital of atom A is nearest to the positive lobe of the p orbital of atom B. It is said that the π* bond is higher in energy, less stable, than the π bond. I would have thought it to be the other way around, due to electrostatic repulsion when two lobes of p orbitals which have the same charge overlap, and the electrostatic attraction when two lobes of p orbitals which have opposite charges overlap.

Why am I wrong? Why is it not this way around?

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Firstly, the + and – signs you see do not indicate charge. They mark the sign of the electron’s wave function. In an atom, the electrons always have a negative charge, and the nucleus always has a positive one.

When a bonding orbital is formed, the electron density is concentrated in the area between the nuclei. The electrostatic attraction between the electrons and two (instead of one) nuclei, lowers the energy of the molecule. Furthermore, the electron density between the nuclei shields the electrostatic repulsion between the two positive nuclei to a certain degree.

In an antibonding orbital, the electron density is moved away from the internuclear area: the electron-nucleus attractions are minimized and the internuclear repulsion maximized. Thus, the molecule is destabilized.

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  • $\begingroup$ When you say the electron density is concentrated in the area between the nuclei, isn't that in a σ bond where the electron density is in the AB axis. I thought in a π bond the electron density is zero along the AB axis and is actually above and below the AB axis. $\endgroup$
    – Saul McShane
    Dec 27 '15 at 13:08
  • $\begingroup$ Yes, I don't mean the axis, I mean the nuclei's neighbourhood: in a π MO, the electrons can easily interact with both nuclei. In a π* MO, this possibility is reduced drastically. $\endgroup$
    – Corundum
    Dec 27 '15 at 13:11
  • $\begingroup$ There's just one thing I don't understand, what is the significance of the sign of an electron's wave function? What does it determine? Why does one lobe of the p orbital have a positive wave function and the other lobe have a negative wave function? Also, in the second example from the top of the page, showing the overlap of a s orbital and a p orbital, why does the s orbital have a positive wave function? $\endgroup$
    – Saul McShane
    Dec 27 '15 at 13:15
  • $\begingroup$ @SaulMcShane chemistry.stackexchange.com/q/35212/16683 $\endgroup$
    – orthocresol ♦
    Dec 27 '15 at 13:20
  • $\begingroup$ The wave function itself is no real physical quantity: what has a significance is its square modulus, which indicates the probability of finding the electron in a specific position (electron density). The sign of the wave function is important when combining orbitals in molecules – if you add two same-signed orbitals, a bonding MO will form; otherwhise, an antibonding MO will form. Regarding your doubt about why the orbitals have their respective signs… well, that's a long story, but you can find all the orbitals on the orbitron website. $\endgroup$
    – Corundum
    Dec 27 '15 at 13:24

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