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MO3. Lessons from Dihydrogen

Molecular Orbitals

In the case of the hydrogen molecule, we took two atomic orbitals and combined them to form two molecular orbitals. These new molecular orbitals had different wavelengths than the two atomic orbitals: one had a longer wavelength and was a little lower in energy, while the other had a shorter wavelength and was a little higher in energy. If we take into account the energy of the two original atomic waves, and compare them to the total energy of the two new molecular waves, there is no change overall.

  • The average energy of the orbitals has remained almost constant.
  • Also, the number of waves has remained constant.

We started with two atomic orbitals, and by combining them we produced two molecular orbitals. Both of these ideas are useful in considering the formation of more complex molecules from individual atoms.

Of course, from the point of view of the two real electrons, some remarkable changes have occurred. Both of these electrons have adopted a longer wavelength and a lower energy and that has made all the difference. There is an occupied molecular orbital and an unoccupied molecular orbital; only the occupied orbital makes a real energetic contribution to the overall stability of the molecule. The unoccupied orbital is completely imaginary.

  • The electrons have lower kinetic energy in the bond than they had before bonding.
  • Electronic energy has decreased. A stable bond has formed.

A bonding picture of He2 would look exactly the same, because it would also involve the overlap of 1s electrons on one atom with 1s electrons on the other atom. There would be a different electronic energy, however. That difference would affect the prospects of helium-helium bond formation.

Problem MO3.1.

Construct molecular orbital diagrams for the following diatomic species and discuss the likelihood of bond formation in each case.

  1. He2
  2. Li2
  3. Be2