The Valence-State Energy of the Bivalent Oxygen Atom

Abstract

A significant simplification in the discussion of the structure of molecules can be achieved by referring the energies of molecules not to the energies of their constituent atoms in any one of the stationary spectroscopic states of the isolated atoms, but instead to a hypothetical valence state of each atom. [1] The valence state of an atom is defined as that state in which it has the same electronic structure as it has in the molecule. Thus in the valence state the spins of the electrons that are used for forming bonds in a molecule are considered to resonate between positive and negative values, corresponding to the resonance of each bofiding electron with another electron of opposite spin in the atom with which the bond is formed. For the bivalent oxygen atom, with six L electrons, two orbitals, one mainly 2s in character and one 2p, are occupied by unshared pairs, and the other two orbitals, mainly 2p but with a small amount of s hybridization, are occupied by one bonding electron apiece, the spins of which are parallel 50 per cent of the time and antiparallel 50 per cent of the time. In the following paragraphs values of the energy of this bivalent state of the oxygen atom relative to the normal state of the isolated atom, 3P, are derived by four different methods. It is found that the four values agree reasonably well with one another, and their average value, 0.74 v.e., is indicated to be reliable to about 0.05 v.e.