LCAO-SCF Computations for Ethylene

A series of LCAO‐SCF calculations with a minimal Slater basis set were carried out for the ethylene molecule and for its positive and negative monovalent ions in a number of electronic states. These computations, which included configurational mixing of the two low ¹A states, were performed for seven values of the twist angle between the two CH₂ groups, from 0° to 90°. For planar ethylene four singlet levels, two triplets, and two doublets of each of the ions, were investigated. The two ${}^1{A}_g$ states, the triplets, and the ionic doublets were individually computed (with open‐shell SCF procedures where applicable), while their energies, as well as those of the other singlets, were also derived from the data of the closed‐shell calculations. This individual SCF optimization was found to have only limited effect in the case of the neutral molecule, but was more significant for the ions. The energy obtained for the planar ground state, though lower than that of a 40‐function Gaussian calculation, is considerably higher than the energies reported for basis sets of 60 Gaussians or of 66 Gaussian lobe functions. The effect of the neglect of $\mathrm{\sigma \hspace{0.17em}-\hspace{0.17em}\pi }$ exchange was examined in a series of calculations for planar ethylene (including the ions). It was found that the errors introduced by this approximation are roughly proportional to the number of π electrons in each state, and are quite significant for the determination of energy differences involving changes in the number of these electrons. The assignment of the so‐called olefin mystery band is considered, and we favor the opinion that this band correlates with the ethylene Rydberg singlet transition at 7.11 eV, with the corresponding upper level probably having some valence‐shell $\mathrm{\pi \to }\mathrm{CH}*$ character. The twisting potential curves are obtained for four neutral and four ionic states. The ionic levels are degenerate in perpendicular configuration, and show a Jahn‐Teller splitting at that point. Similar behavior should be expected of the Rydberg $\mathrm{\pi \to ns}$ states.