Abstract

The spin-orbit contribution to the zero-field splitting in methylene is calculated at bond angles of 135° and 180° using the correct microscopic Hamiltonian. The two lowest states of 1A1 symmetry as well as the lowest states of 1,3A2 and 1,3B2 symmetry are considered in the second-order perturbation treatment. Configuration-interaction wavefunctions obtained using a double-zeta-plus polarization quality basis of contracted Gaussian-lobe functions are used to describe each state. Spin-orbit contributions to the splitting parameter, D, of 0.023 and 0.0245 cm−1 are obtained at 135° and 180°, respectively. Configuration interaction is found to reduce the spin-orbit contribution at 135° by about 50%. Combining these results with the results for the spin-spin dipole contribution to D gives D=0.807 cm−1 and E=0.049 cm−1 at 135° which is slightly higher than the current experimental value of D=0.76±0.02 cm−1. The increase in the value of the parameter D from 0.807 cm−1 at 135° to 0.934 cm−1 at 180° is observed to come almost entirely from increased spin-spin contributions at the larger angle.

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