Abstract

The paramagnetic resonance spectrum of a benzene solution of bis-cyclopentadienyl vanadium shows a g factor equal to 2.00 and a V51 isotropic hyperfine splitting equal to 77 Mc. The small V51 hyperfine splitting demonstrates that none of the three odd electrons in this sandwich compound spends an appreciable part of its time in a 4s atomic orbital of the vanadium atom. The observed spectroscopic splitting factor shows that the orbital ground state is a singlet. This result signifies that one of the three unpaired electrons in this substance must be in an orbital of a1g symmetry, and that the other two electrons must be in orbitals of eg symmetry. This latter conclusion is in good agreement with theoretical orbital energy levels derived by a number of investigators. The paramagnetic resonance spectrum of an air oxidation product of bis-cyclopentadienyl vanadium has been observed. The spectroscopic splitting factor is g=1.99 and the hyperfine splittings are 210 Mc. It is suggested that the oxidation product may be the neutral vanadyl sandwich, VO (C5H5)2.

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