Why Is (SiO)(4) Calculated To Be Tetrahedral, Whereas (CO)(4) Is Square Planar? A Molecular Orbital Analysis Academic Article uri icon


  • Qualitative molecular orbital (MO) theory predicts that square-planar tetrasilacyclobutanetetraone D4h-(SiO)4 should, like D4h-(CO)4, have a triplet ground state, and the results of the (U)CCSD(T)-F12b/cc-pVTZ-F12//(U)B3LYP/6-311+G(2df) calculations, reported here, confirm this expectation. Calculations at the same level of theory find that square-planar tetrasilacyclobutanetetrathione D4h-(SiS)4 also has a triplet ground state. However, these ab initio calculations predict that (SiO)4 and (SiS)4 both have a singlet state of much lower energy, with a tetrahedral (Td) equilibrium geometry and six, electron-deficient, Si–Si bonds. In contrast, the lowest singlet state of (CO)4 and of (CS)4 is calculated to prefer a D4h to a Td geometry. An analysis, based on the second-order Jahn–Teller effect, rationalizes the influence of the electronegativity difference between A and Y in (AY)4 on the energy difference between a D4h and Td geometry. This analysis predicts that (BF)4 and (BCl)4, which are isoelectronic with, respectively, (CO)4 and (CS)4, should both prefer a Td to a D4h equilibrium geometry. These qualitative predictions have been confirmed by our calculations, and (BCl)4 is known experimentally to have a Td equilibrium geometry.

publication date

  • January 1, 2013