Why is SO32- sp3 hybridized even though it forms a double bond with the oxygen (in resonance)?

I thought double bonds needed a p orbital so I would think it would be sp2d so it can still have an unhybridized p orbital for the double bond. Thanks!


1 Answer 1


This is a point of continuing drama. Theoretical calculations as long ago as the 60s (cf. J. Am. Chem. Soc., 1994, 116 (10), pp 4414–4426) strongly suggest there is no d orbital participation in hybridized orbitals for Period 3 "hypervalent" compounds like the sulfite anion. In fact, these calculations appear to suggest the observed bond shortening is probably explained quite simply by a significant ionic component to the bond, which means the best Lewis structures may well be the simple octet-rule following one, with 4 sp3 bonds around the central S, occupied by 3 bonding pairs (giving 3 sigma bonds) and 1 lone pair. No pi bond at all. This gives you higher formal charges than the resonance structure with a single double bond -- but that's what's meant by saying the calculations suggest the best VB description simply has more ionic content to the bonding between the S and O atoms (represented by the higher formal charges), rather than any d mixing.

  • $\begingroup$ Not 4 "sp3 bonds"... Hybridisation should not be applied to bonds and only to orbitals. Be careful of your terminology. $\endgroup$ Commented Jun 3, 2018 at 2:12
  • $\begingroup$ Would be good if u could provide additional insights from such papers, especially those that do calculations on sulfite in particular. Perhaps, show charge distribution for the structure. $\endgroup$ Commented Jun 3, 2018 at 2:14

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