Why dioxovanadium $\ce{[VO2]+}$ is pale yellow, whereas orthovanadate $\ce{[VO4]^3-}$ is colorless?

Both of them should be colorless if we ignore charge transfer for a moment, because in both cases vanadium is at +5 oxidation state.

Now considering the charge transfer is the reason for the pale yellow color of dioxovanadium species, why orthovanadate does not show charge transfer band?


1 Answer 1


The colour will depend on several factors. You've identified the oxidation state (+5 in both cases) as one. Other factors are:

  • Ligands
  • Complex shape

With $\ce{[VO2]}^{+}$, the actual complex ion is $\ce{[VO2(H2O)4]}^{+}$ (Taken from ChemGuide). So our ligand is water, and the shape is tetrahedral (or square-planar).

The $\ce{[VO4]^{3-}}$ ion usually only exists at high pH in solution but polymerises at lower pHs. These polymeric ions can have colour (e.g. $\ce{[V10O28]^{6-}}$ is orange).(Link for colour / polymerisation) However, I could find no details as to the ligands (I would assume that they would be hydroxide ions at high pH) nor shape.

If we are talking about the solid, then the same applies... does the solid have any water of crystallisation. (Think copper sulfate... blue when "wet" and white when "dry").


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