What is the reason that vanadyl sulfate shows blue colour, even in its anhydrous form? If it's d-d transition, can you explain how it works in this compound?


Vanadyl sulfate or $\ce{VOSO4}$ has $\ce{VO^2+}$ ion. It exists as a blue hydrated crystalline salt, known as acid vanadyl sulfate, $\ce{VOSO4 \cdot H2SO4 \cdot xH2O}$. If you're to heat it at $\pu{533 K}$ with concentrated sulfuric acid, you obtain anhydrous vanadyl sulfate, which is rather a greyish-green crystalline powder.¹

The ion $\ce{VO^2+}$ is more accurately written as a complex $\ce{[VO2(H2O)4]+}$.² Inductive effects of ligands and different symmetries of orbitals causes the d-orbitals to split up and thus they become non-degenerate.

The electrons possibly absorb and emit electromagnetic radiation in the visible spectrum. A rough estimate leads me to conclude that this complex would've been absorbing radiations towards the higher-wavelength (possibly red) region of the visible-spectrum thus emitting radiations corresponding to a blueish tone of light.

The colour is possibly due to d-d transition as the central metal ion ($\ce{V+4}$) has one electron in its d-orbital.


  1. B.Sc. Chemistry–II (UGC), R.L. Madan
  2. Chemguide – Vanadium

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.