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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?

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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.


References:

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