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This question already has an answer here:

I was studying about different vitriol and noticed that some hydrates are coloured while some are colourless even after having water in their crystals.

For example; $\ce{FeSO4.7H2O}$ (iron sulfate heptahydrate) has a green colour while anhydrous iron sulfate has no colour, similarly $\ce{CuSO4.5H2O}$ (copper sulfate pentahydrate) has a blue colour while anhydrous copper sulate has no colour. When we come to zinc sulfate,it has no colour in all of its three hydrates (the most common one being the heptahydrate).

Why does zinc sulfate not have any colour like $\ce{FeSO4.7H2O}$ or $\ce{CuSO4.5H2O}$ regardless of whether it has water of crystallisation?

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marked as duplicate by Klaus-Dieter Warzecha, Todd Minehardt, ringo, Wildcat, Jannis Andreska Oct 17 '16 at 12:06

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  • $\begingroup$ i don't know which tags r suitable so please add suitable tag if u know. $\endgroup$ – Vidyanshu Mishra Oct 16 '16 at 17:21
  • $\begingroup$ For $\ce{Zn^2+}$, the answer is simple: Zn(II) is $\mathrm{d^{10}}$, therefore no d-d* transitions. $\endgroup$ – orthocresol Oct 16 '16 at 17:59
  • $\begingroup$ it's either d-d transitions or charge transfer transitions and it all depends on the d-orbital occupancy and the geometry. I feel like this question has likely been answered on here already though. $\endgroup$ – gannex Oct 16 '16 at 18:18
  • $\begingroup$ Is this d-d transition related to d block elements only?? $\endgroup$ – Vidyanshu Mishra Oct 16 '16 at 18:20