# Why is MnSO4 colourless?

I get it that transition metal salts have colour due to d-d transition from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$ orbitals in octahedral and the opposite in tetrahedral but why would $\ce{Mn^2+}$ not have any characteristic colour? Is it so because the orbitals are degenerate in $\ce{MnSO4}$? or because d-d transition only takes place when there is completely empty d orbital?

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• I'm already a member at this site. Thank you @Jan – user29557 Apr 30 '16 at 21:00
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• It does have a colour, but it is very faint. It has to do with the selection rules governing d-d transitions in a metal complex: wwwchem.uwimona.edu.jm/courses/selrules.html in particular the $\Delta S = 0$ rule. – orthocresol Apr 30 '16 at 21:02
• @orthocresol Oh I got the reason. So there must be no electron with same spin in that orbital, to which the electron is being transited to? Right? Because if it were present it should have opposite spin which will violate Hund's rule? – user29557 Apr 30 '16 at 21:08

However, it is true that the colour is rather faint, and you need a white background to properly see it. That’s because manganese(II) is a $\mathrm{d^5}$ system and a high-spin one of those meaning that all five d-orbitals are occupied by a single electron:
• @user29557 Indeed - so an example of an 8-electron ion would be $\ce{Ni^2+}$, and $\ce{NiSO4}$ has what I personally think is a very pretty colour (there's a picture on Wikipedia). – orthocresol Apr 30 '16 at 21:21
• @user29557 $\ce{Ni^2+}$ is a typical $\mathrm{d^8}$ system, so yeah. – Jan Apr 30 '16 at 21:21