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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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  • $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved the formatting of your post by using MathJax for chemical formulae. For more information on how to do so yourself, check out the help center, this meta-post or this one. And there are even more hints available. $\endgroup$ – Jan Apr 30 '16 at 20:59
  • $\begingroup$ I'm already a member at this site. Thank you @Jan $\endgroup$ – user29557 Apr 30 '16 at 21:00
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    $\begingroup$ 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. $\endgroup$ – orthocresol Apr 30 '16 at 21:02
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    $\begingroup$ @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? $\endgroup$ – user29557 Apr 30 '16 at 21:08
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Manganese(II) sulphate is not colourless. It is a faint pink. Compare this image taken from Wikipedia, where a full list of authors is available:

Manganese(II) sulphate tetrahydrate

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:

d5 high-spin

Exciting an electron directly is forbidden due to the spin-retaining rule: The excited electron must occupy an already populated orbital, which is only possible if it flips its spin, which is forbidden by selection rules. Special mechanisms are needed to account for the (obviously possible but faint) transition here; the page Orthocresol linked in the comments names spin-orbit coupling.

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  • $\begingroup$ So basically, the orbitals should be empty for transition to take place,right? $\endgroup$ – user29557 Apr 30 '16 at 21:13
  • $\begingroup$ @user29557 Not necessarily. For example if you added one more spin down electron to the lower level, then that electron could be promoted to the upper level, as long as it retains its spin state in the upper level (no change in spin). $\endgroup$ – orthocresol Apr 30 '16 at 21:14
  • $\begingroup$ @orthocresol Oh. So, if we had 8 electrons, there would be a characteristic colour, right? $\endgroup$ – user29557 Apr 30 '16 at 21:18
  • $\begingroup$ @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). $\endgroup$ – orthocresol Apr 30 '16 at 21:21
  • $\begingroup$ @user29557 $\ce{Ni^2+}$ is a typical $\mathrm{d^8}$ system, so yeah. $\endgroup$ – Jan Apr 30 '16 at 21:21

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