We all know that $\ce{CuSO4.5H2O}$ is blue and it turns to $\ce{CuSO4}$ on heating which is white. I also learnt about coordination complexes and d-d transitions and how transition metal compounds are colored. Recently I came to know that $\ce{CuSO4.5H2O}$ is also a coordination compound and in its structure 4 water molecules are coordinated with $\ce{Cu++}$ and one is hydrogen bonded to $\ce{SO4^{2-}}$. So I thought lets try to apply the theory here and see what causes the blue color of Copper sulphate...

$\ce{CuSO4}$ has $\ce{Cu^{2+}}$ ions and I wrote down the electronic configuration as $3d^94s^0$. Since no d-d transition is possible so it should be white ...

$\ce{CuSO4.5H2O}$ is actually $\ce{[Cu(H2O)4]SO4.H2O}$ and the complex ion is $\ce{[Cu(H2O)4]^{2+}}$. $\ce{H2O}$ is a weak field ligand therefore the electronic configuration is $3d^94s^24p^6$. But still no d-d transition is possible so it should also be white!? What went wrong in my calculation???

  • $\begingroup$ I think I found the answer to this question but still I'll wait for answers just to confirm. . $\endgroup$ – Kartik Apr 6 '17 at 4:29

The Blue color which arises when we Dissolve CuSO4 is indeed due to d-d transition !

Well , the reason behind this is the Splitting Effect ;

The Cu+2 ion is hydrated by four H2O molecules , i.e , it forms a complex [Cu(H2O)4]+2 ;

Now Due to the water molecules , the d-orbitals which were degenerate(of same energy) prior to the hydration , are split up into two discrete groups having different energy levels (they are called t2g and eg Groups );

The Energy difference(i.e the Energy of photon which will be released due to the transiotion) between the groups is corresponding to the Blue Color seen.

Refer to Crystal Field Theory if you are willing to know more.

Hope this Helps !



The d9 configuration is an unsymmetrical field, with the t2g orbitals being completely filled while the eg orbitals have only three electrons. So, according to the Jahn Teller theorem, these octahedral shapes will undergo slight distortions and will not be a perfect octahedron. It is easy to see why it is so. Due to this unsymmetrical filling up of d electrons, some ligands will be repelled more as compared to other when they approach to form the complex. In this case, the eg orbitals(which point directly at the approaching ligands) have two electrons in one while one electron in the second eg orbital. So the ligand along either dz² or dx²-y² will be repelled more as compared to the other. Hence it will experience some sort of elongation towards that side, and more importanrly, to remove the asymmetry, further sub-splitting will occur in the eg level, so even the restrictions of spin will be lifted.Clearly,the Laporte rule is no longer valid, and so, it can undergo d-d transitions. While in the anhydrous form, we can clearly see that a centre of symmetry will be present, and so transitions will be Laporte-forbidden

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  • $\begingroup$ Along with the downvoting,how about somebody actually points out the mistake in my answer? $\endgroup$ – Yusuf Hasan Jan 4 '19 at 2:24

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