# Considering the d-d transition how, does tetracyanidonickelate(II) ion exist as a colored complex?

As I refer on google, the resources indicated that,

the d block elements which have unpaired electrons as in their ions, when they are making complexes with ligands, they can absorb different frequencies in the spectrum due to degenerated d-d orbital transition.

But I am confused in understanding how there is a colour for $\ce{[Ni(CN)4]^2-}$ complex.

My question is: In the $\ce{Ni (II)}$ ion there are two unpaired electrons in the d orbitals. But while they are hybridizing there aren't any unpaired electrons, all are paired. In the solution state, the nickel ions do not exist as $\ce{Ni^2+}$ ions, but rather as $\ce{[Ni(CN)4]^2-}$ ions. But $\ce{[Ni(CN)4]^2-}$ has no unpaired electron. So it is unable to absorb energy from the spectrum and transit between d orbitals. (My logic is that there are no more spot in d orbitals to transit between after the complex occur)

But as the image below this complex is yellow in colour.

I can't find the mistake in my logic. I referred to many resources but I wasn't able to find any satisfactory answer.

• Hi Osal, please cite the source (website, book, paper) of the quoted text. Google is never the source for any information (it's just a search engine). Thank you! – Gaurang Tandon Jun 25 '18 at 6:01
• See the edit now... – Osal Thuduwage Jun 25 '18 at 6:16
• Websites are like Wikipedia, ChemGuide, etc. Please mention one like that if you recall. If you don't recall, no problem, let it be like that. – Gaurang Tandon Jun 25 '18 at 6:23
• chemistry.stackexchange.com/questions/4667/…. : but the quoeted text had been not mentioned directly, I quoted the summary that can get from it. – Osal Thuduwage Jun 25 '18 at 6:33
• Hybridisation is a very flawed theory for metal complexes... Please, stop using it as soon as you can. Look up an MO diagram for a typical square planar complex. – orthocresol Jun 25 '18 at 9:28

• Your answer is regarding the mechanism of showing color of $\ce{[Ni(CN)4]^2-}$ complex. But the question is actually not that. You are describing standing on the Ni2+ ion. But my logic after existing the complex, the nickel will not exist as Ni2+ ion, it is exist as the $\ce{[Ni(CN)4]^2-}$ complex at the compound. As I shown in the figure there are no more d orbitals in the $\ce{[Ni(CN)4]^2-}$ after having the complex. So, as I mentioned above how there is a transition between d orbitals while there is no any space in the d orbitals regarding the complex. That's the relevant. – Osal Thuduwage Jun 26 '18 at 3:03