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.

enter image description here

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.

enter image description here

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

  • 1
    $\begingroup$ 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! $\endgroup$ Jun 25, 2018 at 6:01
  • $\begingroup$ See the edit now... $\endgroup$ Jun 25, 2018 at 6:16
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2018 at 6:23
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    $\begingroup$ 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. $\endgroup$ Jun 25, 2018 at 9:28
  • 1
    $\begingroup$ When the metal complex is formed the symmetry of the ion is no longer spherical but octahedral ( four ligands in a plane and one north and one south) or square planar( four in a plane). The effect this change in symmetry has is to split the energy of the d orbitals into two groups of 3 and 2 (labelled $t_{2g}\,e_g$}. Transitions giving the colour occur between these two sets of d orbitals. You can see examples in the answers below. $\endgroup$
    – porphyrin
    Apr 22, 2019 at 13:38

2 Answers 2


enter image description here

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which are effectively isoelectronic anyways).

As you can see the frontier orbitals of this complex are all mostly on the metal complex , and I'd expect the excitation of electrons from lower-energy orbitals into these anti bonding orbitals to be the reason for the colors.

  • $\begingroup$ How we are sure that the color doesn't come from the transition between $p$ (filled) to $s$ (empty) orbital? $\endgroup$
    – ado sar
    Nov 29, 2020 at 16:34

This can be explained by Crystal Field Theory (CFT) According to CFT, when a strong field ligand brings about pairing on electrons in d orbital, splitting of d-orbitals occurs (loss of degeneracy, i.e. all 5 d orbitals are not of same energy level)

Splitting of d-orbitals in Ni+2

Now as the upper orbitals are not of the same energy as lower ones, transitions take place giving rise to colours. CFT has the capability to explain the rise of colours in coordination compounds.

Read more here: https://en.wikipedia.org/wiki/Crystal_field_theory

  • $\begingroup$ 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. $\endgroup$ Jun 26, 2018 at 3:03
  • 5
    $\begingroup$ Wrong energy scheme for a square planar complex. You’ve illustrated it for an octahedral geometry. $\endgroup$ Jun 26, 2018 at 7:55

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