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.
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.