Home > Community > Considering the d-d transition how, does tetracyanidonickelate(II) ion exist as a colored complex?
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Aaron Kaufman

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

Brett Johnson  Follow

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

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James Leland Harp  Follow
Wrong energy scheme for a square planar complex. You’ve illustrated it for an octahedral geometry.More
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Edward Lewis  Follow
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. Thats the relevant.More
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Brighton Munezi  Follow

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.

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Isaac To  Follow
How we are sure that the color doesnt come from the transition between $p$ (filled) to $s$ (empty) orbital?More
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