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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.

Gaurang Tandon
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Osal Thuduwage
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    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 06:01
  • See the edit now... – Osal Thuduwage Jun 25 '18 at 06:16
  • 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 06:23
  • https://chemistry.stackexchange.com/questions/4667/why-do-transition-elements-make-colored-compounds. : but the quoeted text had been not mentioned directly, I quoted the summary that can get from it. – Osal Thuduwage Jun 25 '18 at 06:33
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    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 09:28
  • @orthocresol♦: Can you say the reason of why does hybridisation a flowed theory for metal complex. Instead of that, is there any mechanism to understand the electron arrangement of the metal complexes. Thanks. – Osal Thuduwage Jun 26 '18 at 03:09
  • https://chemistry.stackexchange.com/q/76726/16683 – orthocresol Jun 26 '18 at 07:56
  • @orthocresol♦: As you indicated in the link that; "All in all, hybridisation theory is simply inadequate - it does not explain the rich chemistry of the transition metals and their complexes. Even something simple, such as the d-d* transitions responsible for the colours of complexes, cannot be explained by it" So is there any accurated way for describe the colours of the complexes? Thanks – Osal Thuduwage Jun 26 '18 at 09:04
  • Molecular orbital theory or at the very least crystal field theory – orthocresol Jun 26 '18 at 10:01
  • I'd suggest you pick up a book on ligand field theory or MO theory; it'll be very helpful. But you're right though- a d8 square planar complex should be diamagnetic. – ANZGC FlyingFalcon Apr 22 '19 at 12:36
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    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. – porphyrin Apr 22 '19 at 13:38
  • I think the key piece that you are missing is that the ligand electrons don't just transfer into the metal orbitals. The metal and ligand orbitals combine to form bonding and antibonding orbitals (you should learn this even in valence bond theory). In transition metal complexes, the metal d electrons generally occupy non-bonding and anti-bonding orbitals, whereas the ligand electrons occupy bonding orbitals. The non-bonding and anti-bonding orbitals closely resemble the original metal d orbitals and so we use the same names for them, which may be where you're getting confused. – Andrew Apr 22 '19 at 18:14

2 Answers2

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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.

ANZGC FlyingFalcon
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  • How we are sure that the color doesn't come from the transition between $p$ (filled) to $s$ (empty) orbital? – ado sar Nov 29 '20 at 16:34
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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

Sreyas Adury
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  • 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 03:03
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    Wrong energy scheme for a square planar complex. You’ve illustrated it for an octahedral geometry. – orthocresol Jun 26 '18 at 07:55