Why do transition metals element make colored compounds both in solid form and in solution? Is it related to their electrons?
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1$\begingroup$ it is due to d-d transition and charge transfer spectra $\endgroup$– user4790Commented Mar 9, 2014 at 17:27
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4 Answers
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You are absolutely correct, it all about the metal's electrons and also about their d orbitals.
Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level.
However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels).
This forms the basis of Crystal Field Theory. How these d orbitals split depend on the geometry of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this:
As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour?
Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of $300$GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white).
However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it.
So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the $d_{yz}$ orbital to the $d_{z^2}$ orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta.
This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal.
answered Nov 8, 2015 at 5:27
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$\begingroup$ The crystal field theory is deprecated, but what you describe is also valid for the ligand field theory, which should be used instead now. $\endgroup$– Martin - マーチン ♦Commented Nov 8, 2015 at 6:54
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$\begingroup$ Then how come KMnO4 has color (a very deep purple)? The manganese would have an oxidation state of 7-, which means the atom would no longer have occupied d orbitals at all. $\endgroup$ Commented Nov 10, 2016 at 6:28
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$\begingroup$ A broader question: How does crystal field theory apply to transition metals that are part of polyatomic ions? $\endgroup$ Commented Nov 10, 2016 at 6:29
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3$\begingroup$ Won't the electrons emit the light in the same frequency as it absorbed when it comes back to the ground state? $\endgroup$ Commented Mar 11, 2018 at 8:56
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$\begingroup$ Nanoputian, the complementary colour explanation works, but the comment above asks a good question. Is it that the electron excitation-deexcitation method of explaining this just doesn't work? $\endgroup$– harryCommented May 18, 2021 at 9:31
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The partially full d-orbitals in transition metals have energy splittings that happen to lie in the visible range. Depending on the arrangement of substituents (known as ligands) that attach to them, the electron energies split according to crystal field theory. Similar splitting in the s or p orbitals produce gaps in the ultraviolet, and any visible light goes right through, so we don't see any color. In transition metals, however, visible light excites the electrons from a lower d orbital to a higher one and only letting some light through.
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3$\begingroup$ A perhaps deeper explanation for the first sentence in your answer can be found in this article. It seems that the fact the 3d orbitals are anomalously compact explains why many electronic transitions in fourth period d-block metal complexes are in the visible region. $\endgroup$ Commented Nov 20, 2013 at 1:23
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$\begingroup$ Why can't electron transition take place between fully filled split d orbitals? $\endgroup$– UserCommented Sep 19, 2017 at 1:47
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1$\begingroup$ Because of the Pauli exclusion principle that says that two electrons can't be in the same place at the same time. For the transition to take place, an electron has to move from a lower split level to a higher split level. If there is no empty spot, no transition can take place. $\endgroup$– craigimCommented Sep 19, 2017 at 2:18
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$\begingroup$ How we are sure that the color doesn't come from the transition between p (filled) to s (empty) orbital? $\endgroup$– ado sarCommented Nov 29, 2020 at 16:36
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Colored compounds of transition elements are associated with partially filled (n-1)d orbitals. Tthe transition metal ions containing unpaired d-electrons undergoes an electronic transition from one d-orbital to another. During this d-d transition process, the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. The color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in the visible region for all transition elements.
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Transition elements have partially filled d orbitals. We also know that when electrons jump from one orbital to another light is emitted due to which the compounds of transition elements seem to be colored compounds.
Note: This can also happen in some organic compounds though in this case, it is p orbitals and not d orbitals.
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$\begingroup$ In case of transition metals ($3d$) why we can't have a transition from $4s$ to $5s$ where would be in the visible range? $\endgroup$– ado sarCommented Nov 29, 2020 at 16:38
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$\begingroup$ Firstly, in complexes usually the 4s electrons are removed to form the central positive charged ion. Secondly, the energy gap between the 4s and 5s orbitals is extremely high that visible light photons does not have enough energy to overcome. Such transition would requre photons of at least ultraviolet light. And we are not supposed to see ultraviolet light. In short, it is theoretically possible for a 4s-5s transfer, but we cannot see it. $\endgroup$ Commented Feb 22, 2021 at 3:02