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For example, elemental iodine is deep violet. Its sigma bond or perhaps the lone pairs are capable of absorbing all visible light frequencies except violet which is why we see it as that color. Lithium, on the other hand, is red which means its valence electrons must be reflecting red frequencies. So, does that mean the valence/bonding electrons of $\ce{I2}$ are higher energy than the lithium valence electrons? In other words, is there a relationship between an orbitals energy and the type of light it can reflect?

You could perhaps compare $\ce{I2, Sn^{II}I2}$, and $\ce{Sn^{IV}I4}$, and which are violet, yellow, orange (in decreasing order of visible light energy).

Perhaps it has to do with symmetry? $\ce{I2}$ is $D_{\infty\mathrm{h}}$, stannous iodide is $C_{2\mathrm{v}}$, and stannic iodide is $T_\mathrm{d}$.

orthocresol
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Nova
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    Bonds do not absorb light, neither do lone pairs. It is a transition between certain orbitals that actually works. – Ivan Neretin Feb 23 '16 at 21:05
  • You could have edited earlier question http://chemistry.stackexchange.com/questions/46785/why-is-tiniv-iodide-bright-orange-while-elemental-iodine-is-violet – Mithoron Feb 23 '16 at 21:26
  • related http://chemistry.stackexchange.com/questions/16633/why-is-gold-golden – Mithoron Feb 23 '16 at 21:34
  • @IvanNeretin: It is even worse, the absorption of light corresponds transition between states. – ssavec Feb 23 '16 at 22:03
  • Would you leave out iodine, there could be some trends observed. Lot of information can be gained from https://en.wikipedia.org/wiki/Ligand_field_theory and subsequently https://en.wikipedia.org/wiki/Tanabe%E2%80%93Sugano_diagram – ssavec Feb 23 '16 at 22:05
  • Probably interesting too, although more organic: http://chemistry.stackexchange.com/questions/9493/what-is-the-origin-of-the-colour-of-azo-dyes/9495 – Klaus-Dieter Warzecha Feb 24 '16 at 06:40
  • I understand there is absorption going on here, but WHICH electrons (as in WHICH orbitals) are the ones doing the absorbing of red, orange, yellow, and green to give Iodine its purple color. Is there any way to know this information? – Nova Feb 24 '16 at 17:36

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Rather than give answers to your specific question I try to explain what happens in general as there is some confusion in your question. You should really consult a book on spectroscopy, of which there are many, for a thorough explanation.

In a molecule the arrangement of electron in orbitals is called a state, hence ground state, first exited state etc. When all the electrons a molecule has fill up all the orbitals with no gaps left, this is the ground state. They will fill up half of all the orbitals, two to an orbital. When a photon is absorbed and it has sufficient energy (usually in the visible or UV part of the spectrum) an electron is promoted from one of the bonding orbital to a anti-bonding orbital which has higher energy. This arrangement of electrons is called an excited state. This transition between states is what gives molecules and atoms their colour.

In iodine the difference between the lowest excited state and the ground state occurs in the visible part of the spectrum. (for simplicity, I'm ignoring any vibrational transitions accompanying the electronic excitation). Benzene has a larger energy gap between ground and excited state than I$_2$ and so absorbs only in the UV and looks transparent to us.

The same basic idea applies to all atoms and molecules. Where the molecules or atoms absorb is a function of the exact electronic energy difference between ground state and excited state of the particular species. An understanding of the nature of the bonding and so energetics allows us to predict trends in absorption spectra, but exact values can only be obtained from quantum calculations.

porphyrin
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