# What does a molecules color have to do with its bond/orbital energies?

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 $T_\mathrm{d}$.

• Bonds do not absorb light, neither do lone pairs. It is a transition between certain orbitals that actually works. Feb 23, 2016 at 21:05
• You could have edited earlier question chemistry.stackexchange.com/questions/46785/… Feb 23, 2016 at 21:26
• Feb 23, 2016 at 21:34
• @IvanNeretin: It is even worse, the absorption of light corresponds transition between states. Feb 23, 2016 at 22:03
• Would you leave out iodine, there could be some trends observed. Lot of information can be gained from en.wikipedia.org/wiki/Ligand_field_theory and subsequently en.wikipedia.org/wiki/Tanabe%E2%80%93Sugano_diagram Feb 23, 2016 at 22:05

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.