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