Simple way to understand why chromophores often have double bonds?
No. Without going into quantum mechanics and molecular orbital theory, there is no way to understand why chromophores typically have double bonds.

Everything related to colour is related to the excitation of electrons from one orbital to another and therefore any explanation that does not rely on orbitals in at least some way is doomed to fail.



If you decide to accept a little MO theory then you can be helped. Double bonds are formed by an underlying &#x3c3; bond (or single bond) and a &#x3c0; bond. Each bond is formed by two atomic orbitals (one from each atom) that combine to give a bonding and an antibonding orbital. The energy difference between the bonding and antibonding orbital corresponds to the excitation energy: a photon that carries exactly that energy will be absorbed and the electron excited to the higher level. The greater the overlap between the atomic orbitals, the higher the energy difference between bonding and antibonding.

Now in &#x3c0; orbitals, the atomic p orbitals are oriented perpendicular to the bond axis, so any overlap is doomed to be much smaller than the corresponding single bond’s &#x3c3; overlap. Therefore, the energy difference from the bonding to the antibonding &#x3c0; orbital is not far into the ultraviolet region; typical values would correspond to a wavelength of $150$ to $200~\mathrm{nm}$. Extending the &#x3c0; network will move those orbitals closer together energetically (again, the complete answer requires much more MO theory) which is why benzene has an energy difference corresponding to a wavelength of $250~\mathrm{nm}$. Thenceforth, it is not a long way until you reach visual wavelengths ($400{-} 700~\mathrm{nm}$).