- TL;DR Summary
- The Fe -phen complex is a redox indicator. In the context of d-d orbital transitions why is the oxidized form blue?
Ferroin is a redox indicator. It is the complexation of phenanthroline and iron. In its reduced state the Fe (II) - phenanthroline complex is red and in its oxidized state the Fe(III)- phenanthroline complex is blue. I am trying to understand in the context of d-d orbital transitions why this is true. I know that phenanthroline is a strong field ligand so the Fe (II) would be low spin and the transition would be in the blue region explaining why the reduced form is red. But why is the oxidized form blue? My only thought is that an occupied orbital is lower in energy than an unoccupied orbital so the paired spin on the reduced form would need even more energy to make it to an unoccupied orbital than an electron from the ground state of the oxidized form, which would all be the same in energy. This thing about the difference in energy of occupied d orbitals vs unoccupied d orbital might me something I made up. I don't quite remember since I took inorganic 4 years ago.