# Why brown ring complex is coloured?

In my textbook it is stated that brown ring complex formed during the test of nitrate ion is coloured due to charged transfer phenomenon but I think it is due to d-d orbital electron transition. Which is correct and why?

• Mar 12 at 7:59
• @NisargBhavsar there is only one iron atom in the complex Mar 12 at 16:19

Monsch, G. and Klufers, P. $$\ce{[Fe(H2O)5(NO)]^2+}$$, the "Brown-Ring" Chromophore. Angew. Chem. Int. Ed. 2019, 58, 8566-8571. DOI: 10.1002/anie.201902374.
As for the origin of the brown color, how one describes that is also somewhat dependent on what the oxidation state assignment is. There are two electronic transitions associated with the color - one being from the Fe-NO $$\pi$$ bond orbitals to completely metal centered d orbitals and the other being from the same starting orbital but going into the Fe-NO $$\pi^*$$ orbital. Whether you describe these as $$d \rightarrow d$$ transitions on the metal or as ligand-metal charge transfer depends whether you assign the $$\pi$$ bond electrons to the metal or the ligand in your determination of the oxidation state.
The reactions involved here are: $$\ce{2 HNO3 + 3H2SO4 + 6FeSO4 -> 3Fe2(SO4)3 + 2NO + 4H2O}$$ $$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$$
In this case, the brown complex is $$\ce{[Fe(H2O)5NO]SO4}$$ and the colour is determined by the excitation of electrons in d orbitals of the $$\ce{Fe+}$$. It's also a Laport allowed transition, because the $$\ce{NO}$$ breaks the central symmetry of the previous complex.