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?


2 Answers 2


There is an excellent discussion of this compound in this paper:

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.

The authors note that this complex is difficult to interpret in the context of the IUPAC definition of formal oxidation state, and that reasonable arguments can be made for Fe(I), Fe(II) or Fe(III) as the proper oxidation state of the central atom, with the NO oxidation state varying accordingly so that the total charge of +2 is retained.

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.


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