When iron(II) sulfate dissolves in water, does iron(II) form a complex ion with water surrounding it?

1 Answer
Oct 9, 2015

Yes. All iron sulfate compounds complex with water until all the binding sites are filled up.

In this case, you get an octahedral complex that can be written as [Fe(H_2O)_6]SO_4[Fe(H2O)6]SO4, or drawn as:

where the oxygens on the waters point towards the iron.

A common name for it is Ferrohexahydrite. The scientific name for this would be either hexaaquairon(II) sulfate or iron(II) sulfate hexahydrate.

You can call Fe^(2+)Fe2+ a d^6d6 transition metal in the context of Crystal Field Theory.

The waters are known as the ligands in this metal-ligand complex. Water is a relatively weak-field ligand, meaning that it generates only small repulsive forces that do not cause much splitting of the dd-orbital energy levels (as a ligand approaches to form a metal-ligand interaction, repulsive forces would decrease the stability of the interaction, thereby creating a difference in the energy levels of the most significantly different orbital shapes).

https://upload.wikimedia.org/https://upload.wikimedia.org/

That means that the energy levels of the d_(x^2 - y^2)dx2y2 and d_(z^2)dz2 orbitals are not very high above the energy levels of the d_(xy)dxy, d_(xz)dxz, and d_(yz)dyz orbitals. Therefore, this octahedral complex gives rise to high-spin state, in which electrons are likely to fill all the dd-orbitals one at a time, then double up.