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QUESTION

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

Yes. All iron sulfate 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, 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+) a d^6 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 d-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).

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