Question

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

Answer 1

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`, 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.