How can I understand the Lewis acid-base interaction of iron(III) and water? The interaction between $\ce{H+}$ and $\ce{NH3}$ as respectively a Lewis acid and a Lewis base is clear for me. $\ce{NH3}$ has a lone electron pair and $\ce{H+}$ has no electrons, or, saying politely, an incomplete duet, and can use nitrogen's lone pair to complete its duet, and a dative (coordinate) bond is formed in the process.

However when I think about $\ce{[Fe(H2O)6]^3+}$, I don't see where these six lone pairs of electrons (which are coming from water molecules) can go in the iron ion. The electron configuration of $\ce{Fe^3+}$ is: $\mathrm{1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5}$. It has five 3d electrons, and all (namely, 2) of its 4s electrons are gone. All five 3d electrons have the same spin, and they occupy five d orbitals. OK, I do remember that 3d orbitals are funny and in the transition metals 3d orbitals get divided into low and high energy level 3d orbitals (hence the color of transitional metal solutions). 

But even with this, how do six water molecules position their lone pairs there? We have space for one pair of electrons (where 4s electron left) and then space for 5 more electrons, so in total, for 7 electrons. 6 molecules of water would give us at least six pairs of electrons, i.e. 12 electrons. So we are 5 places short to place all these electrons. 

So the question is: how are they placed?
So, effectively, the question is: How are electrons from the covalent bond between ligand and central metal atom or ion placed with regard to the electron configuration? 

$\ce{Fe^{3+}}$ has the condensed electron configuration of [Ar]$3d^5$. When the ligands approach the iron ion from the X, Y, and Z, the ion forms six bonds. It would be incorrect to state that the remaining 3d electrons are involved in the bonding with the ligands. Instead, the iron ion uses 6 orbitals from the 4s, 4p, and 4d orbitals to accept the lone pairs of electrons from the ligands. However, before they are used, the orbitals are hybridized in order to create six orbitals of equal energy. 

If you want to dig deeper into the subject, I highly recommend the lecture notes from this webpage here.