How can I understand the Lewis acid-base interaction of iron(III) and water?

Question

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?

Answer 1

Coordination chemistry is that point in time where the simple main-group octet rule starts to loose its practicality. (In reality, not even main-group chemistry is as easy as it is often made in introductory courses, but the fact that d-orbitals of the same period are energetically too far removed to actively take part in bonding and that core orbitals are usually fully populated make things easier.)

The general principle is that electrons are represented by wavefunctions called orbitals that have an intrinsic energy value. These orbitals — being waves — can be mixed in an additive or subtractive way much like you can add or subtract sine waves to each other. When mixing orbitals, you always have one method which creates constructive intereference and one which creates destructive interference — the constructive interference will always have a lower energy than the original and the destructive one will always be higher. The caveat is, that the energy gain from constructive interference will always be less than the energy lost from destructive interference, so that mixing is only overall favourable if one of the resulting orbitals ends up empty (or filled by only one electron).

In the case of Lewis acids and bases, one side (the Lewis base) will always contain a pair of electrons in an orbital of relatively high energy (for filled orbitals) and the other side (the Lewis acid) will usually contain no electrons in an orbital of relatively low energy (for unfilled orbitals). These two can mix favourably: Lowering the populated ligand’s orbital and raising the unpopulated Lewis acid’s orbital creates an overall stabilisation of the system. Typically, for an octahedral complex like $\ce{[Fe(H2O)6]^3+}$, the overall picture will look something like this:

Figure 1: Orbital scheme of a basic octahedral complex. Image originally taken from Prof. Klüfers’ internet scriptum to his coordination chemistry course.

On the right-hand side in figure 1, you can see the six donating ligand orbitals that are each populated with an electron pair. The symbols underneath donate the orbitals’ symmetry point groups. On the left-hand side, you can see a central metal. The lowest-lying set of orbitals corresponds to 3d, the next to 4s and the final to 4p. You can see how mixing with these unoccupied orbitals gives the ligand orbitals an overall stabilisation. It is this that makes $\ce{[Fe(H2O)6]^3+}$ more stable than a ‘naked’ $\ce{Fe^3+}$ ion. The energy gained is largest for the octahedral geometry shown which is why exactly this arrangement is adopted.

There are higher-arching rules, too. For the broad area of metal-carbonyl complexes, one often finds arrangements that correspond to an electron count of 18 overall — and a stable number 18 is also true for many other complexes. 18, of course, corresponds to 10 d-electrons, 2 s-electrons and 6 p-electrons and thus a fully-populated ‘shell’. (Note that the d-electrons belong to the lower shell, but energetically they are all close together; much closer than the d-subshell of the same shell.) Therefore, $\ce{[Fe(CO)5]}$, pentacarbonyliron(0) is stable, iron having eight electrons and each carbonyl donating two more. Cobalt would need nine in addition to the nine it has in its ground state, so it assumes a dimeric structure of $\ce{[Co2(CO)8]}$. Nickel is satisfied with eight additional electrons, so $\ce{[Ni(CO)4]}$. However, you should never just look at the higher-arching rule and thereby decide whether a complex is more stable than another one or not.

Answer 2

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.

Answer 3

I'm gonna give basic results of Crystal Field and related VBT, and I don't think the generalizations made on shape are good in that ppt like Ni+4 Co+3 etc..

Assumptions

1.Electrostatic interaction only between ligand's dipole or charge and positive charge of central metal ion

2.Ligands are point charges

_{(source: pgcc.edu)}

Look up shape of an octahedral, x,y,z axes passing through vertices and central metal at its centre, ligands approach along this axis and increase the energy those orbitals which are along axes which are dx2-y2 and dz2 due to electronic repulsion

so, we have 3 "d" degenerate {equal energy} of one level of energy and two of a higher level, now FILL the orbitals {from start if you want to but just the d subshell is fine} by paulli,hund's rule . When there are 3 unpaired electron in lower energy level, its about to happen, if the energy gap between the levels {increases with strength of field of ligand, look up spectrochemical series for strength [chelating ligands have extra stability and need to be taken care of properly, their strength is more than specified in order as that order doesn't take ring formation entropy increase into consideration [if we made 2 ammonias into ethylene diammine, reaction's products are more so entropy increases and free energy becomes more negative]]} is more than energy needed to piss off hund (pair the electrons) then, pairing will start until the 3 orbitals have 6 electrons, now the next batch starts and start filling like you would normally do, when it's done, start to count orbitals left with no electrons with are lowest in energy and they (6 of them) hybridise to 6 equal energy orbital and as my comment said, each coordinate bond fills one orbital so, we needed 6 orbitals

things are a little complicated for tetrahedral and square planar with the energy gaps not really as simple as described above {to this day i can't convince myself of dz2 in square planar} so, just google it and learn it , and try to derive it by the same method as before

When we write d before, it means inner orbital d orbital is used ex- 3d for Fe+3 and when after one means 4d for Fe+3 i.e. the next shell's d

1. Always start with coordination number i.e. number of ligands [Any one wants to make inner orbital complex as shorter and better bonds]

When it's

4 - Tetrahedral {sp3}[only if we don't get an empty d orbital left , even by hook or crook] or Square Planar {dsp2}

5(rare) - Triagonal Bipyamidal {sp3d} and {dsp3}

6 - octahedral {sp3d2} or {d2sp3} with same logic, latter being preferred if possible

Other things include delocalized (resonance) coordinate bond and it will be specified on how many atoms electron cloud being donated is delocalized , colour [photon accepted and released's wavelength can be correlated withe energy gap between HOMO and LUMO and the electron gets excited to the next lowest energy orbital i.e. LUMO, and then dexcites , photon's energy =hv and can be used to measure crystal field splitting energy with is the energy gap we are discussing], magnetic moment = (n(n+2))^1/2 Bohr Magneton {it's a unit} where n is the number of unpaired electrons in the configuration etc, Also, that energy gap increases rapidly with oxidation number and size and charge density {size becomes smaller as we move down in d block due to lanthanoid/actinoid contraction} so, pretty much any ligand is strong for 4d and 5d metals and for 3d, anything after and including nh3 is strong and anything before and incuding H2O is weak except for Co+3, by strong and weak i mean gap is or is not big enough to cause pairing,

Extra :

1. If there is one unpaired electron and all ligands are strong, then transference will take place if inner orbital complex would be formed {meaning, the electron jumps to a higher level for the good of them team, and the emptied orbital can take part in bonding}

2. If there are 2 unpaired electrons and again, all ligands are strong, again they will be forcefully paired if dsp2 i.e. CN=4 is there. {meaning would be formed if they got paired}

3. Any metal with oxidation number > 4 , All ligands are strong for it like Ni+4. and other factors like number of d electrons, proton number etc also influence it but basically, it increases with positive charge on central metal

from here ligand theory starts