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Why does Fe(III) form octahedral coordination complexes if it has 5 electrons in its d-orbitals? I understand that Fe(II) has 6 electrons in its d-orbitals and 6 lone pairs from 6 ligands as the 12 electrons fill up the 3d, 4s and 4p orbitals (18-electron rule, $6+12=18$). However, Fe(III) has 5 electrons and 6 lone pairs is not enough. I know basic crystal field theory and thought that some of the metal-ligand interactions were covalent bonds, but it doesn't add up for both low and high spin states.

Also, do the ligands need to line up with the orbitals form coordinate bonds? I know it sounds counterintuitive since they repel and that's the basis of the crystal field splitting, but how do coordinate bonds form in coordination complexes then?

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Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ complex).

So how are they still stable? Well, coordination compounds quickly exceed the simplistic bonding pictures that students are taught at the beginning of their chemistry classes. And the ligand field (or crystal field) model—while it does a good job of explaining some observations at a low level—also doesn’t exactly capture what we consider the current knowledge of coordination compounds. A much better description of how they are bonded lies within molecular orbital theory and scheme 1 below.

Molecular orbital scheme of an octahedral complex
Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left of the scheme, the metal’s 3d, 4s and 4p orbitals are placed, while the right-hand half contains the ligand orbitals; simplified as one σ-type orbital per ligand in an octahedral complex. After taking a look at the scheme and finding it perplexing, we can attempt to find the stabilising and destabilising factors in forming a complex.

The most obvious stabilising effect is that all ligand orbitals form bonding interactions that lower their energy. For this, they make use of the typically unoccupied metal’s 4s and 4p orbitals alongside the often occupied 3d orbitals. As we are stabilising occupied orbitals at the expense of unoccupied orbitals, this already results in a net stabilisation.

The second bit of stabilisation results from the interaction of the ligands with parts of the 3d orbitals: precisely the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals, often collectively termed $\mathrm{e_g^*}$ orbitals according to their symmetry group. These may be occupied, but these also have better overlap with the ligand orbitals so the bond formed is stronger than in the case of 4s and 4p.

Finally, the electrons in the other three metal d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$; collectively termed $\mathrm{t_{2g}}$ according to their symmetry group) are, in fact, non-bonding in this simplified picture as their symmetry does not match any ligand orbital. Thus, any electrons in here do not incur any stabilisation but also—arguably more importantly—are not destabilised.

The 18-electron rule derives from this simplified picture and assumes that all bonding and non-bonding orbitals are occupied (that means: all ligand orbitals and the $\mathrm{t_{2g}}$) while all antibonding orbitals (most notably: $\mathrm{e_g^*}$) remain unoccupied. If you count electrons and add up, this gives 18.

However, as I have established the $\mathrm{t_{2g}}$ electrons don’t actually destabilise the complex (under the simplified scheme). Thus, it is relatively irrelevant if there are 0, 2 or 4 of them; it doesn’t change the general scheme. Likewise, even though any $\mathrm{e_g^*}$ orbitals are antibonding, their overall effect is not extreme, so there can be electrons occupying them while still allowing for a stable complex.

The picture on the metal is further complexified (pun intended) by considering high spin versus low spin. In an uncomplexed metal ion, all five d orbitals are energetically equivalent, so they should be filled according to Hund’s rule. In a complex, you have two options:

  • Hund’s rule is observed across all five d orbitals regardless of the energy difference

  • the energy difference is large enough that spin pairing in the $\mathrm{t_{2g}}$ orbitals occurs before the $\mathrm{e_g^*}$ orbials are filled according to Hund’s rule.

The first case is termed high spin and should be treated as the standard case for 3d metals, while the second case is termed low spin and is the standard case in 4d and 5d metals. Note however, that clever choice of ligands may cause a 3d complex to adopt a low-spin configuration and some complexes are even capable of flipping between high and low spin.

While $\ce{Fe(H2O)6]^2+}$ may look like a prime example of the 18 electron rule, it is actually a high-spin complex so the outline I drew above is no longer valid. On the other hand, high-spin $\ce{[Fe(H2O)6]^3+}$ has 5 d-electrons which allows for every d orbital to be populated by exactly one electron—a ‘slightly more favourable’ state. Therefore, high-spin iron(II) complexes are easily oxidised in aquaeous solution but high-spin iron(III) complexes are not as easily reduced. (Obviously, the picture changes when you consider low-spin complexes such as $\ce{[Fe(CN)6]^3-}$, which is a good oxidising agent.)


tl;dr:

  • The 18-electron rule is more of a guideline than an actual rule; it states that more stable adducts can be formed, if, but that’s all;
  • All ligand–metal interactions can and should be considered covalent bonds;
  • Technically, the orbitals and the ligands do need to line up but because electrons are volatile we can assume they have lined up automatically;
  • Essentially, the crystal field model isn’t entirely wrong when it predicts the $\mathrm{e_g^*}$ orbitals to be raised to a higher energy level. It just provides a bad explanation. The orbitals end up being antibonding; that should equally explain.
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