# Advanced Inorganic Chemistry/Electron counting and the 18 electron rule

The 18 electron rule can be used to predict the reactivity of complexes, especially transition metal organometallic complexes. Transition metals contain 1 x s, 3 x p and 5 x d orbitals which can carry up to 2, 6 and 10 electrons respectively. Complexes containing 18 electrons are therefore preferred as this leads to an increased stability resulting from fully filling all the s, p and d orbitals available. This rule could be seen as analogous to the octet rule in organic chemistry with both trying to achieve noble gas configuration, but the addition of d orbitals increases the electron count from 8 to 18. Although having 18 electrons is most stable, this rule is not followed by all complexes.

## Ligand effects on the 18 electron rule

Whether a complex follows the rule or not can depend on whether the ligands attached to the metal are π-acceptors, π-doners or just σ-doners. In addition, the period that the transition metal is in also matters.

### Complexes that follow the rule

With strong field ligands such as CO and CN$^-$ that are σ-doners and π-acceptors the 18 electron rule is followed. π-acceptors have a large octahedral splitting meaning the t$_{2g}$ orbital is a low energy bonding orbital, whilst the e$_g$* is higher in energy and is antibonding. Due to the large difference in energy between the t$_{2g}$ and e$_g$* orbitals, electrons only occupy up to the t$_{2g}$. The e$_g$* is left unoccupied and so the most stable 18 electron configuration is obeyed.

### Exceptions

First row transition metals with medium or weak field ligands often have between 12-22 electrons. Due to a small difference in energy between the t$_{2g}$ and e$_g$* orbital, the t2g orbital becomes non-bonding and the e$_g$* orbital made up of the 3d$_z$2 and 3d$_x$$^2$-$_y$$^2$ becomes only weakly antibonding. This therefore means up to an additional 4 electrons can be added.
Second or third row transition metals usually adopt configurations with 12-18 electrons. As 4d and 5d transition metals have larger orbitals on the metal it means there is an increased metal d orbital-ligand repulsion, which causes a greater octahedral splitting. The t2g is thought to be non-bonding whilst the e$_g$* is a high energy antibonding orbital and so is never occupied. These complexes therefore have 18 electrons or fewer.

## Electron Counting

There are 2 methods for counting electrons the ionic method and the covalent method. In the ionic method, all the ligands are removed from the metal and electrons are given to the ligand in order to fill all its valence orbitals. For example, if CO is removed from a metal, both the C and O have a full octet and the molecule has a neutral charge. This ligand is therefore a 2 electron donor neutral ligand. In contrast, for methyl the octet is filled by adding one electron which gives CH$_3$$^-$. Once the metal forms a bond with the anionic carbon, the methyl acts as a 2 electron donor and the metal is oxidised.
In contrast, in the covalent method all bonds are treated as covalent and the electron count is determined by what makes a ligand neutral. L ligands such as CO or PR$_3$ datively donate two electrons to the metal and are neutral. X ligands also exist which treat the ligands as radicals and donate 1 electron to the metal, examples include H and R. One electron in the covalent bond is provided by the ligand and the other by the metal. These ligands are negatively charged. Finally Lewis acids, such as BF$_3$, are Z ligands where instead the metal datively donates its electrons to the ligands.

### Examples of electron counts using the ionic and covalent method for some key ligands

| Ligand | Covalent | Ionic | Charge |
| M-X, M-R, M-H | 1 | 2 | -1 |
| M-CO | 2 | 2 | 0 |
| M-PR3 | 2 | 2 | 0 |
| η$^2$-C2H4 | 2 | 2 | 0 |
| M-OR | 1 | 2 | -1 |
| M-Ar | 1 | 2 | -1 |
| M-C(O)-R | 1 | 2 | -1 |
| η$^1$-allyl | 1 | 2 | -1 |
| η$^3$-allyl | 3 | 4 | -1 |
| η$^5$-cyclopentadienyl | 5 | 6 | -1 |
| η$^6$-benzene | 6 | 6 | 0 |
| η$^7$-cycloheptatrienyl | 7 | 6 | +1 |
| μ-X (M-X-M) | 3 | 4 | -1 |

### Steps for covalent method

* Identify the group number of the metal.
* Identify the electron counts of each ligand.
* If more than one metal in the complex account for metal-metal bonds (each bond is one electron).
* If the complex has a positive charge subtract an electron from the total count, if it has a negative charge add an electron to the total.
* To obtain the total number of electrons add up the metal group number, electron count of ligands, charges and metal-metal bonds (if present).

### Steps for ionic method

* Work out the d electron count of the metal using the following equation: number of d electrons = group number – oxidation state.
* Identify the electron counts of each ligand.
* If more than one metal in the complex account for metal-metal bonds (each bond is one electron).
* To obtain the total number of electrons add up the metal d electron count, electron count of the ligands, charges and metal-metal bonds (if present).

## Practice Questions

Ru(CO)$_5$:
* Covalent method: Ru (group 8) contributes 8 electrons, each CO 2 electrons: 5 x 2 = 10 electrons, no charge. Total = 10 + 8 = 18 electrons.
* Ionic method: Ru number of d electrons: 8-0 = d$^8$. Each CO is 2 electrons: 5 x 2 = 10 electrons. Total = 10 + 8 = 18 electrons.
Re(CO)$_2$(PR$_3$)$_2$CH$_3$(C$_2$H$_2$):
* Covalent method: Re (group 7) contributes 7 electrons, each CO 2 electrons: 2 x 2 = 4 electrons, each PR$_3$ 2 electrons: 2 x 2 = 4 electrons, CH$_3$ contributes 1 electron and C$_2$H$_2$ contributes 2 electrons. Total = 7 + 4 + 4 + 1 + 2 = 18 electrons.
* Ionic method: Re number of d electrons: 7-1 = d$^6$. Each CO 2 electrons: 2 x 2 = 4 electrons, each PR$_3$ 2 electrons: 2 x 2 = 4 electrons, CH$_3$ contributes 2 electrons and C$_2$H$_2$ contributes 2 electrons. Total = 6 + 4 + 4 + 2 + 2 = 18 electrons.
Re$_2$Cl$_8$$^{2-}$:
For each Re:
* Covalent method: Re (group 7) contributes 7 electrons, each Cl is a 1 electron donor: 4 x 1 = 4 electrons, 4 Re-Re bonds each contributing one electron: 4 x 1 = 4. For each Re the charge can be split so overall -1 charge each so add one electron to each. Total: 7 + 4 + 4 + 1 = 16 electrons.
* Ionic Method: Re number of d electrons: 7-3 = d$^4$. Four Cl each contribute 2 electrons each: 4 x 2 = 8 electrons, 4 x Re-Re bonds contribute 4 electrons. Total 4 + 8 + 4 = 16 electrons.
Ni(Cp)$_2$:
* Covalent method: Ni (group 10) contributes 10 electrons, each Cp contributes 5 electrons: 2 x 5 = 10 electrons. Total 10 + 10 = 20 electrons.
* Ionic method: Ni number of d electrons: 10-2 = d$^8$. Each Cp contributes 6 electrons = 6 x 2. Total = 12 + 8 = 20 electrons.

## References

* ↑ Pfennig, B. (2015). Inorganic Chemistry. New Jersey: John Wiley & Sons. pp. 628.
* ↑ Miessler, G. & Tarr, D. (2014). Inorganic Chemistry. Essex: Pearson. pp. 388.