# How to determine net charge of a polycation using Wade's rules?

In order to determine the net charge of polycations, which adopt a Wade-like structure (octahedral, trigonal bipyramidal, ...), Wade's rules can be used.

For example, the charge $x$ of the octahedral cluster $\ce{[Bi6]^x+}$ should be determined:

• It is known that a Wade-cluster overall "needs" $2n+2$ electrons (to fill the $n+1$ energetically favoured, bonding MOs).
• Bi atoms contain three valence p-electrons (the two s-electrons do not take part in the cluster-bonding). So in a $\ce{Bi_n}$ cluster, there are $3n$ electrons already available for bonding.
• But only $2n+2$ electrons are needed: We need to remove $2n+2-3n=-n+2=-(n-2)$ electrons to obtain the favoured $2n+2$ electron count in the cluster.
• So for example the $\ce{[Bi6]^x+}$ should have $-(6-2)=-4$ electrons "removed". We would arrive at a $\ce{[Bi6]^4+}$ net charge... ...but in reality, there is "only" a $\ce{[Bi6]^2+}$ - with two electrons more than determined according to the Wade's rules.

EDIT: In contrast to that, the rules aboves apply to $\ce{[Bi5]^3+}$ (Thanks to @orthocresol).

Is there a misconception in my above thoughts? I can not find an illogical step - Maybe Wade's rules can not be applied to the polyatomic clusters (my textbooks suggests this nevertheless)? Or I should have take the s-electrons into account?

As far as I can tell, your calculations are correct. A closo structure would require $2n+2$ cluster electrons (or $4n+2$ total electrons), which is satisfied by the formula $[\ce{Bi}_n]^{(n-2)+}$. As such, for $n = 6$, a closo octahedral structure would correspond to the ion $\ce{[Bi_6]^4+}$ as you have outlined.

The contradiction seems to be that the ion $\ce{[Bi_6]^2+}$ is not closo, but rather nido, according to e.g. Hampel & Ruck:[1]

The $\ce{[Bi_6]^2+}$ cation, a nido cluster with 16 skeletal electrons, has the shape of a distorted octahedron with an opened edge.

These require $2n$ cluster electrons for complete population of the bonding molecular orbitals, consistent with the formula $\ce{[Bi_6]^2+}$. I do not currently have access to Holleman-Wiberg (nor would I understand the German version!), so I cannot check what they have written, but it is possible that it is a simple typo.

Interestingly, some of the higher bismuth polycations also display a tendency to have "extra" electrons. For example, Greenwood and Earnshaw[2] describe the following cations:

$$\begin{array}{cccc}\hline \text{Cation} & \text{Formal oxidation state} & \text{Cluster structure} & \text{Point group symmetry} \\ \hline \ce{Bi+} & 1.00 & - & - \\ \ce{Bi3+} & 0.33 & \text{Triangle} & D_\mathrm{3h} \\ \ce{Bi5^3+} & 0.60 & \text{Trigonal bipyramid} & D_\mathrm{3h} \\ \ce{Bi8^2+} & 0.25 & \text{Square antiprism} & D_\mathrm{4h} \\ \ce{Bi9^5+} & 0.56 & \text{Tricapped trigonal prism} & C_\mathrm{3h} (\sim\!D_\mathrm{3h}) \\ \hline \end{array}$$

The rationalisation which they provide for this is

[The application of Wade's rules] would account for the stoichiometries $\ce{Bi3+}$ and $\ce{Bi5^3+}$ but would also lead one to expect $\ce{Bi8^6+}$ and $\ce{Bi9^7+}$ for the larger clusters. However, these charges are very large and it seems likely that the lowest-lying nonbonding orbital would also be occupied in $\ce{(Bi+)_n^2-}$. For $\ce{(Bi+)8^2-}$ this is an $\mathrm e_1$ orbital which can accommodate 4 electrons, thereby reducing the charge from $\ce{Bi8^6+}$ to $\ce{Bi8^2+}$ as observed. In $\ce{(Bi+)9^2-}$ the lowest nonbonding orbital is $\mathrm a_2''$ which can accommodate 2 electrons, thus reducing the charge from $\ce{Bi9^7+}$ to $\ce{Bi9^5+}$ as observed.

### References:

1. Hampel, S.; Ruck, M. Synthesen, Eigenschaften und Kristallstrukturen der Cluster-Salze Bi6[PtBi6Cl12] und Bi2/3[PtBi6Cl12]. Z. Anorg. Allg. Chem. 2006, 632 (7), 1150–1156. DOI: 10.1002/zaac.200500379.

2. Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, pp 590–591.