How do you apply Wade's rules to Zintl polyhedral cluster anions ?

For example, I believe $[\ce{Bi_3}]^{3-}$ is a triangle. My approach to justify would be:

  • 5 electrons per each Bi atom
  • 3 electrons for the charge
  • 18 total electrons

$18 = 4\times3+6 \therefore$ arachno cluster: trigonal bipyramid ($n+2$ polyhedron) with two missing vertices $\therefore$ linear.

If it was a nido cluster $(4n+4\ \ce{e-})$, then it would be a tetrahedron with one missing vertex $\therefore$ a triangle. Am I mistaken about the observed geometry or is my method incorrect?


3 Answers 3


Let the number of atoms be n.

If the number of skeletal p electrons is 2n + 2, the structure is closo (a deltahedron with n vertices).

If the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).

If the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).

If the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).

see Excursions Beyond the Zintl Border and Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary

For example:

Neutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons.

$\ce{Sn_5^{2-}}$, $\ce{Pb_5^{2-}}$, $\ce{Ge_5^{2-}}$ and $\ce{Ge_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).

$\ce{Sn_9^{4-}}$ and $\ce{Ge_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. (gyroelongated square bipyramid with one square pyramid apex missing)

$\ce{Bi_4^{2-}}$ and $\ce{Sb_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).

Also, quoting Farley and Castleman's Observation of Gas-Phase Anionic Bismuth Zintl Ions J. Am. Chem. Soc. 1989, 111, 2734-2735

Since bismuth and antimony are typically trivalent, anionic clusters of these elements very quickly exceed the maximum 2n + 8 electrons accountable under Wades rules for “hypo” compounds. In particular, the previously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and $\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are particularly intriguing due to the inability of current theories to account for them and lack of other corresponding metallic compounds with which they may be compared.

Now going back to the example $\ce{Bi_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:

  1. The structure is isoelectronic with species known to be bent in the ground state including $\ce{O3}$, $\ce{S3}$, $\ce{S2O}$, $\ce{SO2}$ and allyl anion.

  2. Experimental determination published in Stabilization of Ozone-like [Bi3]3- and in a more recent article that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).

  3. Theoretical work in Zintl Anions Analogous to O3.

Though ozone and like species are found to be bent, it is also theorized that cyclic (equilateral triangular ring) ozone represents another higher local potential energy minimum, and this would be the same for $\ce{Bi_3^{3-}}$ too.

$\ce{Bi_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal bipyramid. A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.

  • $\begingroup$ triangular As3 ring was found in organoarsenicum chemistry and in $\ce{[LMAs3ML]^{n+}}; n=1,2; M=Co,Ni$ complexes, implying that triangular structure is acceptable. Bent structure for ozone is preferable because of extra stability of $\pi$-bond for 2nd row elements, however, they are virtually unknown except very special circumstances for 3rd+ rows. Given that, I highly doubt stability of bent geometry. However, relatively high charge density in $\ce{[Bi3]^{3-}}$ makes its formation without external stabilization extremely improbable, so I doubt this consideration are of any use. $\endgroup$
    – permeakra
    Jan 8, 2015 at 18:10
  • $\begingroup$ I understand that ring As3 is known, but I'm not aware of ring $\ce{[As3]^{3-}}$. The relative energies of the ring and bent forms of $\ce{[Sb3]^{3-}}$ are calculated in pubs.acs.org/doi/pdf/10.1021/jp980004e $\endgroup$
    – DavePhD
    Jan 8, 2015 at 18:26
  • $\begingroup$ $\ce{[Sb3]^{3-}}$ has an optimal angle of 131 degrees, even greater than ozone, and the bent form is more stable by 0.5 eV. $\endgroup$
    – DavePhD
    Jan 8, 2015 at 18:45
  • $\begingroup$ $\ce{As3}$ unit in mentioned complex may be considered as $\ce{[As3]^{3-}}$. $\endgroup$
    – permeakra
    Jan 8, 2015 at 23:55
  • $\begingroup$ As for anions, highly-charged anions cannot form without solvation stabilization, and in condensed state the stabilization participates greatly in relative stability. For the matter, I recommend to read the link you provided, as it clearly states that for $\ce{[Na4Sb3]^{+}}$ the geometry with triangular structure is preferable. However, the only state I was able to find with 1:1 K/Bi has $\ce{Bi4}$ and $\ce{Bi2}$ units, and $\ce{Ba2Bi3e}$ phase has complex 3d network of Bi atoms. $\endgroup$
    – permeakra
    Jan 9, 2015 at 0:02

The wiki article contains a note

To predict the structure of an arachno cluster, the closo polyhedron with n + 2 vertices is used as the starting point, and the n+1 vertex nido complex is generated by following the rule above; a second vertex adjacent to the first is removed if the cluster is composed of mostly small atoms, a second vertex not adjacent to the first is removed if the cluster is composed mostly of large atoms.

$\ce{Bi}$ atom is definitely a large atom, so we should remove out non-adjacent vertexes. The only way to choose non-adjacent vertexes in trigonal bipyramid is to take axial vertexes, gaining a triangle.

Also, please not, that removing two adjacent vertexes in trigonal bipyramid may also lead to triangle, if one removes one equatorial and one axial position. removing two equatorial positions will lead not to triangle or line, but to an angle.

However, for $\ce{[Bi3]^{3-}}$ the amount of electrons per atom is exactly six. This means, that 6n rules suite situation better. 6n rule leads to ring in the case.

I recommend to consider Wade's rules as general guidelines, but not a final truth, sometimes they fail spectacularly. For example, consider $\ce{[CH]6}$ cluster. Wade's rule predicts it to be a Trigonal prism. As we know, IRL this molecule prefers to be a ring, with prismane being an esoteric and a quite rare form.

  • $\begingroup$ $\ce{[Bi3]^{3-}}$ is isoelectronic with ozone and allyl anion and bent, not a ring. See etd.nd.edu/ETD-db/theses/available/etd-07072009-102951/… and pubs.acs.org/doi/pdf/10.1021/ja0038732 $\ce{[P3]^{3-}}$ and $\ce{[As3]^{3-}}$ are also bent. $\endgroup$
    – DavePhD
    Jan 7, 2015 at 15:33
  • $\begingroup$ @DavePhD The links you provided are irrelevant as they do not contain a proof that an isolated structure is bent. The ozone-like bent structure is stabilized by means of covalent interactions with transition metal. In fact the links consider mixed clusters, not $\ce{[Bi3]^{3-}}$ $\endgroup$
    – permeakra
    Jan 7, 2015 at 22:23
  • $\begingroup$ Fig.1.2D of the thesis depicts the isolated ion as bent, so I wouldn't say the link is irrelevant, but I agree that it isn't experiment proof. $\endgroup$
    – DavePhD
    Jan 8, 2015 at 1:42
  • $\begingroup$ @DavePhD Fig. 1.2D depicts hypothetical structure, isoelectronic to allyl-anion. This means, that said structure MAY appear in places where normal allyl-anion may appear, but it does not state anything about the free $\ce[Bi3]^{3-}}$ anion itself. I can draw many similarly isoelectronic structures, it does not mean they would appear IRL. $\endgroup$
    – permeakra
    Jan 8, 2015 at 12:23

So-called 2n, 4n, 14n and so forth rules are not necessary at all. Simply, Wade's rule in its original form i.e. 'n' rule is equally applicable to Zintl ions and sufficient to predict their structural type e.g., closo, nido, arachno, hypho and so on. For example, the Skeleton Electron Pair Count becomes six [0.5(3 x 5 - 3)] for [Bi5]3+; indicating (n + 1) formulation and hence it must adopt closo structure. Similarly, the SEPC becomes nine [0.5(3 x 5 + 3)] for [As5]3-, implying its hypho structure from (n + 4) formulation.

Moreover, the contribution of the vertex atom (here it is Bi/As) can be easily calculated from the 'v + x - 2' rule, where v is the # of valance electron(s). The value of x (contribution of exo ligand towards vertex atom) must be zero since there is no exo ligand bonded with the vertex atoms in Zintl ions, Naked Clusters. Jadavpur University, India.


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