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.
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.
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).
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:
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.
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).
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.
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).
Since bismuth and antimony are typically trivalent, anionic clustersof these elements very quickly exceed the maximum 2n + 8 electronsaccountable under Wades rules for “hypo” compounds. In particular, thepreviously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and$\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species areparticularly intriguing due to the inability of current theories toaccount for them and lack of other corresponding metallic compoundswith 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:
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.
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).
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.
@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.More
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.
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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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
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:
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.
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).
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.
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
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:
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.
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).
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.
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