Jemmis mno rules
Jemmis mno rules represent a unified rule for predicting and systematizing structures of compounds, usually clusters. The rules involve electron counting. They were formulated by Eluvathingal Devassy Jemmis to explain the structures of condensed polyhedral boranes such as B20H16, which are obtained by condensing polyhedral boranes, by sharing a triangular face, an edge, a single vertex or even four vertices. These rules are addition and extensions to Wade’s rules and polyhedral skeletal electron pair theory.[1][2] Jemmis mno rule provides the relationship between polyhedral boranes, condensed polyhedral boranes and β-rhombohedral boron.[3][4] This is similar to the relationship between benzene, condensed benzenoid aromatics and graphite, brought by Hückel’s 4n+2 rule, as well as that between tetracoordinate tetrahedral carbon compounds and diamond. The Jemmis mno rule reduces to Hückel rule when restricted to two dimensions and reduces to Wade’s rules when restricted to one polyhedron.[5]
Contents
Electron counting rules
Electron counting rules are the tools to predict the preferred electron count for molecules. Octet rule, 18-electron rule and Hückel’s 4n+2 π-electron rule are proved to be very useful in predicting the stability of molecules. Wade’s rules were formulated to explain the electronic requirement of monopolyhedral borane clusters. Jemmis mno rule is an extension of Wade’s rule, generalized to include condensed polyhedral boranes as well. The first condensed polyhedral borane B20H16, is formed by the sharing of 4 vertices between two icosahedra. According to Wade’s n+1 rule for closo structures, B20H16 should have a charge of +2 (n+1 = 20+1 = 21 pairs required; 16 BH units provide 16 pairs; four shared boron atoms provide 6 pairs; thus 22 pairs are available). To account for the existence of B20H16 as a neutral species, and to understand the electronic requirement of condensed polyhedral clusters, a new variable, “m”, corresponding to the number of polyhedra (sub-clusters) was introduced.[6] In Wade’s n+1 rule, “1” corresponds to the core bonding molecular orbital (BMO) and “n” corresponds to the number of vertices, which in turn is equal to the number of tangential surface BMOs. If m polyhedra condense to form a macropolyhedron, m core BMOs will be formed. Thus the SEP requirement of closo-condensed polyhedral clusters is m+n. Single-vertex sharing is a special case, where each sub-cluster needs to satisfy Wade’s rule separately. Let a and b be the number of vertices in the sub-clusters including the shared atom. The first cage requires a+1 and second cage requires b+1 skeletal electron pairs (SEPs). Thus a total of a+b+2 or a+b+m SEPs are required; but a+b = n+1, as the shared atom is counted twice. Thus the rule can be modified to m+n+1or generally, m+n+o, where “o” corresponds to the number of single-vertex sharing condensations. The rule can be made more general by introducing a variable, “p”, corresponding to the number of missing vertices, and “q”, the number of caps. Thus the generalized Jemmis rule can be stated as follows:
The SEP requirement of condensed polyhedral clusters is m+n+o+p-q, where, “m” is the number of sub-clusters, “n” is the number of vertices, “o” is the number of single-vertex shared condensations, “p” is the number of missing vertices and “q” is the number of caps.[4][7]
Examples
B20H16
m+n+o+p-q = 2+20+0+0+0 = 22 SEPs are required; 16 BH units provide 16 pairs; four shared boron atoms provide 6 pairs; thus 22 pairs are available as required by Jemmis rule. This explains why B20H16 is stable as a neutral species.[7]
B21H181−
Closo-B21H181− is formed by the face-sharing condensation of two icosahedra. The m+n+o+p-q rule demands 23 SEPs; 18 BH units provide 18 pairs, 3 shared boron atoms provide 4.5 pairs; the negative charge provide 0.5 pair. Thus a total of 23 SEPs are available, which is the same as the required count.[8]
B12H16
The bis-nido-B12H16 is formed by the edge-sharing condensation of nido-B8 unit and nido-B6 unit. The m+n+o+p-q count of 16 SEPs are satisfied by ten BH units which provide 10 pairs, two shared boron atoms which provide 3 pairs and six bridging H atoms which provide 3 pairs.[7]
[Cu(B11H11)2]3−
m+n+o+p-q = 26 SEPs. A transition metal with n valence electrons will provide n-6 electrons for skeletal bonding, as 6 electrons occupying the metal-like orbitals don’t contribute much to the cluster bonding. Thus Cu provides 2.5 pairs, 22 BH units provide 22 pairs; three negative charges provide 1.5 pair. Thus 26 SEPs are available as required.[7]
Ferrocene
According to m+n+o+p-q rule, ferrocene requires, 2+11+1+2-0=16 SEPs. 10 CH units provide 15 pairs; Fe provides one pair. Thus a total of 16 SEPs are available, which satisfies m+n+o+p-q rule.[7]
B18H202−
B18H202− is a bis-nido edge-shared polyhedron. Here, m+n+o+p-q = 2+18+0+2-0 = 22; 16 BH units provide 16 pairs; 4 bridging hydrogen atoms provide 2 pairs; two shared boron atoms provide 3 pairs; along with the two negative charges, 22 SEPs are available as required.[7]
Triple decker complexes
Triple decker complexes are known to obey a 30 valence electron (VE) rule. If we subtract 6 pairs of non-bonding electrons from the two metal atoms, it becomes 9 pairs. For a triple decker complex with C5H5 as the decks, m+n+o+p-q = 3+17+2+2-0=24. Subtracting the 15 pairs corresponding to C-C σ bonds, it becomes 9 pairs. Thus Jemmis rule merges with the 30VE rule for triple decker complexes. For example, consider (C5Me5)3Ru2+. 15 C-CH3 groups provide 22.5 pairs. Each Ru provides one pair. Removal of one electron corresponding to the positive charge of the complex leads to a total of 22.5+2-0.5 = 24 pairs.
β-rhombohedral boron
The structure of β-rhombohedral boron is complicated by the presence of partial occupancies and vacancies.[9][10][11] The idealized unit cell, B105 has been shown to be electron-deficient and hence metallic according to theoretical studies; but β-boron is a semiconductor.[12] Application of Jemmis rule, shows that the partial occupancies and vacancies are necessary for electron sufficiency. B105 can be conceptually divided into a B48 fragment and B28-B-B28 (B57) fragment. As per Wade’s rule, the B48 fragment requires 8 electrons (the icosahedron at the centre (green) requires 2 electrons; each of the six pentagonal pyramids (black and red) completes an icosahedron in the extended structure, and hence the electronic requirement of each of them is 1). The B28-B-B28 or B57 is formed by the condensation of 6 icosahedra and two trigonal bipyramids. Here, m+n+o+p-q = 8+57+1+0-0 = 66 pairs are required for stability; but 67.5 are available. Thus B28-B-B28 fragment has 3 excess electrons. Thus the idealized B105 is deficient by 5 electrons. The 3 excess electrons in the B28-B-B28 fragment can be removed by the removal of one B atom, thereby leading to B27-B-B28 (B56). The requirement of 8 electrons by the B48 fragment can be satisfied by 2.66 boron atoms. Thus the unit cell should contain 48+56+2.66=106.66, which is very close to the experimental result.[3]
References
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