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In chemistry, the **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

Electron-counting rules are used to predict the preferred electron count for molecules. The octet rule, the 18-electron rule, and Hückel's 4*n* + 2 pi-electron rule are proven to be useful in predicting the molecular stability. Wade's rules were formulated to explain the electronic requirement of monopolyhedral borane clusters. The Jemmis *mno* rules are an extension of Wade's rules, generalized to include condensed polyhedral boranes as well.

The first condensed polyhedral borane, B_{20}H_{16}, is formed by sharing four vertices between two icosahedra. According to Wade's *n* + 1 rule for *n*-vertex *closo* structures, B_{20}H_{16} 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 B_{20}H_{16} as a neutral species, and to understand the electronic requirement of condensed polyhedral clusters, a new variable, *m*, was introduced and corresponds to the number of polyhedra (sub-clusters).^{[6]} In Wade's *n* + 1 rule, the 1 corresponds to the core bonding molecular orbital (BMO) and the *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 skeletal electron pair (SEP) requirement of closo-condensed polyhedral clusters is *m* + *n*.

Single-vertex sharing is a special case where each subcluster needs to satisfy Wade's rule separately. Let *a* and *b* be the number of vertices in the subclusters including the shared atom. The first cage requires *a* + 1 and the second cage requires *b* + 1 SEPs. Therefore, 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. The rule can be modified to *m* + *n* + 1, or 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. As such, 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 subclusters,*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]}

*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, which describes why B_{20}H_{16} is stable as a neutral species.^{[7]}

*closo*-B_{21}H−18 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 and 3 shared boron atoms provide 4+1⁄2 pairs; the negative charge provides one half pair.^{[8]}

The bis-*nido*-B_{12}H_{16} is formed by the edge-sharing condensation of a *nido*-B_{8} unit and a *nido*-B_{6} 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]}

*m* + *n* + *o* + *p* − *q* = 26 SEPs. A transition metal with *n* valence electrons provides *n* − 6 electrons for skeletal bonding as 6 electrons occupying the metal-like orbitals do not contribute much to the cluster bonding. Therefore Cu provides 2+1⁄2 pairs, 22 BH units provide 22 pairs; three negative charges provide 1+1⁄2 pairs.^{[7]}

According to the *m* + *n* + *o* + *p* − *q* rule, ferrocene requires 2 + 11 + 1 + 2 − 0 = 16 SEPs. 10 CH units provide 15 pairs while Fe provides one pair. ^{[7]}

B_{18}H2−20 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 which provide 1 pair.^{[7]}

Triple-decker complexes are known to obey a 30-valence electron (VE) rule. Subtracting 6 pairs of nonbonding electrons from the two metal atoms brings the number of SEPs to 9 pairs. For a triple-decker complex with C_{5}H_{5} as the decks, *m* + *n* + *o* + *p* − *q* = 3 + 17 + 2 + 2 − 0 = 24. Subtracting the 15 pairs corresponding to C–C sigma bonds, it becomes 9 pairs. For example, consider (C_{5}(CH_{3})_{5})_{3}Ru+2: 15 C–CH_{3} groups provide 22+1⁄2 pairs. Each ruthenium atom provides one pair. Removing the electron corresponding to the positive charge of the complex leads to a total of 22+1⁄2 + 2 − 1⁄2 = 24 pairs.

The structure of β-rhombohedral boron is complicated by the presence of partial occupancies and vacancies.^{[9]}^{[10]}^{[11]} The idealized unit cell, B_{105} has been shown to be electron-deficient and hence metallic according to theoretical studies, but β-boron is a semiconductor.^{[12]} Application of the Jemmis rule shows that the partial occupancies and vacancies are necessary for electron sufficiency.

B_{105} can be conceptually divided into a B_{48} fragment and a B_{28}−B−B_{28} (B_{57}) fragment. According to Wade's rule, the B_{48} 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; as such the electronic requirement for each of them is 1). The B_{28}−B−B_{28} or B_{57} 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 required for stability, but 67+1⁄2 are available. Therefore the B_{28}−B−B_{28} fragment has 3 excess electrons and the idealized B105 is missing 5 electrons. The 3 excess electrons in the B_{28}−B−B_{28} fragment can be removed by removing one B atom, which leads to B_{27}−B−B_{28} (B_{56}). The requirement of 8 electrons by the B_{48} fragment can be satisfied by 2+2⁄3 boron atoms and the unit cell contains 48 + 56 + 2+2⁄3 = 106+2⁄3, which is very close to the experimental result.^{[3]}