Magadh University, Bodh-Gaya
Dr. Partha Pratim Das Assistant Professor P.G. Department of Chemistry Magadh University, Bodh-Gaya E-Content Title: Boranes
Stream: M. Sc. (2nd Semester)
Email id: [email protected] Boranes: Boranes or Boron hydrides are the class of compounds having the general formula BxHy. Boron does not react with hydrogen directly to form any of boron hydrides or boranes, but many of these compounds are easily synthesized under special conditions. Due to high affinity of boron towards oxygen, they readily get oxidized by air. Hence, these compounds do not occur in nature. This whole class is derived from the borane
(BH3) only, which exists as a transient intermediate and dimerizes to form diborane (B2H6) immediately. All higher boranes consist of boron clusters, which are polyhedral in nature. Besides neutral boranes, a large number 2− of anionic boron hydrides also exist. The general formula for single-cluster boron hydrides are BnHn , BnHn+4, BnHn+6, BnHn+8 and BnHn+10 (n = number of boron atoms) for closo-, nido-, arachno-, hypho- and klado- type, respectively. There are also series of substituted neutral hypercloso-boranes, which have theoretical formula
BnHn.
Naming of neutral boranes is done with a Greek prefix showing number of boron atoms and number of hydrogen atoms in brackets, while, for naming of anions, hydrogen count is dictated first followed by the boron count and the overall charge in the bracket in the last, in addition prefixes like closo- nido- etc., which are related to geometry and number of framework electrons of a particular cluster, can also be added.
2− Formula B5H9 B4H10 B6H6 Name (IUPAC) Pentaborane(9) Tetraborane(10) hexahydridohexaborate(2−) Type Nido- Arachno- Closo-
VBT approach for 3-centre-2-electron bond in borane: Boron hydrides are electron-deficient species. Simplest boron hydride with 3-center-2-electron bond is diborane (B2H6), bacially dimeric unit of BH3 molecules. One s and three p orbitals mix and generate four sp3 hybrid orbitals, three half-filled and one empty. How come dimerization occurs then, as it seems three hybrid orbitals would be taken up by three H atoms? Each boron atom in diborane can use two of its sp3 half-filled hybrid orbitals to bind two hydrogen atoms (terminal-H). Remaining two hybrid orbitals (one half-filled and one empty) are then used to bind bridging-H atoms as depicted pictorially. Dotted lobes represent empty orbitals. Valence electron of bridging hydrogen and one electron from one half filled sp3 hydrid orbital of one boron atom gets delocalized over all three atoms (B–H–B), producing banana shaped 3-centre-2-electron bond. This depiction of banana bond is contradictory to VSEPR theory though, as it deals with bounded valence electrons into regions of space that are localized between two participating nuclei only. In addition, the geometry is also unexplainable as two bonds around each bridging hydrogen should be linear i.e. the B–H–B bond angle should be ~ 180° with straight-line coordination around, according to VSEPR theory.
Molecular Orbital approach for 3-Centre-2-Electron Bond in borane: As we want to understand the nature of bonding involved in B–H–B bridges, our focus will be on just this part of the molecule only. Four equivalent sp3 hybrid orbitals of each Boron atoms and two s orbitals of hydrogen atoms construct the symmetry adapted linear combinations (SALCs) of atomic orbitals. Irreducible components for H atom, ΓH = a1g + b3u, and for B atom, ΓB = a1g + b1g + b2u + b3u.
Single-electron wave functions and corresponding molecular orbital shapes for B–H–B portion of diborane are depicted above, focussing mainly only on the largest lobes of the four sp3 hybrid orbitals. Energies two nonbonding molecular orbitals unequal as smaller lobes on the opposite side also have a good possibility to * overlap. Bonding and anti-bonding MOs, a1g and a1g , respectively, are formed by mixing of two SALCs of * a1g symmetry from both H and B atom each. Similarly, bonding and anti-bonding MOs, b3u and b3u , respectively, are formed by mixing of two SALCs of b3u symmetry from both H and B atom each. Shapes of MOs a1g and b3u symmetry clarifies that electron density present in them is delocalized over all three nuclei involved in B–H–B unit of diborane. Some extent of direct B–B bonding, due to small-size of H-atom, is also present in diborane, which helps to stabilize the molecule geometry.
Differrent types of bonds present in higher boranes: Overall bonding of higher boranes are explained by counting “framework electrons” i.e. electrons involved in skeletal structure of Bn cluster, where terminal B– H bonds are discarded and B–H–B bridges are considered as part of the Bn framework. Each B atom uses one electron in traditional 2-center-2-electron terminal B–H bond, whereas, remaining two electrons are used in
Bn framework. Concept of localized 3-centre-2-electron and 2-centre-2-electron orbital approach is used to explain main features of boranes, though, electrons involved in the framework construction of Bn cluster are very much delocalized. Structures of higher boranes have the following different kinds of bonds:
1. Terminal B–H bond: It is a normal 2-center-2-electron covalent bond shown as B–H. 2. Direct B–B bond: It is a 2-center-2-electron bond connecting two boron atoms, represented as B–B. 3. Bridging or open B–H–B bond: It is a 3-center-2-electron bond connecting two boron and one hydrogen atoms, represented as B–H–B. 4. Closed B–B–B bond: It is a 3-center-2-electron bond connecting three boron atoms present at the corners of an equilateral triangle, depicted as B–B–B. 5. Open bridging B–B–B bond: It is a 3-center-2-electron bond connecting three boron atoms to form a B– H–B like bridge, found in some diborane, represented as B–B–B.
Geometry prediction for higher boranes using lipscomb’s model of STYX Numbers: American chemist, W. N. Lipscomb developed a method to determine possible key combinations of bonding profile for a specific boron hydride. His topological model involving STYX numbers and rules was the initial approach to understand the structures of boranes and related species. Each boron atom in neutral boranes or hydroborate anions has at least one H attached by a traditional 2-center-2-electron sigma bond. Hence, one B–H bond is present per boron atom. Further, he proposed that every other bond present in the cluster must belong to any of the following categories: (i) 3-centre-2-electrons B–H–B bond, indicated as “S”. (ii) Closed, open or a mixture of both 3-centre 2-electrons B–B–B bond, indicated as “T”. (iii) 2-center-2-electrons B–B bond, indicated as “Y”.
(iv) 2-center-2-electrons B–H terminal bond (Number of BH2 groups), indicated as “X”.
Prediction of structures of borane clusters by Lipscomb’s model involves following three major components:
(1) Relationship between STYX code and valence electrons: Correlation between borane’s formula and the number and types of bonds in the cluster is given by simple equations of balance, which relate all kinds of sigma bonds to the number of valence electrons available. (a) Three centers orbital balance: Electron deficiency in a borane cluster can be removed only if one 3-center-2-electron bond is created by each boron atom. Hence, sum of the number of 3-centre 2- electron B–H–B bonds and the number of 3-centre 2-electron B–B–B bonds must be equal the number of B–H units (n) i.e. [ = S + T]. (b) Hydrogen balance: Assuming one terminal hydrogen is attached to each boron atom, number of 𝑛𝑛 hydrogen atoms leftover (m), must be distributed among bridges and additional B–H terminal bonds i.e. [ = S + X]. (c) Electron balance: If each BH unit contributes one pair of electrons to the skeleton and each of 𝑚𝑚 the extra hydrogens gives one electron; number of electron pairs contributed by additional hydrogens must be calculated by dividing their number by two. All of these electron pairs must be participating in bonding; and therefore, the total number of bond pairs can be given as: + ( /2) = S + T + Y + X; Hence, Y = ½ (S-X)
𝑛𝑛 𝑚𝑚 (2) Calculation of various STYX possibilities: Upon knowing all correlations between various kinds of bonds with framework electron pairs in boron hydrides, one need to follow the following steps to determine various STYX possibilities.
(a) Note down the general formula of given borane cluster in form like BnHn+m to calculate the values of n and m. (b) Determine the number of B–H–B bridges, represented by S. Value of S must lie within the range: [m/2 to m]. Therefore, value of S must satisfy the condition; m/2 ≤ S ≥ . Validation of the higher limit comes from the fact that, m represents total additional hydrogen atoms, out of which some are present in 𝑚𝑚 B–H–B bond and some are present in B–H terminal bond. Hence, if all the additional hydrogen atoms are present in B–H–B bond, the value of “S” at its maximum can just be equal to m. Validation of the lower limit can be derived as; + ( /2) = S + T + Y + X + ( /2) = ( − X) + T + Y + X as = S + X 𝑛𝑛 𝑚𝑚 = ( /2) + T + Y 𝑛𝑛 𝑚𝑚 𝑚𝑚 𝑚𝑚 S + T = ( /2) + T + Y as = S + T 𝑛𝑛 𝑚𝑚 S = ( /2) + Y 𝑚𝑚 𝑛𝑛 So, S is always equal to or greater than (m/2). 𝑚𝑚 (c) For each different values of S obtained in 2nd step, one need to determine equally possible solutions for the values of T, Y and X. (d) There could be many sets of STYX numbers for a given borane cluster and thus many possible topologies.
(3) Short listing of STYX codes using empirical rules: To pick the correct STYX code among several possibilities, empirical rules have been developed. Such as, (a) STYX number sets with negative value have no physical meaning. They are out of consideration. (b) All boranes have at least a 2-fold symmetry. Hence, it is assumed that any hydride probably should have at least one plane, centre or two-fold axis of symmetry. (c) Only one terminal and no bridging hydrogen may be attached to boron, which is bound to five neighbouring borons, restricting B–H–B bridges and BH2 groups to the open edges of boron frameworks. (d) If a boron atom is bound to four other boron atoms, it will probably not make use of more than one B–H–B bridge. (e) If a boron atom that is bound to only two other boron atoms, it will be involved in at least one B– H–B bridge.
These concepts can be understood and exemplified using the diborane structure. General formula for
diborane can be written as B2H6 i.e. B2H2+4; giving n = 2 and m = 4.
As, m/2 ≤ S ≥ , so 4/2 ≤ S ≥ 4, Hence, possible values for S = 2, 3, 4.
When S = 2, 𝑚𝑚 As, = S + T, T = 0. As, = S + X, X = 2. As, 𝑛𝑛 + ( /2) = S + T + Y + X, Y = 0. 𝑚𝑚 When𝑛𝑛 S =𝑚𝑚 3, As, = S + T, T = -1. As, = S + X, X = 1. As, 𝑛𝑛 + ( /2) = S + T + Y + X, Y = 1. 𝑚𝑚 When𝑛𝑛 S =𝑚𝑚 4, As, = S + T, T = -2. As, = S + X, X = 0. As, 𝑛𝑛 + ( /2) = S + T + Y + X, Y = 2. 𝑚𝑚 If we𝑛𝑛 arrange𝑚𝑚 these values in a table it will look like: Number of B–H–B Number of B–B–B Number of B–B bonds Number of BH2 groups bonds bonds S T Y X 2 0 0 2 3 -1 1 1 4 -2 2 0 So, among the three sets (2,0,0,2), (3,-1,1,1) and (4,-2,2,0), only the first one is physically possible, as the rest possess one negative value each. Experimental finding of the structure supports this set of STYX number, as diborane possess two B–H–B bridges, but no three-center B–B–B bond or two-center B–B bond. Further, two B–H terminal bonds in addition to those already considered i.e. two BH2 groups.
Structure and Bonding of Some Higher Borane Clusters: Tetraborane-10 (B4H10):
Plane projections and three-dimensional structure of B4H10
Four B–H–B bridges, no closed or bridging B–B–B unit, one B–B bond and two terminal BH2 groups are present [STYX code is 4012]. Electron requirement of various kinds of bonds and individual contribution of boron and hydrogen atoms are tabulated below. Four 3-centre 2-electron and 7 2-centre 2-electron bonds present here, require a total 4×2 + 7×2 = 22 electrons. 4 Boron atoms provide 12 valence electrons and 10 electrons are actually contributed by 10 hydrogen atoms.
Nature of the bond Number Total electron required Contribution from Contribution from of bond 4B atoms 10H atoms B–H–B 4 8 4 4 B–B–B 0 0 0 0 B–B 1 2 2 0 B–H 6 12 6 6
Pentaborane-9 (B5H9): Here five atoms of boron arranged in a square pyramidal geometry. Each boron atom has a terminal hydride ligand and four other hydrides span the edges of the base of the pyramid. There are four B–H–B bridges, one closed or triply bridged B–B–B unit, two B–B bond and no terminal BH2 groups. STYX code is 4120. Electron requirement of various kinds of bonds and individual contribution of boron and hydrogen atoms are tabulated below. Five 3-centre 2-electron and 7 2-centre 2-electron bonds present here, require a total 5×2 + 7×2 = 24 electrons. 5 Boron atoms provide 15 valence electrons and 9 electrons are actually contributed by 9 hydrogen atoms in total for both types of bonds.
Nature of the bond Number Total electron required Contribution from Contribution from of bond 4B atoms 10H atoms B–H–B 4 8 4 4 B–B–B 1 2 2 0 B–B 2 4 4 0 B–H 5 10 5 5
Plane projections and three-dimensional structure of B5H9
Pentaborane-11 (B5H11):
Plane projections and three-dimensional structure of B5H11
Pentaborane-11 has unsymmetrical square-pyramidal geometry. Five boron atoms are present at the five corners of square pyramid, similar to B5H9 moiety. It is evident from the structure that there are 3 B–H– B bridges, 2 closed or triply bridged B–B–B unit, zero B–B bond and 3 terminal BH2 groups. Hence, STYX is 3203. 5, 3-Centre 2-electron and 8, 2-centre 2-electron bonds present here require 5×2 + 8×2 = 26 electrons in total. 5 Boron atoms provide 15 valence electrons and 11 electrons are contributed by eleven hydrogen groups for bonding. Nature of the bond Number Total electron required Contribution from Contribution from of bond 4B atoms 10H atoms B–H–B 3 6 3 3 B–B–B 2 4 4 0 B–B 0 0 0 0 B–H 8 16 8 8
Hexaborane-10 (B6H10):
Plane projections and three-dimensional structure of B6H10
Structure suggests that 4 B–H–B bridges, 2 closed or triply bridged B–B–B unit, 2 B–B bond and no terminal BH2 groups are present. STYX code is 4220. 6, 3-Centre 2-electron and 8, 2-centre 2-electron bonds require total 6×2 + 8×2 = 28 electrons. 6 Boron atoms have 18 valence electrons, while 10 electrons are contributed by 10 hydrogen groups that are participating in both types of bonds.
Nature of the bond Number Total electron required Contribution from Contribution from of bond 4B atoms 10H atoms B–H–B 4 8 4 4 B–B–B 2 4 4 0 B–B 2 4 4 0 B–H 6 12 6 6
Decaborane-14 (B10H14)
Plane projections of its three-dimensional structure of B10H14
Three-dimensional structure of B10H14
Here, B10 framework resembles an incomplete octadecahedron in B10H14. Each boron atom has one terminal hydride ligand and the four boron atoms near the open part of cluster, possess extra hydrides. There are 4 B–H–B bridges, 6 B–B–B units (4 B-B-B triple bridge bonds and two B-B-B bent bridge), 2 B–B bond and zero terminal BH2 groups. STYX code is 4620. 10, 3-Centre 2-electron and 12, 2-centre 2- electron bonds require 10×2 + 12×2 = 44 electrons in total. 10 Boron atoms provide 30 valence electrons and 14 electrons are contributed by 14 hydrogen groups for overall bonding. Nature of the bond Number Total electron required Contribution from Contribution from of bond 4B atoms 10H atoms B–H–B 4 8 4 4 B–B–B 6 12 12 0 B–B 2 4 4 0 B–H 10 20 10 10 Deltahedral Cages with five to twelve Vertices:
Wade’s Rules: Lipscomb’s topological model involving STYX numbers and rules was satisfactory towards understanding and rationalization of structures of few the boranes and related species, but it was not enough practical towards higher boranes and carbonyl clusters. More sophisticated and comprehensive method was essential indeed. British chemist, Kenneth Wade provided a straightforward and elegant explanation for the geometries of ‘‘electron-deficient’’ cluster compounds in terms of the number of skeletal electron pairs (SEPs) available, later known as Wade’s rules. This Rule focuses mainly on boranes, carboranes and low-valent transition-metal clusters; whose structures are not explainable by traditional 2-center-2-electron bond. Another British chemist, Michael Mingos, extended the principles of Wade’s rules about counting skeletal electron pairs to electron-precise and electron-rich clusters and gave a more generalized way of calculating skeletal electron contribution. Since then it is known as Wade-Mingos rules or Polyhedral Skeletal Electron Pair Theory (PSEPT). Structure prediction emphasizes upon different sets of rules [4n, 5n or 6n], developed on the basis of the number of electrons present per vertex. 4n rules are very much accurate towards predicting the structures of clusters with 4 electrons per vertex in case of many carboranes or boranes. Clusters following 4n rule are generally classified as closo-, nido-, arachno- or hypho-, depending on whether they represent a complete (closo-) deltahedron, or a deltahedron that is missing one (nido-), two (arachno-) or three (hypho-) vertices. If count of electrons is near to 5 electrons per vertex, structure changes occur, following the 5n rules (based on 3-connected polyhedrons). Similarly, when electron count is further increased, clusters with 5n electron counts become very unstable and 6n rules take the charge (structures are dependent on rings).
Above shown structural relationships explain several geometrical aspects of borane clusters, providing conceptual way to rationalize these systems. Diagonal lines connect species that have same number of skeletal electron pairs (SEP). For Example; structure of dodecahydro-closo-dodecaborate anion 2− [B12H12 ] is a regular icosahedron. Removal of one B–H unit and addition of two H atoms converts nido-B5H9 into a butterfly-shaped arachno-B4H10 borane. Nevertheless, it must be noted that inter- conversion of these structures may or may not be possible chemically and same statement can be made for their chemical formula.
4n Rule: Basis closo-polyhedra for 4n rules are made of triangular faces [n = number of vertices in the cluster]. Number of vertices in cluster prediction and corresponding basis polyhedron is listed below:
Number of Vertex Polyhedron 4 Tetrahedron 5 Trigonal bipyramid 6 Octahedron 7 Pentagonal bipyramid 8 D2d (trigonal) dodecahedron 9 Tricapped trigonal prism 10 Bicapped square antiprism 11 Octadecahedron 12 Icosahedron
Upon counting electron predicted structure can be found. 4n Rules are summarized in the following table:
Electron count Name Predicted structure 4n − 2 Bicapped closo n − 2 vertex closo polyhedron with 2 capped faces 4n Capped closo n − 1 vertex closo polyhedron with 1 face capped 4n + 2 Closo Closo polyhedron with n vertices 4n+ 4 Nido n + 1 vertex closo polyhedron with 1 missing vertex 4n + 6 Arachno n + 2 vertex closo polyhedron with 2 missing vertex 4n + 8 Hypho n + 3 vertex closo polyhedron with 3 missing vertex 4n + 10 Klado n + 4 vertex closo polyhedron with 4 missing vertex
When counting electrons for each cluster, the number of valence electrons is enumerated. For each transition metal or ions present, 10 electrons are subtracted from the total electron count. For example, in Rh6(CO)16 the total number of electrons would be 6×9 + 16×2 − 6×10 = 86–6 × 10 = 26. Therefore, the cluster is a closo polyhedron because n = 6, with 4n + 2 = 26.
The following other rules are considered when predicting the structure of clusters:
1. For clusters consisting mostly of transition metals, any main group elements present are often best counted as ligands or interstitial atoms, rather than vertices. 2. Larger or more electropositive atoms tend to occupy vertices of high connectivity and smaller or more electronegative atoms tend to occupy vertices of low connectivity. 3. In the special case of boron hydride clusters, each boron connected to 3 or more vertices has one terminal hydride, while each boron connected to 2 other vertices has 2 terminal hydrogens. If more hydrogens are present, they are placed in open face positions to even out the coordination number of the vertices. 4. For the special case of transition metal clusters, ligands are added to the metal centers to give the metals reasonable coordination numbers, and if any hydrogen atoms are present they are placed in bridging positions to even out the coordination numbers of the vertices.
In general, closo structures with n vertices are n-vertex polyhedra.
To predict the structure of a nido cluster, the closo cluster with n + 1 vertices is used as a starting point; if the cluster is composed of small atoms a high connectivity vertex is removed, while if the cluster is composed of large atoms a low connectivity vertex is removed.
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.
Example:
2− (1) Pb10 :
Electron count: 10 × number of valence electron of Pb + 2 (for the negative charge) = 10×4 + 2 = 42 electrons. Since n = 10, 4n + 2 = 42, so the cluster is a closo bicapped square antiprism.
2+ (2) S4 :
Electron count: 4× number of valence electron of S – 2 (for the positive charge) = 4×6 – 2 = 22 electrons. Since n = 4, 4n + 6 = 22, so the cluster is arachno. Starting from an octahedron, a vertex of high connectivity is removed, and then a non-adjacent vertex is removed.
4− (3) B5H5 :
Electron count: 5 × number of valence electron of B + 5 × number of valence electron of H + 4 (for the negative charge) = 5×3 + 5×1 + 4 = 24 Since n = 5, 4n + 4 = 24, so the cluster is nido. Starting from an octahedron, one of the vertices is removed. The hydrogen atoms have been omitted for clarity.
(4) C2B7H13:
The rules are useful in also predicting the structure of carboranes. Electron count = 2× number of valence electron of C + 7 × number of valence electron of B + 13 × number of valence electron of H = 2×4 + 3×7 + 13×1 = 42 Since n in this case is 9, 4n + 6 = 42, the cluster is arachno.
(5) C2B5H7: Electron count = 2× number of valence electron of C + 5 × number of valence electron of B + 7 × number of valence electron of H = 2×4 + 3×5 + 7×1 = 30. Since n in this case is 7, 4n + 2 = 30, the cluster is closo. The hydrogens have been omitted for clarity.
(6) Os6(CO)18:
Electron count: 6 × number of valence electron of Os + 18×CO – 6×10 (for 6 osmium atoms) = 6×8 + 18×2 – 60 = 24 Since n = 6, 4n = 24, so the cluster is capped closo. Starting from a trigonal bipyramid, a face is capped. The carbonyls have been omitted for clarity.
The bookkeeping for deltahedral clusters is sometimes carried out by counting skeletal electrons instead of the total number of electrons. The skeletal orbital (electron pair) and skeletal electron counts for the four types of deltahedral clusters are:
• Bicapped closo: n - 1 skeletal orbitals, 2n - 2 skeletal electrons • Capped closo: n skeletal orbitals, 2n skeletal electrons • closo: n + 1 skeletal orbitals, 2n + 2 skeletal electrons • nido: n + 2 skeletal orbitals, 2n + 4 skeletal electrons • arachno: n + 3 skeletal orbitals, 2n + 6 skeletal electrons • hypho: n + 4 skeletal orbitals, 2n + 8 skeletal electrons • Klado: n + 5 skeletal orbitals, 2n + 10 skeletal electrons
The skeletal electron counts are determined by summing the total of the following number of electrons:
• 2 from each BH unit • 3 from each CH unit • 1 from each additional hydrogen atom (over and above the ones on the BH and CH units) • anionic charge as it is.
Example: C2B7H13, n is 9, while electron pairs are 2×CH + 7×BH + 4×H = 6 + 14 + 4 = 24 or 12 pairs. Numbers of framework electron pairs are 9 + 3 = 12; hence, the structure is arachno.
5n Rule: As discussed previously, the 4n rule mainly deals with clusters which have approximately 4 electrons on each vertex. As more electrons are added per vertex, the number of electrons per vertex approaches 5. Rather than adopting structures based on deltahedra, the 5n-type clusters have structures based on a different series of polyhedra known as the 3-connected polyhedra, in which each vertex is connected to 3 other vertices. The 3-connected polyhedra are the duals of the deltahedra. The common types of 3-connected polyhedra are listed below:
Number of Vertex Polyhedron 4 Tetrahedron 6 Trigonal prism 8 Cube 10 Pentagonal prism 12 D2d pseudo-octahedron 14 Dual of triaugmented triangular 16 Square truncated trapezohedron 18 Dual of edge-contracted icosahedron 20 Dodecahedron
The 5n rules are as follows:
Total electron count Predicted structure
5n n-vertex 3-connected polyhedron 5n + 1 n – 1 vertex 3-connected polyhedron with one vertex inserted into an edge 5n + 2 n – 2 vertex 3-connected polyhedron with two vertices inserted into edges 5n + k n − k vertex 3-connected polyhedron with k vertices inserted into edges
Example:
(1) P4: Electron count: 4× number of valence electron of P = 4×5 = 20 It is a 5n structure with n = 4, so it is tetrahedral.
(2) P4S3: Electron count 4× number of valence electron of P + 3× number of valence electron of S = 4×5 + 3×6 = 38 It is a 5n + 3 structure with n = 7. Three vertices are inserted into edges.
(3) P4O6: Electron count 4× number of valence electron of P + 6× number of valence electron of O = 4×5 + 6×6 = 56 It is a 5n + 6 structure with n = 10. Six vertices are inserted into edges.
6n Rule: As more electrons are added to a 5n cluster, number of electrons per vertex approaches 6. Instead of adopting structures based on 4n or 5n rules, the clusters tend to have structures governed by the 6n rules, which are based on rings. The rules for the 6n structures are as follows.
Total electron count Predicted structure k 6n – k n-membered ring with ⁄2 transannular bonds 6n – 4 n-membered ring with 2 transannular bonds 6n – 2 n-membered ring with 1 transannular bond 6n n-membered ring 6n + 2 n-membered chain (n-membered ring with 1 broken bond)
Example:
(1) S8: Electron count = 8 × number of valence electron of S = 8×6 = 48 electrons. Since n = 8, 6n = 48, so the cluster is an 8-membered ring.
(2) C6H14: Electron count = 6× number of valence electron of C + 14× number of valence electron of H = 6×4 + 14×1 = 38 Since n = 6, 6n = 36 and 6n + 2 = 38, so the cluster is a 6-membered chain.