A borane cluster B₆H₆²⁻ has 7 skeletal electron pairs. According to Wade's rules, what geometry does the B₆ framework adopt?
APentagonal bipyramidal (nido, n+1 SEP for a 7-vertex parent)
BOctahedral (closo, n+1 = 7 SEP for n = 6 vertices)
CTrigonal prismatic
DPlanar hexagonal
Wade's rules: for a closo (closed) cluster with n vertices, the number of skeletal electron pairs is n+1. B₆H₆²⁻ has 6 boron atoms, so n = 6. The predicted SEP count for a closo structure is 6+1 = 7. Each BH unit contributes 2 skeletal electrons (B has 3 valence electrons, 1 is used for the B-H bond, 2 remain for the skeleton); 6 BH units give 12 electrons, plus the 2⁻ charge gives 14 electrons = 7 SEP. This matches the closo prediction, so the B₆ framework is a regular octahedron. This is confirmed experimentally: B₆H₆²⁻ is indeed a perfect octahedral cage.
Question 2 True / False
Wade's rules predict that nido clusters have n+2 skeletal electron pairs for n vertices, which corresponds geometrically to a closo polyhedron with one vertex removed.
TTrue
FFalse
Answer: True
Wade's nomenclature classifies clusters by their electron count relative to the number of vertices: closo (n+1 SEP, closed polyhedron), nido (n+2 SEP, one vertex removed from the next-larger closo polyhedron), arachno (n+3 SEP, two vertices removed). B₅H₉ is a classic nido borane: 5 boron atoms with n+2 = 7 SEP. Its B₅ framework is a square pyramid — which is an octahedron (the 6-vertex closo structure) with one vertex removed. This pattern — adding electrons opens the cluster — reflects the antibonding character of the additional electron pairs.
Question 3 True / False
The isolobal analogy allows Wade's rules, originally developed for boranes, to be applied to transition metal clusters by recognizing that a BH fragment is isolobal with certain metal-ligand fragments.
TTrue
FFalse
Answer: True
The isolobal analogy (developed by Roald Hoffmann) identifies molecular fragments that have the same number, symmetry, and approximate energy of frontier orbitals. A BH fragment has two skeletal electrons and three frontier orbitals — it is isolobal with fragments like Fe(CO)₃, Co(CO)₃⁺, and Re(CO)₃(Cp). This means a transition metal cluster compound can be analyzed using Wade's rules by replacing each MLₙ fragment with its isolobal BH equivalent and counting skeletal electrons the same way. Os₆(CO)₁₈ (an octahedral cluster) has the same 7 SEP as B₆H₆²⁻ — both are closo.
Question 4 Short Answer
Explain why adding two electrons to a closo cluster (converting it to nido) causes one vertex to 'open,' and relate this to the electronic structure of the cluster.
Think about your answer, then reveal below.
Model answer: In a closo cluster, all n+1 bonding MOs of the skeletal framework are filled — these are the MOs that hold the cage together. Adding two more electrons (to reach n+2 SEP) must place them in what was the LUMO of the closo cage, which is an antibonding MO concentrated at one vertex. Populating this antibonding orbital weakens the bonding at that vertex, causing it to 'open' — the bonds to that vertex elongate and eventually the atom at that position becomes a bridging or external group rather than a true vertex. The result is a nido geometry: the same basic cage with one vertex removed. Further electron addition (n+3 SEP, arachno) populates another antibonding MO, opening a second vertex.
This is an elegant example of the connection between electron count and structure. The rules are not arbitrary — they arise from the MO energy level pattern of polyhedral clusters, where the number of bonding MOs is always n+1 for an n-vertex cage (a result provable from Euler's theorem for polyhedra).