Boron Hydrides: Classification of Boranes

Take a jar of diborane, B₂H₆, and warm it gently. Instead of behaving like a simple hydride, it rearranges. It forms a whole family of cage compounds — B₄H₁₀, B₅H₁₁, B₆H₁₀, B₁₀H₁₄. Each has its own peculiar geometry. None looks anything like a normal molecule built from two-center two-electron bonds. This is the strange, beautiful chemistry that Alfred Stock spent two decades coaxing out of high-vacuum lines in the 1910s and 1920s. William Lipscomb and later Kenneth Wade turned it into one of the most elegant classification schemes in inorganic chemistry. Boron simply does not have enough electrons to draw ordinary Lewis structures for these molecules. Yet they are stable, isolable, and chemically rich. Understanding why requires abandoning the electron-pair bond as the only currency of bonding. We must accept that boron clusters spend their electrons on binding whole faces and cages, rather than single edges. This matters far beyond a curiosity of the periodic table's most notoriously electron-poor element. The boranes are the historical proving ground for three-center two-electron (3c–2e) bonding. They are also the proving ground for cluster electron-counting rules. These rules now underpin the structural chemistry of transition-metal carbonyl clusters, Zintl ions, metallaboranes, and even some organometallic catalysts. The vocabulary introduced here — closo, nido, arachno, hypho — is not boron-specific jargon. It is a general grammar for cluster shapes. This grammar recurs whenever a set of atoms shares a common electron-deficient bonding budget. Learning to read a borane formula and predict its shape is a genuine skill. This holds true using nothing but its atom count and charge. It is not memorization, once the pattern is seen. The big picture is this: every closed, near-spherical boron cluster (a closo borane) can be thought of as a deltahedron. A deltahedron is a polyhedron whose every face is a triangle. Removing one, two, or three vertices from that parent deltahedron generates new shapes. Patching the resulting hole with extra hydrogens creates them. This generates nido (nest), arachno (web), and hypho (net) clusters respectively. Each has a systematically decreasing symmetry and increasing openness. Each carries a fixed number of skeletal electron pairs relative to the vertex count. The formulas $\text{B}_n\text{H}_n^{2-}$ , $\text{B}_n\text{H}_{n+4}$ , $\text{B}_n\text{H}_{n+6}$ , and $\text{B}_n\text{H}_{n+8}$ are not four unrelated series. They need not be memorized independently. They are one idea, viewed from four different vertex-deficiencies of the same underlying deltahedra. This article builds the classification from structure first. It starts with what these cages actually look like. Then it states the counting relationships that assign a formula to a shape. It works through the reasoning that connects vertex-removal to formula change. Finally, it finishes with worked examples. These span a foundational closo ion, a workhorse nido cluster, and a trickier arachno case with a non-obvious charge. The formal electron-counting machinery of Wade's Rules for Cluster Electron Counting is foreshadowed here. The bonding picture of Three-Center Two-Electron Bonding is also foreshadowed. Both are developed fully elsewhere. Here the focus is on recognizing and classifying the clusters themselves.

Key takeaways

Boron hydrides form polyhedral cage/cluster molecules because boron lacks enough valence electrons for classical 2c–2e bonds throughout.
Closo boranes are closed deltahedra: $\text{B}_n\text{H}_n^{2-}$ , with every vertex occupied and no missing corners.
Nido boranes are closo deltahedra missing one vertex ("nest" shaped): $\text{B}_n\text{H}_{n+4}$ .
Arachno boranes are missing two vertices ("web" or spider shaped, more open): $\text{B}_n\text{H}_{n+6}$ .
Hypho boranes are missing three vertices (very open "net" fragments, usually unstable/rare): $\text{B}_n\text{H}_{n+8}$ .
Key idea: each class is the same parent deltahedron with 0, 1, 2, or 3 vertices removed, and the H-count formula tracks that vertex deficiency exactly.

Where this lives today

Borane cluster chemistry, and the closo/nido/arachno vocabulary specifically, now extends far past its historical roots in Stock's vacuum-line hydrides. The closo dianion $[\mathrm{B}_{12}\mathrm{H}_{12}]^{2-}$ and its carborane cousins ( $\mathrm{C}_2\mathrm{B}_{10}\mathrm{H}_{12}$ isomers — ortho-, meta-, para-carborane) are industrially manufactured, exceptionally robust building blocks used in boron neutron capture therapy (BNCT) for tumor targeting, exploiting boron-10's large neutron-capture cross-section, since the icosahedral cage survives biological conditions that would destroy almost any other boron-containing functional group. Nido-carborane anions such as $[\mathrm{C}_2\mathrm{B}_9\mathrm{H}_{11}}]^{2-}$ (dicarbollide) are the ligand basis of an entire branch of sandwich-complex organometallic chemistry, coordinating to transition metals in a fashion directly analogous to cyclopentadienyl, because the open pentagonal face of a nido cage presents exactly the right symmetry and electron count to bind a metal fragment isolobally. The same closo/nido/arachno logic transfers, almost formula-for-formula, to metallaboranes and metal carbonyl clusters, where a transition-metal vertex fragment can substitute isoelectronically for a BH unit, and to Zintl-ion clusters of heavier p-block elements (Ge₉⁴⁻, Sn₉⁴⁻, Pb₉⁴⁻), which adopt nido or arachno cage geometries by exactly the same vertex-removal logic from a parent deltahedron. Superacid and weakly-coordinating-anion chemistry exploits the extraordinary chemical inertness of closo-borate dianions like $[\mathrm{CB}_{11}\mathrm{H}_{12}]^-$ as some of the least nucleophilic, most oxidatively and thermally robust counterions known, used to stabilize highly reactive cationic species (including silylium and even protonated-benzene cations) that would decompose any conventional counterion. Even in materials science, boron-rich solids and boride ceramics draw on cluster-electron-counting intuition developed first for the molecular boranes catalogued here.

Ask the tutor: How does the skeletal electron pair count from Wade's rules actually predict which specific deltahedron (octahedron vs. trigonal bipyramid vs. icosahedron) a given borane will adopt?

The physical picture — thorough prose

Start with the fact that dominates all of boron cluster chemistry: boron has three valence electrons and four valence orbitals (2s, 2pₓ, 2p_y, 2p_z). A neutral BH₃ fragment uses only three of boron's electrons plus one from hydrogen to make one B–H bond and leaves boron with an empty p orbital — it is isoelectronic with a carbocation fragment, chronically short of electron pairs. When many BH units try to assemble into a polyatomic hydride, there are never enough electron pairs to give every pair of adjacent boron atoms its own two-center two-electron bond, the way carbon manages in alkanes. Something else has to hold the cage together. The resolution, developed by Longuet-Higgins and Lipscomb and confirmed by X-ray and later neutron diffraction, is that boron hydrides use two additional bonding motifs beyond the ordinary 2c–2e bond: bridging B–H–B hydrogens, in which one hydrogen sits between two borons and a single electron pair is smeared over the three-atom unit, and B–B–B three-center bonds, in which a pair of electrons occupies a bonding orbital delocalized over a triangular face of borons with no hydrogen at all. These 3c–2e interactions are individually weaker than a normal 2c–2e bond but there are more of them, and collectively they hold cage frameworks together with real thermodynamic stability. The formal machinery for counting how many of each type a given cluster needs belongs to a separate treatment of Three-Center Two-Electron Bonding; what matters for classification is simply that boron clusters are held together by a shared, delocalized pool of skeletal bonding electrons rather than a fixed set of localized bonds — and that pool has a size, in electron pairs, that depends in a very regular way on the cluster's shape. That regularity is captured by imagining an idealized family of closed polyhedra called deltahedra: solids whose faces are all triangles. The tetrahedron (4 vertices), trigonal bipyramid (5), octahedron (6), pentagonal bipyramid (7), dodecahedron in the boron sense — actually a bisdisphenoid (8), tricapped trigonal prism (9), bicapped square antiprism (10), octadecahedron (11), and icosahedron (12) form the standard series used throughout borane and carborane chemistry. A closo borane occupies every single vertex of one of these deltahedra with a BH unit (each boron also carrying one terminal, non-bridging hydrogen), and it is these fully triangulated, vertex-complete cages that are the reference point for everything else. Now imagine physically plucking one vertex out of a closo deltahedron, BH unit and all, while keeping the remaining framework roughly in the shape it had (with some relaxation). The resulting open-faced cluster, missing one corner, is called nido (Latin for "nest") — it looks like a bowl or basket, an open face surrounded by the rest of the polyhedron. Removing a second, adjacent vertex from that same parent deltahedron opens the structure further into a shallower, more spread-out fragment: this is arachno ("spider's web" or "web-shaped"), typically resembling a butterfly or an open, flattened chain-fused-ring shape. Removing a third vertex yields hypho ("net"), an even more open, floppy fragment that is rarely isolable as a stable, stand-alone neutral hydride (most confirmed hypho examples exist as adducts or in derivatized/metal-complexed forms). Each vertex removed pulls the extra electron density that vertex "used" for bonding into the remaining framework, and the compensations show up as extra hydrogens bonded to the open cluster — this is the concrete, countable signature that lets a chemist look at a formula and immediately place it in a class.

Figure

The closo→nido→arachno progression from a common octahedral parent

Note in the figure how the vertex count n is what shrinks (6 → 5 → 4) as vertices are removed from the same 6-vertex octahedral parent, while the hydrogen count grows relative to n (from Hₙ to Hₙ₊₄ to Hₙ₊₆). This anti-correlation — fewer skeletal atoms, more extra hydrogens per remaining atom — is the entire content of the classification, and it is worth internalizing visually before turning to the formulas. It helps to also see how the bridging hydrogens physically decorate a real nido cage, since these are not drawn arbitrarily but sit over specific edges of the missing face.

Figure

nido-B₅H₉: square pyramidal cage with bridging hydrogens over the open base

This is exactly the known solid-state structure of pentaborane(9): a square pyramid of five borons, each carrying one terminal hydrogen, with four bridging hydrogens spanning the four base edges around the open face where the sixth octahedral vertex would have been. Nine hydrogens total for five borons: $\text{B}_5\text{H}_{5+4} = \text{B}_5\text{H}_9$ , matching the nido formula exactly.

The core ideas — the governing equations/relationships

The classification rests on comparing two numbers for a given cluster: the number of boron (or boron+heteroatom) vertices, $n$ , and the total number of hydrogens (terminal plus bridging), which fixes the class. The four canonical formula families are:

\text{closo: } \mathrm{B}_n\mathrm{H}_n^{2-}

where: $n$ = number of vertices = number of boron atoms, and the deltahedron is fully closed (every vertex present, every face triangular). The $2-$ charge is essential — the neutral parent $\mathrm{B}_n\mathrm{H}_n}$ does not exist as a stable closo species; closo boranes are isolated almost exclusively as dianions.

\text{nido: } \mathrm{B}_n\mathrm{H}_{n+4}

where: $n$ = number of boron vertices actually occupied (one fewer than the parent deltahedron's vertex count), and the cluster is neutral in this, its most common isolable form.

\text{arachno: } \mathrm{B}_n\mathrm{H}_{n+6}

where: $n$ = number of occupied vertices (two fewer than the parent deltahedron), neutral parent hydride form.

\text{hypho: } \mathrm{B}_n\mathrm{H}_{n+8}

where: $n$ = number of occupied vertices (three fewer than the parent deltahedron); such species are electron-richest of all and are rarely stable as free neutral hydrides, usually appearing as anions or Lewis-base adducts.

Reading these relationships. Each successive family adds exactly two more hydrogens per formula step (n → n+4 → n+6 → n+8), and this step of $+2$ H is not arbitrary — it is the hydrogen-counting fingerprint of removing one vertex at a fixed total skeletal electron-pair count, which is the actual underlying invariant (developed fully under Wade's Rules for Cluster Electron Counting). For now, treat the four formulas as an empirical, but exact and diagnostic, ladder: given any boron hydride's molecular formula and charge, compute $n$ (borons) and the "excess hydrogen" index

x = (\text{total H count}) - n

where: $x$ counts how many hydrogens exceed one-per-boron. If the species is a dianion and $x = 0$ , it is closo; if neutral and $x = 4$ , nido; if neutral and $x=6$ , arachno; if neutral (or anionic, with charge accounted for) and $x = 8$ , hypho. Species with intermediate or shifted $x$ due to non-zero charge are handled by folding the charge into an equivalent hydrogen count (each extra negative charge behaves, for this bookkeeping, like one extra H⁻ donating a pair, and each positive charge like removing one). A companion relationship, useful for sanity-checking any classification, links the class directly to how many vertices are missing from an $m$ -vertex closo parent:

n = m - k, \qquad k \in \{0,1,2,3\} \text{ for closo, nido, arachno, hypho}

where: $m$ is the vertex count of the parent deltahedron (the smallest deltahedron with $\geq n$ vertices, in practice $m = n+k$ ), and $k$ is the number of vertices removed. This says nido-B₅H₉'s parent deltahedron has $m = n+1 = 6$ vertices — the octahedron — exactly as drawn above; arachno-B₄H₁₀'s parent has $m = n+2 = 6$ vertices too, the same octahedron, just missing two corners instead of one.

Derivation AWhy removing a vertex adds exactly four hydrogens (nido from closo)

Goal. Show why stripping one BH vertex from a closo cluster $\mathrm{B}_n\mathrm{H}_n^{2-}$ (parent size $n+1$ , since the nido cluster of size $n$ derives from an $(n+1)$ -vertex parent) produces a neutral nido cluster with formula $\mathrm{B}_n\mathrm{H}_{n+4}$ , i.e. an increase of exactly four H atoms (net, after also removing the vertex's own single H and its charge).

Step 1 — Write the closo parent explicitly. Take the $(n+1)$ -vertex closo dianion, $\mathrm{B}_{n+1}\mathrm{H}_{n+1}^{2-}$ . Every boron carries exactly one terminal hydrogen; the whole cluster carries an overall 2− charge distributed over the delocalized skeletal bonding framework.

Step 2 — Remove one BH vertex and its associated bonding contribution. Physically deleting one $\{\mathrm{BH}\}$ unit removes one boron atom and one hydrogen atom from the formula, leaving a fragment $\mathrm{B}_n\mathrm{H}_n^{2-}$ — but this fragment is now an open cage (a missing vertex leaves an exposed face), not a closed deltahedron, and it is drastically electron-poor at that open face because the vertex removed had been contributing orbitals and electrons to several three-center bonds around it.

Step 3 — Electron bookkeeping across the removal. Each BH vertex in the closed parent contributes 2 skeletal electrons to the cluster's shared bonding pool (1 from B beyond its terminal B–H bond, 1 from the associated H order — formally, each BH unit donates a total of 3 orbitals and supplies 2 electrons to skeletal bonding after using one orbital+electron pair for the exo B–H bond). Removing a BH⁻-equivalent unit removes 2 electrons and 3 orbitals from the skeletal count, but the total number of skeletal electron pairs required to hold together an $n$ -vertex open (nido) cluster stays fixed at $n+2$ pairs (this invariant is the heart of Wade's rules and is justified in detail elsewhere). To keep that pair count constant after losing the vertex's own contribution, the fragment must gain compensating electron density from somewhere.

Step 4 — Supply the missing density as extra protons plus electrons, i.e. as H atoms/hydride. The open face left behind is capped, not by a new boron, but by extra hydrogen atoms that each donate one electron into the framework via B–H–B bridging bonds, while simultaneously neutralizing the residual negative charge. Quantitatively: the parent dianion had charge 2− over $n+1$ vertices. Removing one BH vertex leaves charge 2− over $n$ borons and $n$ terminal H — call this hypothetical intermediate $[\mathrm{B}_n\mathrm{H}_n]^{2-}$ . To reach the neutral, isolable nido hydride, this dianion must be protonated twice (each added H⁺ combines with a lone pair from the electron-rich open face to form a new bridging B–H–B bond, consuming one unit of negative charge and adding one H atom):

[\mathrm{B}_n\mathrm{H}_n]^{2-} + 2\,\mathrm{H}^+ \longrightarrow \mathrm{B}_n\mathrm{H}_{n+2}

But this only accounts for +2 H, not +4. The resolution is that the correct comparison is not "remove one vertex from the $n$ -vertex closo parent" but "compare the $n$ -vertex nido cluster to the $(n+1)$ -vertex closo cluster and separately track that the nido cluster's own most stable closo count is $n+1$ , i.e. $m = n+1$ ." Redo the count directly from the skeletal electron pair invariant: a closo cluster of $n$ vertices has $n+1$ skeletal pairs; an isoelectronic nido cluster built on the same $n$ -vertex fragment (one vertex short of its own $(n+1)$ -vertex closo parent) also carries $n+2$ skeletal pairs (one more pair than the analogous closo count for that vertex number), because the open face permits more localized bridging bonds to form without over-filling antibonding cluster orbitals. Each terminal BH unit supplies 2 skeletal electrons (as in Step 3); $n$ BH units supply $2n$ electrons, i.e. $n$ pairs. The required total is $n+2$ pairs $= 2n+4$ electrons. The shortfall of $4$ electrons ( $=2$ pairs) is made up by four additional H atoms, each donating one electron as a bridging/extra hydrogen, giving the neutral closed-shell formula

\mathrm{B}_n\mathrm{H}_{n} \; (\text{skeleton}) + 4\,\mathrm{H} \;\longrightarrow\; \mathrm{B}_n\mathrm{H}_{n+4}

which is exactly the nido formula, with no residual charge.

Reading the result. The "+4" is not a coincidence of nomenclature; it is the electron count needed to raise an $n$ -vertex fragment from "closo-appropriate" bonding ( $n+1$ pairs) up to "nido-appropriate" bonding ( $n+2$ pairs) while also neutralizing charge, and each unit of that raise costs one extra H atom donating one electron, doubled by the pairing requirement and the charge neutralization together. The same argument, run for two and three removed vertices, generates the $+6$ (arachno) and $+8$ (hypho) increments, each successive class needing one additional skeletal electron pair (equivalently two more H atoms) than the last.

Go deeper · Derivation B · 2 minClassifying an unknown borane from its bare formula

Derivation BClassifying an unknown borane from its bare formula

Goal. Given only a molecular formula (and its charge), derive a short algorithm — used constantly by working chemists — that identifies $n$ , computes the excess-hydrogen index, and reads off closo/nido/arachno/hypho without needing to draw the cage first.

Step 1 — Extract $n$ and total H, and fold in the charge. For a borane $\mathrm{B}_n\mathrm{H}_p^{\,z-}$ (or with $z$ positive for a cation, rare), define the charge-adjusted hydrogen count

p' = p + z

where: $z$ is the magnitude of negative charge (each extra negative charge is treated as equivalent, for electron-counting purposes, to one additional bridging hydride donating a pair — consistent with the fact that closo clusters are characteristically dianions of formula $\mathrm{B}_n\mathrm{H}_n^{2-}$ , i.e. $p'=p+2 = n+2$ ... this needs care, shown below).

Step 2 — Compute the excess index $x = p' - n$ . This single number is the entire classification key:

x = 0 \Rightarrow \text{closo}, \qquad x = 4 \Rightarrow \text{nido}, \qquad x = 6 \Rightarrow \text{arachno}, \qquad x = 8 \Rightarrow \text{hypho}

Step 3 — Apply carefully to a charged closo example. Take $\mathrm{B}_6\mathrm{H}_6^{2-}$ : here $n=6$ , $p=6$ , $z=2$ , so $p' = 6+2=8$ , and $x = p'-n = 8-6=2$ . This does not match the pattern $x=0$ for neutral formulas above — because the closo formula is defined directly as $\mathrm{B}_n\mathrm{H}_n^{2-}$ without folding the charge into $x$ at all. The correct, consistent convention (used throughout this classification) is: compare the raw H : B ratio and the raw charge simultaneously against the four reference formulas, rather than trying to collapse charge and H-count into one universal index. That is:

\text{closo test: } p = n \text{ and } z = 2

\text{nido test: } p = n+4 \text{ and } z = 0 \ (\text{most common}), \text{ or } p=n+3,\ z=1 \text{ for } [\mathrm{B}_n\mathrm{H}_{n+3}]^-,\text{ etc.}

Step 4 — Generalize to anionic nido/arachno/hypho members. Many known clusters are anionic derivatives of the neutral reference formulas, formed by loss of H⁺ (deprotonation of a bridging hydrogen) rather than by changing the skeletal electron count. For instance, nido- $\mathrm{B}_5\mathrm{H}_9$ deprotonates to $[\mathrm{B}_5\mathrm{H}_8]^-$ : same $n=5$ , same electron-counting class (nido), but $p=8=n+3$ and $z=1$ . The general rule, robust to charge, is to compute the total skeletal electron pairs implied, $\mathrm{SEP} = \tfrac12\big(3n + p + z - 2n\big) = \tfrac12(n+p+z)$ (each B contributes 3 orbitals/2 electrons after its exo bond, each extra H or negative charge contributes 1 electron beyond the 1-per-B baseline — full justification under Wade's rules), and read off:

\mathrm{SEP} = n+1 \Rightarrow \text{closo}, \quad n+2 \Rightarrow \text{nido}, \quad n+3 \Rightarrow \text{arachno}, \quad n+4 \Rightarrow \text{hypho}

Reading the result. For $[\mathrm{B}_5\mathrm{H}_8]^-$ : $\mathrm{SEP} = \tfrac12(5+8+1) = \tfrac12(14) = 7 = n+2$ with $n=5$ , confirming nido, matching the parent $\mathrm{B}_5\mathrm{H}_9$ 's class even though a bridging proton was removed. This SEP-based test is the rigorous, charge-robust version of the simple formula-matching done in Step 1–2, and it is exactly what the full skeletal-electron-pair (Wade's rules) treatment formalizes; the four boxed reference formulas above remain the fast, correct answer for the overwhelmingly common case of neutral nido/arachno/hypho hydrides and dianionic closo clusters.

Worked examples

Example 1 (foundational): classify B₄H₁₀. Count borons: $n=4$ . Count hydrogens: $p = 10$ . The species is neutral, $z=0$ . Compute the excess index directly against the reference formulas:

p - n = 10 - 4 = 6

An excess of 6 matches the arachno reference formula $\mathrm{B}_n\mathrm{H}_{n+6}$ exactly. So B₄H₁₀ (tetraborane(10)) is arachno, built from an octahedral ( $m=6$ ) parent deltahedron with two vertices removed ( $k=2$ , since $n=m-k=6-2=4$ ). This is consistent with its known "butterfly" structure: four boron atoms in a bent, open rhombus-like arrangement, four terminal B–H, four bridging B–H–B, and two extra "endo" terminal hydrogens on the two wingtip borons, for a hydrogen total of $4+4+2=10$ . Checking the limiting case: if instead only one vertex were missing from the same octahedral parent ( $n=5$ ), the reference formula predicts $\mathrm{B}_5\mathrm{H}_9$ — precisely pentaborane(9), the canonical nido cluster discussed above — confirming that B₄H₁₀ and B₅H₉ are two different vertex-deficiencies of the same six-vertex octahedral ancestor.

Example 2 (applied): classify the borane anion $[\mathrm{B}_{12}\mathrm{H}_{12}]^{2-}$ and explain its exceptional stability. Here $n=12$ , $p=12$ , $z=2$ . Testing against the closo reference $\mathrm{B}_n\mathrm{H}_n^{2-}$ :

p = 12 = n, \qquad z = 2 \ (\text{matches exactly})

So $[\mathrm{B}_{12}\mathrm{H}_{12}]^{2-}$ is closo, built on the 12-vertex icosahedron, the most symmetric and most thermodynamically robust of all deltahedral boranes. It survives boiling in acid or base, resists oxidation, and is the terminus of nearly every borane pyrolysis or hydrolysis sequence — a real chemical fact, not a classification artifact, that follows directly from the closed-shell, maximally symmetric electron count ( $n+1 = 13$ skeletal pairs exactly filling the icosahedral cluster bonding manifold, with no partially-filled or antibonding orbitals engaged). Applying the SEP formula as a cross-check:

\mathrm{SEP} = \tfrac12(n+p+z) = \tfrac12(12+12+2) = \tfrac12(26) = 13 = n+1

confirming closo independently of the fast formula-matching test. As a limiting case, note that shrinking this same icosahedral series down to $n=6$ gives $[\mathrm{B}_6\mathrm{H}_6]^{2-}$ (octahedral closo), the smallest commonly cited closo borate dianion, so the pattern holds from $n=6$ all the way to $n=12$ without modification — closo boranes are known for essentially every $n$ in that range.

Example 3 (exam/competition level): a mixed anionic arachno species. Suppose an exam question gives the formula $[\mathrm{B}_9\mathrm{H}_{14}]^{-}$ and asks for its classification and parent deltahedron. First, $n=9$ , raw $p=14$ , $z=1$ . Testing directly against neutral arachno ( $p=n+6$ ) fails since $14 \ne 9+6=15$ . But charge must be included. Compute SEP:

\mathrm{SEP} = \tfrac12(n+p+z) = \tfrac12(9+14+1) = \tfrac12(24) = 12

Compare to $n+3 = 12$ : this matches exactly, so $[\mathrm{B}_9\mathrm{H}_{14}]^-$ is arachno (SEP $=n+3$ ). Its parent deltahedron has $m = n+2 = 11$ vertices — the 11-vertex octadecahedron — with two vertices removed to expose the open, web-like arachno face. As a check against the simpler charged reference formula: an arachno anion formed by simple deprotonation of the hypothetical neutral $\mathrm{B}_9\mathrm{H}_{15}$ (which would be $n+6=15$ , matching the neutral arachno formula precisely) losing one H⁺ gives exactly $[\mathrm{B}_9\mathrm{H}_{14}]^-$ , so the anion is nothing more exotic than deprotonated arachno-B₉H₁₅, confirming the SEP-based classification by an entirely independent route. This double-check — reference-formula matching and SEP arithmetic agreeing — is exactly the kind of cross-validation a well-prepared exam answer should show.

Limiting cases, scope & common traps

Situation	What happens	Correct classification approach
Neutral hydride, $p=n$	Almost never isolable as a stable neutral molecule	Closo species require the 2− charge; don't force a neutral closo assignment
Very small $n$ (2 or 3)	B₂H₆, B₃H₈⁻ etc. sit outside the clean deltahedral pattern	Treat $n\le4$ cases (diborane, etc.) as special/limiting; classification language is most reliable for $n\ge5$
Large closo dianions ( $n=9$ – $12$ )	Exceptionally robust, aromatic-like 3D delocalization	Confirms closo assignment; chemically these are the most inert boranes known
Anionic nido/arachno species from deprotonation	H count looks "off" from the neutral reference formula by exactly the charge	Always fold charge into the SEP formula, don't pattern-match blindly
Hypho species	Rarely stable as free neutral hydrides	Expect these mostly as base adducts, e.g. $\mathrm{B}_8\mathrm{H}_{12}\cdot\mathrm{L}_2$ -type species, not bare $\mathrm{B}_n\mathrm{H}_{n+8}$
Heteroboranes (carboranes, metallaboranes)	A CH or a transition-metal fragment can substitute isoelectronically for a BH vertex	Same closo/nido/arachno/hypho ladder applies once vertex-electron equivalences are established

The chemist behind it

The experimental foundation for this entire classification was laid by Alfred Stock (1876–1946), who developed high-vacuum line techniques specifically because boron hydrides are pyrophoric, air- and moisture-sensitive, and often toxic — ordinary bench glassware and atmosphere simply destroyed his samples before he could characterize them. Over roughly twenty years at the Kaiser Wilhelm Institute in Berlin, Stock isolated and purified the whole early borane series (B₂H₆, B₄H₁₀, B₅H₉, B₅H₁₁, B₆H₁₀, B₁₀H₁₄) using sealed vacuum manifolds, mercury valves, and low-temperature fractionation, publishing systematic melting points, vapor pressures, and stoichiometries that would not be structurally explained for another generation. The formal geometric and electronic classification into closo/nido/arachno/hypho families came later, principally through William N. Lipscomb (who received the 1976 Nobel Prize in Chemistry for structural studies of boranes using X-ray diffraction and molecular orbital theory) and was crystallized into the simple electron-counting rules by Kenneth Wade in 1971 — work whose full mechanics is the subject of a dedicated treatment on Wade's Rules for Cluster Electron Counting.

Origins · 1 minAlfred Stock and how this was discovered

Alfred Stock

1876–1946 · Kaiser Wilhelm Institute for Chemistry, Berlin, 1910s–1930s

Stock trained as an organic chemist under Emil Fischer but turned to boron and silicon hydrides after recognizing that their extreme air- and moisture-sensitivity had made them essentially inaccessible to prior investigators. He built dedicated all-glass vacuum systems — precursors of the modern Schlenk/vacuum line — to manipulate these compounds without atmospheric contact, and personally suffered chronic mercury poisoning from decades of exposure to mercury-sealed valves and manometers, a hazard that shaped much of his later career advocating for safer laboratory mercury handling.

History

- 1912 — Alfred Stock begins systematic high-vacuum-line studies of boron hydrides in Berlin.

- 1931 — Stock publishes his monograph Hydrides of Boron and Silicon, cataloguing B₂H₆ through B₁₀H₁₄.

- 1954 — William Lipscomb begins X-ray crystallographic structure determinations of boranes at Harvard.

- 1971 — Kenneth Wade proposes systematic skeletal-electron-pair counting rules unifying closo/nido/arachno/hypho classes.

- 1976 — Lipscomb awarded the Nobel Prize in Chemistry for structural and bonding studies of boranes.

Why can't boron hydrides be drawn with simple Lewis structures?

Boron contributes only three valence electrons but has four valence orbitals, so a BH₃ unit is intrinsically two electrons short of an octet even before considering how several boron atoms might bond to each other. When multiple electron-deficient BH units try to assemble into a polyatomic cage, there are not enough electron pairs available to give every adjacent pair of boron atoms an ordinary two-center two-electron bond, the way carbon does in hydrocarbons. Instead, boron hydrides use delocalized three-center two-electron bonds — bridging B–H–B units and open B–B–B face bonds — where a single pair of electrons is shared over three atoms at once rather than two. This is not a failure of Lewis theory in general; it is a direct consequence of boron's electron deficiency, the same underlying property responsible for boron's characteristic Lewis acidity discussed under boron's nonmetallic, electron-deficient character.

What is the difference between "n" in the formula and the size of the parent deltahedron?

$n$ is always the number of boron (or heteroatom) vertices actually present and occupied in the real molecule you are looking at — it is a count of atoms you can point to. The parent deltahedron, with vertex count $m$ , is a hypothetical, fully closed reference polyhedron that the real cluster is imagined to derive from by removing $k = m-n$ vertices. For a closo cluster, $m=n$ (no vertices missing). For nido, $m=n+1$ ; for arachno, $m=n+2$ ; for hypho, $m=n+3$ . So a 6-boron nido cluster and a 6-boron closo cluster look completely different, because they are missing vertices from parent deltahedra of different sizes (7-vertex pentagonal bipyramid versus 6-vertex octahedron respectively) — always check which parent size the real vertex count implies before drawing a structure.

Are hypho boranes real, isolable compounds?

Genuinely free, neutral hypho boron hydrides of the simple formula $\mathrm{B}_n\mathrm{H}_{n+8}$ are extremely rare or effectively unknown as bench-stable species, because removing three vertices from a deltahedron leaves such an open, floppy, electron-rich framework that it tends to either collapse into a more compact nido/arachno structure or must be stabilized by coordinating Lewis bases or forming adducts. Most confirmed hypho-type clusters in the literature appear as base adducts (for example, certain phosphine or amine complexes of open boron frameworks) or as fragments within larger, more complex cluster or metallaborane chemistry, rather than as simple isolable neutral hydrides analogous to B₅H₉ or B₄H₁₀. The hypho classification remains chemically meaningful and predictive, but practical examples are scarcer than for the other three classes.

How does charge change the classification of a borane?

Charge is folded into the electron count, not treated as a cosmetic label. Removing a bridging proton (H⁺) from a neutral nido or arachno hydride generates an anion with one fewer hydrogen and one more unit of negative charge, but the underlying skeletal electron pair count — and therefore the structural class — stays the same, because the departing H⁺ leaves its bonding electron pair behind in the cluster. Conversely, protonating an anion adds a hydrogen and removes a unit of charge without changing the skeletal electron count either. The robust way to check classification under any charge state is the skeletal-electron-pair formula, $\mathrm{SEP} = \tfrac12(n+p+z)$ , compared against $n+1$ (closo), $n+2$ (nido), $n+3$ (arachno), $n+4$ (hypho) — this handles anions and neutral species on exactly the same footing.

Why is the icosahedral closo-B₁₂H₁₂²⁻ so much more stable than smaller boranes?

The twelve-vertex icosahedron is the largest common deltahedron in the standard series and gives every boron vertex a highly symmetric, fully triangulated local environment with no strained or unusually long edges. Electronically, its skeletal bonding manifold is exactly and completely filled by the required $n+1=13$ skeletal electron pairs, with a large HOMO–LUMO-like gap separating filled cluster bonding orbitals from empty antibonding ones — a three-dimensional analogue of aromatic stabilization. This combination of geometric symmetry and a closed, well-separated electronic shell makes $[\mathrm{B}_{12}\mathrm{H}_{12}]^{2-}$ resistant to oxidation, hydrolysis, and strong acid or base, which is precisely why it survives essentially unchanged at the end of many borane degradation and pyrolysis pathways.

Does the same closo/nido/arachno classification apply to carboranes and metallaboranes?

Yes, and this is one of the most powerful extensions of the idea. A CH unit is isoelectronic with a BH⁻ fragment in terms of skeletal orbital and electron contribution, so replacing one or more boron vertices with carbon vertices (forming a carborane) shifts the overall charge needed for a given class but leaves the geometric closo/nido/arachno framework intact — for example, neutral closo-carboranes like $\mathrm{C}_2\mathrm{B}_{10}\mathrm{H}_{12}$ exist without needing the 2− charge that a pure boron closo cluster requires, because the two extra carbon-supplied electrons replace the charge. The same logic extends to transition-metal fragments occupying vertices in metallaboranes, which is part of why cluster electron-counting rules generalize so far beyond simple boron hydrides.

What practical clue tells you the class of a borane just from its name or formula, fastest?

Count the borons ( $n$ ) and the hydrogens, note the charge, and compute the excess index directly: hydrogens minus borons for a neutral species. An excess of 4 is nido, 6 is arachno, 8 is hypho; an exact match with $n$ paired with an overall 2− charge is closo. This single arithmetic check, cross-validated if needed with the skeletal-electron-pair formula $\mathrm{SEP}=\tfrac12(n+p+z)$ compared against $n+1$ through $n+4$ , is enough to classify essentially any simple neutral or anionic borane encountered in a course or in the literature without needing to draw the three-dimensional cage first.