Lesson 3 · 30 min

Boranes and clusters: counting skeletal electrons

Diborane holds 8 atoms together with only 12 valence electrons: three-centre two-electron bonds are the answer. Wade’s rules need only a count of electron pairs to tell whether a borane or carborane cage is closo, nido or arachno — then compare with measured structures to see what the rules can predict.

In this lesson

Explain why diborane is “electron-deficient” and describe its structure with three-centre two-electron bonds
Count the skeletal electron pairs of boranes and carboranes from their formulas and use Wade’s rules to tell closo, nido and arachno cages apart
Compare an idealised polyhedral fragment with a measured structure, and see what the rules can and cannot predict

From 1912 the German chemist Alfred Stock prepared a series of boron hydrides in vacuum apparatus of his own design: B₂H₆, B₄H₁₀, B₅H₉, B₁₀H₁₄ and more. Several of them ignite spontaneously in air, and their formulas made no sense: ethane, C₂H₆, has 14 valence electrons, just enough for 7 two-electron bonds, but diborane, B₂H₆, has only 12. Only in the 1950s, when W. N. Lipscomb solved their structures by X-ray diffraction, did a pattern emerge: hydrogen atoms can bridge two borons, and the boron atoms enclose polyhedral cages — work recognised with the 1976 Nobel Prize in Chemistry. In 1971 K. Wade distilled it into a simple rule that only needs electrons to be counted.

Predict first

B₂H₆ has only 12 valence electrons. Joining 8 atoms with two-electron bonds as in ethane would take 7 bonds and 14 electrons. What is its structure most likely to be?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Like ethane, but with only one electron in the B–B bond Two BH₂ groups bridged by two hydrogen atoms, each bridge holding three atoms together with one pair A straight chain H–B–H–B–H Two separate BH₃ molecules that do not bond

Three-centre two-electron bonds

Each boron is roughly tetrahedral: the two terminal hydrogens lie in a plane with the two borons, and the two bridging hydrogens sit above and below it. In each B–H–B bridge the two boron hybrids and the hydrogen 1s orbital combine into three molecular orbitals: bonding, non-bonding (the two boron orbitals out of phase, with the hydrogen on its node) and antibonding. Only two electrons are available; they fill the bonding orbital, and three atoms are held together by one pair.

Such “electron-deficient” bonding is not rare: [Al(CH₃)₃]₂ dimerises through bridging methyl groups, and metal clusters and many hydrides contain multicentre bonds. What they share is more orbitals than electrons: rather than leave orbitals empty, one pair binds more than two atoms at once.

E_\pm = \frac{\alpha_\mathrm{B} + \alpha_\mathrm{H}}{2} \pm \sqrt{\left(\frac{\alpha_\mathrm{B} - \alpha_\mathrm{H}}{2}\right)^2 + 2\beta^2},\qquad E_0 = \alpha_\mathrm{B}

Hückel orbital energies of a B–H–B bridge. α are Coulomb integrals and β (< 0) the resonance integral between the boron and hydrogen orbitals; the symmetric combination of the boron orbitals mixes with H 1s, the antisymmetric one stays non-bonding.

$\alpha_\mathrm{B}$: Coulomb integral of a boron hybrid
$\alpha_\mathrm{H}$: Coulomb integral of hydrogen 1s
$\beta$: boron–hydrogen resonance integral

Measured gas-phase structures (NIST CCCBDB) · Hückel orbitals of the bridge

Three-centre two-electron bonds

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Click two atoms for a distance, three for an angle.

The three molecular orbitals of a B–H–B bridge

ΔE = αB − αH (in units of |β|): 0.5

B–H (terminal) mean 1.203 Å, B–H (bridge) 1.315–1.484 Å, B–B 1.704 Å

Two electrons fill the bonding orbital: three atoms share one pair.

Tetraborane, B₄H₁₀, has four B–H–B bridges. Click pairs of atoms to measure: the two borons on the central “hinge” are 1.70 Å apart, the “wing” borons 1.86 Å from them, and each bridging hydrogen sits at unequal distances from its two borons.

Wade’s rules: count pairs, find the polyhedron

In the larger boranes the borons enclose polyhedra with triangular faces (deltahedra). Wade noticed that counting the electron pairs that hold the skeleton together tells you which kind it is:

each B–H unit gives the skeleton 2 electrons (one of boron’s 3 valence electrons goes to its terminal B–H bond), each C–H gives 3;
each extra hydrogen (bridging or endo) gives 1, and a negative charge gives its electrons.

With n skeletal atoms and p skeletal electron pairs: p = n + 1 is closo — the n atoms fill every vertex of a deltahedron; p = n + 2 is nido — an (n + 1)-vertex polyhedron missing one vertex; p = n + 3 is arachno — missing two adjacent vertices. Molecules with the same number of pairs are cut from the same parent polyhedron.

p = \tfrac{1}{2}\left(2\,n_{\mathrm{BH}} + 3\,n_{\mathrm{CH}} + n_{\mathrm{H}}^{\prime} - z\right),\qquad p = n + 1\ (\text{closo}),\ n + 2\ (\text{nido}),\ n + 3\ (\text{arachno})

n = n_BH + n_CH is the number of skeletal atoms, n_H′ the extra hydrogens and z the ionic charge (z < 0 for anions).

$p$: skeletal electron pairs
$n$: number of skeletal atoms
$z$: ionic charge

Skeletal electron pairs from the formula · idealised closo polyhedra (edges 1.75 Å)

Wade’s rules: counting skeletal electron pairs

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5 BH × 2 + 0 CH × 3 + 4 extra H × 1 + 0 from the negative charge = 14 skeletal electrons = 7 pairs

5 cage atoms, 7 pairs: p = n + 2, a nido cage cut from a 6-vertex closo polyhedron with 1 vertex removed.

Step through [B₆H₆]²⁻, B₅H₉ and B₄H₁₀: all three have 7 skeletal pairs and all come from the octahedron — closo fills its 6 vertices, nido leaves 1 empty, arachno 2. Then compare [B₁₂H₁₂]²⁻ with 1,2-C₂B₁₀H₁₂: two C–H replace two B–H⁻, the pair count is unchanged, and the cage is still an icosahedron.

Worked example

The skeleton of B₅H₉

Use Wade’s rules to predict the skeleton of pentaborane, B₅H₉.

Concept check

Why is 1,2-C₂B₁₀H₁₂ an icosahedron, just like [B₁₂H₁₂]²⁻?

A C–H gives the skeleton 3 electrons, like a B–H plus a negative charge: both have 13 skeletal pairs and 12 skeletal atoms Carbon and boron have the same atomic radius Neutral molecules are always closo Carbon atoms always sit at polyhedral vertices

Try the calculation

How many skeletal electron pairs does [B₁₀H₁₀]²⁻ have?

Key takeaways

B₂H₆ has only 12 valence electrons: four terminal two-electron B–H bonds plus two three-centre two-electron B–H–B bonds; the bridge bonds are longer (1.32 Å against 1.20 Å).
Wade’s rules: count 2 per B–H, 3 per C–H, 1 per extra hydrogen, plus the negative charge; p = n + 1 closo, n + 2 nido, n + 3 arachno, and equal pair counts share a parent deltahedron.
The rules predict the topology of the skeleton, not its dimensions: measured B₄H₁₀ has a 1.70 Å hinge, 1.86 Å wing edges and a 117° fold, not the equal edges and 109.5° of the ideal fragment.

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Data and references

W. N. Lipscomb, “The boranes and their relatives” (Nobel lecture), Science 196, 1047–1055 (1977).
K. Wade, “The structural significance of the number of skeletal bonding electron-pairs in carboranes, the higher boranes and borane anions, and various transition-metal carbonyl cluster compounds”, J. Chem. Soc. D, Chem. Commun. 1971, 792–793.
NIST Computational Chemistry Comparison and Benchmark Database (CCCBDB), SRD 101, experimental geometries of B₂H₆, B₄H₁₀ and B₅H₉.
C. J. Dain, A. J. Downs, G. S. Laurenson, D. W. H. Rankin, J. Chem. Soc., Dalton Trans. 1981, 472–477 (structure of B₄H₁₀).
H. J. Hrostowski, R. J. Myers, “The microwave spectra, structure, and dipole moment of stable pentaborane”, J. Chem. Phys. 22, 262–265 (1954).