Boranes and Wade’s Rules: Predicting Cluster Shapes

Why boranes need special treatment

Boron has three valence electrons but four valence orbitals, so it cannot form enough conventional two-centre two-electron bonds to satisfy every atom. The result is electron deficiency, and boron solves it by forming three-centre two-electron bonds, in which one electron pair holds three atoms together.

The simplest example is diborane, where two hydrogen atoms bridge the two borons. Each bridge is a single pair shared across B–H–B. Drawing diborane with a conventional B–B bond and terminal hydrogens is wrong, and it is a standard question to explain why.

Wade's rules

Count the skeletal electron pairs — those involved in holding the cluster framework together, excluding terminal bonds to hydrogen.

Counting, step by step

For a neutral borane B_nH_m:

1. Each BH unit contributes 2 skeletal electrons — boron's 3 electrons plus hydrogen's 1, minus the 2 used in the terminal B–H bond.
2. Each extra hydrogen beyond the n terminal ones contributes 1 electron.
3. Add 1 electron for each negative charge; subtract 1 for each positive charge.
4. Divide the total by 2 to get skeletal pairs.

Worked example: B₅H₉

Five BH units give 5 × 2 = 10 electrons. Four extra hydrogens give 4 more, so 14 electrons, which is 7 pairs. With n = 5, seven pairs is n + 2, so the structure is nido. The parent closo polyhedron has 6 vertices — an octahedron — and removing one vertex gives a square pyramid, which is the observed shape.

Worked example: B₆H₆²⁻

Six BH units give 12 electrons, plus 2 for the charge, giving 14 electrons or 7 pairs. With n = 6, that is n + 1, so it is closo — a complete octahedron.

Extending to other clusters

The rules generalise well beyond boranes. Carboranes replace a BH unit with CH, which contributes three skeletal electrons rather than two, since carbon has one more electron than boron. Metallaboranes replace a vertex with a transition metal fragment, whose contribution is calculated from its d electron count and its ligands.

This generality is why the topic is examined: one counting scheme covers boranes, carboranes and metal clusters, and questions frequently mix them to test whether the principle is understood rather than the boron cases memorised.

Properties and reactivity

• Lower boranes are highly reactive and many ignite in air; the closo dianions are much more stable.
• Closo clusters are generally the most stable of the three types, being closed and fully bonded.
• Nido and arachno clusters, having open faces, are more reactive at those open positions.
• The bonding is delocalised over the cluster, which is why localised two-centre structures fail to describe them.

Frequently asked questions

What exactly is a three-centre two-electron bond?

A single electron pair delocalised over three atoms rather than two. In diborane a bridging hydrogen shares one pair with both borons, which is how the molecule holds together despite not having enough electrons for conventional bonds.

How do I know the parent polyhedron?

From the pair count, not the vertex count. A cluster with p skeletal pairs derives from the closo polyhedron with p − 1 vertices, then removes vertices until the actual number is reached.

Do the rules ever fail?

Yes, for some larger clusters and for certain metal-rich systems where alternative bonding schemes are needed. They are reliable for the range asked at entrance level.

Why are the dianions more stable than neutral boranes?

Because the added electrons complete the closo bonding requirement, giving a closed shell with no open faces and no unsatisfied bonding capacity.