Clusters, extended structures and unusual bonding

A three-center bridge is not two independent pair bonds

Diborane B₂H₆ has 2× 3+6× 1=12 valence electrons. Its selected bonding description uses four terminal ordinary B–H pair bonds, consuming eight electrons, and two bridging B–H–B three-center/two-electron units, consuming four. Each bridge shares one pair over three centers.

Counting all eight B–H contacts as independent two-center/two-electron bonds would demand 16 electrons, more than the available 12. A contact count is therefore not a localized pair-bond count. These bridges are not automatically ordinary intermolecular hydrogen bonds. The electron-allocation model itself gives no measured geometry, partial charges or individual bond-order value.

Terminal and skeletal electron pools are separate

For the simple closo borane B₆H₆²⁻, total valence electrons are 6× 3+6× 1+2=26. Six terminal B–H pairs use 12, leaving 14 skeletal electrons, or seven pairs. In this restricted borane class, the closo count is n+1 skeletal pairs for n vertices; n=6 gives seven, consistent with the selected octahedral closed cage.

An octahedral graph has six vertices, twelve edges and eight triangular faces. Its twelve drawn edges do not mean twelve independently localized two-center pair bonds. Nor does this cage have the same bridging-H arrangement as diborane: the selected B₆H₆²⁻ has terminal hydrogens. More general clusters require their own electron-count conventions, not automatic application of this restricted rule.

A conventional unit cell counts shares rather than molecules

For conventional cubic rock salt, represent chloride at eight corners and six faces, sodium at twelve edges and the body center. Periodic sharing gives chloride count 8×(1/8)+6×(1/2)=4, sodium count 12×(1/4)+1=4. The cell thus contains four NaCl formula units and net charge zero under Na⁺/Cl⁻ assignments.

Each ion has six opposite-ion neighbors in this stated phase, including neighbors across cell boundaries. The unit cell is not an isolated Na₄Cl₄ molecule, and coordination count differs from formula ratio. No lattice length, radius or rule that every one-to-one salt must adopt rock salt follows from this count.

Three sites can give bonding, nonbonding and antibonding combinations

For an original equal-site open-chain toy, use an orthonormal three-site basis and matrix rows (0,−1,0), (−1,0,−1) and (0,−1,0) in arbitrary energy units. Diagonal energies are all zero; adjacent couplings are equal and nonadjacent coupling is zero. Eigenvalues are −√2, 0 and +√2.

Normalized eigenvectors can be written (1/2,1/√2,1/2), (1/√2,0,−1/√2) and (1/2,−1/√2,1/2). Four independent electrons occupy the lower and middle levels with two each, leaving the upper level empty. The middle combination has zero coefficient at the central site. Summing occupied squared coefficients gives toy site populations 1.5, 1 and 1.5, conserving four electrons.

This is an illustrative three-center/four-electron orbital pool, not a complete real molecule. XeF₂ is heteronuclear and has other orbitals, overlap, correlation and ionic contributions. The equal-site toy does not predict its actual population charges, energies, distances or bond strengths. An expanded Lewis drawing is not evidence that literal d-orbital hybridization is required.

Choose a representation that answers the stated question

Ordinary pair ledgers, multicenter descriptions, skeletal counting, periodic composition and orbital models each expose different constraints. A graph can record connectivity; a cage edge records topology; an electron count records a chosen pool. None should be silently substituted for the others.

Assess the question, phase and evidence before extending a familiar model. Actual unusual bonding needs structural and electronic support. The examples here establish internally consistent constrained descriptions, with the real-system limits kept beside their calculations.

Worked examples

Try it yourself

1. Does one B–H–B three-center/two-electron bridge contain two independent ordinary pair bonds?

   Show the answer and reasoning

   No.

   Its one pair is shared across three centers in the selected description.

2. In the same simple B₆H₆²⁻ closo ledger, subtract six terminal B–H pairs from 26 total valence electrons. What remains?

   Show the answer and reasoning

   14 skeletal electrons, or seven pairs.

   26−6× 2=14, and 14/2=7=n+1 for n=6 in the restricted cluster class; cage edges do not count localized pairs.

3. For the stated four-electron three-site toy, both occupied eigenvectors have central coefficients 1/√2 and 0. Find the central site population.

   Show the answer and reasoning

   1.0 electron in the stipulated basis-population model.

   Two electrons per occupied orbital give 2(1/√2)²+2× 0²=1. This is a toy basis population, not a real molecular atomic charge.

4. Do equal-site toy populations 1.5,1,1.5 predict real XeF₂ atomic charges?

   Show the answer and reasoning

   No.

   The heteronuclear molecule and its full electronic structure are outside the equal-site model.

What to carry forward

• Distinguish contacts from independent pair bonds.
• Separate terminal and skeletal electrons in the stated cluster class.
• Use fractional shares for periodic composition.
• Keep orbital-pool assumptions visible.
• Do not turn a representation into unmeasured real-system properties.