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An alkyl group attached across metal centres. A bridging methyl can form a three-centre two-electron bond, rather than two full independent two-electron metal–carbon bonds.

A hydrogen ligand bonded to more than one skeletal centre. A B–H–B bridge in a borane is commonly represented as a three-centre two-electron bond; it is not two independent ordinary B–H single bonds.

The four terminal methyl groups form ordinary Al–C sigma bonds, with formal bond order one. Each of the two bridging methyl carbons connects the two aluminium centres through a three-centre two-electron bond. One occupied bonding combination extends over Al–C–Al; the other combinations are unoccupied in the simple electron-deficient bridge model. A single shared pair therefore supplies one bond in total across the two contacts, giving formal Al–C bridge bond order $1/2$ per contact, compared with one for terminal Al–C.

Each aluminium has approximately tetrahedral, $s{p}^3$-like orbital directions. A terminal methyl carbon uses $s{p}^3$-like orbitals for its three C–H bonds and its Al–C bond. A bridging methyl carbon still uses three directed orbitals for C–H bonding, while its remaining approximately $s{p}^3$ hybrid overlaps with orbitals on both aluminium atoms. This shared bridge avoids placing five independent ordinary two-electron bonds around carbon. The hybrid labels are idealized and the bridge geometry is distorted; the four occupied C–H/bridge orbital directions are not equivalent to five localized carbon bonds.

The total valence count provides a useful check: $2(3)+6(4)+18(1)=48$ electrons. Eighteen C–H bonds use 36, four terminal Al–C bonds use eight, and two three-centre bridges use four. A description assigning a full two-electron bond to each of the four bridging contacts would require too many electrons.

Two electrons can occupy a fully bonding orbital delocalized equally over all three hydrogen atoms. Three $1s$ atomic orbitals on an equilateral triangle form one totally symmetric $a_1^{\prime }$ combination and a degenerate pair $e^{\prime }$ of higher orbitals. In an orthonormal equal-interaction model with diagonal energy $\alpha$ and off-diagonal $\beta <0$,The lower orbital is proportional to $1s_A+1s_B+1s_C$, with constructive overlap along every edge. Its two electrons give a three-centre two-electron bond and equivalent interactions between every pair of nuclei. At the equilibrium distance, this bonding competes with nuclear Coulomb force to favour the symmetric triangular arrangement. The occupied shell is nondegenerate and closed, so there is no first-order Jahn-Teller incentive to distinguish the three edges. A straight-line or localized single-bond description would miss that equally shared bonding pattern.

For the CO and NCO cases, the added fragment acts as a Lewis base at boron, rather than becoming a new skeletal vertex. Both have a triangular ${\mathrm{B}}_3$ framework. A useful localized drawing has three $\mathrm{B}{\mathrm{H}}_2$ groups, one B–H–B bridge between two of them, and the donor attached to the third boron. The boron–boron framework includes three-centre two-electron bonds; its edges should not all be read as ordinary independent two-electron single bonds.

For the neutral carbonyl adduct, attach CO through carbon, giving $\mathrm{B}\leftarrow \mathrm{C}\equiv \mathrm{O}$ at the donor-bearing $\mathrm{B}{\mathrm{H}}_2$ site. For the anionic isocyanate adduct, attach $\mathrm{NC}{\mathrm{O}}^{-}$ through nitrogen, giving a B–N=C=O connectivity and the overall $-1$ charge. In both drawings the hydrogen count is $3(2)+1=7$. Weak or asymmetric additional B–H contacts can distort this idealized drawing; a hydrogen-rich borane is not fully characterized by a polygon alone. The carbonyl triborane structure was established in the original borane study.^[ref]

The rhodium fragment is instead a skeletal building block. Its neutral electron count is $9+2(2)+1=14$, so it contributes two skeletal electrons, like a BH vertex. Combining it with ${\mathrm{B}}_3{\mathrm{H}}_7$ gives $3(2)+4+2=12$ skeletal electrons, or six pairs, for four vertices. A consistent low-vertex structure is a tetrahedral $\mathrm{Rh}{\mathrm{B}}_3$ skeleton, with the $\mathrm{RhH}(\mathrm{PP}{\mathrm{h}}_3{)}_2$ fragment capping the boron triangle. Keep the terminal Rh–H ligand distinct from the seven hydrogens on the borane portion. The latter can again be represented by three terminal $\mathrm{B}{\mathrm{H}}_2$ groups and a B–H–B bridge. The total formula then contains eight hydrogens outside the phenyl groups. Treating Rh as a pendant one-centre Lewis base would miss its skeletal role.

Trialkylaluminium compounds commonly form dimers with two bridging alkyl groups and four terminal alkyl groups. Each electron-deficient bridge is a three-centre two-electron bond spanning Al–C–Al; the carbon–metal bond electrons are shared with the second aluminium center. A localized ordinary two-center bond to each Al would overcount the electrons.

For $[{\mathrm{R}}_2\mathrm{AlH}{]}_2$, hydride bridges are expected to be especially favorable, giving a four-membered Al–H–Al–H core with two terminal alkyl groups on each Al. The Al–H–Al links are again three-centre two-electron bonds. Hydride is small and can bridge without the steric cost of a bulky alkyl bridge. Aggregation still depends on substituent size, donor solvent and concentration; this is the natural dimeric model, not a statement that all dialkylaluminium hydrides are dimers under every condition.