= Solution
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 \b[formal Al–C bridge <bond order> $1/2$ per contact], compared with one for terminal Al–C.
Each <aluminium> has approximately tetrahedral, $sp^3$-like orbital directions. A terminal methyl <carbon> uses $sp^3$-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 $sp^3$ 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.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-aluminium-bridges.png]
{title=Original comparison of chloride lone-pair bridges and electron-deficient methyl bridges, with their different electron accounting}
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