 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 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.