 Solution

The requested equality is $32$ electrons if each unsaturated $C=CHPh$ unit is counted as a four-electron donor to the chosen extended cluster framework. It needs that additional convention; a count confined to the localized ${\mathrm{Al}}_7{\mathrm{C}}_5$ skeleton gives $28$ instead.

For $B$, an $\mathrm{AlR}$ fragment contributes two skeletal electrons, a singly substituted cap $\mathrm{CR}$ contributes three, and the bridging hydrogen contributes one:For the second formula, let $x$ be the contribution assigned to each unsaturated unit. The comparison isThus $N=N_B$ requires $x=4$. A bookkeeping partition that includes the two alkene $\pi$ pairs in the extended cluster electron pool givesIn contrast, assigning the $C=C$ double bond entirely to the exoskeletal substituent leaves each corresponding cage carbon with two electrons for the cage:The difference is exactly the two $\pi$ pairs, four electrons, rather than an arithmetic discrepancy. These are different electron partitions: do not compare $28$ from one partition with $32$ from the other as though both were the same skeletal count, and do not infer a structural classification merely by inserting the extended count into Wade's rules for twelve localized skeletal vertices.

Both the original and revised question pages print the same neutral molecular formula; the revised copy does not supply a charge, a drawing, or an explicit instruction about these $\pi$ electrons.^[ref] The primary study identifies the open cluster with the two genuine $C=C$ groups and reports their hydroalumination to give the capped aluminium cube. Its accessible abstract does not establish the orbital partition needed to select the four-electron convention.^[ref]

Conditional conclusion: the advertised $32$-electron equality follows under the explicit extended-framework convention above. Without evidence that those $\pi$ pairs belong in the cluster electron pool, retain the $28$-electron localized count and this qualification. The source checks do not justify asserting a typo or inventing a charge to force agreement.

TODO: confirm the intended orbital partition from a complete primary orbital analysis or an authoritative clarification of the question.