Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 9 / a

An isolobal analogy compares fragment frontier orbitals, not atomic sizes or total electron counts. The neutral BH fragment has four valence electrons; two make the external B–H bond, leaving two skeletal electrons in three frontier orbitals directed towards possible cluster neighbours. For the neutral $\mathrm{CpCo}$ fragment, $\mathrm{Cp}={\mathrm{C}}_5{\mathrm{H}}_5$ contributes five electrons and cobalt nine, giving 14. Removing the 12 electrons assigned to filled, predominantly non-skeletal fragment orbitals again leaves two electrons in three cluster-directed frontier orbitals.

The corresponding frontier sets have comparable numbers, spatial symmetries and occupations, allowing replacement of a BH vertex by CpCo in a suitable cluster-electron count. This is why the metal fragment can cap an open carborane face without adding an arbitrary new set of skeletal electrons. Exact orbital energies, ligand preferences and bond lengths need not be identical.^[ref]