= Solution
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 \b[two skeletal electrons in three frontier orbitals] directed towards possible cluster neighbours. For the neutral $\mathrm{CpCo}$ fragment, $\mathrm{Cp=C_5H_5}$ contributes five electrons and <cobalt> nine, giving 14. Removing the 12 electrons assigned to filled, predominantly non-skeletal fragment orbitals again leaves \b[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.<Building Bridges Between Inorganic and Organic Chemistry (Nobel Lecture) by Roald Hoffmann>[1982, isolobal fragment correspondence and its use across organic/inorganic bonding]{ref}
Back to article page
