 Solution

Use Wade's rules with the metal included as a skeletal vertex. Each BH vertex supplies two skeletal electrons, each CH supplies three, and the two additional hydrogen atoms supply two more. The $\mathrm{Ru}(\mathrm{CO}{)}_3$ fragment supplies $8+3(2)-12=2$ skeletal electrons after the metal's non-skeletal orbitals have been accounted for. ThusThe six-vertex cluster is therefore nido, with a pentagonal-pyramidal skeleton: Ru is the apex, bonded to the five vertices of the planar ${\mathrm{B}}_3{\mathrm{C}}_2$ base, and the three carbonyl ligands project away from that base. This is the six-vertex fragment obtained by removing an equatorial vertex from a seven-vertex pentagonal bipyramid. The five ring atoms carry their terminal B–H or C–H bonds; the two additional hydrogens belong to the hydrogen-rich open-face bonding framework. The skeletal electron count does not, by itself, locate every bridging hydride, so a unique detailed hydrogen arrangement should not be inferred from it.