Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 9 / a / Solution

For the CO and NCO cases, the added fragment acts as a Lewis base at boron, rather than becoming a new skeletal vertex. Both have a triangular ${\mathrm{B}}_3$ framework. A useful localized drawing has three $\mathrm{B}{\mathrm{H}}_2$ groups, one B–H–B bridge between two of them, and the donor attached to the third boron. The boron–boron framework includes three-centre two-electron bonds; its edges should not all be read as ordinary independent two-electron single bonds.

For the neutral carbonyl adduct, attach CO through carbon, giving $\mathrm{B}\leftarrow \mathrm{C}\equiv \mathrm{O}$ at the donor-bearing $\mathrm{B}{\mathrm{H}}_2$ site. For the anionic isocyanate adduct, attach $\mathrm{NC}{\mathrm{O}}^{-}$ through nitrogen, giving a B–N=C=O connectivity and the overall $-1$ charge. In both drawings the hydrogen count is $3(2)+1=7$. Weak or asymmetric additional B–H contacts can distort this idealized drawing; a hydrogen-rich borane is not fully characterized by a polygon alone. The carbonyl triborane structure was established in the original borane study.^[ref]

The rhodium fragment is instead a skeletal building block. Its neutral electron count is $9+2(2)+1=14$, so it contributes two skeletal electrons, like a BH vertex. Combining it with ${\mathrm{B}}_3{\mathrm{H}}_7$ gives $3(2)+4+2=12$ skeletal electrons, or six pairs, for four vertices. A consistent low-vertex structure is a tetrahedral $\mathrm{Rh}{\mathrm{B}}_3$ skeleton, with the $\mathrm{RhH}(\mathrm{PP}{\mathrm{h}}_3{)}_2$ fragment capping the boron triangle. Keep the terminal Rh–H ligand distinct from the seven hydrogens on the borane portion. The latter can again be represented by three terminal $\mathrm{B}{\mathrm{H}}_2$ groups and a B–H–B bridge. The total formula then contains eight hydrogens outside the phenyl groups. Treating Rh as a pendant one-centre Lewis base would miss its skeletal role.