= Solution
For the CO and NCO cases, the added fragment acts as a <Lewis base> at <boron>, rather than becoming a new skeletal vertex. \b[Both have a triangular $\mathrm{B_3}$ framework]. A useful localized drawing has three $\mathrm{BH_2}$ 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 $\mathrm{B{\leftarrow}C{\equiv}O}$ at the donor-bearing $\mathrm{BH_2}$ site. For the anionic isocyanate adduct, attach $\mathrm{NCO^-}$ 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.<Boranes. XXXVII. Reactions of triborane(7) and the structure of triborane(7) carbonyl by James D. Glore and others>[1973, pp.2175–2178, structural assignment of the <carbonyl> adduct]{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 $\mathrm{B_3H_7}$ gives $3(2)+4+2=12$ skeletal <electrons>, or six pairs, for four vertices. A consistent low-vertex structure is \b[a tetrahedral $\mathrm{RhB_3}$ skeleton], with the $\mathrm{RhH(PPh_3)_2}$ 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 $\mathrm{BH_2}$ 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.
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