 Solution

Use ionic electron counting, counting each ordinary donor atom once and keeping ligand charge distinct from denticity. A is 16-electron, B is 14-electron and C is 16-electron, so none is an 18-electron complex under this convention.

A contains a modified, monoanionic PNP pincer: the exocyclic double bond and altered pyridine-ring bond pattern in the drawing matter. Its two phosphines and its central N donor provide six electrons altogether, while neutral N${}_2$ provides two. Neutrality then gives Co(I), $d^8$, and $8+6+2=16$. Treating the modified pincer as a neutral intact pyridine ligand would incorrectly produce Co(0) and a 17-electron count. A four-coordinate approximately square planar molecular geometry in a $d^8$ complex can leave the strongly sigma-antibonding orbital empty and be stable at 16 electrons. The underlying modified-pincer chemistry has been examined using structure, spectroscopy and electronic-structure calculations.^[ref]

B has a neutral tertiary amine donor at the centre of its tripod and three anionic amido arms. Their combined charge is $-3$, so Re is Re(III), $d^4$. The four N donors give eight electrons and N${}_2$ gives two: $4+8+2=14$. Such an electron count is not inherently unexpected in an amido-supported complex: strong amido pi donation and the imposed five-coordinate geometry alter the orbital energies, and the empty orbitals are not nine equivalent slots that must all be filled. Steric protection and chelation also help stabilize this coordinatively restricted structure. The 14-electron figure uses the usual two-electron ionic count for each amido donor; pi bonding should not be appended to it inconsistently.

C has an anionic acyl carbon donor and a neutral phosphine in the chelate, together with two neutral PMe${}_3$ donors. Thus Co(I), $d^8$, receives $2+2+4=8$ ligand electrons, giving $8+8=16$. This is again a reasonable four-coordinate $d^8$ structure. The acyl C=O is part of a carbon-bound acyl ligand, not an additional free CO donor.