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2-Picolylamine is pyridin-2-ylmethanamine, with a pyridine nitrogen and an aminomethyl nitrogen adjacent on the ring. It binds as an unsymmetrical neutral bidentate ligand, forming a five-membered chelate ring.

A chelate ring is a closed bonded path containing the central atom and a ligand with multiple attachments to it. A tethered Cp–pyridyl ligand can close such a path when the pyridyl N coordinates.

Chelation binds multiple donor sites of one ligand to the same central atom, forming a chelate ring. It can retain a donor near a metal even when one attachment temporarily opens.

Treat the two donors of 2-Picolylamine as distinct: $P$ denotes pyridyl N and $A$ denotes amino N. Each ligand must join two adjacent octahedral positions because its five-membered chelate ring cannot span a trans pair. The three $P$ sites can either occupy a triangular face, giving a facial isomer, or a meridian containing a trans pair, giving a meridional isomer. Each geometry has a pair of nonsuperposable mirror images. Therefore there are four configurational stereoisomers: fac-$\Delta$, fac-$\Lambda$, mer-$\Delta$ and mer-$\Lambda$. Rapid ring puckering or amine inversion does not introduce additional persistent configurations in this count.

The mer geometry is the reasonable steric prediction for the more stable form. It spreads the bulky pyridyl rings farther apart, allowing one pyridyl–pyridyl trans relationship, whereas the fac geometry groups all three pyridyl donors on one face. This is a qualitative preference, not a proof independent of counterions and solvent: hydrogen bonding and packing can favor the alternative. Primary structures of different salts contain both geometries, so neither should be claimed impossible.^[ref]

The readily displaced acetonitrile sites have both been exhausted. A second chloride would have to displace a strongly bound carbonyl ligand, open the chelate ring, or associate to an already 18-electron center to make an unfavorable crowded 20-electron species. The product is neutral, so further chloride uptake also lacks the charge-neutralization advantage of $D\to E$ and would generate an anionic complex if substitution occurred. The chelate effect disfavors loss of the tethered pyridine, while strong pi backbonding helps retain CO under the stated conditions. These combined kinetic and bonding arguments explain the absence of further reaction. The 18-electron rule alone would not forbid replacing CO by chloride, since that replacement is isoelectronic; it therefore cannot be the whole explanation.

Write $L^{-}$ for the cyclopentadienyl ligand bearing its $\mathrm{C}{\mathrm{H}}_2$–2-pyridyl arm; ${\eta }^5:\kappa N$ means that its ring and pyridyl nitrogen both bind the same metal. $D=[\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2(\mathrm{NCMe}){]}^{+}$. The base removes the pyridinium N–H proton in $C$, restoring a pyridine lone pair. Intramolecular nitrogen coordination then replaces one acetonitrile ligand and closes a chelate ring. Both CO ligands are retained; the molybdenum remains Mo(II), $d^4$.

This is an proton transfer followed by chelation, with metal electron count $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{NCMe}}=18$. The chelate effect, favorable formation of the Mo–N bond, and release of a solvent molecule explain the driving force. Merely invoking a 16-to-18-electron change would be misleading because $C$ is already 18-electron. In the preceding acid treatment, protonation of the ${\eta }^3$ allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for $C$'s overall $2+$ charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.