Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 2 / B8 / b / ii

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, with chloride replacing the remaining acetonitrile. The pendant pyridine remains attached. The substitution removes the positive electric charge on $D$ but does not change the Mo(II) oxidation state. In ionic electron counting, both neutral acetonitrile and anionic chloride ligand donate two electrons, giving $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{Cl}}=18$ in $E$.

acetonitrile is the remaining weak, exchangeable monodentate donor. Its displacement forms a favorable Mo–Cl bond at a cationic center, retains the intramolecular Mo–N bond, and restores an 18-electron product after substitution. The following formulas specify the coordination changes directly:Tetramethylammonium is a spectator counterion; it is not a carbon donor to Mo.