= Solution
\b[$E=\mathrm{Mo}(\eta^5\!:\!\kappa N\text{-}L)(\mathrm{CO})_2\mathrm{Cl}$], 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:
$$
\underbrace{[\mathrm{Mo}(\eta^5\!:\!\kappa N\text{-}L)(\mathrm{CO})_2(\mathrm{NCMe})]^+}_{D}
\xrightarrow{\mathrm{Cl^-}}
\underbrace{\mathrm{Mo}(\eta^5\!:\!\kappa N\text{-}L)(\mathrm{CO})_2\mathrm{Cl}}_{E}+\mathrm{MeCN}.
$$
<Tetramethylammonium> is a spectator counterion; it is not a <carbon> donor to Mo.
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