To determine which compound has the longest C–O bond length, we need to consider the electronic structure and bonding in each complex.$\newline$The C–O bond length is influenced by the back-donation of electrons from the metal to the $\pi^*$ orbitals of CO.$\newline$More back-donation results in a weaker C–O bond and thus a longer bond length.$\newline$Let's analyze each option:$\newline$• Option 1: $Ni(CO)_4 \newline$ - Nickel is in the zero oxidation state.$\newline$ - It has a full d-shell, which allows for significant back-donation.$\newline$ - This results in a moderate C–O bond length.$\newline$• Option 2: $[Co(CO)_4]^- \newline$ - Cobalt is in the -1 oxidation state.$\newline$ - The negative charge increases electron density on the metal, enhancing back-donation.$\newline$ - This leads to a longer C–O bond compared to neutral complexes.$\newline$• Option 3: $[Fe(CO)_4]^{2-} \newline$ - Iron is in the -2 oxidation state.$\newline$ - The high negative charge greatly increases electron density on the metal.$\newline$ - This results in the strongest back-donation and the longest C–O bond length among the options.$\newline$• Option 4: $[Mn(CO)_6]^+ \newline$ - Manganese is in the +1 oxidation state.$\newline$ - The positive charge reduces electron density, decreasing back-donation.$\newline$ - This results in a shorter C–O bond length.$\newline$Therefore, the complex with the longest C–O bond length is $[Fe(CO)_4]^{2-},$ corresponding to Option 3.