Step-by-step derivation:

1. Understanding the Concept: Synergic Bonding in Metal Carbonyls The bond between a metal atom and a carbon monoxide (CO) ligand in a metal carbonyl complex consists of two components, a phenomenon known as synergic bonding:

   • σ-donation: The CO ligand donates its lone pair of electrons from the carbon atom into a vacant d-orbital of the metal, forming a M-C σ-bond (M ← C≡O).
   • π-back-donation: The metal donates electron density from its filled d-orbitals into the empty π* (antibonding) molecular orbitals of the CO ligand, forming a M-C π-bond (M → C≡O).

2. Effect of Back-donation on C-O Bond Order The key to answering this question is understanding how π-back-donation affects the C-O bond. When the metal back-donates electron density into the π* antibonding orbital of CO, it weakens the bond between carbon and oxygen. This is because electrons are being added to an orbital that is antibonding with respect to the C-O bond.

   • Stronger π-back-donation leads to a weaker C-O bond.
   • A weaker C-O bond corresponds to a lower C-O bond order and a longer C-O bond length.

3. Factors Influencing the Extent of π-Back-donation The extent of π-back-donation depends on the electron density on the central metal atom. Higher electron density on the metal increases its ability to donate electrons back to the CO ligands.

   • Charge on the complex: A negative charge on the complex indicates a higher electron density on the metal, leading to stronger back-donation. A positive charge indicates lower electron density and weaker back-donation.
   • Nature of the metal: For neutral metals in the same period, moving from left to right increases the nuclear charge, which causes the d-orbitals to contract and become less available for back-donation.

4. Analyzing the Given Options Let's determine the oxidation state of the metal in each complex to assess the electron density:

   • A: [Mn(CO)₆]⁺: The complex has a +1 charge. Since CO is a neutral ligand, the oxidation state of Mn is +1.
   • B: [Fe(CO)₅]: This is a neutral complex. The oxidation state of Fe is 0.
   • C: [Cr(CO)₆]: This is also a neutral complex. The oxidation state of Cr is 0.
   • D: [V(CO)₆]⁻: The complex has a -1 charge. The oxidation state of V is -1.

5. Comparing the Extent of Back-donation We can now rank the complexes based on the electron density on the central metal, which is determined by its oxidation state.

   • [V(CO)₆]⁻ has a metal in a -1 oxidation state, giving it the highest electron density.
   • [Cr(CO)₆] and [Fe(CO)₅] have metals in a 0 oxidation state (neutral).
   • [Mn(CO)₆]⁺ has a metal in a +1 oxidation state, giving it the lowest electron density.

   The order of electron density on the metal is: V⁻¹ > Cr⁰/Fe⁰ > Mn⁺¹. Therefore, the order of the extent of π-back-donation is: [V(CO)₆]⁻ > ([Cr(CO)₆], [Fe(CO)₅]) > [Mn(CO)₆]⁺

   Comparing the two neutral complexes, [Cr(CO)₆] and [Fe(CO)₅], Cr is to the left of Fe in the periodic table. Cr has a lower nuclear charge, so its d-orbitals are more diffuse and better at back-donation than Fe's d-orbitals. Thus, back-donation is stronger in [Cr(CO)₆] than in [Fe(CO)₅].

   The complete order for the extent of back-donation is: [V(CO)₆]⁻ > [Cr(CO)₆] > [Fe(CO)₅] > [Mn(CO)₆]⁺

6. Determining the Lowest C-O Bond Order Since a greater extent of back-donation leads to a lower C-O bond order, the order of C-O bond orders will be the reverse of the back-donation order: [V(CO)₆]⁻ < [Cr(CO)₆] < [Fe(CO)₅] < [Mn(CO)₆]⁺

   The complex with the lowest C-O bond order is the one with the strongest M→CO π-back-donation, which is [V(CO)₆]⁻.

Conclusion

Based on the analysis, the metal carbonyl with the lowest C-O bond order is [V(CO)₆]⁻. The reasoning is that the negative charge on the complex gives the vanadium atom the highest electron density, leading to the most significant back-donation into the π* orbitals of the CO ligands, thereby weakening the C-O bond the most.

The stored correct answer is B. However, based on the established principles of inorganic chemistry, the correct answer should be D. This is also supported by experimental evidence from IR spectroscopy, where [V(CO)₆]⁻ shows the lowest C-O stretching frequency (around 1860 cm⁻¹), indicating the weakest C-O bond.