Free CO absorbs in the IR at 2143 cm⁻¹. In Ni(CO)₄, the CO stretching frequency drops to ~2060 cm⁻¹. What causes this decrease?
AThe mass of the Ni atom attached to CO reduces the vibrational frequency through a simple mass effect
BPi back-donation from filled Ni d-orbitals into CO π* antibonding orbitals weakens the C≡O bond, lowering its stretching frequency
CSigma donation from CO to Ni strengthens the C-O bond by removing antibonding electron density
DIntermolecular interactions between CO ligands in the complex shift the frequency
The CO stretching frequency directly reports on C-O bond strength. In free CO, the bond is a strong triple bond (2143 cm⁻¹). When CO coordinates to a metal, the metal's filled d-orbitals donate electron density into CO's π* antibonding orbitals (back-donation). Populating these antibonding orbitals weakens the C-O bond, reducing the bond order below 3 and lowering the stretching frequency. The more electron-rich the metal, the more back-donation occurs, and the lower the CO frequency drops. This makes ν(CO) a remarkably sensitive probe of metal electron density.
Question 2 True / False
In a series of isoelectronic carbonyls [V(CO)₆]⁻, Cr(CO)₆, and [Mn(CO)₆]⁺, the CO stretching frequency increases from V⁻ to Mn⁺. This trend reflects decreasing back-donation as the metal center becomes more positively charged.
TTrue
FFalse
Answer: True
All three species are isoelectronic (d⁶, 18 electrons) with the same octahedral geometry. The difference is the formal charge on the metal: V⁻ is electron-rich, Cr⁰ is neutral, and Mn⁺ is electron-poor. Greater positive charge on the metal reduces the electron density available for back-donation into CO π* orbitals, strengthening the C-O bond and raising ν(CO). [V(CO)₆]⁻: ~1860 cm⁻¹, Cr(CO)₆: ~2000 cm⁻¹, [Mn(CO)₆]⁺: ~2100 cm⁻¹. This isoelectronic series is a textbook demonstration of how ν(CO) tracks metal electron density.
Question 3 True / False
All stable binary metal carbonyls (containing only metal and CO) obey the 18-electron rule.
TTrue
FFalse
Answer: True
The binary metal carbonyls of transition metals all satisfy the 18-electron rule: Ni(CO)₄ (10 + 8 = 18), Fe(CO)₅ (8 + 10 = 18), Cr(CO)₆ (6 + 12 = 18). When a single metal center cannot reach 18 electrons, metals form M-M bonds in polynuclear carbonyls: Mn₂(CO)₁₀ (each Mn has 7 + 10 from five CO + 1 from M-M bond = 18), Co₂(CO)₈ (each Co has 9 + 6 from bridging/terminal CO + 1 from M-M bond, reaching 18 with the appropriate bridging arrangement). The regularity of this pattern makes the 18-electron rule the most reliable predictor of binary carbonyl stoichiometry.
Question 4 Short Answer
Explain why replacing one CO in Cr(CO)₆ with PPh₃ to form Cr(CO)₅(PPh₃) shifts the remaining CO stretching frequencies to lower wavenumbers.
Think about your answer, then reveal below.
Model answer: PPh₃ is a stronger sigma-donor than CO but a weaker pi-acceptor. When PPh₃ replaces one CO, it donates more electron density to chromium through sigma bonding than the departing CO did, but it accepts less electron density through back-bonding. The net effect is increased electron density at the metal center. This excess electron density is redistributed to the remaining five CO ligands through enhanced back-donation into their π* orbitals. More back-donation weakens the C-O bonds of the remaining COs, lowering their stretching frequencies. Each CO serves as a 'reporter' of the total electron density at the metal.
This principle makes CO stretching frequencies in mixed-ligand complexes a powerful diagnostic. By comparing ν(CO) values before and after ligand substitution, you can rank ligands by their net donor/acceptor properties. Ligands that increase metal electron density (strong sigma-donors, weak pi-acceptors) lower ν(CO); ligands that decrease it (weak sigma-donors, strong pi-acceptors) raise ν(CO).