In ligand field theory, how does a pi-acceptor ligand like CO increase Δ_oct compared to a pure sigma-donor like NH₃?
ACO forms stronger electrostatic interactions with the metal due to its dipole moment
BCO has empty pi-antibonding orbitals that accept electron density from the filled metal t₂g orbitals (back-bonding), lowering the t₂g energy and increasing the gap to the eg set
CCO is a stronger sigma-donor than NH₃, pushing the eg orbitals to higher energy
DCO reduces the electron-electron repulsion in the eg orbitals by withdrawing charge from them
The key mechanism is pi-back-bonding. The metal's filled t₂g orbitals have the correct symmetry to overlap with CO's empty π* orbitals. Electron density flows from metal to ligand through this overlap, stabilizing (lowering the energy of) the t₂g set. Since Δ_oct is the gap between t₂g and eg, lowering t₂g while eg stays roughly the same (or rises slightly from sigma interactions) increases Δ. This is why CO sits at the far end of the spectrochemical series. NH₃ has no empty pi-orbitals, so it cannot accept back-donation — it only interacts through sigma-donation, giving a moderate Δ.
Question 2 True / False
Ligand field theory predicts that halide ligands (F⁻, Cl⁻, Br⁻, I⁻) are weak-field ligands because their filled p-orbitals act as pi-donors, raising the energy of the metal t₂g orbitals and decreasing Δ.
TTrue
FFalse
Answer: True
Halides have filled p-orbitals perpendicular to the metal-ligand bond axis. These orbitals have the correct symmetry to overlap with the metal t₂g orbitals. Since the ligand p-orbitals are filled, electron density is donated from ligand to metal t₂g, raising the t₂g energy. This decreases the t₂g-eg gap (Δ), making halides weak-field ligands. The effect is strongest for larger, more polarizable halides (I⁻ > Br⁻ > Cl⁻ > F⁻), which is why the spectrochemical series places I⁻ at the very bottom.
Question 3 True / False
Ligand field theory reduces to crystal field theory when all metal-ligand interactions are treated as purely electrostatic.
TTrue
FFalse
Answer: True
LFT is a more general framework that includes CFT as a limiting case. When you remove all covalent interactions (sigma and pi bonding between metal and ligand orbitals) and treat ligands as point charges, LFT reproduces exactly the CFT predictions: the d-orbital splitting pattern depends only on geometry, and Δ is determined solely by electrostatic parameters. The power of LFT is that it adds covalency without discarding the intuitive d-orbital splitting picture — you can still talk about t₂g and eg sets, high-spin and low-spin, and CFSE, while also explaining trends (like the spectrochemical series) that CFT cannot.
Question 4 Short Answer
Explain why CO is a stronger-field ligand than CN⁻, even though CN⁻ is negatively charged and should interact more strongly with a positively charged metal ion in an electrostatic model.
Think about your answer, then reveal below.
Model answer: In the electrostatic model of CFT, CN⁻ should produce a stronger field because its negative charge creates a stronger point-charge interaction with the metal cation. But LFT shows that field strength is dominated by covalent interactions, not electrostatics. CO is a superior pi-acceptor: its π* orbitals are lower in energy and better matched to the metal t₂g orbitals than those of CN⁻, making back-bonding more effective. CO is also an excellent sigma-donor through its carbon lone pair. The combination of strong sigma-donation (raising eg) and strong pi-acceptance (lowering t₂g) produces the largest Δ of any common ligand. CN⁻, while also a good pi-acceptor, is slightly less effective because its negative charge raises the energy of its π* orbitals, making them a less favorable target for back-donation.
This example perfectly illustrates why CFT alone is insufficient: the relative field strengths of CO and CN⁻ cannot be rationalized without invoking covalent pi-interactions. LFT resolves this and many similar puzzles in the spectrochemical series.