Practice problems drawn from Housecroft, Atkins, JD Lee, and others. Select an answer or click “Show Answer” to reveal the explanation.
In an octahedral crystal field, the five d orbitals split into two sets. Which orbitals are raised in energy (the eg set) and which are lowered (the t2g set)?
In an octahedral field, the six ligands lie along the x, y, and z axes. The dz² and dx²−y² orbitals point directly at the ligands, so they experience greater electrostatic repulsion and are raised in energy (the eg set). The dxy, dxz, and dyz orbitals point between the ligand axes, experience less repulsion, and are stabilized (the t2g set). The energy gap between these two sets is Δoct.
The crystal field splitting parameter Δoct measures the energy separation between the t2g and eg sets. Which combination of factors leads to the largest Δoct?
Three factors increase Δoct: (1) higher metal oxidation state increases the effective nuclear charge, pulling ligands closer; (2) moving down a group (first → second → third row) increases the spatial extent of d orbitals, enhancing metal–ligand overlap; (3) stronger-field ligands (e.g. CN−, CO) interact more strongly. Combining all three — high charge, third-row metal, strong-field ligand — gives the largest Δoct.
In a tetrahedral field, Δtet ≈ 4⁄9 Δoct. Which statement best explains why Δtet is smaller?
Two factors make Δtet smaller. First, there are only four ligands instead of six, reducing the total electrostatic interaction. Second, in a tetrahedron, no ligand points directly along the d orbital axes — the interaction is indirect. Combined, these effects give Δtet ≈ 4⁄9 Δoct for the same metal–ligand combination. The splitting pattern is also inverted: the t2 set is raised and the e set is lowered.
Calculate the crystal field stabilization energy (CFSE) in terms of Δoct for a d6 ion in (i) high-spin and (ii) low-spin octahedral environments. Which answer is correct?
For d6 high-spin: t2g4eg2. CFSE = 4(−0.4Δ) + 2(+0.6Δ) = −0.4Δoct. For d6 low-spin: t2g6eg0. CFSE = 6(−0.4Δ) = −2.4Δoct, but three extra electron pairings cost 2P (there is already one forced pairing in high-spin, giving a net cost of 2P additional). The low-spin configuration is favored when Δoct is large enough to offset the pairing energy penalty.
Whether an octahedral complex adopts a high-spin or low-spin configuration depends on the competition between two energy quantities. Which are they?
When Δoct < P (pairing energy), electrons prefer to occupy the higher-energy eg orbitals rather than pair up in t2g, giving a high-spin configuration. When Δoct > P, the energy cost of pairing is less than the cost of promoting an electron to eg, so electrons fill t2g first — a low-spin configuration. This competition applies to d4–d7 octahedral complexes.
[CoF6]3− is high-spin but [Co(CN)6]3− is low-spin. Both contain Co3+ (d6). Using the spectrochemical series, explain this difference.
The spectrochemical series ranks ligands by their ability to split d orbitals: I− < Br− < Cl− < F− < H2O < NH3 < en < NO2− < CN− < CO. F− is near the weak-field end, producing a small Δoct that cannot overcome the pairing energy P — hence high-spin (4 unpaired electrons). CN− is near the strong-field end, generating a large Δoct > P, so all six d electrons pair in t2g — low-spin (0 unpaired electrons).
A d5 metal ion is placed in an octahedral field. How many unpaired electrons does it have in the high-spin and low-spin configurations, respectively?
In high-spin d5: all five d orbitals are singly occupied (t2g3eg2), giving 5 unpaired electrons and maximum spin multiplicity. In low-spin d5: electrons fill t2g first, giving t2g5eg0. Two of the three t2g orbitals hold paired electrons and one holds a single electron, yielding 1 unpaired electron. This dramatic difference (5 vs. 1) is directly detectable by magnetic susceptibility measurements.
Tetrahedral complexes are almost always high-spin. Why is a low-spin tetrahedral configuration essentially never observed?
The tetrahedral crystal field splitting Δtet is inherently only about 4/9 the magnitude of Δoct for the same metal and ligand. This reduced splitting is almost never large enough to exceed the electron pairing energy P. Since Δtet < P in virtually all cases, electrons occupy the higher e set before pairing in the lower t2 set, producing a high-spin configuration. Low-spin tetrahedral complexes would require an unusually large Δtet that is rarely achievable.
The Jahn-Teller theorem states that any non-linear molecule with an orbitally degenerate ground state will undergo a geometric distortion. Which octahedral d-electron configurations show the strongest Jahn-Teller distortions?
Strong Jahn-Teller distortions occur when the eg orbitals are unevenly occupied, because these orbitals point directly at the ligands and unequal occupation creates a large driving force for distortion. High-spin d4 (t2g3eg1) and d9 (t2g6eg3) both have one eg orbital singly occupied and the other empty or doubly occupied. Low-spin d7 also shows a strong effect. Uneven t2g occupation produces weaker distortions.
Cu2+ complexes (d9) typically display tetragonal elongation, where the two axial bonds are longer than the four equatorial bonds. Use the Jahn-Teller effect to explain this observation.
Cu2+ is d9 with configuration t2g6eg3. The three electrons in the doubly degenerate eg set cannot occupy both orbitals equally. In a tetragonal elongation, the axial ligands move away, lowering dz² relative to dx²−y². The extra electron occupies the now-lower dz², removing the degeneracy and lowering the total energy. This is a textbook example of a strong Jahn-Teller distortion.
Square planar geometry can be viewed as the limiting case of tetragonal distortion of an octahedral complex. What happens in this extreme limit?
Starting from a regular octahedron, tetragonal elongation progressively lengthens the two axial bonds. In the extreme limit, the axial ligands are effectively removed, leaving four ligands in the xy plane — a square planar geometry. This causes extensive further splitting of the d orbitals: dx²−y² becomes highest in energy (pointing directly at the four remaining ligands), while dz² drops significantly. The resulting d orbital order from low to high energy is approximately dxz,dyz < dz² < dxy < dx²−y².
Most Pd2+ and Pt2+ complexes adopt square planar geometry rather than tetrahedral. Why is this the case?
Pd2+ and Pt2+ are d8 ions from the second and third transition rows, which means their d orbitals are spatially extended and produce very large crystal field splittings. In square planar geometry, the dx²−y² orbital is pushed so high in energy that all eight d electrons fill the four lower orbitals, leaving dx²−y² empty. This is strongly stabilized when Δ is large. First-row d8 ions like Ni2+ can be tetrahedral or square planar depending on the ligand field strength.
Ligand field theory (LFT) improves on crystal field theory (CFT) in several ways. What is the key conceptual advance of LFT over CFT?
Crystal field theory treats ligands as point charges or dipoles — a purely electrostatic model that ignores covalency entirely. Ligand field theory (LFT) instead constructs molecular orbitals from metal d orbitals and ligand σ/π orbitals, capturing covalent bonding, π back-donation, and the true electronic structure of the complex. LFT explains the spectrochemical series, magnetic properties, and spectroscopic data that CFT cannot fully account for.
In ligand field theory, σ-donor ligands alone produce a baseline Δoct. How do π-acceptor ligands (such as CO and CN−) increase Δoct beyond this baseline?
π-Acceptor ligands like CO and CN− possess low-energy empty π* orbitals that interact with the filled metal t2g orbitals. This π back-bonding delocalizes electron density from metal to ligand, stabilizing (lowering) the t2g set. Since the eg level is unaffected by this interaction, the gap Δoct increases. This is why π-acceptor ligands appear at the strong-field end of the spectrochemical series, and conversely, π-donor ligands (like halides) raise t2g and reduce Δoct.
Using ligand field theory, explain why CO produces a much larger Δoct than H2O when coordinated to the same metal ion.
CO acts as both a strong σ-donor through its lone pair on carbon and a strong π-acceptor through its low-lying π* orbitals. The σ-donation pushes the eg set up, while the π-acceptance stabilizes the t2g set through back-bonding — both effects increase Δoct. In contrast, H2O is a modest σ-donor with no significant π-acceptor capability; its lone pairs act as weak π-donors that slightly raise t2g, reducing Δoct. This dual σ/π character is why CO sits at the very top of the spectrochemical series.
Crystal Field & Ligand Field Theory