In an octahedral crystal field, which d-orbitals are raised to higher energy?
Ad_xy, d_xz, and d_yz (the t₂g set)
Bd_z² and d_x²−y² (the eg set)
CAll five d-orbitals are raised equally
DOnly d_z² is raised; the other four remain degenerate
In an octahedral field, six ligands approach along the x, y, and z axes. The d_z² and d_x²−y² orbitals (eg set) point directly at the ligands and experience strong electrostatic repulsion, raising their energy. The d_xy, d_xz, and d_yz orbitals (t₂g set) point between the ligand axes and experience less repulsion, placing them at lower energy. The energy gap between these two sets is the octahedral crystal field splitting parameter, Δ_oct (or 10Dq).
Question 2 Multiple Choice
A d⁶ metal ion in an octahedral field can be either high-spin (4 unpaired electrons) or low-spin (0 unpaired electrons). What determines which configuration is adopted?
AThe atomic number of the metal ion
BThe relative magnitudes of the crystal field splitting energy (Δ_oct) and the electron pairing energy (P)
CWhether the complex is charged or neutral
DThe principal quantum number of the d-orbitals
The competition between Δ_oct and the pairing energy P determines spin state. If Δ_oct > P (strong-field ligands), electrons pair in the lower t₂g set before occupying the higher eg set — producing a low-spin configuration (t₂g⁶ eg⁰, zero unpaired electrons for d⁶). If Δ_oct < P (weak-field ligands), electrons maximize spin by occupying all five d-orbitals singly before pairing — giving a high-spin configuration (t₂g⁴ eg², four unpaired electrons for d⁶). The metal identity matters insofar as it affects Δ_oct (higher oxidation states and heavier metals give larger Δ), but the direct determinant is the Δ vs P comparison.
Question 3 True / False
Tetrahedral crystal field splitting (Δ_tet) is approximately 4/9 of octahedral splitting (Δ_oct) for the same metal-ligand combination.
TTrue
FFalse
Answer: True
Two factors make tetrahedral splitting smaller: there are only four ligands instead of six (reducing total electrostatic interaction), and the ligands do not point directly at any d-orbital (they approach between the axes). The combined geometric and numerical effect gives Δ_tet ≈ (4/9)Δ_oct. Because tetrahedral splitting is inherently smaller, it almost never exceeds the pairing energy — which is why tetrahedral complexes are nearly always high-spin.
Question 4 True / False
In crystal field theory, the total energy of the d-electrons in a complex is always lower than in the free (spherically symmetric) ion, regardless of the electron configuration.
TTrue
FFalse
Answer: False
CFT splitting is a redistribution of energy: the t₂g set drops by 0.4Δ_oct per electron while the eg set rises by 0.6Δ_oct per electron (in an octahedral field). The crystal field stabilization energy (CFSE) — the net energy lowering — depends on how electrons distribute. For configurations like d⁵ high-spin (t₂g³ eg²), CFSE is 3(−0.4Δ) + 2(+0.6Δ) = 0 — no net stabilization. For d¹⁰, CFSE is similarly zero. So not all configurations gain stabilization from the crystal field.
Question 5 Short Answer
Explain why Cr³⁺ (d³) octahedral complexes are exceptionally kinetically inert, connecting your answer to crystal field stabilization energy.
Think about your answer, then reveal below.
Model answer: Cr³⁺ has the configuration t₂g³ in an octahedral field, with one electron in each of the three lower-energy orbitals. This gives a large CFSE of 3 × (−0.4Δ_oct) = −1.2Δ_oct. For any ligand substitution to occur, the complex must pass through a transition state with different geometry (five-coordinate or seven-coordinate), which necessarily changes the d-orbital splitting pattern and partially loses this CFSE. The large CFSE of t₂g³ means the activation energy for any geometry change is high, making ligand exchange slow — hence kinetic inertness. This is why Cr(III) complexes can be isolated as specific isomers that persist indefinitely at room temperature.
Kinetic inertness is distinct from thermodynamic stability. A complex can be thermodynamically unstable (the products are lower energy) but kinetically inert (the barrier to reaching those products is high). CFSE contributes to this barrier by stabilizing the ground-state geometry relative to any transition-state geometry.