The Barycentre Rule in Octahedral Fields

Introduction

An octahedral diagram has three lower d orbitals and two higher ones. It is tempting to place the upper pair Δₒ above an arbitrary zero line and call the lower triplet zero. That is fine for transition energies but wrong for crystal field stabilisation energy. The conventional zero is the average energy of the five orbitals before directional splitting, called the barycentre.

Core explanation

Imagine a spherical ligand field that repels all five d orbitals equally. Their common energy is the reference barycentre. Changing to six ligands on the x, y and z axes splits the five orbitals because their shapes point in different directions. The d(x²−y²) and d(z²) orbitals face ligands and form the higher e g set; d(xy), d(xz) and d(yz) point between axes and form the lower t₂g set.

Let the lower set shift down by a units of energy per electron and the upper set shift up by b. The separation is a+b=Δₒ. Because there are three lower orbitals and two upper orbitals, preserving the mean energy requires 3(−a)+2b=0. Solve the simultaneous equations: 2b=3a and a+b=Δₒ, so a=2Δₒ/5=0.4Δₒ and b=3Δₒ/5=0.6Δₒ. Thus each t₂g electron contributes −0.4Δₒ and each e g electron contributes +0.6Δₒ relative to the barycentre.

The coefficients are weighted by orbital degeneracy , not by how many electrons a particular complex happens to have. The barycentre rule describes the level positions before filling. For a d¹ electron in t₂g, orbital energy is lowered by 0.4Δₒ. For a d³ configuration t₂g³, the lowering is 1.2Δₒ. For a d⁵ high-spin configuration t₂g³e g², the weighted sum is −1.2Δₒ+1.2Δₒ=0. No orbital CFSE remains, even though the five orbitals are still split and light absorption may still occur.

The signs matter. A negative calculated CFSE means lower orbital energy than the hypothetical unsplit reference. Some textbooks report stabilisation magnitude as a positive number, such as “1.2Δₒ stabilisation” for d³. State which convention you use. Pairing energy is another term and does not alter the barycentre coefficients. Its contribution matters when comparing configurations with different numbers of electron pairs.

The barycentre is an idealised common reference, not the energy of one actual d orbital in the octahedral complex. Nor does its preservation imply that ligand binding has no energy change overall. Ligand–metal bonding, electrostatic attraction and repulsion, and covalency contribute to total energy; CFSE isolates one directional splitting contribution.

Step-by-step reasoning

Mark a horizontal reference line for five equally energetic orbitals. Place three t₂g boxes below and two e g boxes above. Write 3E(t₂g)+2E(e g)=0 and E(e g)−E(t₂g)=Δₒ. Solve to obtain −0.4Δₒ and +0.6Δₒ, then multiply each by the actual occupancy and add.

Visual explanation

Draw three boxes below a dashed barycentre and two boxes above it. The lower arrow from the line has length 0.4Δₒ; the upper arrow length 0.6Δₒ. Three short downward shifts balance two larger upward shifts, so the average of all five positions stays on the dashed line.

Real-world analogy

Five passengers share a fixed average seat height. Three seats move down by four steps while two move up by six steps. The total downward movement is 3×4=12 steps and upward movement is 2×6=12, so the average height is unchanged despite a ten-step gap between groups.

Real-world example

Spectra of an octahedral transition-metal complex can reveal an energy separation associated with d-level excitation. When that Δₒ is reported in kJ mol⁻¹ or cm⁻¹, the barycentre coefficients allow its electron configuration to be converted into an orbital stabilisation estimate.

Why?

Why is the upper shift larger than the lower shift? Only two orbitals are raised, while three are lowered. Each raised orbital must move further than each lowered orbital so that the weighted average of five orbital energies remains fixed.

Common misconception

“The three lower orbitals are −0.5Δₒ and the two higher orbitals are +0.5Δₒ because they are separated by Δₒ.” Equal shifts would move the weighted average downward: 3(−0.5)+2(+0.5)=−0.5, not zero.

Worked example

Calculate orbital CFSE for t₂g⁴e g¹. Four electrons in lower orbitals contribute 4(−0.4Δₒ)=−1.6Δₒ, while one upper electron contributes +0.6Δₒ. The total is −1.0Δₒ. If Δₒ=20,000 cm⁻¹, the CFSE is −20,000 cm⁻¹ in energy-equivalent wavenumber units; pairing-energy accounting is a separate step.

Quick check

1. What relation preserves the octahedral d-orbital barycentre? Answer: 3E(t₂g)+2E(e g)=0 when energies are measured from the barycentre. 2. Does high-spin octahedral d⁵ have zero splitting? Answer: No. It has zero net orbital CFSE in the simple model, although Δₒ remains nonzero.

Exam focus

Label the fivefold degeneracy of the reference and the 3:2 degeneracies after splitting. Use occupancy times level energy; avoid dividing Δₒ equally above and below zero.

Advanced insight

The barycentre rule is a trace argument: splitting redistributes the energies of a five-dimensional orbital set under an idealised angular interaction while preserving their mean. It provides a consistent reference for comparing fillings, not an exact prediction of total complex formation enthalpy.

Summary

Octahedral t₂g orbitals lie 0.4Δₒ below and e g orbitals 0.6Δₒ above the barycentre. Their 3:2 degeneracies make the weighted mean zero. CFSE sums these contributions for occupied orbitals.

Practice questions

1. Derive the upper level position if the lower level is −0.4Δₒ. Answer: The gap is Δₒ, so E(e g)=−0.4Δₒ+Δₒ=+0.6Δₒ. Equivalently, 3(−0.4)+2x=0 gives x=+0.6. 2. Calculate orbital CFSE for t₂g²e g⁰ and t₂g³e g². Answer: The first is 2(−0.4Δₒ)=−0.8Δₒ. The second is 3(−0.4Δₒ)+2(+0.6Δₒ)=0. 3. Does a zero CFSE prove that no d–d transition is possible? Answer: No. CFSE compares total ground-state orbital energy with a reference; a nonzero splitting can still permit an excitation between lower and higher levels, subject to selection rules.