Pairing Energy and Its Origins

Introduction

Electrons prefer low-energy orbitals, but putting two electrons into one orbital is not free. They repel one another, and an arrangement with parallel spins in different orbitals can gain exchange stabilisation. Pairing energy P packages these effects into a useful comparison. It is the reason an electron may occupy a higher e g orbital even while a lower t₂g orbital still has room for a second electron.

Core explanation

The Pauli exclusion principle allows two electrons in one spatial orbital only when their spins are opposite. Bringing their charge distributions into the same region increases Coulombic repulsion. Conversely, distributing same-spin electrons among separate degenerate orbitals is favoured by Hund’s rule; exchange effects contribute to that preference. The effective pairing energy represents the energetic penalty of choosing a new pair over an arrangement with electrons in separate orbitals. It is a simplified parameter, not a literal bill for only Coulomb repulsion.

In an octahedral complex, filling t₂g before e g saves Δₒ for each electron moved from an upper to a lower orbital. For d⁴, high-spin t₂g³e g¹ has four unpaired electrons and no electron pair; low-spin t₂g⁴e g⁰ has two unpaired electrons and one pair. The low-spin arrangement gains one Δₒ in orbital energy, since its orbital CFSE changes from −0.6Δₒ to −1.6Δₒ. It incurs one pairing cost P. A simple comparison therefore favours low spin if the energy gained by dropping an electron to t₂g exceeds the effective pairing cost.

For d⁵, high-spin t₂g³e g² has five unpaired electrons and no pairs; low-spin t₂g⁵e g⁰ has one unpaired electron and two pairs. The orbital benefit is 2Δₒ, but two pairs are added. For d⁶, high spin has one pair and low spin three; again two additional pairs enter the comparison. Counting all low-spin pairs without subtracting the pairs already present in high spin shifts the reference arbitrarily.

P is not exactly the same for every metal and electron configuration. Electron–electron repulsion depends on how diffuse the orbitals are and how much covalent mixing with ligands occurs. Consequently, the textbook rule “Δₒ greater than P gives low spin” is a compact guide for first-pass comparisons, not a universal numerical law for every real compound. Temperature and entropy can also matter near a spin crossover.

Tetrahedral splitting is usually smaller than octahedral splitting for corresponding ions and ligands, so the orbital benefit of pairing early is generally too small to outweigh P. That is why familiar tetrahedral transition-metal complexes are ordinarily high spin. The statement reflects common chemistry, not an algebraic impossibility of other electronic structures.

Step-by-step reasoning

Draw both plausible fillings for a given d count. Calculate orbital CFSE for each. Count the number of doubly occupied d orbitals in each configuration, then subtract high-spin pair count from low-spin pair count. Compare the orbital-energy change with the added pairing cost; only then predict which filling the simple model favours.

Visual explanation

For d⁴, draw three lower t₂g boxes with one upward arrow each. In the high-spin drawing, put the fourth upward arrow in e g. In the low-spin drawing, place a downward arrow beside one lower upward arrow instead. Mark the move downward as a saving of Δₒ and the new pair as a cost P.

Real-world analogy

Four travellers can pay for an extra hotel room or share a crowded room. The extra room has a price, while crowding has a separate discomfort cost. Choosing the cheaper arrangement requires both values, just as electrons balance orbital splitting against pairing cost.

Real-world example

Water often gives Fe²⁺ octahedral complexes a comparatively smaller Δₒ and a high-spin d⁶ state. Cyanide produces a much larger splitting and low-spin d⁶ hexacyanoferrate(II). The metal d count is unchanged; the ligand-dependent orbital-versus-pairing balance changes.

Why?

Why does Hund’s rule not force high spin in every complex? Hund’s rule favours separate occupancy among orbitals of the same energy. In a ligand field, t₂g and e g are no longer equal, and a sufficiently large gap can compensate for the cost of pairing below.

Common misconception

“P is simply the energy of electrostatic repulsion between two electrons.” It is an effective comparison that also reflects lost exchange stabilisation and can depend on the electronic environment. It should not be treated as a universal fixed constant.

Worked example

For octahedral d⁴ with Δₒ=18,000 cm⁻¹ and P=22,000 cm⁻¹, low spin saves one Δₒ but pays P. Its energy relative to high spin is −18,000+22,000=+4,000 cm⁻¹, so high spin is favoured in this simple model. If Δₒ instead rises to 27,000 cm⁻¹ with the same assumed P, the difference becomes −5,000 cm⁻¹ and low spin is favoured.

Quick check

1. How many extra pairs does low-spin d⁶ have compared with high-spin d⁶? Answer: Two extra pairs: three in t₂g⁶ versus one in t₂g⁴e g². 2. Why are tetrahedral complexes generally high spin? Answer: Their splitting is commonly too small to reward early pairing enough to overcome pairing cost.

Exam focus

Write both configurations and count pair differences . Use an explicit energy difference, such as low minus high, and interpret its sign. Do not apply Hund’s rule across nondegenerate levels without an energy comparison.

Advanced insight

Actual multiplet energies depend on several electron–electron repulsion parameters, often represented by Racah parameters, rather than a single perfectly transferable P. Ligand-field theory and spectroscopy resolve those details; the one-P model is most useful for transparent qualitative and introductory quantitative reasoning.

Summary

Pairing costs energy through electron repulsion and exchange effects. Spin state follows a competition between that cost and the orbital saving from moving electrons into lower split levels. Compare configurations with the same pair-count reference.

Practice questions

1. Count unpaired electrons and pairs in high-spin and low-spin octahedral d⁵. Answer: High-spin t₂g³e g² has five unpaired and zero pairs. Low-spin t₂g⁵ has one unpaired and two pairs. 2. For d⁴, Δₒ=24,000 cm⁻¹ and P=20,000 cm⁻¹. Which spin state does the one-P model favour? Answer: Low spin gains one Δₒ but pays one P, giving −4,000 cm⁻¹ relative to high spin. It is favoured in this simplified comparison. 3. Why does a low-spin d⁶ diagram not require adding 3P if comparing against high-spin d⁶? Answer: High-spin d⁶ already contains one pair. The low-spin diagram has three pairs, so only the difference of two pairs changes the comparison.