Octahedral CFSE for d⁴ to d⁷: Low-Spin Cases

Introduction

When the octahedral gap grows large, putting an electron in e g can cost more than pairing it in t₂g. The resulting low-spin d⁴ through d⁷ configurations have fewer unpaired electrons and more negative orbital CFSE than their high-spin counterparts. The comparison is meaningful only when the extra pairs are included in the energy difference.

Core explanation

Low-spin d⁴ fills t₂g⁴e g⁰: one lower orbital contains a pair and the other two contain one electron each. Orbital CFSE is 4(−0.4Δₒ)=−1.6Δₒ, with two unpaired electrons and one pair. Low-spin d⁵ is t₂g⁵e g⁰, giving −2.0Δₒ, one unpaired electron and two pairs. Low-spin d⁶ fills all three lower orbitals twice, t₂g⁶e g⁰, giving −2.4Δₒ and no unpaired electrons. Its three pairs make it diamagnetic in a simple spin model.

The seventh electron cannot enter a t₂g orbital already holding two electrons because of the Pauli principle. Low-spin d⁷ therefore has t₂g⁶e g¹, orbital CFSE 6(−0.4Δₒ)+0.6Δₒ=−1.8Δₒ, one unpaired electron and three pairs. “Low spin” describes the lowest feasible spin for that count within the usual octahedral levels; it does not require zero unpaired electrons for every dⁿ.

Compare d⁴ with high-spin t₂g³e g¹, which has −0.6Δₒ and no pairs. The low-spin form gains 1.0Δₒ in orbital energy but adds one pair; its simplified energy relative to high spin is −Δₒ+P. For d⁵, low spin gains 2Δₒ and adds two pairs: −2Δₒ+2P. For d⁶, the same difference arises because low-spin t₂g⁶ has two more pairs than high-spin t₂g⁴e g². For d⁷, low spin again gains Δₒ and adds one pair compared with high-spin t₂g⁵e g². These neat differences motivate the familiar Δₒ-versus-P rule, within the simplified one-P model.

The d⁶ pair comparison is a common source of error. Low spin contains three electron pairs, but high spin already contains one, so the additional cost is 2P. If energies are measured against a d⁶ arrangement with no pairs, that hypothetical reference is not a valid five-orbital filling; directly compare the actual two arrangements instead.

Low-spin configurations are often associated with stronger-field ligands such as CN⁻ or CO, and with many 4d or 5d metal ions, where splitting is commonly larger. This is a tendency rather than an automatic classification: oxidation state, geometry and precise bonding also matter. A measured moment and absorption spectrum can test whether the assigned state is plausible.

Step-by-step reasoning

Fill each t₂g box singly, then pair in the three lower boxes before placing electrons in e g. Stop at the required d count. Compute CFSE from the level occupancies, count paired boxes and unpaired arrows, and subtract the high-spin energy on the same pair-count convention.

Visual explanation

Draw a larger vertical Δₒ than in the high-spin diagram. For d⁴, the fourth arrow points down beside a lower arrow, not into e g. Continue filling t₂g to six; only d⁷ sends its seventh arrow to an upper box.

Real-world analogy

Suppose upper-floor workstations become very costly to rent, while sharing a lower-floor workstation has a fixed inconvenience. Workers now share downstairs until the downstairs desks reach their legal capacity. The capacity limit mirrors two opposite-spin electrons per orbital.

Real-world example

Hexacyanoferrate(II), [Fe(CN)₆]⁴⁻, contains Fe²⁺ and is d⁶. In an octahedral strong field it is low spin, t₂g⁶, with no unpaired d electrons in the simple model. Its formal d count matches high-spin aqueous Fe²⁺ even though their magnetic behaviour differs.

Why?

Why does low-spin d⁷ contain an e g electron despite a large gap? The three t₂g orbitals each hold at most two electrons. Once six electrons fill them, the seventh has no allowed lower state and must enter e g.

Common misconception

“Low-spin d⁵ has no unpaired electrons.” Five electrons distributed among three lower orbitals leave one unpaired electron. Only low-spin d⁶ among these four counts has a completely paired t₂g shell.

Worked example

An octahedral d⁵ ion has Δₒ=25,000 cm⁻¹ and effective P=18,000 cm⁻¹. High spin has t₂g³e g², orbital CFSE zero and no pairs. Low spin has t₂g⁵, orbital CFSE −2(25,000)=−50,000 cm⁻¹ and two pairs costing 36,000 cm⁻¹. Its relative energy is −14,000 cm⁻¹, so the simple comparison favours low spin; one electron remains unpaired.

Quick check

1. What is orbital CFSE for low-spin octahedral d⁶? Answer: t₂g⁶ gives 6(−0.4Δₒ)=−2.4Δₒ. 2. How many extra pairs does low-spin d⁷ have relative to high-spin d⁷? Answer: One extra pair: three versus two.

Exam focus

Do not write “d⁷ all in t₂g.” Include the Pauli limit, show the correct pair differences, and distinguish a signed orbital CFSE from total relative energy.

Advanced insight

Observed spin states reflect free energy, not merely a static two-term orbital calculation. Entropy, vibrational structure and covalent changes can make a complex near the crossover switch spin population with temperature, pressure or light.

Summary

Low-spin d⁴–d⁷ configurations are t₂g⁴, t₂g⁵, t₂g⁶ and t₂g⁶e g¹. Their orbital CFSE values are −1.6, −2.0, −2.4 and −1.8Δₒ; additional pairing must be included when comparing with high spin.

Practice questions

1. List unpaired counts for low-spin octahedral d⁴, d⁵, d⁶ and d⁷. Answer: They are two, one, zero and one, respectively, from t₂g⁴, t₂g⁵, t₂g⁶ and t₂g⁶e g¹. 2. Calculate the low-minus-high orbital CFSE difference for d⁷. Answer: Low-spin d⁷ has −1.8Δₒ and high-spin d⁷ has −0.8Δₒ. The difference is −1.0Δₒ before one additional pairing cost is added. 3. Explain why a measured near-zero magnetic moment supports low-spin d⁶. Answer: Its t₂g⁶ configuration has no unpaired electrons, so it is diamagnetic in the simple model. A high-spin d⁶ alternative has four unpaired electrons and a much larger spin contribution.