Octahedral CFSE for d¹ to d³ Ions

Introduction

The first three d electrons provide the cleanest CFSE examples. An octahedral t₂g set has three equal-energy orbitals, so d¹, d² and d³ ions can place every electron in a different lower orbital. There is no competition between pairing below and occupying e g above. That makes these ions useful reference points before tackling d⁴ through d⁷ spin choices.

Core explanation

An octahedral field places t₂g at −0.4Δₒ per electron and e g at +0.6Δₒ. For d¹, put one electron in any t₂g orbital: CFSE = −0.4Δₒ, with one unpaired electron. For d², Hund’s rule places the second electron in a different t₂g orbital, yielding −0.8Δₒ and two unpaired electrons. For d³, each of the three t₂g orbitals receives one electron, yielding −1.2Δₒ and three unpaired electrons. These are signed orbital energies relative to the barycentre; their stabilisation magnitudes are 0.4, 0.8 and 1.2 times Δₒ.

Trying to make a “low-spin d²” arrangement by pairing both electrons in one t₂g orbital does not save any crystal-field splitting energy: both t₂g orbitals have the same energy. It merely introduces pairing cost and loses the preferred separate-spin arrangement. Likewise, pairing d³ electrons before filling the third t₂g orbital offers no orbital benefit. Conversely, promoting any of the first three electrons to e g costs Δₒ without reducing an existing pair. Thus the simple model gives one ground-state filling pattern for d¹–d³, although their precise electronic states have further multiplet detail.

The d count must be established from oxidation state rather than from ligand number. Ti³⁺ is d¹, V³⁺ is d² and Cr³⁺ is d³. In six-coordinate aqua ions such as [Ti(H₂O)₆]³⁺, the six waters donate to the metal, but they do not turn d¹ into d¹³. For an octahedral d³ ion, the calculated spin-only moment uses n=3: √[3(3+2)] = √15 ≈3.87 BM. A real measured moment can differ because of orbital and other effects.

Because all three electron counts have unpaired electrons, their simple spin-only model predicts paramagnetism. They may absorb visible or near-visible light through electronic transitions, but observed spectra are not determined by CFSE alone. Transition selection rules, electron repulsion and covalency affect band intensities and positions. Do not equate −1.2Δₒ with a unique absorption wavelength for every d³ complex.

The increasing CFSE across d¹ to d³ can influence hydration and lattice-energy trends, but a trend in measured thermodynamic quantities includes many other terms. Compare closely related ions before attributing an entire stability difference to CFSE.

Step-by-step reasoning

Calculate oxidation state and d count. For d¹, d² or d³, draw three lower t₂g boxes and place one arrow in each box before any pairing. Leave e g empty. Multiply the number of occupied t₂g orbitals by −0.4Δₒ, and count the arrows without partners for magnetism.

Visual explanation

Place three boxes below a dashed line and two above. Under d¹, draw one upward arrow in the first lower box; under d², arrows in two different lower boxes; under d³, one in each. No arrow enters the upper row, and no box has a paired up/down set.

Real-world analogy

Three identical low-cost rooms can each hold one student comfortably. The first three students each take a separate cheap room. Paying for an expensive room or crowding two students into one cheap room brings no benefit while an empty cheap room remains available.

Real-world example

The octahedral [Cr(H₂O)₆]³⁺ ion contains Cr³⁺, which is d³. Its three electrons occupy separate t₂g orbitals and give a substantial −1.2Δₒ orbital stabilisation. This stable filling is one reason Cr(III) complexes feature prominently in coordination chemistry examples.

Why?

Why is there no ordinary high-spin versus low-spin d³ decision? All three electrons fit in three degenerate lower orbitals without pairing. Any alternative either pairs at no splitting-energy gain or moves an electron into a higher level at an energy cost.

Common misconception

“Three unpaired electrons means three e g electrons.” The lower t₂g group itself has three orbitals, each able to carry one unpaired electron. Octahedral d³ has t₂g³e g⁰.

Worked example

Suppose octahedral [V(H₂O)₆]³⁺ has Δₒ=15,000 cm⁻¹. Vanadium is group 5, so V³⁺ is d². Fill t₂g²e g⁰ with one electron in each of two lower orbitals. CFSE = 2(−0.4)(15,000)=−12,000 cm⁻¹ in energy-equivalent units. Two electrons remain unpaired; the spin-only magnetic moment estimate is √8≈2.83 BM.

Quick check

1. What is the octahedral orbital CFSE of d³? Answer: Three t₂g electrons give −1.2Δₒ. 2. How many electron pairs are present in the d² ground-state diagram? Answer: Zero; the two electrons occupy separate t₂g orbitals.

Exam focus

For d¹–d³, use t₂g¹, t₂g² and t₂g³ directly, with one, two and three unpaired electrons. Avoid inventing a high/low-spin split where no pair-versus-promotion decision exists.

Advanced insight

Even with a unique elementary filling, terms arising from interelectronic repulsion and spin–orbit coupling can split spectroscopic states further. CFSE provides an orbital-energy baseline, while interpreting fine spectral structure requires a richer ligand-field treatment.

Summary

Octahedral d¹, d² and d³ have orbital CFSE values −0.4, −0.8 and −1.2Δₒ, respectively. Their electrons remain separate in t₂g and give one, two and three unpaired electrons in the simple ground-state picture.

Practice questions

1. Find the CFSE and unpaired count for octahedral Ti³⁺. Answer: Ti³⁺ is d¹, so its configuration is t₂g¹. CFSE is −0.4Δₒ and one electron is unpaired. 2. Why does putting the second electron of d² into an occupied t₂g orbital not improve CFSE? Answer: All t₂g orbitals have the same energy. Pairing in one instead of using an empty t₂g orbital leaves orbital CFSE unchanged but adds a pairing penalty. 3. If Δₒ doubles for a d³ ion, what happens to its simple orbital CFSE coefficient and magnitude? Answer: The coefficient stays −1.2, while the numerical magnitude doubles because CFSE=−1.2Δₒ. The three unpaired electrons remain in the same basic filling.