Predicting Spin State from Δo and P

Introduction

Ligand names provide clues about spin state, but a numerical question is settled by an energy comparison. For octahedral d⁴ through d⁷, placing an electron into e g costs Δₒ compared with t₂g, while pairing it below costs an effective P. The useful shortcut “compare Δₒ with P” follows from explicit occupancies and pair counts; it should not be memorised without those conditions.

Core explanation

Choose high spin and low spin fillings for the same oxidation state and geometry. Compute each orbital CFSE from t₂g at −0.4Δₒ and e g at +0.6Δₒ. Count pairs in each diagram. In a simple model, E(low)−E(high) equals the difference in orbital CFSE plus P times the difference in pair count. A negative result favours low spin; a positive result favours high spin at the level of this model.

For d⁴, high spin t₂g³e g¹ has orbital CFSE −0.6Δₒ and no pairs. Low spin t₂g⁴ has −1.6Δₒ and one pair. Their difference is −Δₒ+P. For d⁵, high spin t₂g³e g² has zero CFSE and no pairs; low spin t₂g⁵ has −2Δₒ and two pairs. Difference: −2Δₒ+2P. For d⁶, high spin t₂g⁴e g² has −0.4Δₒ and one pair; low spin t₂g⁶ has −2.4Δₒ and three pairs, again giving −2Δₒ+2P. For d⁷, high spin t₂g⁵e g² has −0.8Δₒ and two pairs; low spin t₂g⁶e g¹ has −1.8Δₒ and three pairs, giving −Δₒ+P.

The common crossover at Δₒ≈P is a result of treating every added pair with the same P. Real electron-repulsion energies vary between electronic terms, and ligand covalency changes them. Near equality, a difference of a few hundred cm⁻¹ cannot reliably prove one state from the simple rule alone. Spin entropy and vibrational effects can change which state is most populated at a particular temperature. Stronger evidence comes from magnetic moment, spectra, bond lengths and a more detailed ligand-field model.

Keep units consistent. Spectroscopic splittings are often expressed as wavenumbers in cm⁻¹; P may be given in kJ mol⁻¹. Convert before comparing. One cm⁻¹ corresponds to a molar energy of N Ahc(100 m⁻¹), about 0.01196 kJ mol⁻¹. For example, 20,000 cm⁻¹ is about 239 kJ mol⁻¹. Since CFSE is an energy difference , either unit works when both inputs use the same scale.

This comparison applies only after the metal d count and geometry are known. d¹–d³ have no ordinary high/low-spin option because their electrons fit in separate t₂g orbitals; d⁸–d¹⁰ have the same basic octahedral occupancy regardless of this choice. A tetrahedral or square-planar complex has a different splitting diagram and cannot be judged by an octahedral Δₒ alone.

Step-by-step reasoning

Verify d⁴–d⁷ and octahedral geometry. Draw both electron fillings, then compute signed orbital CFSE for each. Count pairs and form low minus high. Substitute numerical Δₒ and P in common units. Interpret the sign and mention uncertainty if the difference is small compared with model limitations.

Visual explanation

Make a two-column energy ledger, high spin on the left and low spin on the right. One row records t₂g and e g occupancies, a second orbital CFSE, and a third pair count. Subtract the left column from the right before substituting numbers.

Real-world analogy

Choosing between two train routes requires both fare and transfer cost. Looking only at the cheaper fare can select the worse journey. Orbital CFSE is like fare; pairing energy is like a transfer cost that may reverse the preference.

Real-world example

A chemist comparing related Fe(II) complexes can combine a measured absorption energy with magnetic data. A large splitting and near-zero moment support low-spin d⁶, whereas a smaller splitting and roughly four unpaired electrons support high spin. Neither ligand colour nor name alone is a complete energy calculation.

Why?

Why does a simple Δₒ>P boundary emerge for several d counts? Each electron moved from e g down to t₂g saves one Δₒ and creates one extra pair in these idealised comparisons. The two changes therefore balance when their per-event costs match.

Common misconception

“Any positive CFSE number proves high spin.” Some tables state positive stabilisation magnitudes, while signed CFSE is negative. Compare actual energy differences and pair counts; the sign convention of a tabulated stabilisation magnitude is not a spin rule.

Worked example

An octahedral d⁶ ion has Δₒ=21,000 cm⁻¹ and P=17,000 cm⁻¹. Low minus high equals −2Δₒ+2P = −42,000+34,000=−8,000 cm⁻¹. The elementary model favours low spin, t₂g⁶, with zero unpaired electrons. If P instead were 24,000 cm⁻¹, the difference would be +6,000 cm⁻¹ and high spin, t₂g⁴e g², would be favoured.

Quick check

1. What does a negative E(low)−E(high) mean? Answer: Low spin is lower in the stated energy model. 2. For d⁴, how many extra pairs does the low-spin form contain? Answer: One extra pair relative to high-spin t₂g³e g¹.

Exam focus

Use one sign convention and one unit throughout. Write both configurations and their pair counts, then show the subtraction. Do not apply the Δₒ-versus-P shortcut to the wrong d range or geometry.

Advanced insight

Near a spin crossover, free energy includes a temperature term −TΔS. A high-spin state has more spin microstates, which can increase its entropy and population at higher temperature even if a zero-temperature energy comparison slightly favours low spin.

Summary

For ideal octahedral d⁴–d⁷ comparisons, low minus high includes an orbital benefit of one or two Δₒ and an added pairing cost of one or two P. Negative difference favours low spin; measured properties test the model.

Practice questions

1. For d⁷, Δₒ=14,000 cm⁻¹ and P=19,000 cm⁻¹. Predict the simple-model spin state. Answer: E(low)−E(high)=−Δₒ+P=+5,000 cm⁻¹, so high-spin t₂g⁵e g² is favoured, with three unpaired electrons. 2. Why is a d³ ion not assigned low spin merely because Δₒ>P? Answer: Its three electrons already occupy separate lower t₂g orbitals. Pairing them gives no splitting-energy gain, so the usual pair-versus-promotion decision does not arise. 3. A d⁵ ion has Δₒ=P in the one-P model. What can be concluded? Answer: The simple orbital-plus-pairing comparison gives equal energies because −2Δₒ+2P=0. Real spin state needs further energetic terms or experimental evidence.