Octahedral CFSE for d⁴ to d⁷: High-Spin Cases

Introduction

The fourth d electron creates the first genuine octahedral spin choice. It can pair with a t₂g electron or occupy an empty e g orbital. When Δₒ is small compared with the effective pairing cost P, the higher orbital is cheaper than a new pair. This weak-field choice produces the high-spin configurations from d⁴ through d⁷.

Core explanation

Always begin with three parallel-spin electrons in the three lower t₂g orbitals. For high-spin d⁴, place the fourth in one e g orbital: t₂g³e g¹. Its orbital CFSE is 3(−0.4)+1(+0.6)=−0.6Δₒ, with four unpaired electrons and no pairs. For high-spin d⁵, the fifth fills the other e g orbital: t₂g³e g². The lower and upper contributions cancel, giving orbital CFSE zero, five unpaired electrons and no pairs.

The sixth electron has no empty d orbital left, so pairing becomes unavoidable. High-spin d⁶ is t₂g⁴e g²: one t₂g orbital is doubly occupied, four electrons are unpaired, and CFSE is 4(−0.4)+2(+0.6)=−0.4Δₒ. The seventh electron makes a second pair in t₂g, giving t₂g⁵e g², three unpaired electrons and CFSE 5(−0.4)+2(+0.6)=−0.8Δₒ. Thus the high-spin sequence of signed orbital CFSE values for d⁴, d⁵, d⁶ and d⁷ is −0.6, 0, −0.4 and −0.8 times Δₒ.

“High spin” does not mean that every electron stays unpaired indefinitely. It means as many as reasonably possible remain unpaired under the specified split and pairing competition. After five d electrons, any further electron must join an occupied orbital. It also does not mean CFSE is always zero or that the complex has no colour. A d⁵ high-spin ion has zero orbital CFSE relative to the barycentre, yet its orbitals are separated by Δₒ and its spectroscopy can still be studied; spin-forbidden transitions may be weak.

For comparisons with low-spin configurations, a consistent energy ledger is essential. The high-spin d⁶ diagram already contains one pair and high-spin d⁷ contains two. If you model pairing cost explicitly, either add P for every pair in both candidate diagrams or add only the difference in pair counts when subtracting their energies. The orbital CFSE values alone do not prove the high-spin state; the assumption Δₒ<P motivates the filling.

Water and fluoride often produce high-spin first-row complexes when an appropriate d⁴–d⁷ ion is present, whereas strong-field ligands such as cyanide more often favour low spin. Metal oxidation state also changes Δₒ, so ligand identity is not the only determinant. Experimental magnetic moments can help confirm the actual unpaired count.

Step-by-step reasoning

Draw three t₂g and two e g boxes. Fill one parallel arrow per box before making a pair, then place sixth and seventh electrons into t₂g as paired arrows. Record both occupancies, use −0.4Δₒ and +0.6Δₒ for CFSE, and count only arrows lacking opposite-spin partners as unpaired.

Visual explanation

Picture five boxes in two rows, with three lower and two upper. Label successive snapshots d⁴ through d⁷. The first five arrows each occupy a different box. Arrows six and seven point downward beside existing arrows in two lower boxes, making one then two pairs.

Real-world analogy

Five workers prefer separate desks even if two desks are on a higher floor, because sharing a desk is inconvenient. Once all five desks are occupied, new workers must share. The extra-floor cost resembles Δₒ and desk-sharing inconvenience resembles pairing cost.

Real-world example

The aqueous hexaaquamanganese(II) ion contains Mn²⁺, a d⁵ centre. With water in an octahedral weak-field setting it is high spin, has five unpaired electrons and zero orbital CFSE in this simple calculation. Its existence shows why zero CFSE is not zero bond stability.

Why?

Why does high-spin d⁶ return to a negative CFSE after d⁵ gave zero? The sixth electron must pair somewhere, and the lower t₂g level is cheaper than e g. Its −0.4Δₒ contribution makes the sum negative again.

Common misconception

“High-spin d⁷ has seven unpaired electrons.” There are only five d orbitals. Its configuration t₂g⁵e g² has two pairs and three unpaired electrons.

Worked example

For high-spin octahedral Fe²⁺, the formal count is d⁶. Fill t₂g⁴e g². CFSE=4(−0.4Δₒ)+2(+0.6Δₒ)=−0.4Δₒ. The t₂g fourth electron makes one pair; the other four electrons are unpaired. Its spin-only moment estimate is √[4(4+2)]=√24≈4.90 BM, before orbital contributions.

Quick check

1. What is the high-spin d⁵ orbital CFSE? Answer: Zero, because three t₂g electrons contribute −1.2Δₒ and two e g electrons contribute +1.2Δₒ. 2. How many pairs are in high-spin d⁷? Answer: Two pairs, both in t₂g orbitals.

Exam focus

Memorise the filling logic, not a bare table. Show each level occupancy and calculate CFSE so that an accidental d-count or pair-count mistake is visible.

Advanced insight

High-spin d⁵ is especially symmetric in the elementary orbital picture: one electron occupies each d orbital with parallel spins. Its zero orbital CFSE and large spin moment help make it a useful reference for trends, although real spectral term energies require electron-repulsion and selection-rule analysis.

Summary

High-spin octahedral d⁴–d⁷ fill all five d orbitals singly before further pairing. Their CFSE sequence is −0.6, 0, −0.4 and −0.8Δₒ; their unpaired counts are four, five, four and three.

Practice questions

1. Give configuration, CFSE and unpaired count for high-spin d⁴. Answer: t₂g³e g¹, CFSE −0.6Δₒ, and four unpaired electrons. There is no pair yet. 2. Give the same quantities for high-spin d⁷. Answer: t₂g⁵e g², CFSE −0.8Δₒ, with three unpaired electrons and two electron pairs. 3. Why does pairing begin at the sixth electron even in a weak field? Answer: The five d orbitals can each hold one electron without pairing; after that all are occupied, so a sixth electron must share an orbital. It occupies a lower t₂g orbital to minimise orbital energy.