Octahedral CFSE for d⁸ to d¹⁰ Ions

Introduction

Once six d electrons fill t₂g, additional electrons must enter e g. For octahedral d⁸ through d¹⁰, a large or small Δₒ no longer creates the familiar pair-versus-promotion choice of d⁴ through d⁷. These configurations finish the octahedral CFSE table and illustrate why a change of geometry can matter more than changing an octahedral spin label.

Core explanation

Octahedral d⁸ has t₂g⁶e g². Six t₂g electrons contribute −2.4Δₒ; two e g electrons add +1.2Δₒ. Its orbital CFSE is therefore −1.2Δₒ. Hund’s rule places the two e g electrons in separate equal-energy orbitals with parallel spin in the simple ground-state picture, leaving two unpaired electrons. Pairing both in one e g orbital would not lower orbital energy because the two e g levels are degenerate, but would introduce a pairing penalty.

For d⁹, t₂g⁶e g³ contributes −2.4Δₒ+1.8Δₒ=−0.6Δₒ. One e g orbital is paired and the other holds one electron, so one electron is unpaired. For d¹⁰, t₂g⁶e g⁴ contributes −2.4Δₒ+2.4Δₒ=0. All five d orbitals are filled and no d electron is unpaired. A zero CFSE here is expected because a completely filled set has the same occupancy-weighted total energy before and after idealised splitting.

In d⁸, d⁹ and d¹⁰ octahedral ions, “strong ligand” does not change how many electrons occupy t₂g versus e g under the basic one-electron filling rules. Stronger splitting scales the CFSE magnitude for d⁸ and d⁹, but does not yield a separate low-spin octahedral d⁸ form analogous to low-spin d⁶. A complex can of course adopt a different geometry with new orbital ordering. Four-coordinate Ni²⁺ is d⁸ and can form tetrahedral or square-planar complexes; square-planar splitting may produce a diamagnetic configuration, unlike the simple octahedral d⁸ diagram. Do not transplant the octahedral electron count into another geometry without redrawing its levels.

Octahedral d⁹ is especially associated with Jahn–Teller distortion. An uneven occupancy of degenerate e g orbitals can make an ideal octahedron unstable toward elongation or compression, removing the exact degeneracy. A real Cu²⁺ aqua complex may therefore have unequal metal–ligand distances. This does not change the formal d⁹ count, but it means the ideal −0.6Δₒ formula is an approximation to a more detailed energy pattern.

The d¹⁰ case also clarifies colour arguments. Many d¹⁰ ions lack ordinary d–d transitions because all d levels are occupied, but a d¹⁰ compound can still be coloured through charge-transfer or other electronic transitions. “d¹⁰ means colourless” is a useful starting tendency, not a theorem.

Step-by-step reasoning

Place six electrons into t₂g, making three pairs. Put the remaining two, three or four electrons into two e g orbitals using Hund filling. Calculate (−0.4n t+0.6n e)Δₒ, then count unpaired arrows. Check geometry before applying this octahedral procedure.

Visual explanation

Draw three filled lower boxes and two upper boxes. For d⁸ place one arrow in each upper box; for d⁹ pair one of those; for d¹⁰ pair both. The lower row stays identical in all three sketches, making the CFSE difference easy to see.

Real-world analogy

The cheaper three rooms each hold two people and are now full. Every new arrival must use the two more expensive rooms. Different prices change the total bill, but no amount of price difference creates extra capacity in the full rooms.

Real-world example

Octahedral Ni²⁺ is d⁸ and commonly has two unpaired electrons. A four-coordinate nickel(II) complex can instead be square planar and diamagnetic if its ligand field supports that geometry. The oxidation state and d count are the same; geometry reorganises orbital energies.

Why?

Why does d¹⁰ have zero orbital CFSE despite a nonzero gap? Each of the three lower orbitals and two upper orbitals holds two electrons. The total is 6(−0.4)+4(+0.6)=0, so downward and upward shifts exactly cancel.

Common misconception

“All d⁸ complexes are paramagnetic because octahedral d⁸ has two unpaired electrons.” Square-planar d⁸ can have a different level ordering and be diamagnetic. State the geometry before predicting magnetism.

Worked example

An ideal octahedral Cu²⁺ ion is d⁹. Write t₂g⁶e g³; CFSE=6(−0.4Δₒ)+3(+0.6Δₒ)=−0.6Δₒ. One e g electron remains unpaired. If Δₒ=16,000 cm⁻¹, the idealised orbital CFSE is −9,600 cm⁻¹. A Jahn–Teller-distorted real complex requires a refined splitting pattern.

Quick check

1. What are the octahedral d⁸ and d¹⁰ CFSE values? Answer: d⁸ gives −1.2Δₒ; d¹⁰ gives zero. 2. How many unpaired d electrons does ideal octahedral d⁹ have? Answer: One electron remains unpaired in the e g set.

Exam focus

Do not invent low-spin octahedral d⁹ by forcing electrons beyond the two-electron capacity of an orbital. Distinguish an octahedral spin assignment from a comparison between octahedral, tetrahedral and square-planar geometries.

Advanced insight

Jahn–Teller distortion in d⁹ changes the degeneracy of the e g set and can create a sizable structural response. This is one route from an ideal textbook splitting diagram to observed elongated Cu(II) coordination environments.

Summary

Ideal octahedral d⁸, d⁹ and d¹⁰ have t₂g⁶e g², t₂g⁶e g³ and t₂g⁶e g⁴. Their orbital CFSE values are −1.2, −0.6 and 0Δₒ, with two, one and zero unpaired electrons in the simple diagram.

Practice questions

1. Calculate the signed orbital CFSE for octahedral d⁹. Answer: t₂g⁶e g³ gives −2.4Δₒ+1.8Δₒ=−0.6Δₒ. 2. Why does increasing Δₒ not create a low-spin octahedral d⁸ arrangement in the usual diagram? Answer: The lower t₂g set is already full at six electrons. The remaining two must occupy e g, and placing them in separate degenerate e g orbitals avoids an unnecessary pair. 3. A d⁸ metal complex is diamagnetic. Must its oxidation state be wrong? Answer: No. The complex may be square planar, where orbital energies differ from octahedral splitting and the eight electrons can be paired.