CHEM 548: Materials Chemistry

Lecture 13: Molecular Orbital Theory, Spin States, and Jahn–Teller Distortions

Mon 10/19/2026 · Meeting 15

Molecular orbitals in transition-metal complexes, spin states and Jahn–Teller distortions

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 5 §5.3

Learning Objectives: MO Theory Part II

Understandhow molecular orbital theory applies to transition metal complexes with octahedral and tetrahedral symmetry

Identifywhich atomic orbitals on the metal and ligands participate in bonding interactions

Constructligand group orbitals (SALCs) for octahedral and tetrahedral complexes, and determine which metal orbitals they can mix with

Explainthe formation of $\sigma$-bonding, $\pi$-bonding and nonbonding molecular orbitals in metal–ligand systems

Interprethow energy matching between metal and ligand orbitals governs bonding strength and orbital participation

Describethe splitting pattern of d orbitals (t_2g/e_g in octahedral, e/t₂ in tetrahedral), and relate it to bonding interactions

Comparethe magnitude and direction of orbital splitting in octahedral vs. tetrahedral geometries

Connectmolecular orbital diagrams to crystal field theory concepts, and identify the frontier orbitals relevant to reactivity

Octahedral Complex: CrCl₆³⁻

Per Solid State Materials Chemistry, §5.3.8.

Let's take chromium hexachloride, CrCl₆³⁻, as our model system. The first step is to work out which atomic orbitals combine to form molecular orbitals. As in the hydrogen-helium example, orbitals interact most strongly when their energies are similar. The chlorine 3s orbital lies around minus 29 eV, while the 3p orbitals are at about minus 13 to minus 14 eV. Chromium falls between titanium and zinc, and interpolating between them puts its 3d orbitals near minus 13 eV, close in energy to the chlorine 3p, while its 4s and 4p orbitals are higher. That alignment lets us simplify the problem and focus on the interaction between the chlorine 3p and the chromium 3d.

[advance: count the chlorine orbitals]

We're going to treat the chlorine 3s orbitals as core orbitals and ignore them. That leaves the 3p orbitals, three on each chlorine and six chlorines, so 18 chlorine 3p orbitals go into the molecular orbital diagram.

[advance: count the chromium orbitals]

For chromium we include all nine valence orbitals: five 3d, one 4s and three 4p.

Electron Counting: CrCl₆³⁻

Cl [Ne] 3s²3p⁵6 × 5 3p e⁻ = 30 e⁻

Cr [Ar] 4s¹3d⁵6 e⁻

3− charge30 + 6 + 3 = 39 valence e⁻

CrCl₆³⁻: Bonding and Antibonding MOs

Per Solid State Materials Chemistry, §5.3.8, Fig. 5.21.

Now we take the nine valence orbitals on chromium and find the ligand SALC that matches each one by symmetry. Each row pairs a chlorine SALC with the chromium orbital of the same symmetry, and gives one bonding and one antibonding molecular orbital.

[advance: add the a₁g row]

The a₁g SALC has every chlorine p orbital pointing in phase at the metal, and it meets the chromium 4s orbital, giving a σ bonding and a σ antibonding orbital. This image is showing the atoms that are in the xy-plane of the molecule so we're taking off the chlorines that would be above and below just so we can see it better but if i were to include those all six chlorines would have the same lobe pointing in at the central chromium.

[advance: add the e_g row]

The e_g row is the chromium dx²−y² and dz² pair, doubly degenerate. Those two point straight at the chlorines, so they mix with a SALC whose p orbitals point back at the metal, and they form σ bonding and σ antibonding orbitals.

[advance: add the t₁u row]

The t₁u SALC points along an axis, with the middle orbitals pointing straight at the metal, where they meet the chromium 4p orbitals. Because the ligand orbitals point directly at the metal, this is σ overlap, so we get a σ bonding and a σ antibonding orbital, three of each.

[advance: add the t₂g row]

The t₂g row is also triply degenerate. Its chlorine SALC has the p orbitals side-on to the bonds, and it meets the chromium dxy, dxz and dyz set. In phase we get a π bonding orbital, out of phase we get a π antibonding orbital, and both carry a nodal plane through the bond axis. We have used nine chromium orbitals and nine ligand SALCs. Chlorine gave us 18 SALCs in total, so half of them are still unused.

CrCl₆³⁻: Nonbonding MOs

Two $t_{1u}$ SALC sets can mix with the Cr $4p$ orbitals; the bonding set overlaps each $4p$ both $\sigma$ and $\pi$ and dominates, leaving this second $t_{1u}$ set nonbonding. Per Solid State Materials Chemistry, §5.3.8, fn. 22.

CrCl₆³⁻: The Full MO Diagram

Per Solid State Materials Chemistry, §5.3.8, Fig. 5.22.

Here is the whole diagram. On the left are the 18 chlorine 3p orbitals.

[advance: add the chromium orbitals]

On the right are the nine orbitals that come from the chromium.

[advance: add the bonding MOs]

Down here are the nine bonding MOs.

[advance: add the nonbonding MOs]

In the middle are the nine nonbonding MOs.

[advance: add the antibonding MOs]

And up here are the nine antibonding MOs. This looks like a complicated diagram, but most of the time we only care about the frontier orbitals, and when the metal has d electrons, almost all the action is in this box. You'll recognize the t₂g and e_g splitting from crystal field theory. The t₂g set, dxy, dxz and dyz, is π antibonding with chlorine. The e_g set, dx²−y² and dz², points straight at the ligands. We'll come back to this diagram over and over through the course.

Tetrahedral Complex: CoCl₄²⁻

Per Solid State Materials Chemistry, §5.3.8.

Electron Counting: CoCl₄²⁻

Cl [Ne] 3s²3p⁵4 × 5 3p e⁻ = 20 e⁻

Co [Ar] 4s²3d⁷9 e⁻

2− charge20 + 9 + 2 = 31 valence e⁻

CH₄ Revisited

CoCl₄²⁻: Co 4s ↔ a₁, Co 4p ↔ t₂

Per Solid State Materials Chemistry, §5.3.6.

The t₂ and e Orbitals in a Tetrahedron

The ligands sit on alternating cube corners: the e pair stays farthest from them and interacts least, so it lies lowest; the t₂ set is closer and mixes more strongly.

Per Solid State Materials Chemistry, §5.3.8, Fig. 5.20.

A tetrahedron is a little trickier to picture than an octahedron. The way I find most intuitive is to draw the tetrahedron inside a cube. The green spheres are the ligands. Across the top are the t₂ orbitals, dxy, dxz and dyz. They don't point directly at the ligands, but they don't point directly between them either.

[advance: add the e orbitals]

Below them are the e orbitals, dx²−y² and dz². These two point directly at the edges of the tetrahedron. If you wanted to get as far away from the ligands as you can, the edges are where you would go.

[advance: the comparison]

So the e orbitals have the smallest interaction with the ligands, and they lie lowest. The t₂ orbitals feel more repulsion from the ligands, so they mix more with the chlorides and lie higher.

CoCl₄²⁻: The Full MO Diagram

Per Solid State Materials Chemistry, §5.3.8, Fig. 5.23.

That gives us an MO diagram that looks like this. On the left are the 12 chlorine 3p orbitals.

[advance: add the cobalt orbitals]

On the right are the same nine orbitals on the cobalt.

[advance: add the bonding MOs]

Down here are the nine bonding MOs.

[advance: add the nonbonding MOs]

Now there are only three nonbonding MOs, because there are fewer ligands.

[advance: add the antibonding MOs]

And up here are the nine antibonding orbitals. Once again, the part we usually care about is this box, the antibonding interactions between the ligands and the metal d orbitals. Compared with the octahedron, the splitting is reversed. The e set is now lower, still doubly degenerate, and the t₂ set is higher. The gap, Δ_tet, is also only about half as big. In the octahedron the d orbitals split cleanly into σ antibonding and π antibonding, but here no orbital is purely σ antibonding, so the gap is smaller.

Summary — MO Theory in Transition-Metal Complexes

	[CrCl₆]³⁻ — octahedral	[CoCl₄]²⁻ — tetrahedral
Orbitals	9 metal (5 d, 1 s, 3 p) + 18 ligand (Cl 3p)	9 metal + 12 ligand
Bonding / antibonding / nonbonding	9 / 9 / 9	9 / 9 / 3
Frontier set	t_2g (d_xy, d_yz, d_xz) = $\pi$ antibonding, lower E e_g (d_x²−y², d_z²) = $\sigma$ antibonding, higher E	e (d_x²−y², d_z²) lower E t₂ (d_xy, d_xz, d_yz) higher E — reversed
Splitting	$\Delta_o$	$\Delta_t \approx \tfrac{1}{2}\Delta_o$ (no pure $\sigma$ antibonding)

Geometry sets symmetry, and symmetry sets both the order and the magnitude of $\Delta$: t_2g below e_g in O_h, e below t₂ in T_d, with the frontier d orbitals governing color and reactivity.

Check Your Understanding: Field Splitting

For chloride ligands (weak field), compare $[\mathrm{FeCl}_6]^{4-}$ (octahedral) and $[\mathrm{FeCl}_4]^{2-}$ (tetrahedral). Determine Fe oxidation state, d-electron count, orbital splitting pattern, and number of unpaired electrons. Identify which has the larger splitting and explain why.

Each Cl⁻ = −1$[\mathrm{FeCl}_6]^{4-}$: $x - 6 = -4 \rightarrow \mathrm{Fe}^{2+} \rightarrow 3d^6$

$[\mathrm{FeCl}_4]^{2-}$$x - 4 = -2 \rightarrow \mathrm{Fe}^{2+} \rightarrow 3d^6$

Both complexes have Fe²⁺ (d⁶).

Octahedral field$t_{2g}$ (lower), $e_g$ (higher). Weak field → high spin → $t_{2g}^{4} e_g^{2}$ → 4 unpaired electrons.

Tetrahedral field$e$ (lower), $t_2$ (higher). Weak field → high spin → $e^{3} t_2^{3}$ → 4 unpaired electrons.

Splitting$\Delta_t \approx \tfrac{1}{2}\Delta_o$: octahedral ligands point straight at the metal orbitals, a stronger σ interaction.

AnswerBoth high-spin $\mathrm{Fe}^{2+}$ ($d^6$), 4 unpaired; $[\mathrm{FeCl}_6]^{4-}$ splits more.

Try this one. Compare [FeCl₆]⁴⁻, which is octahedral, with [FeCl₄]²⁻, which is tetrahedral. Chloride is a weak-field ligand. For each one, work out the iron oxidation state, the d-electron count, the splitting pattern and the number of unpaired electrons. Then say which has the larger splitting, and why.

[advance: reveal the oxidation states]

Each chloride is minus one, so in both complexes iron is 2+, and both are d⁶.

[advance: reveal the splitting patterns and the answer]

In the octahedron, t₂g is lower and e_g is higher. Chloride is weak field, so it's high spin, t₂g⁴ e_g², with four unpaired electrons. In the tetrahedron the order flips, e below t₂, and it's high spin again, e³ t₂³, also four unpaired. The octahedral ligands point straight at the metal orbitals, which is a stronger σ interaction, so Δ_t is only about half of Δ_o. [FeCl₆]⁴⁻ has the larger splitting.

Spin States & Jahn–Teller Distortions

Explain spin-state configurations

Definehigh-spin and low-spin complexes.

Relatecrystal field splitting energy ($\Delta$) and spin pairing energy ($P$).

Predictspin states from ligand type (spectrochemical series) and metal identity (3d vs 4d/5d).

Compare structural preferences of d⁸ ions

Describewhy Ni²⁺ forms octahedral (rock salt) structures while Pt²⁺ forms square planar (Cooperite-type).

Interpretorbital splitting diagrams for square planar fields.

Apply the Jahn–Teller theorem

Identifycases where electronic degeneracy causes structural distortion (e.g., high-spin d⁴, d⁹).

Distinguishbetween elongated and compressed octahedra and the electronic rationale for each.

Describe second-order Jahn–Teller distortions

Explainhow HOMO–LUMO mixing stabilizes asymmetric geometries (e.g., NH₃, PbO).

Recognize“stereoactive lone pair” distortions in s² cations (Pb²⁺, Bi³⁺).

In this lecture we are going to talk about a couple of bonding issues that do not fit cleanly anywhere else, the spin-state configurations of transition-metal ions and Jahn-Teller distortions.

By the end you should be able to define a high-spin and a low-spin complex and relate the crystal field splitting energy, Δ, to the spin pairing energy, P. You should be able to predict a spin state from the ligand, through the spectrochemical series, and from the metal, because 4d and 5d ions have a larger Δ than 3d ions. You should be able to say why nickel oxide takes the rock-salt structure, with octahedral nickel, while platinum oxide takes the cooperite structure, with square-planar platinum, and read the orbital splitting diagram for a square-planar field. You should be able to apply the Jahn-Teller theorem to say when electronic degeneracy distorts a structure, as it does for high-spin d⁴ ions and for d⁹ ions, and tell an elongated octahedron from a compressed one and give the electronic reason for each. And you should be able to describe second-order Jahn-Teller distortions, explain how mixing the HOMO with the LUMO stabilizes an asymmetric geometry in ammonia and in lead oxide, and recognize a stereoactive lone pair in an s² cation like Pb²⁺ or Bi³⁺.

We start with the two ways electrons can fill the d orbitals in an octahedral field.

Low-Spin and High-Spin Configurations

Δ = Crystal field splitting energy

Favors filling the lower energy set (t_2g) of orbitals completely before adding electrons to the higher energy set (e_g)

P = Spin pairing energy

Favors spreading the electrons out across all five d orbitals before placing two electrons in the same orbital (strong Hund’s rule coupling)

When transition-metal ions form complexes, the frontier orbitals that govern their properties are usually the metal's d orbitals. In an octahedral field those split into the lower triply degenerate t₂g set, dxy, dyz and dxz, and the higher doubly degenerate e_g set, dx²−y² and dz². Electrons fill t₂g first, one to an orbital with parallel spins, following Hund's rule. From the fourth electron on there are two options: pair it inside t₂g, or put it in one of the higher e_g orbitals. Pairing costs the spin pairing energy, P, and promoting costs the crystal field splitting energy, Δ. Δ favors filling the lower t₂g set completely before any electron goes into e_g. P favors spreading the electrons across all five d orbitals before two of them share one orbital, which is strong Hund's rule coupling.

Take a d⁶ ion and fill it both ways.

[advance: show the low-spin and high-spin configurations]

With P smaller than Δ, pairing is the cheaper move, so the electrons pair in t₂g before any goes into e_g. Three spin up and three spin down fill the three t₂g orbitals, e_g stays empty, and no electron is left unpaired. That is the low-spin configuration. With P larger than Δ, promoting is cheaper than pairing, so every one of the five d orbitals takes one electron before any orbital takes a second. Three go into t₂g, two into e_g, and only the sixth electron pairs, which it does in t₂g. That leaves four unpaired electrons, and that is the high-spin configuration.

[advance: show the spin-up and spin-down channels]

Textbooks often compare Δ and P using diagrams with two sets of five d-orbital lines, one for spin up and one for spin down. Δ is the split inside a channel, and P is the lift between the two channels. When P is smaller than Δ, the electrons pair in the lower t₂g orbitals before any goes into e_g, which is the low-spin case. When P is larger than Δ, the electrons fill all the spin-up t₂g and e_g orbitals before any spin-down pairing, which is the high-spin case. Whichever of the two is smaller is the price the electrons pay, and that is what fixes the spin state.

High spin vs. Low spin

2nd and 3rd rowTransition-metal ions adopt LS configurations: the 4d and 5d orbitals interact more strongly with the ligands, increasing $\Delta$, plus their larger size decreases $P$.

CovalencyIncreasing the covalency of the metal–ligand bonds increases $\Delta$, thereby favoring LS configurations — either by increasing the metal oxidation state or by moving from left to right across the transition-metal series.

Tetrahedral fieldsTetrahedrally coordinated ions are nearly always high spin: $\Delta_\mathrm{tetr} < \Delta_\mathrm{oct}$.

High field ligandsHigh field ligands (see spectrochemical series) favor LS configurations.

I⁻<Br⁻<S²⁻<Cl⁻<F⁻<C₂O₄²⁻<H₂O<NH₃<CN⁻<CO

weak field strong field

Whether a complex is high spin or low spin comes down to the relative size of Δ and P, and four things move them. Second-row and third-row transition-metal ions take low-spin configurations: their 4d and 5d orbitals interact more strongly with the ligands, which raises Δ, and they are larger, which lowers P. Covalency does the same thing, so raising the metal oxidation state, or moving left to right across the transition-metal series, raises Δ and favors low spin. Tetrahedrally coordinated ions are nearly always high spin, because Δ in a tetrahedral field is smaller than in an octahedral one. And high field ligands favor low spin.

[advance: add the spectrochemical series]

This is the spectrochemical series, weak field on the left and strong field on the right. The halides and sulfide sit at the weak end and give a small Δ, so those complexes are usually high spin. Water and ammonia sit in the middle. Cyanide and carbon monoxide sit at the strong end, and they have empty π* orbitals that stabilize the t₂g levels, which makes Δ large and the complex low spin.

Crystal Chemistry of Group 10 Monoxides

The rock-salt structure of nickel oxide: nickel and oxygen spheres alternating on a cubic lattice, each nickel octahedrally surrounded by six oxygens.

NiO (rock salt)

The cooperite structure of platinum oxide: each platinum sits in a square-planar group of four oxygens.

PtO (cooperite)

Why do NiO and PtO adopt different structure types?

Nickel oxide and platinum oxide have the same stoichiometry, and both metals belong to group 10, yet they crystallize in different structures. Nickel oxide takes the rock-salt structure, so there are six oxygens around each nickel, an octahedron. Platinum oxide takes the cooperite structure, where each platinum sits in a square-planar group of four oxygens. So the same stoichiometry and the same group give us six-coordinate nickel and four-coordinate platinum. Size does not explain it, because platinum is the larger of the two and size would predict the higher coordination number, not the lower one. The cause is electronic.

[advance: put the question up]

Why do nickel oxide and platinum oxide adopt different structure types? The answer is in how the d electrons fill in each geometry.

Octahedral vs. Square Planar

Start with the octahedral case, Ni²⁺ in nickel oxide, which is d⁸. Six of the eight electrons fill the lower t₂g set, and the last two go into the higher e_g pair, one in each with parallel spins. Δ is the gap between the two sets.

[advance: correlate to the square-planar splitting]

When we take away the two ligands along z, the four ligands in the xy plane move closer in to keep the bonding. With no ligands along z, dxz and dyz become less antibonding and drop. dxy rises, because its lobes lie in the xy plane, where the four ligands have moved closer. The e_g pair splits even more. dz² comes down by a lot, below dxy, and dx²−y², which points straight at the four remaining ligands, rises sharply. What we end up with is four low orbitals and one high one. A d⁸ ion fills the four low orbitals and leaves dx²−y² empty. Emptying dx²−y² is a large stabilization, but now every electron is paired, and pairing costs energy.

[advance: the nickel verdict]

Nickel uses 3d orbitals, so Δ is small, and the square-planar gain does not cover the pairing energy. So nickel stays high spin and octahedral, with its two e_g electrons unpaired, in the rock-salt structure.

[advance: the platinum verdict]

Platinum uses 5d orbitals, which interact much more strongly with the ligands, so Δ is large and the complex is low spin. Now the square-planar distortion pays, because it lowers dz² and empties dx²−y². That is why platinum oxide is cooperite. Square-planar coordination shows up this way for d⁸ ions in general, and sometimes for d⁹ ions, wherever the ligand field is strong.

That electronic reason for distorting is what the Jahn-Teller theorem describes.

Jahn–Teller Distortions

Jahn–Teller theoremAn incompletely filled set of otherwise degenerate MOs will undergo a structural distortion that removes the degeneracy and lowers the energy of the occupied orbital.

1^st order distortionFollows the letter of the JT theorem. A partially filled HOMO is needed.

2^nd order distortionInvolves mixing between filled and empty orbitals (that should be close in energy) that lowers the energy of the filled orbitals.

The Jahn-Teller theorem says that an incompletely filled set of otherwise degenerate molecular orbitals will undergo a structural distortion that removes the degeneracy and lowers the energy of the occupied orbital. The system trades symmetry away for energy, and it comes out ahead, because the orbital that goes down is occupied and the one that goes up is not.

[advance: the first-order distortion]

A first-order distortion follows the letter of the theorem, and it needs a partially filled HOMO. Only a partly filled set splits into an occupied level and an empty one. A full or empty set gains nothing, because the split sends as many electrons up as down.

[advance: the second-order distortion]

A second-order distortion involves mixing between filled and empty orbitals that should be close in energy, and that mixing lowers the energy of the filled orbitals. Nothing has to be partially filled, which is why it reaches cases the first-order theorem cannot.

First-Order Jahn–Teller Distortion

A first-order Jahn-Teller distortion turns up most often in octahedral complexes that are high-spin d⁴, like Mn³⁺, or d⁹, like Cu²⁺. The middle column is the undistorted octahedron with a high-spin d⁴ ion in it.

[advance: show the compressed octahedron]

Compressing means shortening the two bonds along z and lengthening the four in the xy plane. The orbitals that point along z become more antibonding, so dz² rises above dx²−y², and the dxz and dyz pair rises above dxy. The orbitals in the plane become less antibonding, so dx²−y² comes down, and the single e_g electron drops into it. Now the degeneracy is gone and the total energy is lower.

[advance: show the elongated octahedron]

Elongating means stretching the two bonds along z and pulling the four in the xy plane in. Everything reverses. Now dx²−y² rises above dz², dxy rises above the dxz and dyz pair, and the single e_g electron drops into dz² instead. Both distortions remove the degeneracy and lower the energy by about the same amount, so from the orbitals alone we cannot tell which distortion the complex takes. But in real structures with high-spin d⁴ ions or d⁹ ions, the octahedron is almost always elongated, and the compressed form is rare. So something we have not drawn yet is stabilizing the elongated structure.

Distortions in d⁹ and d¹⁰ Halides

Copper(II) bromide ribbons: each copper carries four short bonds to bridging bromides, with two longer axial contacts.

CuBr₂ distances
4×2.41 Å, 2×3.15 Å

Mercury(II) bromide: each Hg has two short Hg–Br bonds (2.45 Å, thick sticks) making a linear Br–Hg–Br molecule and four long Hg···Br contacts (3.24 Å, thin lines), a strongly compressed octahedron.

HgBr₂ distances
2×2.45 Å, 4×3.24 Å

Cu²⁺ (d⁹) ions almost always take the 2 long + 4 short distortion (elongated octahedron)

d¹⁰ ions, such as Hg²⁺, adopt very large 2 short + 4 long distortions (compressed octahedron).

Why is this so? Why do d¹⁰ ions distort at all?

Here are two example bromides. - CuBr₂, d⁹: four bonds at 2.41 Å and two at 3.15 Å - HgBr₂, d¹⁰: two bonds at 2.45 Å and four at 3.24 Å - Hg²⁺ coordination often looks linear rather than tetrahedral or octahedral. Shorten two bonds and lengthen four far enough and what is left looks like a linear molecule.

[advance: name the copper distortion]

Copper is d⁹, and each copper has four bonds at 2.41 Å and two much longer ones at 3.15 Å. Four short and two long is the elongated octahedron.

[advance: name the mercury distortion]

The heavy d¹⁰ ions like Hg²⁺ do the opposite, with two short bonds and four long ones, a compressed octahedron.

- 2 short + 4 long = compressed

[advance: put the question up]

So we have two things to explain: why copper elongates, and why a d¹⁰ ion like mercury distorts at all. A filled d¹⁰ shell has no partially filled degenerate orbital, so the first-order theorem says mercury should sit in an undistorted octahedron, but it does not.

Jahn–Teller Distortions: s–d_z² Mixing, d⁹

Levels drawn to scale from a two-level model with illustrative values.

The empty ns orbital can mix with the (n−1)d_z² orbital, but not with the (n−1)d_x²−y² orbital.

Splitting alone moves one e_g level up by δ and the other down by δ.

Mixing with s lowers d_z² by ε and raises the empty s orbital. ε is larger when d_z² sits closer to s: ε_c > ε_e.

d⁹ case (Cu²⁺): the s–d_z² mixing favors preferential occupation of the d_z² orbital (2 long + 4 short favored)

The answer is the empty s orbital just above the d levels, which is the 4s in copper. - Middle: regular octahedron, 3 e⁻ in dz² and dx²−y², same energy - Compress or elongate: s can mix with dz², not dx²−y² .

[advance: split e_g by δ]

Compressing or elongating splits e_g, so one level goes up by δ and the other comes down by δ. - Compress: dz² up, dx²−y² down; elongate: the other way round

- Either way 2 e⁻ in the lower level, 1 in the upper → both lower E by the same δ

- Splitting alone can't tell which way copper goes

For that we need the s orbital.

[advance: mix dz² with s]

Now we let dz² mix with s. - dz² down by ε, which is the distance from the dashed line to solid; s up by the same ; - s is empty → pushing it up costs nothing - Closer to s means more mixing, and since compressing lifts dz² toward s → mixing causes its level to go down ε_c > ε_e Now let's add it up.

[advance: add up the totals]

On the compressed side, dz² has one electron, so the total is minus δ minus ε_c. - Elongated: dz² has two → −δ − 2ε_e - Both get the δ → ε_c against 2ε_e - Compressed has the bigger push but 1 electron, while elongated has two electrons to push down, and that wins; - Copper elongates: 4 short + 2 long; Mercury, with ten d electrons, goes the other way.

Jahn–Teller Distortions: s–d_z² Mixing, d¹⁰

Levels drawn to scale from a two-level model with illustrative values.

The empty ns orbital can mix with the (n−1)d_z² orbital, but not with the (n−1)d_x²−y² orbital.

Splitting alone moves one e_g level up by δ and the other down by δ.

Mixing with s lowers d_z² by ε and raises the empty s orbital. ε is larger when d_z² sits closer to s: ε_c > ε_e.

d¹⁰ case (Hg²⁺): the s–d_z² mixing is largest when the energy separation between the two is minimized (2 short + 4 long favored)

Mercury is d¹⁰, so both of its e_g orbitals are full, with four electrons. - Empty 6s above; Let's compress and elongate it again.

[advance: split e_g by δ]

Compressing or elongating splits e_g the same way, one level up by δ and the other down by δ. But now both levels are full, so two electrons go up by δ and two come down by δ, and the total is zero either way. The splitting gains mercury nothing, so you would think that mercury would just stay an octahedron.

[advance: mix dz² with s]

But now we let dz² mix with s. Again dz² comes down by ε, from the dashed line to the solid one, and again ε is bigger on the compressed side, because compressing puts dz² closer to s. Now let's add it up.

[advance: add up the totals]

The δs cancel on both sides, so compression gives minus 2ε_c and elongation gives minus 2ε_e. Both sides have two electrons in dz², so the bigger push wins, and that's compression. So mercury compresses, with two short bonds and four long ones. This is a second-order Jahn-Teller distortion, because the filled dz² mixes with the empty 6s.

Why s Mixes with d_z² Only

No I just want to clarify why s mixes with z2 but not x2y2. Mixing means the d takes s character and the s takes on d character. Whether that is allowed depends on the symmetry of the whole complex. Start with dx²−y² and rotate the complex 90 degrees around z.

[advance: rotate s and dx²−y²]

The positive and negative lobes of dx²−y² swap places, whereas the s is the same. These orbitals behave differently under a symmetry operation so they cannot mix.

[advance: compare s and dz²]

Now look at dz². That rotation leaves it unchanged, just like s, but we also have to check every other symmetry operation of the complex.

[advance: show which symmetries the distortion removes]

In the regular octahedron, rotations can exchange the z axis with x or y. Those rotations change dz², while s stays unchanged, so mixing is forbidden. Stretching or compressing along z removes those symmetry operations. Since s and dz² behave alike under every remaining symmetry operation, mixing is allowed.

Second-Order Jahn–Teller: NH₃

A good example of a second-order Jahn-Teller distortion is ammonia. On the left is the MO diagram for ammonia in a hypothetical trigonal planar geometry. 1a₁′ is the bonding interaction between the nitrogen 2s and the hydrogen 1s orbitals, and 2a₁′ is the antibonding combination of the same orbitals. 1e′ is the bonding interaction between the nitrogen 2pₓ and 2p_y and the hydrogen 1s orbitals, and the highest MO, 2e′, is their antibonding combination. Finally there is a₂″, the nitrogen 2p_z. In planar ammonia the 2p_z cannot mix with any of the hydrogens, so it is nonbonding, and it is the HOMO.

[advance: add the pyramidal column]

But ammonia is actually trigonal pyramidal, not trigonal planar, and the second-order Jahn-Teller distortion explains why.

[advance: why the filled level falls]

Going from planar to pyramidal changes the symmetry so that the nitrogen 2p_z has a₁ symmetry, the same as the 2s. Now the HOMO, the nonbonding 2p_z, can mix with the LUMO, the antibonding nitrogen 2s. The mixing gives a filled orbital that is mostly nonbonding and an empty orbital that is more strongly antibonding. Mixing the 2p_z with the 2s puts a large lobe of electron density on the side of the pyramid away from the hydrogens, and that makes the filled orbital weakly bonding. The empty orbital goes up in energy, but it has no electrons, so there's no penalty. So a second-order Jahn-Teller distortion happens when the HOMO-LUMO gap is small and a distortion gives the two orbitals the same symmetry. They hybridize, and that stabilizes the filled HOMO.

Pb²⁺ in an Octahedral Field

Per Solid State Materials Chemistry, §5.3.10.

Another classic second-order Jahn-Teller distortion, one we see a lot in solid-state chemistry, comes from a main-group ion with electrons in its valence s orbital, like Pb²⁺. For some materials, ferroelectrics for example, it is a very important distortion.

[advance: add the bonding orbitals]

So let's take lead in an octahedron of chlorides and build its bonding MO diagram.

[advance: add the nonbonding and antibonding orbitals]

Then we add the nonbonding and antibonding levels.

[advance: name the HOMO]

The key thing to focus on is the HOMO, which is the lead 6s antibonding with the chlorides, and the LUMO, which is the lead 6p antibonding with the chlorides. If we could distort the molecule so that the s and the p could mix, the molecule would be stabilized by a second-order Jahn-Teller distortion.

The Stereoactive Lone Pair

Per Solid State Materials Chemistry, §5.3.10.

PbO: From CsCl to Litharge

CsCl-type lead oxide: an oxide cube with one lead at the centre, bonded to all eight, every Pb–O distance the same. Key: Pb, O.

CsCl structure

SOJT
distortion →

Litharge lead oxide: each lead sits off centre with four short Pb-O bonds on one side, leaving two stacked layers with a gap between them where the lone pairs sit. Key: Pb, O.

Litharge structure

The PbO structure is distorted from the high-symmetry CsCl structure by a 2^nd order Jahn–Teller distortion, leading to a stereoactive lone pair on Pb²⁺.

If you look at the crystal chemistry of lead, you find a lot of these stereoactive lone pair distortions, which are really second-order Jahn-Teller distortions. One example is PbO. You might expect it to take something like the CsCl structure, and in some ways its structure is related to CsCl.

[advance: add the distorted structure]

But the lead has shifted off center toward one square face of its cube, and that gives it four short bonds to the oxygens on that face and very long bonds to the oxygens on the top face.

[advance: the closing line]

That leaves a space in the crystal with room for the lone pair. In MO terms the lone pair is the HOMO, and it comes from the filled antibonding lead 6s mixing with the empty 6p_z. The other place we see second-order Jahn-Teller distortions a lot is in d⁰ transition metals with a relatively high valence, like Mo⁶⁺, Nb⁵⁺, Ta⁵⁺ and Ti⁴⁺. They also shift the cation out of the center of its coordination polyhedron, and they are discussed in chapter eight.

Summary: Spin States & Jahn–Teller

Spin States

Two energies$\Delta$ = crystal-field splitting; $P$ = spin-pairing energy.

Low spin ($\Delta > P$)Pair up in t_2g.

High spin ($\Delta < P$)Spread over all five d.

$\Delta$ increases4d/5d, covalent bonds, strong field (CO > NH₃ > H₂O > Cl⁻).

Tetrahedral$\approx \tfrac{1}{2}\Delta_\mathrm{oh}$ → almost always high spin.

Octahedral → Square Planar (d⁸)

Ni²⁺ (3d⁸)Weak field → octahedral, 2 unpaired e_g.

Pt²⁺ (5d⁸)Strong field → square planar (low spin).

Large $\Delta$Stabilizes d_z², empties d_x²−y².

Jahn–Teller Distortion (1st order)

TriggerPartly filled degenerate HOMO → symmetry lowering.

WhereHigh-spin d⁴, d⁹ (Mn³⁺, Cu²⁺).

OutcomeElongation (4 short + 2 long) most stable; compression rare.

2nd order / pseudo Jahn–Teller

MechanismFilled HOMO mixes with a near-empty LUMO of the same symmetry.

NH₃p_z ↔ 2s → trigonal pyramidal.

Pb²⁺, Sn²⁺Pb²⁺ 6s/6p, Sn²⁺ 5s/5p → stereoactive lone pair.

d⁰ ionsTi⁴⁺, Nb⁵⁺, Mo⁶⁺ shift off-center.

Core Idea: Electronic degeneracy and Δ vs P competition drive spin state and geometry distortions.

Check Your Understanding: Jahn–Teller

Why do $\mathrm{Cu}^{2+}$ ($d^9$) complexes typically show elongated octahedral geometry rather than compressed?

Uneven fillingIn $d^9$, the $e_g$ orbitals are unevenly filled.

Elongation winsSplitting $e_g$ gains the same δ either way. Mixing with the empty 4s lowers $d_{z^2}$ in both: elongated, $d_{z^2}$ is the lower level with 2 e⁻ (−2ε_e); compressed, it is the upper level with 1 e⁻ (−ε_c). Even though ε_c > ε_e, two electrons win, so elongation is lower in energy.

Why do Cu²⁺ complexes, which are d⁹, usually show an elongated octahedral geometry rather than a compressed one?

[advance: reveal the filling]

In d⁹ the e_g orbitals are unevenly filled, one of them holding two electrons and the other one. That is a partially filled degenerate HOMO, so the complex distorts.

[advance: reveal which distortion wins]

Either distortion splits e_g, and either way the three electrons gain the same δ from the splitting. The difference comes from the empty 4s, which mixes with dz² and lowers it in both cases. When the octahedron elongates, dz² is the lower e_g level and holds two electrons, so both of them gain from the mixing. When it compresses, dz² is the upper level and holds only one. Compression brings dz² closer to the 4s and mixes it a little more, but two electrons gaining beats one, so the net energy is lower for the elongated octahedron and Cu²⁺ elongates.

Practice questions

Work the Lecture 13 practice questions before the next class.

Open the Lecture 13 practice questions