CHEM 548: Materials Chemistry

Lecture 14: Spin States, Jahn-Teller Distortions, and Bond Valence

Wed 10/21/2026 · Meeting 16

Crystal-field spin states, Jahn-Teller distortions, and the bond valence method

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 5 §§5.3.9–5.3.10, §5.4

Learning Objectives: Spin States & Jahn–Teller

Spin States & Jahn–Teller Distortions

Explain spin-state configurations
- Define high-spin and low-spin complexes.
- Relate crystal field splitting energy (Δ) and spin pairing energy (P).
- Predict spin states from ligand type (spectrochemical series) and metal identity (3d vs 4d/5d).
Compare structural preferences of d⁸ ions
- Describe why Ni²⁺ forms octahedral (rock salt) structures while Pt²⁺ forms square planar (Cooperite-type).
- Interpret orbital splitting diagrams for square planar fields.
Apply the Jahn–Teller theorem
- Identify cases where electronic degeneracy causes structural distortion (e.g., high-spin d⁴, d⁹).
- Distinguish between elongated and compressed octahedra and the electronic rationale for each.
Describe second-order Jahn–Teller distortions
- Explain how HOMO–LUMO mixing stabilizes asymmetric geometries (e.g., NH₃, PbO).
- Recognize “stereoactive lone pair” distortions in s² cations (Pb²⁺, Bi³⁺).

Low-Spin and High-Spin Configurations

Low-spin and high-spin d-electron configurations compared, then the textbook two-channel spin-up and spin-down diagrams.

low-spin configuration (P < Δ) · high-spin configuration (P > Δ)

Δ = Crystal field splitting energy

Favors filling lower energy set (t_2g) of orbitals completely before adding electrons to 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 Hunds rule coupling)

When transition metal ions form complexes, the frontier orbitals governing their properties are usually the metal’s d orbitals.

Octahedral field: d orbitals split into lower-energy triply degenerate t₂g (dxy, dyz, dxz) and higher-energy doubly degenerate e_g (dx²–y², dz²)
Electrons fill t₂g first, each singly with parallel spins according to Hund’s rule
Fourth electron onward, two options: pair within t₂g or occupy the higher e_g
Pairing costs spin-pairing energy (P); promoting costs crystal field splitting energy (Δ)
P < Δ: pair in t₂g before e_g → low-spin; P > Δ: occupy all five d orbitals singly before pairing → high-spin

[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 electrons, one for spin-down
P < Δ: pair in the lower-energy t₂g before e_g — low-spin case; P > Δ: fill all spin-up t₂g and e_g before any spin-down pairing — high-spin case

Next, let's look at the trends that control whether delta or P wins.

High Spin vs. Low Spin: The Trends

High spin vs. Low spin

2^nd and 3^rd row transition-metal ions adopt LS configurations
- The 4d and 5d orbitals interact more strongly with the ligands, increasing Δ, plus their larger size decreases P.
Increasing the covalency of the metal–ligand bonds increases Δ thereby favoring LS configurations
- Either by increasing the metal oxidation state or by moving from left to right across the transition-metal series
Tetrahedrally coordinated ions are nearly always high spin
- Δ_tetr < Δ_oct
High field ligands (see spectrochemical series) favor LS configurations

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

weak fieldstrong field

Whether a complex adopts a high- or low-spin configuration depends on the relative magnitudes of crystal field splitting (Δ) and spin-pairing energy (P).

4d and 5d ions: larger d orbitals, more overlap with ligands — larger Δ and smaller P — these ions are always low spin
Increasing metal–ligand covalency (shorter bond distance, higher metal electronegativity) also increases Δ, favoring low-spin states
Tetrahedral fields: Δ smaller, about half that of octahedral — almost always high spin
Ligand identity: in the spectrochemical series, weak-field ligands like halides give small Δ (high spin)
Strong-field ligands such as CO or CN⁻, with empty π* orbitals stabilizing t₂g levels, give large Δ (low spin)

Now let's watch these rules play out in two group-10 oxides, nickel oxide and platinum oxide.

Crystal Chemistry of Group 10 Monoxides

NiO in the rock-salt structure next to PtO in the cooperite structure, asking why the two group-10 monoxides adopt different structure types.

Octahedral vs. Square Planar

Octahedral d8 splitting diagram for NiO, then the correlation to the square-planar splitting that stabilizes PtO.

To understand this, examine how the 3d orbitals fill.

Octahedral field — such as Ni²⁺ in nickel oxide: eight 3d electrons
These occupy the lower-energy t₂g orbitals first, then the higher-energy e_g, as shown
Now, consider how this filling changes in a square-planar field

[advance: show the square-planar splitting]

Remove the z-axis ligands: the x–y ligands move closer to maintain bonding — d_xz, d_yz drop, becoming less antibonding; d_xy rises, its lobes in the ligand plane
The e_g set splits further — d_z² lowers significantly, even below d_xy; d_x²–y², pointing directly at ligands, rises sharply
Result: one high-energy antibonding d_x²–y² orbital and four lower-energy orbitals
A d⁸ configuration stabilizes the square-planar field by filling the low orbitals and leaving d_x²–y² empty — though all electrons are now paired, increasing repulsion
Ni²⁺ (3d⁸): small splitting favors octahedral to avoid pairing energy; Pt²⁺ (5d⁸): strong metal–ligand interactions produce large splitting, stabilizing the square-planar distortion that lowers d_z² and depopulates d_x²–y²
Square-planar coordination thus commonly appears for d⁸ and sometimes d⁹ ions with strong ligand fields

That electronic driving force for distortion has a general name: the Jahn-Teller theorem.

Jahn–Teller Distortions

The Jahn-Teller theorem with definitions of first-order and second-order distortions.

First-Order Jahn–Teller Distortion

A first-order Jahn–Teller distortion commonly occurs in octahedral complexes with high-spin d⁴ (e.g., Mn³⁺) or d⁹ (e.g., Cu²⁺) configurations.

The e_g orbitals are unevenly occupied — such as one electron shared between dz² and dx²–y² — creating a degenerate partially filled HOMO
This degeneracy drives the system to distort and lower its symmetry to stabilize one orbital and reduce total energy

[advance: show the compressed octahedron]

Distortion removes the degeneracy, as predicted — though the theorem doesn’t specify how distortion occurs; two main distortions are possible
Compressed octahedron: z-axis bonds shorten, xy-plane bonds lengthen — dz² rises, dx²–y² lowers
Both are σ-antibonding, but compression reduces the antibonding character of dx²–y², stabilizing it and lowering the overall energy

[advance: show the elongated octahedron]

We could also elongate the octahedron — stretch the bonds along the z-axis, compress those in the xy-plane — the opposite distortion of the compressed case
Both lower the energy by a similar amount, so from the orbital diagram alone it’s unclear which will occur
In practice, high-spin d⁴ or d⁹ complexes almost always adopt the elongated octahedral geometry; the compressed form is rare
The question: what stabilizes this elongated structure

The real copper and mercury halides tell us which distortion nature picks.

Distortions in d⁹ and d¹⁰ Halides

A CuBr2 chain fragment with four short and two long bonds beside HgBr2 molecules with two short and four long.

CuBr₂ distances
4×2.41 Å, 2×3.15 Å 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?

Jahn–Teller Distortions: s–d_z² Mixing

Molecular-orbital diagram for s and d z squared mixing in distorted octahedra, then the d9 copper case, then the d10 mercury case.

To explain this, we need to include the nearby empty s orbital—4s for Cu²⁺ or 6s for Hg²⁺.

Undistorted octahedron: the metal’s e₉ orbitals (dz², dx²–y²) are degenerate with E_g symmetry; the s orbital has A₁g symmetry
Symmetries differ — no mixing or hybridization occurs
Distortion (elongate or compress) lowers the symmetry from O_h to D₄h (Hermann–Mauguin 4/mmm) — previously forbidden mixing between orbitals of compatible symmetry becomes allowed, changing their relative energies

[advance: show the d⁹ case]

With lowered symmetry, dz² gains A₁g symmetry, matching the s orbital — same-symmetry orbitals can now mix (hybridize), just like σ(s) and σ(p) mixing in O₂
The interaction raises the s orbital and lowers dz² in energy
d⁹: elongation lowers dz² enough that it becomes doubly occupied while dx²–y² remains singly occupied — greater stabilization than compression would give
So s–dz² hybridization under elongation explains the preferred distortion in d⁹ systems

[advance: show the d¹⁰ case]

d¹⁰ (e.g., Hg²⁺): both dz² and dx²–y² are doubly occupied — favorable to raise dz² closer to the s orbital, increasing s–dz² hybridization
Greater hybridization strengthens orbital mixing and stabilizes dz² overall
The opposite distortion would widen the dz²–s gap, reducing hybridization and providing less stabilization
Thus compression increases s–dz² overlap and lowers total energy

The cleanest example of this second-order mixing is ammonia.

Second-Order Jahn–Teller: NH₃

Molecular-orbital correlation diagram for trigonal planar versus trigonal pyramidal ammonia.

A second-order Jahn–Teller distortion is exemplified by ammonia (NH₃).

In a hypothetical trigonal planar NH₃, the molecular orbitals are:
1A₁′ — bonding between N(2s) and H(1s); 2A₁′ — antibonding counterpart
1E′ — bonding between N(2pₓ, 2p_y) and H(1s); 2E′ — antibonding counterpart
A₂″ — nonbonding N(2p_z) orbital, which cannot mix with H orbitals and serves as the HOMO
This configuration sets up conditions for a distortion that mixes the filled nonbonding orbital with nearby empty orbitals

[advance: bring up the full molecular-orbital diagram]

Ammonia is not trigonal planar but trigonal pyramidal due to a second-order Jahn–Teller distortion
Planar → pyramidal: p_z and s both gain A₁ symmetry, allowing mixing between the HOMO (nonbonding p_z) and LUMO (antibonding 2s)
Hybridization creates a filled orbital that is mostly nonbonding but slightly bonding, and an empty orbital that is strongly antibonding
The filled orbital gains stability: electron density shifts toward the pyramid apex, forming a weakly bonding region
The antibonding orbital rises but is unoccupied, so there’s no penalty
Occurs when the HOMO–LUMO gap is small and symmetry lowering enables mixing, thereby stabilizing the filled orbital

The same second-order mixing shows up in the solid state with lead.

Pb²⁺ in an Octahedral Field

Molecular-orbital diagram of the PbCl6 4- octahedron built from lead 6s and 6p orbitals and chlorine 3p SALCs.

The Stereoactive Lone Pair

Energy diagram comparing the octahedral m3m geometry with distortions along the fourfold 4mm and threefold 3m axes that create the stereoactive lone pair.

Lowering the octahedral symmetry allows mixing between Pb 6s and 6p orbitals.

If Pb²⁺ shifts along the fourfold axis — one short and one long Pb–Cl bond — the symmetry drops to D₄h (4mm)
This distortion stabilizes the antibonding Pb 6s–Cl interaction and destabilizes the mainly p_z–Cl interaction
Result: an occupied orbital that is almost nonbonding and localized toward one side of the octahedron
This is the stereoactive lone-pair distortion — analogous to how VSEPR theory explains the pyramidal shape of NH₃ through a filled sp³ hybrid orbital
The molecular orbital picture provides a deeper, quantitative version of that concept

In lead oxide itself, that off-center shift reshapes the entire structure.

PbO: From CsCl to Litharge

PbO distorting from the CsCl structure to the litharge structure by a second-order Jahn-Teller distortion.

Summary: Spin States & Jahn–Teller

Spin States & Jahn–Teller Distortions

Spin States

Δ = crystal-field splitting; P = spin-pairing energy.
Low spin (Δ > P): electrons pair in T₂g.
High spin (Δ < P): electrons spread across all d orbitals.
Δ↑ for 4d/5d metals, short/covalent bonds, strong-field ligands (CO > NH₃ > H₂O > Cl⁻).
Tetrahedral fields ≈ ½ Δ_oh → almost always high spin.

Octahedral → Square Planar (d⁸)

Ni²⁺ (3d⁸): weak field → octahedral (high spin).
Pt²⁺ (5d⁸): strong field → square planar (low spin).
Large Δ stabilizes dz², empties dx²–y².

Octahedral → Square Planar (d⁸)

Ni²⁺ (3d⁸): weak field → octahedral (high spin).
Pt²⁺ (5d⁸): strong field → square planar (low spin).
Large Δ stabilizes dz², empties dx²–y².

Jahn–Teller Distortion (1st Order)

Partially filled degenerate HOMO → symmetry lowering.
Seen in high-spin d⁴ (Mn³⁺) and d⁹ (Cu²⁺).
Elongated octahedra (4 short + 2 long bonds) most stable.
Compression rare.

2nd Order / Pseudo Jahn–Teller

Mixing between filled HOMO and nearby empty LUMO of same symmetry.
Examples:
- NH₃ – pz ↔ 2s mix → trigonal pyramidal.
- Pb²⁺, Sn²⁺ – 6s/6p mix → off-center “stereoactive lone pair.”
- d⁰ ions (Ti⁴⁺, Nb⁵⁺, Mo⁶⁺) – cation shifts off-center.

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

Okay, let’s summarize what we’ve seen.

Spin states: a balance between the crystal-field splitting, Δ, and the spin-pairing energy, P — large Δ pays to pair in the lower t₂g (low-spin); small Δ makes it cheaper to spread across all five d orbitals (high-spin)
Δ gets larger for 4d and 5d metals, for short and more covalent bonds, and for strong-field ligands like CO and CN⁻; tetrahedral splitting is only about half as big — almost always high spin
d⁸ metals: nickel has a small Δ and stays octahedral; platinum has a large Δ and distorts to square planar, pushing dx²–y² way up and stabilizing the rest
Jahn–Teller effect: a partially filled degenerate orbital — like high-spin d⁴ or d⁹ — lowers its energy by distorting, usually by elongating the octahedron, removing the degeneracy and stabilizing one orbital
Second-order Jahn–Teller: mixing between a filled and an empty orbital of the same symmetry — why ammonia isn’t flat, why Pb²⁺ and Sn²⁺ shift off-center to make a stereoactive lone pair, and the same in d⁰ oxides like Ti⁴⁺ or Nb⁵⁺
One principle: the system always finds a way to lower its energy by changing how the d orbitals are occupied or mixed

Before we leave spin states, try the first check-yourself question.

Check Your Understanding: Spin States

Question 1 asking how the spin state of an Fe2+ d6 ion differs between the hexaaqua and hexacyanide complexes, then its answer.

Check Your Understanding: Jahn–Teller

Question 2 asking why Cu2+ d9 complexes show elongated rather than compressed octahedral geometry, then its answer.

Learning Objectives: Bond Valence Method

Learning Objectives

Define bond valence and compute from distances: $s = \exp[(R_0 − R)/B]$ with $B = 0.37$; use tabulated $R_0$
Enforce the valence-sum rule $\sum s_i \approx$ oxidation state to assign oxidation states and validate structures
Convert valence↔distance to predict/compare lengths: $R = R_0 − B \ln s$ (e.g., SrTiO₃, Cr⁶⁺O₄)
Apply to distorted/mixed coordination and locate light atoms via valence deficits

Electrostatic Bond Valence

The Bond Valence Method builds on Linus Pauling’s early work on ionic bonding.

In his 1929 papers, Pauling identified several key balances in ionic compounds
Electroneutrality balance: the total positive charge of the cations must equal the total negative charge of the anions
Connectivity balance: every bond from a cation must terminate at an anion — knowing the stoichiometry and coordination number of one species fixes that of the other
Bond-valence balance, the focus here: the strength of each cation–anion bond must be consistent in both directions
The bond valence s equals the cation charge divided by its coordination number, and this value must also equal the anion charge divided by its coordination number

[advance: highlight the lithium oxide bond graph]

Li₂O, antifluorite structure: oxide ions form a cubic close-packed array, lithium occupies tetrahedral holes
Lithium (+1) has four bonds → s = +1/4 per Li–O bond; oxygen (–2) is coordinated by eight Li → s = –2/8 = –1/4 — the equal magnitudes confirm the bond-valence balance
So when we draw our bond graph here, what we can say is that each of these lines represents a bond with a valence of 1 quarter

[advance: clear the highlight]

SrTiO₃, cubic perovskite: each Ti⁴⁺ is surrounded by six oxygens → each Ti–O bond has a valence of 4/6 = 2/3
Sr²⁺ is coordinated by twelve oxygens → each Sr–O bond 2/12 = 1/6
Balance at oxygen: each O²– must receive two strong bonds from Ti (2 × 2/3 = 4/3) and four weaker bonds from Sr (4 × 1/6 = 2/3) — the total, 4/3 + 2/3 = 2, matches oxygen’s oxidation state
What we’re going to do in this lecture is show you how we can become quantitative in our use of these valences

Pauling returned to this idea in 1947, this time with organic molecules.

Bond Order in Organic Molecules

Pauling's 1947 paper relating carbon-carbon bond order to bond length in butane, butene, and butyne.

Bond Order and Bond Distance

Pauling's logarithmic relation rewritten to connect bond length and bond order in both directions.

Bond Valences, Quantitative

Exponential bond valence versus bond length curve with the v ij equals exp of R0 minus d over B expression.

$v_{ij} = \exp\!\left[(R^{0}_{ij} - d_{ij})/B\right]$, $B = 0.37$ Å. Per Solid State Materials Chemistry, §5.4.

To connect this to inorganic compounds like strontium titanate or lithium oxide, we first recognize that bond valence—analogous to bond order—can take non-integer values.

Following Pauling’s idea, we adopt a similar functional form but use an exponential relationship instead of a logarithmic one
The bond valence between atoms i and j: s_ij = exp((R₀ − R_ij)/b)
R_ij is the observed bond length, R₀ the length of a unit-valence (single) bond, b an empirical constant, typically 0.37 Å
Bond valence decreases exponentially as bond length increases — mirroring Pauling’s covalent trend, generalized for ionic and mixed compounds
Ok, how are we going to use this?

The way we use it is through the valence sum rule.

The Valence Sum Rule

Valence sum rule

$v_i = \sum_j v_{ij}$

The valence sum $v_i$ of an atom is equal to the sum of bond valences $v_{ij}$ around it

The valence sum of each atom, $v_i$, should be equal to the oxidation state of the atom

Bond graph of SrTiO3 marking the Sr-O and Ti-O bond valences, beside the perovskite unit cell.

Sr = 12(0.167) = 2Ti = 6(0.667) = 4O = 2(0.667)+4(0.167) = 2

Fractional Bond Valences

Bond valences in SrTiO3 computed from measured distances: the strontium calculation, then the titanium calculation, then the valence sums.

For strontium in SrTiO₃, the measured Sr–O bond length is 2.761 Å.

Using the bond valence expression, we need the empirical R₀ parameter specific to Sr–O bonds, analogous to looking up an ionic radius. From tabulated data, R₀ = 2.118 Å and b = 0.37 Å.

Substituting: s_ij = exp((2.118 − 2.761)/0.37) = 0.176.

This calculated bond valence (0.176) is slightly higher but close to the idealized value of 1/6 (≈0.167), confirming consistency between measured and theoretical bond strengths.

[advance: add the titanium calculation]

Now the same calculation for titanium: the Ti–O bond length is 1.953 Å, and with R₀ = 1.813 Å we get a bond valence of 0.685.

[advance: show the bond valence sums]

Using the valence-sum rule, we add up the total bond valences around each atom.

For strontium, 12 Sr–O bonds each contribute 0.176, giving 2.11, close to the expected +2. For titanium, 6 Ti–O bonds yield 4.11, near the ideal +4. For oxygen, 4 Sr–O bonds and 2 Ti–O bonds sum to 2.07, matching its –2 oxidation state.

These small deviations confirm that the measured bond lengths and derived bond valences are consistent with the ideal oxidation states, validating the structure as chemically reasonable. And if you didn’t know if you’re structural measurements were correct on a compound, this would be one way to check.

Now for a case where the oxidation states are genuinely uncertain: ilmenite.

Bond Valences in FeTiO₃

Ilmenite FeTiO3 structure of face-sharing octahedra with four distinct bond distances, asking whether it is Fe2+Ti4+ or Fe3+Ti3+.

Bond Valence Sums in FeTiO₃

Bond valence sums for FeTiO3: the iron calculation giving 2.03, then the titanium calculation giving 3.99.

So for iron, we have three iron oxygen bonds, which are 2.078 angstroms in length.

They have a valence of 0.395 valence units.

We have three Feobons that are 2.201 angstroms in length. They have a valence of 0.283 valence units.

Adding these up, three of each bond, we come up with a bond valence sum of 2.03.

So this is suggesting that in this compound we have iron 2 plus.

So by inference, the titanium should be 4 plus.

[advance: add the titanium sums]

Let's look at its bond valence sum to see if that works out.

We have three titanium oxygen bonds that are quite long, 2.089 angstroms. Those have a valence of 0.477.

Then we have three bonds that are much shorter, 1.874 angstroms. Those have a valence of 0.853 valence units.

Like iron, the titanium is 6-coordinate, 3 of each kind of bond. Adding up all of those 6 bonds gives a valence sum of 3.99. Very close to 4, as expected.

Now let's collect the iron and titanium sums and settle the assignment.

FeTiO₃: The Verdict

FeTiO3 verdict: valence sums of 2.03 for iron, 3.99 for titanium, and about 2 for oxygen confirm Fe2+Ti4+O3.

Using Bond Valences to Find H Atoms

Diaspore AlOOH bond valence sums identifying the oxygen site that carries the hydrogen atom.

Predicting Bond Lengths

Inverted bond valence equation predicting distances, then the SrTiO3 worked example giving 2.78 and 1.965 angstroms.

We can also use bond valences predictively in the following sense.

If we were to go back to our earlier equation that related the valence of a bond to its distance, and we did some algebraic manipulation of that equation to solve for the distance, we would get this equation.

Armed with this equation, if we have a valence that we want to see, we can predict what the distance should be.

[advance: work the strontium titanate example]

Let’s return to strontium titanate (SrTiO₃) and its cubic perovskite structure.

From the idealized bond valences in our bond graph, we can now calculate the corresponding bond distances using R = R₀ − b ln(s).

For Sr–O, with R₀ = 2.118 Å, b = 0.37 Å, and s = 0.167, we get R = 2.78 Å.

For Ti–O, substituting the appropriate R₀ value and s = 0.667, we find R = 1.965 Å.

These predicted bond lengths define the idealized SrTiO₃ structure.

Because the cube edge is the only variable in the geometry, the Sr–O bond must be √2 times longer than the Ti–O bond, a relationship that holds almost exactly.

Thus, SrTiO₃ is nearly a perfect cubic perovskite.

Now it's your turn, with chromium six oxide.

Bond Valences Predictively: CrO₃

The CrO3 chain structure drawn as corner-connected tetrahedra.

Bond Valences Predictively

The structure of CrO₃ consists of infinite chains of corner connected tetrahedra.

(a) Use either a bond graph or a Niggli formula to determine the number of bonds to each chemically distinct atom.

(b) Use the bond valence balance to assign ideal valences to each type of bond.

(c) Given the Cr(VI)–O bond valence parameters R₀ = 1.79 Å and b = 0.37 Å estimate the Cr–O bond lengths in this compound.

CrO₃: The Worked Answer

CrO3 worked answer: the bond graph with terminal valence 2 and bridging valence 1, then predicted versus observed bond lengths.

Let’s start with the bond graph.

We have one chromium and three oxygens, and chromium must make four bonds.

The only way to achieve this is with two terminal oxygens, each bonded to one chromium, and one bridging oxygen shared between two tetrahedra. So we have two one-coordinate (terminal) oxygens and one two-coordinate (bridging) oxygen.

To find the bond valences, it’s easiest to sum around the oxygens.

The terminal oxygens each form one bond and must reach a total valence of 2, so each Cr–O bond = 2. The bridging oxygen has two bonds and must also reach a total valence of 2, so each Cr–O bond = 1.

Checking chromium: two bonds of valence 2 and two of valence 1 sum to 6, matching chromium’s +6 oxidation state.

[advance: compare predicted and observed distances]

If we use the bond valences that we're given on the last slide we can predict that the valence of the bond to the terminal oxygen will be two when the length is 1.54 angstroms and the valence of the bond between the chromium and the bridging oxygen will be one when its length is 1.79 angstroms.

In the experimental structure we see bond lengths of 1.57 and 1.75. Not an exact match but not very far off either.

You might ask why we bother with bond valences when ionic radii also predict lengths.

Bond Valences vs. Ionic Radii

Table comparing bond valence and ionic radii predictions for Ca, Mg, Zn, Al, and Fe oxide bonds.

Bond valence parameters can be used to predict bond lengths, just as ionic radii can—making the two approaches partly redundant.

Shannon–Prewitt radii for Mg–O: 4-coordinate 1.95 Å, 6-coordinate 2.10 Å, 8-coordinate 2.25 Å — the values depend on coordination number, requiring separate radii for each case
Bond valences need only one parameter set: for Mg²⁺, 4 bonds → s = 0.5 → R = 1.95 Å; 6 bonds → s = 0.33 → 2.10 Å; 8 bonds → s = 0.25 → 2.21 Å
The results match or closely approximate the ionic-radius predictions
Unlike ionic radii, the bond valence method handles distorted environments naturally
Cr⁶⁺ compounds: ionic radii would predict equal Cr–O distances, but bond valences easily capture the variation among different bond lengths
Bond valences: a more flexible and generally superior way to analyze and predict bond lengths

Let me read back through the bond valence summary.

Summary: The Bond Valence Method

Bond Valence Method — Summary

Core rule:

$s_{ij} = \exp[(R_0 − d_{ij})/b], \quad \sum s_{ij} = V_i$

Purpose:

Links bond lengths ↔ oxidation states; works with distorted structures.

Advantages:

One $R_0$ per atom pair covers all coordination numbers.
Handles mixed bond lengths; ionic radii cannot.

Uses:

Verify oxidation states.
Predict reasonable bond distances.
Locate light atoms (e.g., H).

Practice questions

Work the Lecture 14 practice questions on spin states, the Jahn-Teller theorem, second-order Jahn-Teller distortions, and the bond valence method.

Open the Lecture 14 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.