CHEM 548: Materials Chemistry

Lecture 14: The Bond Valence Method

Wed 10/21/2026 · Meeting 16

Bond valences, the valence sum rule, and predicting bond lengths

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

Learning Objectives: Bond Valence Method

Bond Valence Method

Define bond valenceCompute it 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.

Assign oxidation states and validate structures.

Convert valence ↔ distancePredict and compare lengths: $R = R_0 - B \ln s$.

e.g. SrTiO₃, Cr⁶⁺O₄.

Apply to distorted and mixed coordinationLocate light atoms via valence deficits.

Electrostatic Bond Valence

(Pauling 1929 JACS)

$\mathrm{C}_m^{\,c+[N;]}\;\mathrm{A}_n^{\,a-[M;]}$

$c$, $a$ — charges · $N$, $M$ — coordination numbers

electroneutrality balance$m \times c = n \times a$

connectivity balance$m \times N = n \times M$

bond valence balance$c/N = a/M$

The antifluorite unit cell of Li₂O: a cubic close-packed array of large pale oxide spheres with the smaller black lithium atoms in the tetrahedral holes inside it, one shaded tetrahedron drawn around a lithium to show its four oxide neighbours, beside a leader label reading Li.

The cubic perovskite unit cell of SrTiO₃: black titanium atoms at the eight cube corners, red oxide ions at the twelve edge midpoints, and one large pale strontium at the body centre.

Bond valence balance, $c/N = a/M$: in Li₂O every Li–O bond carries $v = 0.25$ read from either end, and in SrTiO₃ each oxygen receives $2(0.667) + 4(0.167) = 2$.

The bond valence method builds on Pauling's 1929 paper, where he wrote down the balances that any ionic compound has to satisfy. Given the crystal chemical formula with each ion's charge and coordination number, we know three balances have to hold. Electroneutrality says the total charge on the cations cancels the total charge on the anions, so m times c equals n times a. Connectivity says every bond that leaves a cation lands on an anion, so m times N equals n times M. The third is the bond valence balance, which we skipped in chapter one but we'll cover now. A bond between a cation and an anion has the same strength in both directions, so the cation's charge divided by its coordination number equals the anion's charge divided by its coordination number.

[advance: highlight the lithium oxide bond graph]

As an example, lithium oxide takes the antifluorite structure, so the oxide ions are cubic close packed and the lithiums sit in the tetrahedral holes. Lithium is plus one with four bonds, so each lithium-oxygen bond carries a valence of one quarter. Oxygen is minus two with eight lithium neighbours, so from the oxygen end each bond is two over eight, one quarter again. Both ends give the same number.

[advance: clear the highlight and show the verdict]

Strontium titanate is the more interesting graph, because two different cations bond to the same oxygen. Its cubic perovskite, so titanium four plus has six oxygens and each titanium-oxygen bond is four over six, which is 0.667. Strontium two plus has twelve oxygens and each strontium-oxygen bond is two over twelve, 0.167. Now go to an oxygen and add up what arrives: two strong bonds from titanium and four weak ones from strontium, two times 0.667 plus four times 0.167, and that comes to 2, which is the oxidation state of oxygen. What we are going to do next is show how to get valences from measured bond lengths instead of from idealized coordination numbers.

Bond Order in Organic Molecules

A scan of Pauling's 1947 paper on atomic radii and interatomic distances in metals, the instructor's yellow highlight over the sentence giving the carbon-carbon single, double, and triple bond distances and his red margin bars beside the equation that makes the bond-type correction proportional to the logarithm of the bond number.

Bond Order and Bond Distance

$-\Delta R(n) = 0.353 \log n$

$-\bigl(d(n) - d(1)\bigr) = 0.353 \log n$

$d(n)$length of the bond

$d(1)$length of a single bond

$n$bond order

$n = 10^{[d(1) - d(n)]/0.353}$

Bond Valences, Quantitative

$v_{ij} = \exp\!\left[(R^{0}_{ij} - d_{ij})/B\right]$, $B = 0.37$ Å.

Per Solid State Materials Chemistry, §5.4.

The 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

Sr$12(0.167) = 2$

Ti$6(0.667) = 4$

O$2(0.667) + 4(0.167) = 2$

Fractional Bond Valences

Sr$12(0.167) = 2$

Ti$6(0.667) = 4$

O$2(0.667) + 4(0.167) = 2$

bond	$d$ (Å)	$v = e^{(R^0 - d)/0.37}$
Sr–O	2.761	$e^{(2.118 - 2.761)/0.37} = 0.176$
Ti–O	1.953	$e^{(1.813 - 1.953)/0.37} = 0.685$

bond valence sums
Sr	$12 \times 0.176 = 2.112$
Ti	$6 \times 0.685 = 4.110$
O	$4 \times 0.176 + 2 \times 0.685 = 2.074$

The measured strontium-oxygen bond is 2.761 ångströms. The tabulated R zero for a strontium-oxygen bond is 2.118, so the valence is the exponential of 2.118 minus 2.761 over 0.37, which is 0.176. Compare that with the idealized 0.167 from the bond graph.

[advance: add the titanium row]

The titanium-oxygen bond is 1.953 ångströms and R zero for titanium-oxygen is 1.813, so the valence is the exponential of 1.813 minus 1.953 over 0.37, which is 0.685. The idealized value was 0.667, so again it's a little high.

[advance: show the bond valence sums]

So then we can use this valence sum rule and we can add up the total valences of each ion. Strontium has twelve bonds at 0.176, which is 2.112, against an expected plus two. Titanium has six at 0.685, which is 4.110, against an expected plus four. Oxygen takes four strontium bonds and two titanium bonds, four times 0.176 plus two times 0.685, which is 2.074 against an expected minus two. Everything lands within a few percent, which says the measured distances and the assigned oxidation states are consistent. If you did not know whether your structural measurements on a compound were right, this is one way to check them.

Bond Valences in FeTiO₃

Ilmenite FeTiO3: alternating layers of TiO6 and FeO6 face-sharing octahedra on the left, and one octahedron enlarged on the right with its four distinct metal-oxygen distances labelled 1.874, 2.089, 2.201 and 2.078 angstroms.

What are the correct oxidation state assignments in this compound, Fe²⁺Ti⁴⁺O₃ or Fe³⁺Ti³⁺O₃?

Bond Valence Sums in FeTiO₃

$v_{ij} = \exp[(R^{0}_{ij} - d_{ij})/B]$, $B = 0.37$ Å.

$v_{\mathrm{Fe-O}}(1)$	$= \exp[(R^{0}_{\mathrm{Fe-O}} - d_{\mathrm{Fe-O}})/B] = \exp[(1.734 - 2.078)/0.37] = 0.395$
$v_{\mathrm{Fe-O}}(2)$	$= \exp[(R^{0}_{\mathrm{Fe-O}} - d_{\mathrm{Fe-O}})/B] = \exp[(1.734 - 2.201)/0.37] = 0.283$
$v_{\mathrm{Fe}}$	$= 3v_{\mathrm{Fe-O}}(1) + 3v_{\mathrm{Fe-O}}(2) = 3(0.395) + 3(0.283) = 2.03$
$v_{\mathrm{Ti-O}}(1)$	$= \exp[(R^{0}_{\mathrm{Ti-O}} - d_{\mathrm{Ti-O}})/B] = \exp[(1.815 - 2.089)/0.37] = 0.477$
$v_{\mathrm{Ti-O}}(2)$	$= \exp[(R^{0}_{\mathrm{Ti-O}} - d_{\mathrm{Ti-O}})/B] = \exp[(1.815 - 1.874)/0.37] = 0.853$
$v_{\mathrm{Ti}}$	$= 3v_{\mathrm{Ti-O}}(1) + 3v_{\mathrm{Ti-O}}(2) = 3(0.477) + 3(0.853) = 3.99$

FeTiO₃: The Verdict

bond valence Fe$= 2.03$

bond valence Ti$= 3.99$

bond valence O$= 0.395 + 0.283 + 0.477 + 0.853 = 2.008$

Fe²⁺Ti⁴⁺O₃

Using Bond Valences to Find H Atoms

Diaspore, AlOOH: a block of edge-sharing aluminium-oxygen octahedra drawn in blue with the oxygen atoms as red spheres at their corners.

AlOOH (Diaspore)

Bond Distances
O1–Al	1.85, 1.85, 1.86
O2–Al	1.97, 1.97, 1.98

Bond Valence Sums
Al	2.75
O1	1.60
O2	1.16

O2 ← OH group

Predicting Bond Lengths

$d_{ij} = R^{0}_{ij} - B \ln v_{ij}$

$v_{\mathrm{Sr-O}}$$= 0.167$

$d_{\mathrm{Sr-O}} = R^{0}_{\mathrm{Sr-O}} - B \ln v_{\mathrm{Sr-O}}$

$d_{\mathrm{Sr-O}} = 2.118 - 0.37 \ln (0.167)$

$d_{\mathrm{Sr-O}} = 2.78$ Å

$v_{\mathrm{Ti-O}}$$= 0.667$

$d_{\mathrm{Ti-O}} = R^{0}_{\mathrm{Ti-O}} - B \ln v_{\mathrm{Ti-O}}$

$d_{\mathrm{Ti-O}} = 1.815 - 0.37 \ln (0.667)$

$d_{\mathrm{Ti-O}} = 1.965$ Å

Cubic perovskite, cube edge $a$: $d_{\mathrm{Ti-O}} = a/2$, $d_{\mathrm{Sr-O}} = a/\sqrt{2}$

Geometry: $d_{\mathrm{Sr-O}}/d_{\mathrm{Ti-O}} = \sqrt{2} = 1.414$ Predicted: $2.78/1.965 = 1.415$

We can also use bond valences predictively. Solve the bond valence expression for the distance and we get the distance equals R zero minus B times the natural log of the valence. So if we know the valence we expect a bond to have, we can predict its length.

[advance: work the strontium titanate example]

Let's return to strontium titanate and turn the idealized valences from its bond graph into distances. For strontium-oxygen the valence is 0.167, R zero is 2.118, and the predicted distance is 2.78 ångströms. For titanium-oxygen the valence is 0.667, R zero is 1.815, and the predicted distance is 1.965 ångströms.

[advance: show the root-two check]

Recall that in a cubic perovskite the cube edge is the only free parameter. Titanium-oxygen is half the cube edge and strontium-oxygen is half the face diagonal, so the strontium-oxygen bond has to be the square root of two times the titanium-oxygen bond, 1.414 times. The ratio of our two predictions is 1.415, so strontium titanate is close to a perfect cubic perovskite.

Bond Valences Predictively: CrO₃

CrO3: corner-connected chromium-oxygen tetrahedra drawn in blue, running as infinite chains with red oxygen atoms at their corners.

CrO₃

chainThe 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

Box: one CrO₃ unit, 1 Cr, its 2 O_t (each bonded to 1 Cr), and half of each of 2 O_b (each bonded to 2 Cr).

Niggli formula: $\mathrm{CrO}_{2/2+2/1}$

2/2: two O_b, each shared by 2 Cr, count as 1. 2/1: two O_t, each on 1 Cr, count as 2. Total: 3 O per Cr.

Crystal-chemical formula: $\mathrm{Cr}^{[4t]}\,\mathrm{O_b}^{[2]}\,(\mathrm{O_t})_2^{[1]}$

Cr: 4 O in a tetrahedron. O_b: 2 Cr. O_t: 1 Cr. Bonds balance: 4 = 2 + 1 + 1.

Bond graph: one node per atom in the formula; the lines at a node are its coordination number. The two lines to O_b are its two Cr neighbours, not a double bond.

bond	$v_{ij}$	$d$ predicted (Å)	$d$ observed (Å)
Cr–O_b	1.0	1.79	1.75
Cr–O_t	2.0	1.54	1.57

O_t: one bond carries all 2, the Cr=O double bond. O_b: 2 × 1 = 2. At Cr: 2 + 2 + 1 + 1 = 6.

Let's start with the chain. Each chromium sits in a tetrahedron of four oxygens, two bridging and two terminal.

[advance: show the Niggli formula]

If we write that with the Niggli formula, the two over two means two oxygens, each shared by two chromiums, so together they count as one. Two over one means two oxygens, each on one chromium, so they count as two. That gives three oxygens per chromium, CrO3.

[advance: show the crystal-chemical formula]

The crystal-chemical formula says the same thing in a different way. Chromium is four-coordinate in a tetrahedron, the one bridging oxygen in the formula is two-coordinate, and the two terminal oxygens are one-coordinate.

[advance: show the bond graph]

The bond graph has a node for every atom in the formula, and the number of lines = the coordination number. So this shows one Cr coordinated with 4 oxygens, two terminal oxygens with one line each, and one bridging oxygen with two lines because its coordinated to two chromiums.

[advance: show the bond valences]

A terminal oxygen has one line and needs a total valence of two, so that one bond has a valence of two, which chemically is the chromium-oxygen double bond. The bridging oxygen has two bonds and needs a total of two, so each has a valence of one. Check it at chromium: two bonds of valence two and two of valence one add to six, which is the oxidation state of chromium.

[advance: compare predicted and observed distances]

Now we can put those valences through the inverted equation. The bond to the terminal oxygen, valence two, is predicted at 1.54 ångströms, and the bond to the bridging oxygen, valence one, at 1.79. In the real structure of chromium six oxide we measure 1.57 and 1.75. So not an exact match, but not far off.

Bond Valences vs. Ionic Radii

	$R_0$ (Å)	$CN$	$v_{ij}$	$d_{\mathrm{BV}}$ (Å)	$d_{\mathrm{IR}}$ (Å)
Ca²⁺ – O	1.967	6	1/3	2.37	2.38
		8	1/4	2.48	2.50
		10	1/5	2.56	2.61
Mg²⁺ – O	1.693	4	1/2	1.95	1.95
		6	1/3	2.10	2.10
		8	1/4	2.21	2.25
Zn²⁺ – O	1.704	4	1/2	1.96	1.98
		6	1/3	2.11	2.12
Al³⁺ – O	1.62	4	3/4	1.73	1.77
		6	1/2	1.88	1.915
Fe²⁺ – O	1.734	4	1/2	1.99	2.01
		6	1/3	2.14	2.16
Fe³⁺ – O	1.759	4	3/4	1.87	1.87
		6	1/2	2.02	2.025

Per Solid State Materials Chemistry, §5.4.

So the last point I want to make is that these bond valence calculations of bond length can be used to replace the ionic radii. Take magnesium-oxygen. With Shannon-Prewitt radii you predict 1.95 ångströms for four-coordinate, 2.10 for six-coordinate, and 2.25 for eight-coordinate, and you need a separate tabulated radius for each of those coordination numbers. With bond valences you need one number, R zero of 1.693. Four bonds gives each of them a valence of a half, and the predicted length is 1.95. Six bonds gives a third, and 2.10. Eight bonds gives a quarter, and 2.21. Read down the rest of the table and the two columns agree. But ionic radii give one distance per coordination number, so in a chromium six compound they predict all the chromium-oxygen bonds to be equal, and we just saw that they are not. So bond valences are generally the way to go.

Summary: The Bond Valence Method

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).

The valence of a bond is the exponential of R zero minus the distance, over b, and the valences around an atom sum to the oxidation state of that atom.

[advance: show the purpose]

The method links bond lengths to oxidation states. Given the distances, we get the oxidation states, as we did in ilmenite. Given the oxidation states, we get the distances, as we did in chromium six oxide. And it works in a distorted structure, where every bond is a different length.

[advance: show the advantages]

Compared with ionic radii, one R zero per atom pair covers every coordination number. There is no separate parameter for four-, six- and eight-coordinate magnesium, and the method handles a site with mixed bond lengths, which ionic radii cannot.

[advance: show the uses]

There are three uses. You can verify an oxidation state assignment, as we did for iron two plus and titanium four plus in ilmenite. You can predict reasonable bond distances for a structure you are building or refining. And you can locate a light atom, hydrogen most of all, from the valence it leaves missing.

Practice questions

Work the Lecture 14 practice questions before the next class.

Open the Lecture 14 practice questions