Using Tanabe–Sugano Diagrams to Find Δo and B

Introduction

A Tanabe–Sugano diagram is most useful when it converts observed bands into physically interpretable parameters. Its axes are ratios, so an observed ratio of two transition energies can locate the horizontal coordinate without knowing B. One absolute band then fixes B and the field splitting Δₒ. The method is simple algebraically but demanding chemically: the starting bands must belong to the same species and have the correct transition assignments.

Core explanation

At a chosen horizontal coordinate x=Δₒ/B, each assigned excited-state curve has ordinate y i(x)=E i/B. An observed band at wavenumber ν̃ i corresponds to E i/(hc), and spectroscopists express both E i and B in cm⁻¹, so ν̃ i=B y i(x). For two assigned bands, r obs=ν̃₂/ν̃₁=y₂(x)/y₁(x). Find the value of x where the diagram's ratio matches r obs. Then B=ν̃₁/y₁(x) and Δₒ=xB. This ordering matters: dividing one band by a guessed B first can conceal the degree of freedom that the ratio is designed to remove.

Band assignment should come from electron count, geometry, spin multiplicity, expected state order and intensity. A weak shoulder may be spin-forbidden; a very strong band may be charge transfer. Using a charge-transfer band as y₂ creates an apparently precise but meaningless Δₒ and B. Broad unresolved bands make maxima uncertain, so one should estimate band positions and uncertainties rather than present many unjustified significant figures.

The ratio function need not be one-to-one over the entire plotted x range. Two x positions can occasionally yield similar ratios, or nearly parallel curves can make r insensitive to x. Additional observed bands, magnetic spin-state information and reasonable parameter ranges resolve ambiguity. A third independent band gives a prediction ν̃₃,pred=B y₃(x); its residual ν̃₃,obs−ν̃₃,pred tests the fit. Several bands can be fitted by minimising weighted residuals instead of relying on hand-read intersections, but the chemistry of assignment remains the limiting step.

Because the diagrams are scaled by B, different complexes of the same d count can lie at different x values for two reasons: Δₒ changes and B changes. A larger x is not by itself proof that Δₒ is larger. Compare the recovered absolute Δₒ values before ranking field strengths. A more covalent complex may have reduced B through the nephelauxetic effect and therefore a larger Δₒ/B even if Δₒ changes less dramatically.

The method is usually semi-quantitative. Published diagrams often assume a particular C/B ratio; singlet and spin-forbidden energies can depend strongly on C. Low symmetry, strong spin–orbit coupling, band overlap and charge-transfer mixing move real peaks away from idealised octahedral predictions. If bands are assigned correctly but residuals remain systematic, the model may need refinement rather than a forced alternative x coordinate.

Check units carefully. Wavelength λ in nanometres converts to ν̃(cm⁻¹)=10⁷/λ(nm); a larger wavelength means lower energy. B and Δₒ must share cm⁻¹ units before x=Δₒ/B is formed. If a plotted y=E/B is 25 and a measured transition is 20,000 cm⁻¹, B is 800 cm⁻¹, not 25/20,000. This arithmetic is a common exam trap.

Step-by-step reasoning

Establish oxidation state, d count, geometry and likely spin state. Assign at least two bands to curves on the matching diagram and convert wavelengths to cm⁻¹. Form the observed energy ratio and scan the diagram for an x where the curve-ordinate ratio agrees. Read y₁ at that x, calculate B=ν̃₁/y₁ and Δₒ=xB. Predict other bands and assess residuals, uncertainties, covalency and whether the selected spin region is chemically plausible.

Visual explanation

Draw two target curves y₁(x) and y₂(x) above the baseline. Beneath them sketch a smaller plot of their ratio R(x)=y₂(x)/y₁(x). Draw a horizontal line at r obs; its crossing gives x. Return to the main plot and mark y₁ at that x, then write B=ν̃₁/y₁ and Δₒ=xB beside the graph. Add a third curve as an independent check.

Real-world analogy

A photograph of two known landmarks provides their apparent height ratio before the camera's absolute scale is known. Matching that ratio can locate the viewpoint; one known real height then fixes the scale. Spectral band ratios locate the diagram coordinate, and one absolute energy fixes B.

Real-world example

Two octahedral Cr³⁺ complexes may show different peak spacings when H₂O is replaced by NH₃. Analysing both on the d³ diagram gives separate Δₒ and B values. Comparing just the first peak can suggest a change in splitting, but the multi-band ratio reveals whether interelectronic repulsion and covalency changed as well.

Why?

Why does taking a band ratio eliminate B? Both measured transition energies contain the same multiplicative B for a single complex in the model: ν̃₁=B y₁ and ν̃₂=B y₂. Dividing cancels B, leaving a function only of x and the chosen curves.

Common misconception

“A larger Δₒ/B always means a stronger absolute ligand field.” It may reflect a smaller B from covalency rather than a larger Δₒ. Recover and compare Δₒ in cm⁻¹ before making that claim.

Worked example

An illustrative spectrum has assigned bands at 18,000 and 27,000 cm⁻¹, giving r obs=1.50. Suppose a published diagram at x=24.0 shows ordinates y₁=22.5 and y₂=33.75, whose ratio is also 1.50. Then B=18,000/22.5=800 cm⁻¹ and Δₒ=24.0×800=19,200 cm⁻¹. A third assigned curve at that x with y₃=48.0 predicts 38,400 cm⁻¹. If the third measured feature lies near 38,000 cm⁻¹, the residual is −400 cm⁻¹, small compared with a broad band; if it lies at 55,000 cm⁻¹, revisit the assignment. The ordinates here are explicitly illustrative readings, not a tabulation for a particular ion.

Quick check

1. A plot gives y₁=20 at x=25 and a band is at 16,000 cm⁻¹. Find B and Δₒ. Answer: B=16,000/20=800 cm⁻¹ and Δₒ=25×800=20,000 cm⁻¹.

Exam focus

Write ν̃ i=B y i(Δₒ/B) first. Show ratio cancellation, then calculate B and Δₒ in that order. Label illustrative diagram values clearly, keep energy units consistent, and do not overstate precision from broad experimental peaks.

Advanced insight

Numerical fitting can treat x and B as parameters and minimise a weighted sum of squared residuals across several bands. Weights should reflect experimental uncertainty and the confidence of each assignment. Because state curves can exchange character near avoided crossings, following energy order alone may give a false match; symmetry and parentage also matter.

Summary

Two assigned band energies yield a ratio that locates Δₒ/B on the correct Tanabe–Sugano diagram. One absolute energy then gives B and Δₒ. Additional bands, magnetic data and uncertainty checks test whether the derived values represent the complex rather than an accidental ratio match.

Practice questions

1. Observed bands are 20,000 and 30,000 cm⁻¹. What plotted-ordinate ratio must match them? Answer: y₂/y₁=30,000/20,000=1.50; no value of B is needed to make this ratio. 2. At the matching x=28, a first target has y₁=25 and its band is 20,000 cm⁻¹. Find B and Δₒ. Answer: B=20,000/25=800 cm⁻¹; Δₒ=28×800=22,400 cm⁻¹. 3. Give two reasons a fitted B might be untrustworthy despite exact arithmetic. Answer: A selected peak may be charge transfer or an overlapping transition rather than the assigned d–d target, and the model's assumed symmetry or C/B ratio may not fit the real complex. 4. Why is a third band useful after fitting two bands? Answer: It gives an independent predicted energy; agreement supports the assignments and model, whereas a large residual exposes an incorrect transition or missing physics.