Chemistry · Lesson

Colour, magnetism and spectroscopic behaviour

Connect light absorption and magnetism to electronic states while keeping intensity, energy and concentration distinct. Use stipulated photon, spin and absorption calculations to test those distinctions.

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By the end, you can

Convert a stated photon wavelength into energy and wavenumber.
Calculate selected spin-only moments and multiplicities.
Distinguish transition energy from transition intensity.
Use Beer–Lambert absorption with its species, path and concentration assumptions.

Colour involves both energies and intensities

Absorption occurs when light couples suitable initial and final states with the corresponding energy difference. Some metal complexes have transitions with substantial d-character; others show ligand-centred or charge-transfer absorption. The presence of colour is not a unique test for an incomplete d shell, and a d¹⁰ entity can have non-d–d absorption.

A perceived colour depends on which wavelengths and intensities reach the observer, together with illumination, concentration, path length and other optical effects. One splitting Δ or one absorption maximum does not by itself specify a complete spectrum. In particular, the energy of a transition and the probability of its absorption are different properties.

Convert photon units before interpreting the transition

For a stipulated wavelength 600. nm in vacuum, use vacuum light speed c=299792458 m s⁻¹ and Planck constant h=6.62607015× 10⁻³⁴ J s. Frequency c/λ is approximately 4.99654× 10¹⁴ Hz and photon energy hc/λ approximately 3.31074× 10⁻¹⁹ J. With elementary charge e=1.602176634× 10⁻¹⁹ C and Avogadro constant N_A=6.02214076× 10²³ mol⁻¹, this is 2.06640 eV per photon or 199.378 kJ per mole of photons. The stated SI constants are exact.

Wavenumber 1/λ is approximately 16666.7 cm⁻¹ after converting 600. nm to 6.00× 10⁻⁵ cm. Values rounded to the wavelength’s three significant figures are 1.67× 10⁴ cm⁻¹, 2.07 eV and 199 kJ mol⁻¹. This invented photon input does not identify a real complex or prove that a single transition energy equals its crystal-field splitting.

A selected unpaired count gives a limited magnetic model

For a simple spin-only estimate with n unpaired electrons, μ_eff≈√[n(n+2)] μ_B, where μ_B is the Bohr magneton. It assumes an appropriate spin description and neglects the orbital contribution and other complications. A paramagnetic response involves unpaired-spin contributions; a complete measurement also includes other magnetic terms and depends on conditions.

Selected Fe(III) d⁵ high spin has n=5 and moment√35≈ 5.916 μ_B, while low spin has n=1 and√3≈ 1.732 μ_B. Selected Fe(II) d⁶ high spin has n=4 and√24≈ 4.899 μ_B, while low spin has n=0 and spin-only moment 0. With parallel unpaired spins in these selected occupations, S=n/2 and multiplicity 2S+1 give 6, 2, 5 and 1 respectively.

A zero spin-only estimate does not erase diamagnetism or establish every temperature-dependent response. Orbital effects, spin–orbit coupling, magnetic interactions and population of excited states can invalidate a simple spin-only identification. An open-shell singlet can also have S=0: counting singly occupied orbitals without their spin coupling is insufficient in a more general state.

Selection rules refer to specified states and symmetry

In the ideal electric-dipole approximation without spin mixing, the spin rule is ΔS=0. For a system with inversion symmetry, opposite parity g↔u is allowed by the parity rule, whereas g↔g or u↔u is forbidden in that ideal approximation. The g/u labels mean even/odd under inversion and apply only where that symmetry description is valid.

An S=0,g→S=0,u transition satisfies both selected rules. S=2,g→S=2,g satisfies spin but fails parity; S=0,g→S=1,u satisfies parity but fails spin. These are classifications of invented state labels, rather than measured intensities. Forbidden does not mean that every real absorption is exactly zero: distortion, vibrational coupling, spin–orbit interactions and other processes can relax a simplified rule.

Many ideal centrosymmetric d–d transitions are weak under the parity rule, while charge-transfer transitions can be much more intense. Actual spectra require the relevant states, symmetries, environment and transition couplings. An absorption band’s intensity cannot be inferred from Δ alone.

Absorbance is logarithmic, rather than a transmitted fraction

For a stipulated single absorbing species, use A=εcℓ and A=−log₁₀(T), where T is transmitted/incident intensity under the selected optical assumptions. ε is the decadic molar absorption coefficient, c amount concentration and ℓ path length. Appropriate monochromatic light, a stable species and controlled optical effects are needed for the simple comparison.

In an invented system at one fixed wavelength, ε=100 L mol⁻¹ cm⁻¹, c=0.00100 mol L⁻¹ and ℓ=1.00 cm give A=0.100 and T=10⁻⁰·¹≈ 0.794328, or 79.4% transmission. Fourfold dilution with no species change gives A=0.0250 and T≈ 0.944061. Dilution changes transmission nonlinearly even when absorbance is linear in concentration.

Alternatively, c=0.000500 mol L⁻¹ and ℓ=2.00 cm give the same A=0.100. One absorbance does not uniquely determine both concentration and path length or identify a complex. Scattering, mixtures, association and changes of chemical speciation require a more complete model.

Transmission in one stipulated Beer–Lambert system

Transmitted intensity fraction T

Stipulated concentration / mmol L⁻¹

Selected model concentrations
— Fixed-species Beer–Lambert model

Underlying data

Stipulated concentration / mmol L⁻¹	Transmitted intensity fraction T
0	1 · Selected model concentrations
0.25	0.9440608762859234 · Selected model concentrations
0.5	0.8912509381337456 · Selected model concentrations
1	0.7943282347242815 · Selected model concentrations

Original model at a fixed wavelength with ε=100 L mol⁻¹ cm⁻¹, path 1.00 cm and no concentration-dependent change of species or optical behavior. The dashed curve is T=10^(−εcℓ); markers select four concentrations. These invented inputs are not a measured spectrum or a unique complex identification.

Worked examples

Keep wavelength and wavenumber units explicit

Use the invented 600. nm photon.

λ=6.00× 10⁻⁷ m=6.00× 10⁻⁵ cm.
1/λ≈ 16666.7 cm⁻¹.
hc/λ≈ 3.31074× 10⁻¹⁹ J=2.06640 eV.
Multiplying by N_A gives 199.378 kJ per mole of photons before rounding.

A photon-energy scale, without identifying a real band or assigning it automatically to Δ_o.

Use the selected occupation, rather than the metal name alone

Compare the high-spin and low-spin Fe(II) d⁶ occupations.

High spin has n=4, so μ_spin=√24≈ 4.899 μ_B.
Low spin has n=0, so the spin-only value is 0.
Selected S values 2 and 0 give multiplicities 5 and 1.

Different restricted spin predictions for the same formal d count, with orbital and magnetic-interaction effects omitted.

Convert the same absorption arithmetic to transmission

Use ε=100 L mol⁻¹ cm⁻¹, c=0.00100 M and ℓ=1.00 cm, then dilute fourfold.

Initial A=100× 0.00100× 1.00=0.100.
T=10⁻⁰·¹≈ 0.794328.
Diluted A=0.0250 gives T≈ 0.944061.

Absorbance quarters, while the transmitted fraction changes from about 0.794 to 0.944.

Apply both ideal rules separately

Classify S=2,g→S=2,g and S=0,g→S=1,u under the stated electric-dipole rules.

The first has ΔS=0 but unchanged parity.
The second changes parity but also changes total spin.
Each fails one selected ideal rule.

Neither is allowed by both ideal rules; no exact real intensity or absolute absence has been predicted.

Try it yourself

A selected spin-only state has three parallel unpaired electrons. Find its multiplicity and estimated moment.
Show the answer and reasoning
Multiplicity 4; μ_spin=√15≈ 3.87 μ_B.
S=3/2 gives 2S+1=4; use n=3 in√[n(n+2)], retaining the spin-only assumptions.
Keep ε=100 L mol⁻¹ cm⁻¹ and c=0.00100 M, but use path 2.00 cm. Find A and T.
Show the answer and reasoning
A=0.200; T≈ 0.631.
A=εcℓ doubles; T=10⁻⁰·²≈ 0.630957.
Under the selected inversion-symmetric electric-dipole rules, does S=1,u→S=1,g satisfy the spin and parity tests?
Show the answer and reasoning
Yes, both selected tests.
ΔS=0 and a parity change satisfy these ideal rules; that alone does not determine a measured intensity.
Why can two different concentration/path pairs give the same absorbance?
Show the answer and reasoning
The simple model fixes their product cℓ.
Without other information, one A cannot determine both variables or identify the absorbing entity.

What to carry forward

Separate transition energy from intensity.
Convert wavelength units before photon arithmetic.
Use spin-only moments within their stated limits.
Selection rules require specified symmetry and states.
Absorbance and transmitted fraction have different mathematical forms.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.