A colourless solution, a brightly coloured dye and the orange pigment in carrots can all be understood using the same basic idea: molecules interact with light because their electrons occupy molecular orbitals with different energies. If light supplies exactly the right amount of energy, an electron can be promoted from one molecular orbital to another.
This is the basis of UV–visible spectroscopy. We therefore need to connect three ideas: the molecular orbitals available to the electrons, the energy required to move between those orbitals, and the wavelength of radiation corresponding to that energy. Once these connections are clear, ideas such as chromophores, conjugation, absorption intensity and the Beer–Lambert law fit naturally into the same picture.
Try these five questions before reading the chapter. Focus on the relationship between molecular-orbital energies, wavelength and the structural features of a molecule.
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1. Molecular orbitals provide the energy levels
Before considering an electronic transition, we first need to know where the electron starts and where it can go. When atomic orbitals combine, molecular orbitals are formed. For UV–visible spectroscopy, five labels appear repeatedly: σ, π, n, π* and σ*.
The σ and π orbitals are bonding molecular orbitals. A non-bonding orbital, written n, usually contains a lone pair located mainly on an atom such as oxygen, nitrogen or a halogen. The π* and σ* orbitals are antibonding molecular orbitals and are normally unoccupied in the electronic ground state.
A useful schematic energy order for many organic molecules is: σ < π < n < π* < σ*. The exact orbital energies depend on the molecule, so this should be understood as a qualitative model rather than an absolute universal ordering.
Bonding, non-bonding and antibonding orbitals
Bonding orbitals, σ and π: these are relatively low in energy and are occupied in the ground state.
Non-bonding orbitals, n: these contain lone-pair electrons that are not part of a conventional bonding molecular orbital.
Antibonding orbitals, π* and σ*: these lie at higher energy and provide possible destinations for excited electrons.
The molecular-orbital energy ladder
The following diagram is intentionally schematic. The important feature is the vertical separation between the orbitals, because that separation determines the photon energy required for excitation.
Notice that every excitation arrow points upward. Absorption promotes an electron from a lower-energy occupied orbital to a higher-energy unoccupied orbital.
2. Absorption of light causes an electronic transition
Light consists of photons. A molecule can absorb a photon when the energy of that photon matches the energy difference between an occupied molecular orbital and an available higher-energy orbital.
Here, ΔE is the energy difference between the two molecular orbitals, h is the Planck constant, ν is the frequency of the radiation, c is the speed of light and λ is its wavelength.
Why wavelength tells us about the energy gap
Because photon energy is inversely proportional to wavelength, a high-energy electronic transition requires shorter-wavelength radiation. Conversely, a small orbital-energy gap can be crossed using longer-wavelength radiation.
Large ΔE → high-energy photon → short λ
Small ΔE → low-energy photon → long λ
Ground state and excited state
Before absorption, the molecule is normally in its electronic ground state. After a suitable photon has been absorbed, one electron occupies a higher-energy molecular orbital and the molecule is in an excited electronic state.
The excitation does not mean that an electron is removed completely from the molecule. It remains part of the molecule but occupies a higher-energy molecular orbital.
3. The four common types of electronic transition
The notation for an electronic transition simply identifies the orbital from which the electron starts and the orbital into which it is promoted. Four transitions are particularly important in organic molecules.
| Transition | Where it is commonly encountered | Energy and wavelength | Typical spectral character |
|---|---|---|---|
| σ → σ* | Saturated σ-bonded systems such as alkanes | Very large energy gap; very short wavelength | Usually in the far-UV region |
| n → σ* | Molecules containing lone pairs, such as alcohols, amines and halogeno compounds | Lower energy than σ → σ* | Usually UV absorption |
| π → π* | Alkenes, carbonyl compounds, aromatic systems and other π systems | Lower energy than σ → σ*; strongly affected by conjugation | Often relatively intense |
| n → π* | Unsaturated groups containing a heteroatom lone pair, especially carbonyl groups | Often a relatively small energy gap | Often weak compared with π → π* |
σ → σ* transitions
A σ bond represents a large stabilisation relative to its corresponding σ* antibonding orbital. The energy difference is therefore large. Consequently, σ → σ* excitation requires high-energy, short-wavelength ultraviolet radiation.
Molecules containing only σ bonds may therefore show little or no absorption in the wavelength region normally used for routine UV–visible measurements even though they can absorb at sufficiently short UV wavelengths.
n → σ* transitions
Molecules containing atoms with lone pairs possess non-bonding orbitals. An electron from such an n orbital can be promoted to a σ* orbital. Because the n orbital lies above the σ-bonding orbital, the energy required is lower than for the corresponding σ → σ* excitation.
π → π* transitions
A π → π* transition is possible when the molecule contains a π bond. Alkenes and aromatic compounds therefore have characteristic UV absorptions associated with their π-electron systems.
The π → π* transition becomes especially important when several π bonds are conjugated because conjugation changes the spacing between the π molecular orbitals.
n → π* transitions
An n → π* transition requires both a non-bonding electron pair and an accessible π* orbital. The carbonyl group is the classic example: oxygen provides non-bonding electrons while the C=O bond provides the π* orbital.
π → π* in a carbonyl: an electron is promoted from the C=O π-bonding orbital to the π* antibonding orbital.
n → π* in a carbonyl: an electron from a non-bonding oxygen orbital is promoted into the C=O π* orbital. This usually requires less energy and therefore occurs at a longer wavelength.
4. Allowed and forbidden electronic transitions
Absorption intensity depends not only on whether the photon has the correct energy. Quantum-mechanical selection rules also determine how probable a transition is.
A transition that has a high probability is described as allowed and normally gives a relatively intense absorption band. A transition with a much lower probability is described as forbidden and gives a much weaker absorption.
π → π* transitions are commonly relatively intense, whereas n → π* transitions are commonly much weaker because the latter are restricted by selection rules.
Forbidden does not mean impossible
The word forbidden can be misleading. It does not necessarily mean that the transition can never occur. Rather, its probability is much lower, so the corresponding absorption band has a much smaller intensity.
If a spectrum of a carbonyl compound shows one strong absorption and another much weaker absorption, intensity provides useful evidence: the strong band may correspond to π → π*, while the weaker, longer-wavelength band may correspond to n → π*.
5. Chromophores are the parts of molecules that absorb light
A chromophore is the structural part of a molecule responsible for a characteristic UV or visible absorption. Chromophores contain electrons that can undergo electronic transitions at wavelengths accessible to the spectrometer.
| Structural feature | Important electrons | Possible transition |
|---|---|---|
| C=C | π electrons | π → π* |
| C≡C | π electrons | π → π* |
| C=O | π electrons and oxygen lone pairs | π → π* and n → π* |
| Aromatic ring | Delocalised π electrons | π → π* |
| Heteroatom attached to a suitable unsaturated system | Lone-pair and π electrons | Transition energies depend on the complete chromophore |
Why many saturated molecules appear colourless
A saturated hydrocarbon contains σ bonds but no π electrons or non-bonding electron pairs. Its main electronic excitation is therefore σ → σ*, which generally lies at very short wavelength.
This is why compounds such as simple alkanes do not normally absorb visible light. They may absorb in the far-UV, but visible photons do not possess enough energy to drive the relevant σ → σ* transitions.
6. Conjugation decreases the electronic energy gap
Conjugation occurs when adjacent p orbitals can overlap across several atoms, allowing π electrons to become delocalised over a larger part of the molecule. A familiar pattern is alternating double and single bonds.
As conjugation becomes more extensive, a larger set of closely spaced π molecular orbitals is formed. The energy difference between the highest occupied molecular orbital, HOMO, and the lowest unoccupied molecular orbital, LUMO, generally becomes smaller.
The bathochromic or red shift
A shift of an absorption band towards a longer wavelength is called a bathochromic shift, commonly called a red shift. Increasing conjugation is an important cause of such a shift.
Why extensive conjugation can produce colour
If the energy gap becomes sufficiently small, the molecule may begin to absorb light within the visible region rather than only in the ultraviolet region. The remaining transmitted or reflected visible wavelengths then determine the colour observed.
β-Carotene is a familiar example. Its long conjugated π system gives a much smaller relevant electronic energy gap than an isolated C=C bond, moving its strong absorption into the visible region.
Do not memorise “more conjugation means more colour” as an isolated rule. The underlying reasoning is: more extensive delocalisation → smaller HOMO–LUMO gap → lower-energy absorption → longer wavelength .
7. HOMO and LUMO provide a useful way to describe excitation
The HOMO is the highest-energy molecular orbital occupied by electrons in the ground state. The LUMO is the lowest-energy molecular orbital that is unoccupied.
The lowest-energy electronic excitation is often discussed in terms of the HOMO–LUMO energy gap because this is the smallest separation between an occupied and an available unoccupied molecular orbital.
Large HOMO–LUMO gap: more energy is required, so absorption occurs at shorter wavelength.
Small HOMO–LUMO gap: less energy is required, so absorption occurs at longer wavelength.
HOMO and LUMO do not always have the same orbital labels
In a simple alkene, the important frontier orbitals may be π and π*. In a carbonyl compound, the highest occupied orbital may instead have substantial non-bonding character, so an important low-energy excitation can be n → π*.
This is why it is better to identify the actual orbitals involved rather than assuming that every electronic transition must simply be called “HOMO → LUMO”.
8. Reading a UV–visible absorption spectrum
A UV–visible spectrum normally plots absorbance against wavelength. An absorption band appears because photons over a range of closely related energies can be absorbed, rather than producing an infinitely narrow line.
What is λmax?
The wavelength at which an absorption band reaches its maximum absorbance is written λmax. Comparing λmax values can therefore reveal changes in the energy of an electronic transition.
If λmax increases, the corresponding transition energy has decreased.
Intensity gives different information from wavelength
The position of an absorption tells us about the energy difference between molecular orbitals. The intensity of the absorption tells us about how strongly the species absorbs at that wavelength and is related to the probability of the transition.
λmax: mainly tells us about the energy required for the transition.
Absorbance or molar absorptivity: tells us how strongly the radiation is absorbed.
9. The Beer–Lambert law connects absorbance with concentration
UV–visible spectroscopy is not only useful for identifying electronic transitions. If a species absorbs at a suitable wavelength, the amount of light absorbed can also be used to determine its concentration.
Here, A is absorbance, I₀ is the incident light intensity, I is the transmitted light intensity, ε is the molar absorptivity, c is the concentration of the absorbing species and l is the path length through the sample.
| Quantity | Meaning | Common units |
|---|---|---|
| A | Absorbance | No units |
| ε | Molar absorptivity at the chosen wavelength | dm³ mol⁻¹ cm⁻¹ when c is in mol dm⁻³ and l is in cm |
| c | Concentration of absorbing species | mol dm⁻³ |
| l | Optical path length through the sample | cm |
Direct proportionality
At a fixed wavelength, ε is constant for a given absorbing species. If the same cuvette is used, l is also constant. The Beer–Lambert law then reduces to a simple proportionality:
This means that doubling the concentration doubles the absorbance, provided the Beer–Lambert relationship is obeyed under the experimental conditions.
Worked example
A solution has ε = 1.50 × 10⁴ dm³ mol⁻¹ cm⁻¹ at the wavelength being used. Its absorbance in a 1.00 cm cuvette is 0.450.
Using A = εcl:
c = A/(εl) = 0.450 ÷ [(1.50 × 10⁴)(1.00)] = 3.00 × 10⁻⁵ mol dm⁻³.
Why measurements are often made at λmax
At λmax, the species absorbs particularly strongly. This generally gives greater sensitivity to changes in concentration, making λmax a convenient wavelength for quantitative measurements.
10. Bringing the whole topic together
An electronic-transition problem can usually be solved by moving through a consistent chain of reasoning. Start with molecular structure, identify the available electrons and orbitals, determine the likely transition, and then connect its energy to the wavelength observed.
Structure → molecular orbitals → electronic transition → ΔE → wavelength → absorption spectrum
A useful examination approach
| Step | Question to ask | Chemical reasoning |
|---|---|---|
| 1 | Does the molecule contain only σ bonds? | Expect mainly high-energy σ → σ* absorption. |
| 2 | Are lone pairs present? | n → σ* or, where a π* orbital is available, n → π* may occur. |
| 3 | Are π bonds present? | π → π* excitation becomes possible. |
| 4 | Is the π system conjugated? | Greater conjugation generally decreases ΔE and increases λmax. |
| 5 | Is the absorption strong or weak? | Consider whether the transition is relatively allowed or forbidden. |
| 6 | Is concentration required? | Use A = εcl with consistent units. |
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Thinking longer wavelength means higher-energy radiation | Photon energy is inversely proportional to wavelength. | Use ΔE = hc/λ: longer λ means smaller ΔE. |
| Drawing an absorption transition arrow downwards | Absorption promotes an electron to a higher-energy orbital. | Draw excitation arrows upwards from occupied to unoccupied levels. |
| Assuming every molecule absorbs visible light | Many molecules absorb only in the ultraviolet because their electronic energy gaps are too large. | Ask whether the relevant transition energy corresponds to the UV or visible region. |
| Assuming every molecule containing a lone pair shows n → π* | An n → π* transition also requires a suitable π* orbital. | Look for both a non-bonding electron pair and an unsaturated group with a π* orbital. |
| Saying a forbidden transition cannot occur at all | Forbidden normally means low transition probability, not absolute impossibility. | Expect a weaker absorption band rather than automatically predicting no absorption. |
| Saying conjugation simply “adds colour” | Colour is a consequence of changing electronic energy levels, not a separate effect. | Explain that conjugation decreases the relevant energy gap and shifts absorption to longer wavelength. |
| Using A = εcl without checking units | The units of ε depend on those chosen for c and l. | Keep c, l and ε in a mutually consistent set of units. |
What you should be able to do
Recognise the important molecular orbitals
Distinguish σ, π, n, π* and σ* orbitals and identify which are normally occupied or unoccupied in the ground state.
Identify electronic transitions
Recognise σ → σ*, n → σ*, π → π* and n → π* transitions from molecular structure.
Connect energy with wavelength
Use ΔE = hν = hc/λ to explain why smaller energy gaps correspond to longer wavelengths.
Explain conjugation and red shifts
Relate increasing conjugation to a smaller HOMO–LUMO gap and an increase in λmax.
Interpret absorption intensity
Distinguish relatively intense allowed transitions from weaker transitions restricted by selection rules.
Apply the Beer–Lambert law
Use A = εcl to calculate absorbance, concentration or another unknown quantity with consistent units.
These ten questions test whether you can apply electronic-transition ideas to molecular structure, wavelength, conjugation and Beer–Lambert calculations.
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Frequently asked questions
Electronic transitions: questions students often ask
The details of UV–visible spectroscopy become much easier once the same energy-level argument is applied consistently. These questions tie together the points that students most often mix up.
Why must an electronic-transition arrow point upwards?
During absorption, the molecule takes in energy from a photon. The electron must therefore move from a lower-energy occupied orbital into a higher-energy unoccupied orbital. A downward arrow would instead represent a loss of electronic energy.
Is σ < π < n < π* < σ* always the exact orbital-energy order?
No. It is a useful schematic ordering for discussing many common organic molecules, but actual orbital energies depend on molecular structure and substituents. The diagram should therefore be used for qualitative reasoning rather than treated as a universal numerical ordering.
Why does σ → σ* require such a short wavelength?
The σ-bonding orbital is strongly stabilised while the σ* antibonding orbital is high in energy. The large separation means a high-energy photon is needed. Since ΔE = hc/λ, high energy corresponds to short wavelength.
What kinds of molecules can undergo n → σ* transitions?
The molecule must contain non-bonding electrons, usually a lone pair on a heteroatom such as oxygen, nitrogen or a halogen. An electron from this n orbital can then be promoted into an available σ* orbital.
What kinds of molecules can undergo π → π* transitions?
A π → π* transition requires a π system. Alkenes, carbonyl compounds and aromatic molecules are common examples because they contain occupied π orbitals and corresponding unoccupied π* orbitals.
Why are n → π* transitions especially associated with carbonyl compounds?
A carbonyl group contains both ingredients needed for the transition: non-bonding electrons on oxygen and an unoccupied C=O π* orbital. An oxygen lone-pair electron can therefore be promoted into π*.
Why is an n → π* absorption often weaker than a π → π* absorption?
Transition intensity depends on quantum-mechanical selection rules. π → π* transitions are commonly relatively allowed and therefore intense, while n → π* transitions are commonly restricted and have lower probability, producing weaker absorption.
What exactly is meant by a red shift?
A red shift, or bathochromic shift, means that an absorption band moves to a longer wavelength. Because wavelength is inversely related to photon energy, the transition responsible for the band now requires less energy.
Why does increasing conjugation cause a red shift?
Greater conjugation delocalises π electrons over a larger system and creates more closely spaced π molecular orbitals. The relevant HOMO–LUMO energy gap generally becomes smaller, so a lower-energy, longer-wavelength photon can cause excitation.
Does every conjugated molecule have to be coloured?
No. Conjugation shifts absorption towards longer wavelength, but the shift may still leave λmax within the ultraviolet region. A molecule becomes visibly coloured only when its electronic absorptions extend into the visible region strongly enough to remove part of the visible spectrum.
What is the difference between a chromophore and a HOMO?
A chromophore is a structural region of a molecule responsible for characteristic light absorption. The HOMO is an individual molecular orbital: specifically, the highest occupied molecular orbital. A chromophore can contain several molecular orbitals, including the HOMO and LUMO involved in an important transition.
What does λmax tell us?
λmax is the wavelength at which a particular absorption band has maximum absorbance. Its position provides information about the energy of the associated electronic transition: a larger λmax corresponds to a smaller transition energy.
Why does absorbance have no units?
Absorbance can be written as A = log₁₀(I₀/I). The ratio I₀/I is dimensionless because both intensities have the same units. Consequently, absorbance itself is dimensionless.
Why is absorbance directly proportional to concentration?
From the Beer–Lambert law, A = εcl. At a fixed wavelength for a given species, ε is constant, and if the same cuvette is used, l is also constant. Therefore A is directly proportional to c.
Why is λmax often chosen for quantitative analysis?
The absorbing species has a relatively large molar absorptivity at λmax, so changes in concentration produce readily measurable changes in absorbance. This generally improves the sensitivity of the measurement.
Electronic transitions are ultimately an energy-level story. Molecular structure determines the orbitals available, those orbitals determine ΔE, and ΔE determines the wavelength that can be absorbed. Conjugation, chromophores, absorption intensity and the Beer–Lambert law then build on that same foundation, allowing UV–visible spectroscopy to connect molecular electronic structure with both qualitative and quantitative chemical information.