GATE UV-Visible Spectroscopy — Transitions, Selection Rules and λmax
UV-visible questions in GATE Chemistry split cleanly into two families. One family is quantitative concentration work built on the Beer–Lambert law, which has its own dedicated article in this knowledge base. The other family — the one covered here — asks why a band appears where it does and with the intensity it has: which orbitals are involved, whether the transition is allowed, how a solvent or a substituent moves it, and what a molar absorptivity of 10 versus 10 000 is telling you. That reasoning is worth more marks than the arithmetic, and it is where most candidates are weakest.
Converting a wavelength into an energy
Every UV-visible question is secretly an energy question, so learn the three conversions and never derive them under exam pressure.
A band appears at λmax = 300 nm. Express its energy three ways.
Per photon: E = hc/λ = (6.626 × 10−34 × 2.998 × 108) / (3.00 × 10−7) = 1.9865 × 10−25 / 3.00 × 10−7 = 6.62 × 10−19 J
Per mole: 6.62 × 10−19 × 6.022 × 1023 = 3.99 × 105 J mol−1 = 399 kJ mol−1. Cross-check with the shortcut: 1.196 × 105 / 300 = 398.7 ✓
In electronvolts: 6.62 × 10−19 / 1.602 × 10−19 = 4.13 eV. Shortcut: 1 239.8 / 300 = 4.133 ✓
In wavenumbers: 107 / 300 = 33 333 cm−1
Notice that 399 kJ mol−1 is comparable with a covalent bond energy. That is the physical reason UV light causes photochemistry while infrared light only makes bonds vibrate.
The transitions, in energy order
| Transition | Energy | Typical λmax | Typical ε (L mol−1 cm−1) | Found in |
|---|---|---|---|---|
| σ → σ* | Highest | Below ~150 nm (vacuum UV) | large | Saturated alkanes |
| n → σ* | High | ~150–250 nm | 102–103 | Alcohols, amines, halides |
| π → π* | Moderate | ~180 nm upward; longer with conjugation | 103–105 | Alkenes, carbonyls, aromatics |
| n → π* | Lowest of the organic set | ~270–350 nm | 10–100 | Carbonyls, nitro, azo |
| Charge transfer | Variable | Often visible | 103–104 | MnO4−, CrO42− |
| d → d | Low | Visible | 1–100 (often < 20 for octahedral) | Transition-metal complexes |
The single most useful diagnostic on this page: ε tells you which family you are looking at. An intense visible band (ε ≈ 104) in a metal compound is charge transfer, not d–d, no matter how strongly coloured the sample looks. A pale band with ε ≈ 10 in the same region is d–d.
Selection rules — why weak bands are weak
Both rules are strictly true only for an idealised centrosymmetric molecule that does not vibrate. Real molecules relax them, and how much they relax explains the observed intensities.
| Case | Spin | Laporte | Consequence |
|---|---|---|---|
| π → π* in butadiene | allowed | allowed | ε ≈ 104, strong band |
| n → π* in a ketone | allowed | forbidden by symmetry overlap | ε ≈ 10–100, weak but visible |
| d–d in octahedral [Ti(H2O)6]3+ | allowed | forbidden (centrosymmetric) | ε of order 1–10; only vibronic coupling gives intensity |
| d–d in tetrahedral [CoCl4]2− | allowed | relaxed — no centre of symmetry, d–p mixing | ε an order or two higher; deep blue |
| d–d in [Mn(H2O)6]2+ (high-spin d5) | forbidden | forbidden | ε of order 0.01–1; very pale pink |
| LMCT in MnO4− | allowed | allowed | ε of order 103–104; intense purple |
The Mn(II) versus Mn(VII) contrast is a classic exam pair. Permanganate has no d electrons at all, so its colour cannot be d–d; it is a ligand-to-metal charge transfer, oxygen lone pair into empty metal orbital. Manganese(II) has five d electrons and is almost colourless, because its transitions break both selection rules at once.
Colour: what you see is not what is absorbed
| Absorbed λ (nm) | Colour absorbed | Colour observed |
|---|---|---|
| 400–435 | Violet | Yellow-green |
| 435–480 | Blue | Yellow |
| 480–490 | Green-blue | Orange |
| 500–560 | Green | Purple / red |
| 560–580 | Yellow | Violet-blue |
| 595–605 | Orange | Green-blue |
| 605–750 | Red | Blue-green |
A question that says "the complex is yellow" is telling you it absorbs in the blue, near 435–480 nm, and therefore has a moderately large splitting. Read the complementary colour, never the observed one, into an energy.
From λmax to the crystal-field splitting
[Ti(H2O)6]3+ has a single broad band with λmax near 500 nm. Find Δo.
Ti3+ is d1, so the one band is the single t2g → eg excitation and its energy is Δo.
ν̄ = 107 / 500 = 20 000 cm−1
In kJ mol−1: 1.196 × 105 / 500 = 239 kJ mol−1
In eV: 1 239.8 / 500 = 2.48 eV
Absorbing green light at 500 nm leaves the transmitted light purple, which is exactly the colour of the ion. The band is also visibly asymmetric, with a shoulder, because the excited eg1 state is Jahn–Teller distorted — a favourite follow-up question.
Only for d1 (and by symmetry d9) does one band equal Δo directly. For d2, d3, d7 and d8 there are several bands and Δo must be extracted from a Tanabe–Sugano analysis, not simply read off.
Substituent and solvent effects — the four "-chromic" words
| Term | Direction | Cause |
|---|---|---|
| Bathochromic (red shift) | λmax increases | Extended conjugation; auxochrome with a lone pair; polar solvent on π → π* |
| Hypsochromic (blue shift) | λmax decreases | Polar / hydrogen-bonding solvent on n → π*; protonation of an amine auxochrome |
| Hyperchromic | ε increases | Substituent that improves orbital overlap; relaxation of a selection rule |
| Hypochromic | ε decreases | Steric twisting that breaks planarity and shortens effective conjugation |
The solvent rule is the one GATE tests most often, and it has a clean physical explanation. Going to a more polar, hydrogen-bonding solvent stabilises the n lone pair strongly by hydrogen bonding, which raises the n → π* gap — a blue shift. The same solvent stabilises the more polar π* excited state of a π → π* transition, which lowers that gap — a red shift. Opposite directions, same cause, and the direction identifies the transition type.
Woodward–Fieser rules — predicting λmax for a diene
These empirical rules let you predict λmax by adding increments to a base value. Different textbooks quote slightly different base values — the acyclic diene base is given as 214, 215 or 217 nm depending on the edition. In an exam, use the table printed in the question; if none is printed, state which base value you used. Never present one edition's number as the only correct one.
| Contribution | Increment (nm) |
|---|---|
| Base: heteroannular (transoid) diene | 214 |
| Base: homoannular (cisoid) diene | 253 |
| Each alkyl substituent or ring residue | +5 |
| Each exocyclic double bond | +5 |
| Each double bond extending the conjugation | +30 |
| Auxochrome: –OR | +6 |
| Auxochrome: –Cl, –Br | +5 |
| Auxochrome: –SR | +30 |
| Auxochrome: –NR2 | +60 |
| Auxochrome: –OC(O)CH3 (acetate) | 0 |
(a) A heteroannular diene carries four ring residues and one exocyclic double bond. Predict λmax.
214 (base) + 4 × 5 (ring residues) + 1 × 5 (exocyclic) = 214 + 20 + 5 = 239 nm
(b) A homoannular diene carries five ring residues.
253 + 5 × 5 = 278 nm
(c) Take the diene in (a) and extend the conjugation by one more double bond that itself carries an –OR group.
239 + 30 (extension) + 6 (–OR) = 275 nm
The rules apply only to conjugated systems. Cross-conjugated and non-planar systems fall outside them, and predictions typically agree with experiment to within a few nanometres rather than exactly.
Instrumental limits worth remembering
Every solvent absorbs below some wavelength, its cut-off, and you simply cannot record a spectrum there. Water and acetonitrile are transparent far into the UV; acetone, with its own n → π* band, is useless below roughly 330 nm. Ordinary glass cells absorb in the UV, so quartz cells are required below about 350 nm. A question that asks "why was this band not observed?" is often asking about the solvent or the cell, not the molecule.
Common mistakes
- Reading the observed colour as the absorbed colour. A red solution absorbs green, not red.
- Calling every coloured metal compound a d–d absorber. Permanganate and chromate have d0 metals; their colour is charge transfer.
- Expecting d–d bands to be intense. In a centrosymmetric octahedral complex they are Laporte-forbidden and weak; that weakness is the answer, not a problem.
- Getting the solvent shift backwards. Polar solvent: n → π* blue-shifts, π → π* red-shifts.
- Setting one absorption band equal to Δo for any dn. That shortcut is safe only for d1 and d9.
- Applying Woodward–Fieser to a cross-conjugated or non-planar system. The rules were fitted to planar conjugated dienes and enones.
- Quoting a base value as universal. Editions differ; say which you used.
- Forgetting that λ and energy are inversely related. A red shift is a decrease in transition energy.
Where this appears in GATE Chemistry
| Question type | What is tested | Usual form |
|---|---|---|
| λ → energy conversion | E = hc/λ and the cm−1, eV, kJ mol−1 shortcuts | Numerical answer |
| Identify the transition | ε magnitude plus λ region | Multiple choice |
| Selection-rule reasoning | Spin and Laporte rules, vibronic relaxation | Explain the intensity |
| Δo from a spectrum | d1/d9 special case | Numerical answer in cm−1 |
| Solvent shift direction | Excited-state versus lone-pair stabilisation | Conceptual |
| Woodward–Fieser addition | Careful increment bookkeeping | Numerical answer in nm |
How much weight instrumental methods carry changes between syllabus revisions, so confirm the current scope from the official GATE notification for your exam year rather than from any secondary summary.
Do the photon arithmetic properly. Every one of these questions starts by turning a wavelength into an energy, and that single step — powers of ten in metres, then a multiplication by Avogadro's number — is where marks quietly disappear. The photon and de Broglie calculator does E = hc/λ directly, so you can check your value in cm−1, joules or electronvolts before building the rest of the answer on it.
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