Intuition The one-sentence coreA complex looks coloured because it absorbs a specific slice of visible light when an electron jumps between energy levels — most often between the split d-orbitals (t 2 g ↔ e g t_{2g}\leftrightarrow e_g t 2 g ↔ e g complementary colour of what was absorbed.
Definition Colour from absorptionWhen white light passes through a complex, photons whose energy equals an allowed electronic energy gap Δ E \Delta E Δ E complementary colour .
WHAT links energy to colour? The photon must satisfy:
Δ E = h ν = h c λ \Delta E = h\nu = \frac{hc}{\lambda} Δ E = h ν = λ h c
WHY this exact equation? Energy is quantised. Only a photon carrying exactly the gap energy can promote the electron. Too little → no jump; too much → no jump (it has to match an actual level difference).
Worked example Complementary colour logic[ Ti ( H 2 O ) 6 ] 3 + [\text{Ti}(\text{H}_2\text{O})_6]^{3+} [ Ti ( H 2 O ) 6 ] 3 + Why purple? Remove green from white → the eye sums the rest → purple/violet .
Absorbed λ (nm)
Absorbed colour
Seen colour
400–450
violet
yellow-green
490–500
green
red/purple
580–600
yellow
blue
650–700
red
green
In an octahedral field the five d-orbitals split into lower t 2 g t_{2g} t 2 g e g e_g e g Δ o \Delta_o Δ o t 2 g t_{2g} t 2 g e g e_g e g is the d–d transition.
HOW to predict the colour wavelength? The absorbed energy equals the gap:
Δ o = h ν abs = h c λ abs \Delta_o = h\nu_{\text{abs}} = \frac{hc}{\lambda_{\text{abs}}} Δ o = h ν abs = λ abs h c
Spectroscopists report this as a wavenumber ν ˉ = 1 / λ \bar\nu = 1/\lambda ν ˉ = 1/ λ ν ˉ ∝ E \bar\nu \propto E ν ˉ ∝ E Δ o = h c ν ˉ \Delta_o = hc\,\bar\nu Δ o = h c ν ˉ
[ Ti ( H 2 O ) 6 ] 3 + [\text{Ti}(\text{H}_2\text{O})_6]^{3+} [ Ti ( H 2 O ) 6 ] 3 + d 1 d^1 d 1 Δ o \Delta_o Δ o Step 1: Absorption peak at λ = 500 nm = 500 × 10 − 9 m \lambda = 500\text{ nm} = 500\times10^{-9}\text{ m} λ = 500 nm = 500 × 1 0 − 9 m Why? Measured from the spectrum.
Step 2: ν ˉ = 1 λ = 1 500 × 10 − 7 cm = 20000 cm − 1 \bar\nu = \dfrac{1}{\lambda} = \dfrac{1}{500\times10^{-7}\text{ cm}} = 20000\text{ cm}^{-1} ν ˉ = λ 1 = 500 × 1 0 − 7 cm 1 = 20000 cm − 1 Why this step? Convert to wavenumber, the standard unit of Δ o \Delta_o Δ o Step 3: Convert to energy: Δ o = h c ν ˉ \Delta_o = hc\bar\nu Δ o = h c ν ˉ h c = 1.986 × 10 − 23 J cm hc = 1.986\times10^{-23}\text{ J cm} h c = 1.986 × 1 0 − 23 J cm Δ o = 1.986 × 10 − 23 × 20000 = 3.97 × 10 − 19 J ≈ 2.5 eV \Delta_o = 1.986\times10^{-23}\times 20000 = 3.97\times10^{-19}\text{ J} \approx 2.5\text{ eV} Δ o = 1.986 × 1 0 − 23 × 20000 = 3.97 × 1 0 − 19 J ≈ 2.5 eV Why? Confirms it lies in the visible range — hence colour.
Step 4: Per mole: × N A = 239 kJ/mol \times N_A = 239\text{ kJ/mol} × N A = 239 kJ/mol Why? Compare with the spectrochemical-series ranking of ligand strength.
Δ o \Delta_o Δ o Δ o \Delta_o Δ o Δ o \Delta_o Δ o spectrochemical series in action.
Definition Charge-transfer transitionAn electron jumps between the metal and the ligand , not within the d-set.
LMCT (Ligand→Metal): electron moves from a filled ligand orbital to an empty/half-filled metal d. Common when metal is in a high oxidation state (easily reduced).MLCT (Metal→Ligand): metal d-electron moves to an empty ligand π ∗ \pi^* π ∗ π \pi π
MnO 4 − \text{MnO}_4^- MnO 4 − intensely purple
Mn is +7 → d 0 d^0 d 0 There are no d-electrons to do a d–d jump! The deep colour comes from LMCT : an O lone-pair electron jumps onto Mn. Why intense? CT transitions are fully allowed (no selection-rule penalty), so ε \varepsilon ε
Feature
d–d band
Charge-transfer band
Electron moves
within d-orbitals
metal ↔ ligand
ε \varepsilon ε small (1–100)
huge (10³–10⁴)
Selection rules
usually forbidden
usually allowed
Example
[ Cu ( H 2 O ) 6 ] 2 + [\text{Cu}(\text{H}_2\text{O})_6]^{2+} [ Cu ( H 2 O ) 6 ] 2 + MnO 4 − \text{MnO}_4^- MnO 4 −
Definition Two key selection rules
Laporte (orbital) rule: In a centrosymmetric molecule, transitions must change parity (g ↔ u g\leftrightarrow u g ↔ u d → d d\to d d → d g → g g\to g g → g Laporte-forbidden . (Δ l = ± 1 \Delta l = \pm1 Δ l = ± 1 Spin rule: Δ S = 0 \Delta S = 0 Δ S = 0 spin-forbidden transitions are even weaker.
HOW do d–d bands appear at all if forbidden?
Intuition Relaxing the Laporte rulePerfect symmetry forbids it, but molecules vibrate . An asymmetric vibration momentarily destroys the centre of symmetry, mixing in some p-character → the transition becomes partly allowed. This is vibronic coupling . It explains why d–d bands are weak but visible (low ε \varepsilon ε
[ Mn ( H 2 O ) 6 ] 2 + [\text{Mn}(\text{H}_2\text{O})_6]^{2+} [ Mn ( H 2 O ) 6 ] 2 + d 5 d^5 d 5 Any d–d transition must flip a spin (an electron going t 2 g → e g t_{2g}\to e_g t 2 g → e g doubly forbidden (Laporte and spin).
Why pale? ε ≈ 0.01 \varepsilon \approx 0.01 ε ≈ 0.01 0.1 0.1 0.1
Forbidden by
Effect on ε \varepsilon ε
Example
Laporte only
ε ∼ 10 \varepsilon \sim 10 ε ∼ 10 [ Ti ( H 2 O ) 6 ] 3 + [\text{Ti}(\text{H}_2\text{O})_6]^{3+} [ Ti ( H 2 O ) 6 ] 3 +
Laporte + spin
ε ∼ 0.1 \varepsilon \sim 0.1 ε ∼ 0.1 [ Mn ( H 2 O ) 6 ] 2 + [\text{Mn}(\text{H}_2\text{O})_6]^{2+} [ Mn ( H 2 O ) 6 ] 2 +
Neither (CT)
ε ∼ 10 4 \varepsilon \sim 10^4 ε ∼ 1 0 4 MnO 4 − \text{MnO}_4^- MnO 4 −
No d-electrons at all
colourless
[ Zn ( H 2 O ) 6 ] 2 + [\text{Zn}(\text{H}_2\text{O})_6]^{2+} [ Zn ( H 2 O ) 6 ] 2 + d 10 d^{10} d 10
d 0 d^0 d 0 d 10 d^{10} d 10 d 0 d^0 d 0 d 10 d^{10} d 10 e g e_g e g no d–d transition → colourless (CT may still occur if oxidation state is high).
Common mistake "Seen colour = absorbed colour."Why it feels right: if a tomato looks red we say it "is" red. The truth: you see the complementary of the absorbed colour for transmitted/transparent samples. A complex absorbing green looks red. Fix: absorb → subtract → see the remainder.
Common mistake "MnO₄⁻ is coloured because of d–d transitions."Why it feels right: it's a transition-metal complex, so "must be d–d." The truth: Mn is +7 = d 0 d^0 d 0 no d-electrons. The colour is LMCT . Fix: check the d-count first; d 0 / d 10 d^0/d^{10} d 0 / d 10
Common mistake "Forbidden means it never happens, soε = 0 \varepsilon = 0 ε = 0 Why it feels right: "forbidden" sounds absolute. The truth: symmetry-breaking vibrations and spin-orbit coupling relax the rules, giving small-but-nonzero ε \varepsilon ε Fix: forbidden ⇒ weak , not zero .
Δ o \Delta_o Δ o Why it feels right: "bigger" intuitively pairs with "bigger λ." The truth: Δ o = h c / λ \Delta_o = hc/\lambda Δ o = h c / λ shorter λ (higher energy, bluer). Fix: remember energy and wavelength are inversely related.
Recall Quick self-test (hide and answer)
Why is [ Sc ( H 2 O ) 6 ] 3 + [\text{Sc}(\text{H}_2\text{O})_6]^{3+} [ Sc ( H 2 O ) 6 ] 3 + d 0 d^0 d 0
Why is a CT band ~1000× stronger than a d–d band? → CT is Laporte-allowed; d–d is Laporte-forbidden.
What relaxes the Laporte rule for octahedral complexes? → asymmetric vibrations (vibronic coupling).
Δ o = 20000 cm − 1 \Delta_o = 20000\text{ cm}^{-1} Δ o = 20000 cm − 1 λ = 1 / 20000 cm = 500 nm \lambda = 1/20000\text{ cm} = 500\text{ nm} λ = 1/20000 cm = 500 nm
Recall Feynman: explain to a 12-year-oldImagine the metal atom's electrons sitting on a low shelf and a high shelf. A photon of light is like a coin of exactly the right value — only then can an electron buy a ticket to jump to the high shelf. The colour of light it "spends" is missing from what comes out, so we see the leftover colours. If both shelves are empty or both are full, no jumping happens, and the thing looks colourless. Sometimes the electron is shy (the jump is "forbidden"), so it only jumps a tiny bit when the atom wiggles — that gives pale colours. But when an electron leaps all the way from the ligand onto the metal (charge transfer), that's a big bold jump → super bright colour, like in purple potassium permanganate.
Mnemonic Remember the rules"L aporte L oses light L ow" — L aporte-forbidden ⇒ l ow ε \varepsilon ε "CT = Can-Transfer = Crazy-Tint" — charge-transfer bands are crazily intense.
"d 0 d^0 d 0 d 10 d^{10} d 10 — no d–d colour.
Crystal Field Theory — origin of Δ o \Delta_o Δ o t 2 g / e g t_{2g}/e_g t 2 g / e g Spectrochemical Series — how ligands tune Δ o \Delta_o Δ o Magnetic Properties of Complexes — high/low spin links to the spin selection ruleBeer–Lambert Law — ε \varepsilon ε Oxidation States of Transition Metals — high state ⇒ LMCTElectronic Spectra & Term Symbols — multiplet transitions (d n d^n d n
Why are most coordination compounds coloured? A d-electron absorbs a visible photon and jumps from
t 2 g t_{2g} t 2 g to
e g e_g e g (gap
Δ o \Delta_o Δ o ); we see the complementary colour of what's absorbed.
Relation between absorbed energy and wavelength? Δ E = h c / λ = h c ν ˉ \Delta E = hc/\lambda = hc\bar\nu Δ E = h c / λ = h c ν ˉ ; larger gap ⇒ shorter wavelength (bluer) absorbed.
Why are d 0 d^0 d 0 d 10 d^{10} d 10 d 0 d^0 d 0 has no electron to excite;
d 10 d^{10} d 10 has no empty d-level to receive it ⇒ no d–d transition.
State the Laporte selection rule. In centrosymmetric molecules transitions must change parity (
g ↔ u g\leftrightarrow u g ↔ u );
d → d d\to d d → d is
g → g g\to g g → g ⇒ forbidden.
State the spin selection rule. Δ S = 0 \Delta S = 0 Δ S = 0 ; transitions changing the number of unpaired electrons are spin-forbidden (very weak).
How do d–d bands appear despite being Laporte-forbidden? Asymmetric molecular vibrations momentarily break the centre of symmetry (vibronic coupling), making them weakly allowed.
What is LMCT vs MLCT? LMCT: ligand→metal electron jump (high oxidation-state metal); MLCT: metal→ligand
π ∗ \pi^* π ∗ jump (π-acceptor ligands).
Why is MnO 4 − \text{MnO}_4^- MnO 4 − d 0 d^0 d 0 Colour is from a Laporte-allowed LMCT (O lone pair → Mn), giving huge
ε \varepsilon ε (~10³–10⁴).
Why is [ Mn ( H 2 O ) 6 ] 2 + [\text{Mn}(\text{H}_2\text{O})_6]^{2+} [ Mn ( H 2 O ) 6 ] 2 + High-spin
d 5 d^5 d 5 : every d–d transition is both Laporte- and spin-forbidden ⇒
ε ∼ 0.01 \varepsilon\sim0.01 ε ∼ 0.01 .
Typical molar absorptivity: d–d vs charge transfer? d–d: 1–100; charge transfer: 10³–10⁴ (because CT is fully allowed).
If Δ o = 20000 cm − 1 \Delta_o = 20000\text{ cm}^{-1} Δ o = 20000 cm − 1 λ = 1 / ( 20000 cm − 1 ) = 5 × 10 − 5 cm = 500 nm \lambda = 1/(20000\text{ cm}^{-1}) = 5\times10^{-5}\text{ cm} = 500\text{ nm} λ = 1/ ( 20000 cm − 1 ) = 5 × 1 0 − 5 cm = 500 nm .
Does "forbidden" mean ε = 0 \varepsilon = 0 ε = 0 No — symmetry-breaking and spin-orbit coupling relax the rules; forbidden means weak, not zero.
transmitted light missing wavelengths
ranked by spectrochemical series
weak field H2O and halides
Complementary colour seen
d-orbital splitting t2g and eg
Large delta-o, blue light
Intuition Hinglish mein samjhoDekho, coordination compound colour kyun dikhate hain? Simple baat hai: jab white light complex pe padti hai, to ek particular wavelength ki light absorb ho jaati hai, kyunki uska energy exactly d-orbitals ke beech ke gap Δ o \Delta_o Δ o t 2 g t_{2g} t 2 g e g e_g e g d–d transition . Jo colour absorb hua, uska complementary colour humari aankh dekhti hai. Isliye Ti(H₂O)₆³⁺ green absorb karke purple dikhta hai.
Ab ek important point: agar metal me d-electron hi nahi hai (jaise d 0 d^0 d 0 d 10 d^{10} d 10 possible hi nahi . Phir bhi MnO₄⁻ itna gehra purple kyun? Kyunki wahan charge transfer (LMCT) ho raha hai — electron oxygen ke ligand se metal pe jump karta hai. Ye transition fully allowed hota hai, isliye bahut intense colour (ε ∼ 10 4 \varepsilon \sim 10^4 ε ∼ 1 0 4 ε ∼ 1 \varepsilon \sim 1 ε ∼ 1 100 100 100
Selection rules samajhna zaroori hai. Laporte rule kehta hai d → d d\to d d → d Spin rule kehta hai unpaired electrons ka number change nahi hona chahiye. Mn²⁺ (d 5 d^5 d 5 Δ E = h c / λ \Delta E = hc/\lambda Δ E = h c / λ Δ o \Delta_o Δ o
