Colour and spectra — d-d transitions, charge transfer; selection rules: Question bank

Before you start, keep these anchors in mind (all built in the parent note):

$Δ_{o}$ = the gap between the lower $t_{2 g}$ orbitals and upper $e_{g}$ orbitals in an octahedral field (Crystal Field Theory). See the picture below.
d–d transition = an electron hops across that gap, absorbing a photon of energy $Δ_{o}$ .
CT (charge-transfer) = an electron hops between metal and ligand, not within the d-set. It comes in two directions (defined next).
$ε$ = molar absorptivity, the "loudness" of an absorption band (Beer–Lambert Law).
Laporte rule = a centrosymmetric molecule forbids transitions that keep the same parity — defined carefully just below.

Definition The other symbols the answers lean on — define them before use

$S$ (total spin quantum number) = a single number that summarises the electron spins. In simple counting terms, $S = \frac{1}{2} \times (number of unpaired electrons)$ . So 5 unpaired electrons → $S = \frac{5}{2}$ ; all paired → $S = 0$ . The spin rule demands $Δ S = 0$ , i.e. the number of unpaired electrons cannot change during the jump.
LMCT (Ligand→Metal Charge Transfer) = an electron jumps from a filled ligand orbital onto the metal. Favoured when the metal is easily reduced (high oxidation state), e.g. $MnO_{4}^{-}$ , $CrO_{4}^{2 -}$ .
MLCT (Metal→Ligand Charge Transfer) = a metal d-electron jumps out onto an empty ligand orbital, usually a $π^{*}$ (antibonding) orbital of a $π$ -acceptor ligand, e.g. CO, CN⁻, bipyridine.
$π^{*}$ = an empty, high-energy "antibonding" orbital on the ligand that can receive an electron.

Formula Three mechanisms that let a "forbidden" d–d band still be visible

A perfectly symmetric octahedral d–d jump is Laporte-forbidden ( $g \to g$ ), yet $ε \sim 1$ – $100$ because three intensity-lending mechanisms leak through:

Vibronic coupling — asymmetric vibrations momentarily break the centre of symmetry, so for an instant the orbitals are no longer purely $g$ and a little $u$ (p-character) mixes in (main contributor to d–d intensity).
Magnetic-dipole-allowed character — some d–d transitions are allowed as magnetic-dipole (not electric-dipole) transitions; this is intrinsically weak but nonzero and adds to the small $ε$ .
Spin–orbit coupling — the coupling between an electron's spin and its orbital motion; it slightly blurs the "spin" and "orbital" labels, so a strictly spin-forbidden jump borrows a little intensity from a spin-allowed one. It is what lets otherwise doubly-forbidden bands (e.g. in $d^{5}$ ) show up faintly rather than not at all. Together these explain why d–d bands are pale but real.

The $t_{2 g} / e_{g}$ split (what $Δ_{o}$ means visually) — read this first: In figure s01, height on the page is energy (up = more energy). The single flat line on the left is the free-ion d level before any ligands arrive. Once six ligands surround the metal, that one level splits into the three lower $t_{2 g}$ orbitals (lavender) and the two higher $e_{g}$ orbitals (coral); the vertical distance between them is exactly $Δ_{o}$ (mint arrow). The butter arrow shows the d–d transition: an electron absorbs a photon of energy $Δ_{o}$ and climbs from $t_{2 g}$ up to $e_{g}$ . That is the whole basis for every "which colour is absorbed" question on this page.

Figure — Colour and spectra — d-d transitions, charge transfer; selection rules

The complementary-colour wheel (absorb one slice → see the opposite):

The spectrochemical series (why CN⁻ splits harder than F⁻):

True or false — justify

The colour you see is the same colour the complex absorbs.

False. For a transmitted/transparent sample you see the complementary colour — absorb green, subtract it from white, the eye sums the leftover and reads it as red/purple (see the colour wheel, figure s02).

A larger

Δ_{o}

shifts the absorption to a longer wavelength.

False. Since

Δ_{o} = h c / λ

, a bigger gap needs a higher-energy photon, which means a shorter wavelength (bluer), not longer.

A "forbidden" transition has

ε = 0

and never happens.

False. Forbidden means weak, not impossible; vibronic coupling, magnetic-dipole character and spin–orbit coupling (all three defined above) relax the rules, so

ε

is small but nonzero (that's why d–d bands are pale but visible).

Charge-transfer bands are generally more intense than d–d bands.

True. CT is usually Laporte-allowed, giving

ε \sim 1 0^{3}

–

1 0^{4}

, whereas Laporte-forbidden d–d bands sit at

ε \sim 1

–

100

All transition-metal complexes are coloured.

False.

d^{0}

(Sc³⁺, Ti⁴⁺) and

d^{10}

(Zn²⁺, Cu⁺) ions have no available d–d jump, so they are colourless unless charge transfer occurs.

Vibronic coupling makes a d–d transition fully allowed.

False. It only partially relaxes the Laporte rule by momentarily breaking the centre of symmetry, mixing in a little

u

(p-character) — the band stays weak, just visible.

Absorbing a photon of any energy above the gap can drive a d–d transition.

False. Energy is quantised: the photon must match the gap exactly — too little or too much both fail to promote the electron.

Strong-field ligands make a complex absorb at shorter wavelengths.

True. Strong-field ligands (CN⁻, CO) give a large

Δ_{o}

, so the complex absorbs high-energy, short-wavelength light — this is the Spectrochemical Series in action (figure s03).

The spin selection rule cares about how many unpaired electrons change.

True.

Δ S = 0

, where

S

is the total spin quantum number (

S = \frac{1}{2} \times

number of unpaired electrons), means the number of unpaired electrons must stay the same; a transition that forces a spin flip is spin-forbidden and extra weak.

A tetrahedral complex has a strict Laporte rule like an octahedral one.

False. Tetrahedral complexes lack a centre of symmetry, so there is no clean

g

u

label to enforce — the Laporte rule is inherently relaxed and their d–d bands are noticeably more intense than octahedral ones.

Only vibrations give d–d bands their small intensity.

False. Vibronic coupling dominates, but magnetic-dipole character and spin–orbit coupling also contribute to the

ε \sim 1

–

100

range — all three lend intensity to an electric-dipole-forbidden jump.

Spot the error

"MnO₄⁻ is deep purple because of a d–d transition."

Error: Mn is +7, i.e.

d^{0}

— there are no d-electrons to jump. The intense colour is LMCT (an oxygen lone-pair electron jumps onto Mn), which is fully allowed.

"[Zn(H₂O)₆]²⁺ is colourless because its

Δ_{o}

is too small to absorb visible light."

Error: The reason isn't a small gap — Zn²⁺ is

d^{10}

, so the

e_{g}

set is full and there is no empty d-level to receive an excited electron. No d–d transition can occur at all.

"[Mn(H₂O)₆]²⁺ is pale because water is a weak-field ligand."

Error: Field strength sets the wavelength, not the faintness. It is pale because high-spin

d^{5}

makes every d–d transition both Laporte- and spin-forbidden, so

ε \approx 0.01

–

0.1

"Since CN⁻ is a strong-field ligand, its complexes are always brightly coloured."

Error: Field strength sets where absorption lands, not intensity. A CN⁻ d–d band is still Laporte-forbidden and weak; only a CT band (e.g. MLCT — a metal electron into a

π^{*}

of CN⁻) would be intense.

d^{1}

ion like Ti³⁺ should show many sharp d–d peaks."

Error: With a single d-electron there is essentially one

t_{2 g} \to e_{g}

excitation, giving a single broad band. It is broad, not sharp, because vibronic coupling and Jahn–Teller distortion (defined above) smear the excited level — not because there are many transitions.

"LMCT is common for metals in low oxidation states."

Error: LMCT (ligand→metal) needs the metal to accept an electron easily, i.e. be easily reduced — that means a high oxidation state (as in MnO₄⁻, CrO₄²⁻).

"MLCT and LMCT are just two names for the same thing."

Error: They run in opposite directions — MLCT is metal→ligand (electron leaves the metal for a ligand

π^{*}

; favoured by

π

-acceptor ligands), LMCT is ligand→metal (favoured by an easily reduced, high-oxidation-state metal).

Why questions

Why is a spin-forbidden d–d band weaker than one that is only Laporte-forbidden?

Because it violates two rules at once (Laporte and spin, i.e.

Δ S \neq = 0

), each of which suppresses intensity; the combined penalty pushes

ε

down to ~0.1 or lower.

Why does changing the ligand change the colour of the same metal ion?

A different ligand changes

Δ_{o}

(its position in the Spectrochemical Series, figure s03), which shifts the absorbed wavelength and hence the complementary colour we see.

Why do we report

Δ_{o}

in wavenumbers (cm⁻¹) rather than wavelength?

Wavenumber

\overset{ν}{ˉ} = 1/ λ

is directly proportional to energy (

Δ_{o} = h c \overset{ν}{ˉ}

), so equal energy gaps map to equal wavenumber differences — convenient and additive, unlike wavelength.

Why can a tiny amount of KMnO₄ deeply stain a solution while a strong solution of [Cu(H₂O)₆]²⁺ is only pale blue?

MnO₄⁻'s LMCT band is allowed (

ε \sim 1 0^{4}

), so few molecules absorb a lot; Cu²⁺'s d–d band is Laporte-forbidden (

ε \sim 10

), so even many molecules absorb little (see Beer–Lambert Law).

Why is high-spin

d^{5}

(like Mn²⁺) special for the spin rule?

All five electrons are unpaired and parallel (

S = \frac{5}{2}

), so any

t_{2 g} \to e_{g}

jump must land in an orbital with the opposite spin — forcing a spin flip that changes

S

, hence every d–d transition is spin-forbidden.

Why does having a high oxidation state favour LMCT rather than MLCT?

A high oxidation state means the metal has empty/half-empty low-lying d-orbitals and is easily reduced, so it readily accepts a ligand electron — that is precisely a ligand→metal (LMCT) transfer (Oxidation States of Transition Metals).

Why do molecular vibrations help a "forbidden" band appear at all?

An asymmetric vibration momentarily destroys the molecule's centre of symmetry, so the orbitals are no longer purely

g

and a little

u

(p-character) mixes in — the transition briefly satisfies the parity requirement (vibronic coupling).

Why does spin–orbit coupling let strictly spin-forbidden bands show up faintly?

Because coupling between spin and orbital motion blurs the pure "spin" label, a spin-forbidden jump can borrow a little intensity from a spin-allowed one — small but nonzero, so the band is faint rather than absent.

Why do d–d bands keep a small intensity even in a perfectly rigid, symmetric picture?

Because some d–d transitions are allowed as magnetic-dipole transitions (not the usual electric-dipole ones); this contribution is intrinsically weak but nonzero, adding to the

ε \sim 1

–

100

range.

Edge cases

Is [Cu(H₂O)₆]²⁺ (

d^{9}

) coloured, and why?

Coloured (pale blue).

d^{9}

has a hole in the

e_{g}

set, so one d–d transition is possible; it's Laporte-forbidden, giving a weak band and a faint colour.

Can a

d^{0}

ion ever be coloured?

Yes — but only via charge transfer. It cannot do a d–d transition, yet in a high oxidation state an LMCT band (like CrO₄²⁻, MnO₄⁻) can give intense colour.

Both

d^{0}

and

d^{10}

are colourless for d–d transitions — is the reason the same?

No.

d^{0}

has no electron to promote;

d^{10}

has no empty upper d-level to receive one. Same outcome, opposite cause.

What colour would you predict if a complex absorbs across the whole visible range weakly and none strongly?

Nearly colourless / grey-white, because no single band removes a distinct slice of white light — nothing dominant is subtracted.

If a complex shows a very intense band, does that alone prove it isn't a d–d transition?

Effectively yes for octahedral cases — Laporte-forbidden d–d bands are weak (

ε ≲ 100

), so an intense band (

ε \sim 1 0^{3}

–

1 0^{4}

) points to charge transfer.

Is a colourless

d^{10}

complex necessarily diamagnetic, and does that matter for colour?

It is diamagnetic (all d-electrons paired,

S = 0

), but that's a magnetic fact (Magnetic Properties of Complexes); the colourlessness comes specifically from the full

e_{g}

leaving no d–d transition available.

Why might two complexes of the same metal ion have different band positions but similar band intensities?

Different ligands change

Δ_{o}

(band position via the Spectrochemical Series), yet if both remain octahedral and Laporte-forbidden, their intensities (

ε

) stay similarly small.