Colour of complexes — d-d transitions: Question bank

Refresher — every symbol used on this page

Before the traps, here is the minimum vocabulary, each anchored to the picture below.

Definition The five terms you must own

d-orbitals — the five "shelves" an electron can sit on in a transition-metal ion. In a free ion they all have the same energy (degenerate).
$Δ$ (crystal-field splitting energy) — when ligands surround the ion, the five shelves split into two energy groups. $Δ$ is the energy gap between them. Big $Δ$ = big jump needed. Built in full in Crystal Field Theory.
$t_{2 g}$ and $e_{g}$ (octahedral case) — the two groups the shelves split into. The lower group of three orbitals ( $d_{x y}, d_{y z}, d_{z x}$ , pointing between the ligands) is called $== t_{2 g} ==$ ; the upper group of two ( $d_{x^{2} - y^{2}}, d_{z^{2}}$ , pointing at the ligands) is called $== e_{g} ==$ . An electron absorbs a photon to jump from $t_{2 g}$ up to $e_{g}$ — that is the d-d transition.
$h$ (Planck's constant) $\approx 6.626 \times 1 0^{- 34} J s$ — the fixed number that converts a light wave's frequency into its photon energy. See Planck's Equation E=hν.
$c$ (speed of light) $\approx 3.0 \times 1 0^{8} m s^{- 1}$ — how fast light travels; it links frequency $ν$ to wavelength $λ$ via $c = λ ν$ .

Figure 1 shows the octahedral split (left), why $Δ_{t} = \frac{4}{9} Δ_{o}$ for tetrahedral (middle), and the complementary-colour wheel (right) — the three pictures every question here refers back to.

Figure — Colour of complexes — d-d transitions

True or false — justify

A complex looks the same colour as the light it absorbs.

False. A solution shows transmitted light = white minus absorbed = the complementary colour; e.g.

[T i (H_{2} O)_{6}]^{3 +}

absorbs green-yellow but looks purple.

Every ion with electrons in d-orbitals is coloured.

False. You need both a d-electron AND an empty d-slot to jump into;

d^{10}

(like

Z n^{2 +}

) has electrons but no empty d-orbital, so it is colourless.

A larger crystal-field splitting

Δ

always means the complex absorbs a longer wavelength.

False. Since

λ = h c /Δ

, wavelength is inversely proportional to

Δ

; a bigger gap needs a more energetic (shorter) photon.

Swapping

H_{2} O

for

C N^{-}

around the same metal ion makes the absorbed wavelength shorter.

True.

C N^{-}

sits far to the strong-field end of the Spectrochemical Series, so

Δ

grows and

λ = h c /Δ

shrinks.

Tetrahedral complexes generally show colours from smaller energy gaps than octahedral ones with the same metal and ligand.

True.

Δ_{t} = \frac{4}{9} Δ_{o}

, so tetrahedral splitting is smaller and the absorbed photon is lower-energy (longer

λ

). See Octahedral vs Tetrahedral Splitting.

The colour we see comes from the electron emitting light as it falls back down.

False. The visible colour is due to absorption of a photon on the way up (the missing colour from white light), not emission.

A strong-field ligand can turn a would-be high-spin ion into a low-spin one.

True. A strong field makes

Δ

exceed the spin-pairing energy

P

, so electrons pair up low rather than occupy the upper

e_{g}

set — changing spin, magnetism and the available d-d jumps.

C u^{+}

(

d^{10}

) and

C u^{2 +}

(

d^{9}

) should have the same colour behaviour.

False.

C u^{2 +}

(

d^{9}

) has one empty d-slot and is coloured;

C u^{+}

(

d^{10}

) is full and colourless — the same element, different oxidation state, opposite result.

Doubling

Δ

doubles the frequency of the absorbed photon.

True.

Δ = h ν

, so

ν = Δ/ h

; frequency is directly proportional to the gap (see Planck's Equation E=hν).

Spot the error

" $[Z n (H_{2} O)_{6}]^{2 +}$ is colourless because water is a weak-field ligand."

Wrong cause. The ligand strength is irrelevant here —

Z n^{2 +}

d^{10}

, so no ligand (weak or strong) can create a possible d-d transition; the d-set is full.

" $S c^{3 +}$ is colourless because its $Δ$ is too small to absorb visible light."

Wrong cause.

S c^{3 +}

d^{0}

— there is no d-electron to promote at all, so the size of

Δ

never enters the argument.

" $[T i (H_{2} O)_{6}]^{3 +}$ looks green because it absorbs at ~500 nm."

Confused absorbed with seen. It absorbs green-yellow (~500 nm), so green is removed; the eye sees the complement, purple.

"Adding $N H_{3}$ to $[C u (H_{2} O)_{6}]^{2 +}$ deepens the blue because $N H_{3}$ absorbs blue itself."

Wrong mechanism.

N H_{3}

doesn't absorb the light; it raises

Δ

(stronger field), shifting the metal's absorption to shorter

λ

, which changes the transmitted (seen) colour.

"Charge-transfer bands and d-d transitions are the same thing, just different names."

Different processes. A d-d transition moves an electron within the metal's d-set; a charge-transfer moves it between ligand and metal — much more intense (see Charge-Transfer Spectra (why KMnO4 is intensely coloured)).

"Because $Δ_{t} = \frac{4}{9} Δ_{o}$ , tetrahedral complexes are always colourless."

Overstated. A smaller gap still absorbs some wavelength (often in the visible/near-IR); it doesn't vanish — the colour just tends toward the red/longer end.

"A high-spin and a low-spin form of the same ion must absorb the same colour."

Wrong. Low-spin means a larger

Δ

(strong field), so

λ = h c /Δ

is shorter — the two forms absorb at different wavelengths and can look strikingly different.

Why questions

Why must the metal ion be partially filled (

d^{1}

d^{9}

) to show a d-d colour?

You need an electron in a lower set to move and an empty higher slot for it to land in;

d^{0}

lacks the electron,

d^{10}

lacks the empty slot.

Why does a higher oxidation state on the metal usually increase

Δ

A higher positive charge pulls the ligands closer, so their electron pairs repel the d-orbitals more strongly, widening the gap between the

t_{2 g}

(lower) and

e_{g}

(upper) sets.

Why do we rewrite

E = h ν

Δ = h c / λ

instead of leaving it in frequency?

Because we describe colour by wavelength; substituting

ν = c / λ

(where

c

is the speed of light) lets us name the actual absorbed colour directly (see Planck's Equation E=hν).

Why is the complementary colour, not the absorbed colour, what we see through the solution?

Light passes through (transmits); the absorbed colour is subtracted from white, and what remains is the complement on the colour wheel.

Why can two complexes of the same metal ion show completely different colours?

Different ligands give different field strengths, so different

Δ

, so different absorbed

λ

— the metal's electron count is the same but the gap it jumps is not.

Why does whether a complex is high-spin or low-spin depend on comparing

Δ

with

P

Climbing to

e_{g}

costs

Δ

; pairing up in

t_{2 g}

costs the spin-pairing energy

P

. Electrons take the cheaper option, so the sign of

(Δ - P)

decides the filling.

Why doesn't a full

d^{10}

ion simply absorb light and jump to the next shell up?

That would be a much larger energy jump (out of the d-set entirely), needing UV rather than visible light — so no visible colour arises from it.

Edge cases

d^{0}

(e.g.

T i^{4 +}

): is it coloured, and why?

Colourless — no d-electron exists to be promoted, so a d-d transition is impossible regardless of

Δ

d^{10}

(e.g.

Z n^{2 +}

C u^{+}

): is it coloured, and why?

Colourless — every d-orbital is full, so there is no empty slot for an electron to jump into.

d^{5}

high-spin (e.g.

M n^{2 +}

[M n (H_{2} O)_{6}]^{2 +}

): why is it only very pale pink?

With

Δ < P

all five d-electrons stay unpaired (one per orbital), so any d-d jump must flip an electron's spin — a "spin-forbidden" move that absorbs only faintly, leaving it almost colourless.

A colourless solution suddenly becomes intensely purple (

K M n O_{4}

d^{0}

M n^{7 +}

) — is that a d-d transition?

No.

M n^{7 +}

d^{0}

with no d-electron; the intense colour comes from a charge-transfer transition, not d-d (see Charge-Transfer Spectra (why KMnO4 is intensely coloured)).

Δ

grows so large the absorbed photon falls into the UV, what happens to the visible colour?

The complex absorbs no visible light and appears colourless — the "missing colour" is now outside the eye's range.

Δ

shrinks so the absorbed photon lies in the infrared, what colour is seen?

Again colourless to the eye for the d-d band, since infrared is invisible — the visible spectrum passes through untouched by that transition.

Connections

Crystal Field Theory
Spectrochemical Series
Octahedral vs Tetrahedral Splitting
Complementary Colours & the Colour Wheel
Charge-Transfer Spectra (why KMnO4 is intensely coloured)
Magnetic Properties of Complexes
Planck's Equation E=hν