Colour and spectra — d-d transitions, charge transfer; selection rules: Worked examples

First, meet the vocabulary once so nothing appears unexplained:

== $λ$ == (lambda) — the wavelength of light, the length of one ripple, measured in nanometres (nm, a billionth of a metre) or centimetres (cm).
== $\overset{ν}{ˉ}$ == (nu-bar) — the wavenumber, literally "how many wave-ripples fit in one centimetre," $\overset{ν}{ˉ} = 1/ λ$ . Bigger wavenumber = more ripples per cm = higher energy.
== $Δ_{o}$ == (delta-oh) — the energy gap between the lower $t_{2 g}$ shelf and upper $e_{g}$ shelf in an octahedral complex (see Crystal Field Theory).
== $ε$ == (epsilon) — the molar absorptivity, a number saying "how greedily this substance drinks light" (see Beer–Lambert Law). Big $ε$ = intense colour.

The three constants we reuse: $h = 6.626 \times 1 0^{- 34} J s, c = 3.00 \times 1 0^{10} cm/s, h c = 1.986 \times 1 0^{- 23} J cm$

The scenario matrix

Every question this topic can throw is one of these cells. Each worked example below is tagged with the cell it fills.

#	Case class	What's special	Example
A	Large $Δ_{o}$ (strong-field)	short λ absorbed, blue end	Ex 1
B	Small $Δ_{o}$ (weak-field)	long λ absorbed, red end	Ex 2
C	$d^{0}$ input (zero d-electrons)	no d–d possible → colour from CT or none	Ex 3
D	$d^{10}$ input (full d-shell)	no empty d-level → colourless	Ex 3
E	Doubly-forbidden ( $d^{5}$ high-spin)	Laporte + spin → tiny $ε$	Ex 4
F	Charge-transfer (allowed)	huge $ε$ , high oxidation state	Ex 5
G	Complementary-colour reasoning	absorbed → seen colour	Ex 6
H	Real-world word problem	concentration from absorbance	Ex 7
I	Exam twist (compare two complexes)	which absorbs bluer & why	Ex 8

Case A — large gap, strong-field ligand

Worked example Ex 1 —

$[Fe (CN)_{6}]^{3 -}$ absorbs at $λ = 24, 000 cm^{- 1}$ (wavenumber given). Find $Δ_{o}$ in J, eV, and kJ/mol. Forecast: CN⁻ is at the strong-field end of the Spectrochemical Series. Guess: is $Δ_{o}$ big (>200 kJ/mol) or small? Guess before reading on.

Step 1. The wavenumber is already given: $\overset{ν}{ˉ} = 24, 000 cm^{- 1}$ . Why this step? The energy gap equals the photon energy, and wavenumber is directly proportional to energy — no wavelength juggling needed.

Step 2. Energy in joules: $Δ_{o} = h c \overset{ν}{ˉ} = (1.986 \times 1 0^{- 23}) (24, 000)$ . $Δ_{o} = 4.766 \times 1 0^{- 19} J$ Why this step? $h c$ has units J·cm, and $\overset{ν}{ˉ}$ has cm⁻¹; multiply and the cm cancels, leaving joules per photon.

Step 3. In electron-volts: divide by $1.602 \times 1 0^{- 19} J/eV$ : $Δ_{o} = \frac{4.766 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}} = 2.98 eV$ Why this step? eV is the natural size-unit for one electron's jump; ~3 eV confirms it's a visible-light-sized gap.

Step 4. Per mole: $\times N_{A} = 6.022 \times 1 0^{23}$ : $Δ_{o} = 4.766 \times 1 0^{- 19} \times 6.022 \times 1 0^{23} = 287 kJ/mol$ Why this step? Chemists compare ligand strengths in kJ/mol; this large value confirms CN⁻ is strong-field.

Verify: 287 kJ/mol is far above H₂O's ~240 kJ/mol — consistent with CN⁻ outranking H₂O in the spectrochemical series. ✓ (Forecast: big gap — correct.)

Case B — small gap, weak-field ligand

Worked example Ex 2 —

$[CrF_{6}]^{3 -}$ absorbs at $λ = 671 nm$ . Find $\overset{ν}{ˉ}$ and $Δ_{o}$ (kJ/mol), and predict where it sits versus Ex 1. Forecast: F⁻ is weak-field. Do you expect the absorption at longer or shorter wavelength than the CN⁻ complex? Guess.

Step 1. Convert nm to cm: $671 nm = 671 \times 1 0^{- 7} cm = 6.71 \times 1 0^{- 5} cm$ . Why this step? Wavenumber is per centimetre, so λ must be in centimetres.

Step 2. Wavenumber: $\overset{ν}{ˉ} = 1/ λ = 1/ (6.71 \times 1 0^{- 5}) = 14, 903 cm^{- 1}$ . Why this step? This is the standard reporting unit and lets us reuse $h c$ directly.

Step 3. Energy per mole: $Δ_{o} = h c \overset{ν}{ˉ} N_{A} = (1.986 \times 1 0^{- 23}) (14, 903) (6.022 \times 1 0^{23})$ . $Δ_{o} = 178 kJ/mol$ Why this step? Same recipe as Ex 1 — combine the two conversions into one line.

Verify: 178 kJ/mol < 287 kJ/mol from Ex 1. Longer wavelength (671 nm > the ~417 nm of Ex 1) ↔ smaller gap. F⁻ weak-field beats CN⁻ strong-field the wrong way? No — the smaller gap belongs to the weaker ligand, exactly as predicted. ✓ (Forecast: longer λ — correct.)

Cases C & D — zero and full d-shells

Worked example Ex 3 — Classify each ion and say if a d–d transition is possible: (i)

$[Sc (H_{2} O)_{6}]^{3 +}$ , (ii) $[Zn (H_{2} O)_{6}]^{2 +}$ , (iii) $VO_{4}^{3 -}$ . Forecast: Which of these three can show colour, and by what mechanism? Guess before reading.

Step 1. Count d-electrons (see Oxidation States of Transition Metals).

Sc is group 3; Sc³⁺ removes all three valence electrons → $d^{0}$ . (Case C)
Zn is group 12; Zn²⁺ → $[Ar] 3 d^{10}$ → $d^{10}$ . (Case D)
V in $VO_{4}^{3 -}$ is +5; V⁵⁺ → $d^{0}$ . (Case C) Why this step? Colour source is decided by d-count first, before any splitting talk.

Step 2. Apply the two forbidden-by-emptiness rules.

$d^{0}$ : no electron sitting in $t_{2 g}$ to be lifted → no d–d transition.
$d^{10}$ : $e_{g}$ is completely full → no empty upper level to receive a jump → no d–d transition. Why this step? A jump needs both a full starting shelf and an empty landing shelf.

Step 3. Check for charge transfer.

$[Sc (H_{2} O)_{6}]^{3 +}$ : Sc³⁺ is not strongly oxidising, H₂O has no low π*; no easy CT → colourless.
$[Zn (H_{2} O)_{6}]^{2 +}$ : same → colourless.
$VO_{4}^{3 -}$ : V is +5, a high oxidation state (easily reduced) with O donors → LMCT possible → can be pale yellow. Why this step? $d^{0} / d^{10}$ can still be coloured if CT is available; you must not stop at "no d–d."

Verify: Real data — $Sc^{3 +}$ and $Zn^{2 +}$ aqua ions are indeed colourless; orthovanadate is pale yellow from LMCT. ✓

Case E — doubly forbidden

Worked example Ex 4 —

$[Mn (H_{2} O)_{6}]^{2 +}$ has $ε \approx 0.02$ . Explain why it is both Laporte- and spin-forbidden, and estimate how much fainter it is than a Laporte-only band ( $ε \approx 10$ ). Forecast: Will this complex look boldly coloured or almost white? Guess.

Step 1. d-count: Mn²⁺ → $d^{5}$ , high-spin → one electron in every d-orbital, all spins parallel (↑↑↑↑↑). See Magnetic Properties of Complexes. Why this step? The electron arrangement decides whether a spin must flip.

Step 2. Laporte check: a $t_{2 g} \to e_{g}$ jump is $g \to g$ (both d-parity), which the Laporte rule forbids in a centrosymmetric octahedron. Why this step? Every octahedral d–d transition carries this penalty.

Step 3. Spin check: the target $e_{g}$ orbital already holds an ↑ electron. An incoming ↑ can't join it (Pauli), so the jumping electron must arrive ↓ — that means a spin flip, violating $Δ S = 0$ . Why this step? This second violation is what makes it doubly forbidden, unique to $d^{5}$ high-spin.

Step 4. Intensity estimate: $ε \approx 0.02$ vs $\approx 10$ gives a ratio $10/0.02 = 500$ . Why this step? Quantifies "much weaker" — about 500× fainter.

Verify: $[Mn (H_{2} O)_{6}]^{2 +}$ is famously pale pink — barely coloured — matching $ε \sim 0.01$ – $0.1$ . ✓ (Forecast: almost white — correct.)

Case F — charge transfer

Worked example Ex 5 —

$CrO_{4}^{2 -}$ (chromate) is intense yellow. Identify the transition type and explain the large $ε \sim 1 0^{4}$ . Forecast: Is this d–d or charge transfer? What is Cr's d-count? Guess.

Step 1. Oxidation state: in $CrO_{4}^{2 -}$ , Cr is +6 → $d^{0}$ . Why this step? $d^{0}$ instantly rules out any d–d transition.

Step 2. With no d-electrons yet a strong colour, the source must be LMCT: an oxygen lone-pair electron jumps onto the easily-reduced Cr⁶⁺. Why this step? High oxidation state + $d^{0}$ + colour is the fingerprint of ligand→metal charge transfer.

Step 3. Selection rule: LMCT moves an electron between different orbitals of different parity (ligand p → metal d), so it is Laporte-allowed — no penalty. Why this step? Allowed transitions give the huge $ε$ .

Step 4. Consequence: $ε \sim 1 0^{4}$ means it absorbs ~ $1 0^{4} /10 = 1000 \times$ more strongly than a Laporte-only d–d band. Why this step? Explains why chromate stains intensely at tiny concentrations.

Verify: Chromate/permanganate are textbook intense-CT ions; $d^{0}$ rules out d–d, matching the deep colour. ✓

Case G — complementary colour

Worked example Ex 6 — A complex absorbs strongly at 605 nm. What colour does it appear? Show the reasoning on the colour wheel.

Forecast: 605 nm is orange light. Will the complex look orange? Guess.

Step 1. Locate 605 nm: this is orange light being removed from the white beam. Why this step? You must know which colour is absorbed before finding what's seen.

Step 2. The eye sums the remaining wavelengths — everything except orange. On the wheel, the colour directly opposite orange is blue. Why this step? Removing one colour leaves its complementary partner; opposite-on-the-wheel = complementary.

Step 3. So the transmitted light looks blue. Why this step? This is the whole "absorb → subtract → see the remainder" logic from the parent note.

Verify: From the parent table, absorbing 580–600 nm (yellow-orange) → seen blue. 605 nm sits right there. ✓ (Forecast "looks orange" was the classic trap — it looks blue.)

Case H — real-world word problem

Worked example Ex 7 — A

$[Ti (H_{2} O)_{6}]^{3 +}$ solution in a 1.00 cm cell gives absorbance $A = 0.150$ at the peak where $ε = 5 L mol^{- 1} cm^{- 1}$ . Find the concentration. Forecast: With such a small $ε$ (Laporte-forbidden d–d), do you expect a big or small concentration to reach $A = 0.15$ ? Guess.

Step 1. Use the Beer–Lambert Law: $A = ε c l$ , where $l$ is the path length in cm and $c$ the concentration in mol/L. Why this step? This is the only equation linking measured absorbance to concentration.

Step 2. Rearrange for $c$ : $c = \frac{A}{ε l}$ . Why this step? We know $A$ , $ε$ , $l$ and want $c$ — isolate it.

Step 3. Plug in: $c = \frac{0.150}{5 \times 1.00} = 0.030 mol/L$ . Why this step? Direct substitution; units L mol⁻¹cm⁻¹ × mol L⁻¹ × cm cancel to a pure number (absorbance is dimensionless).

Verify: Units: $(L mol^{- 1} cm^{- 1}) (mol L^{- 1}) (cm) =$ dimensionless ✓. Because $ε$ is tiny (forbidden band), you need a large 0.03 M concentration for a modest absorbance — matching the forecast that weak absorbers need concentrated solutions. ✓

Case I — exam twist

Worked example Ex 8 —

$[Co (H_{2} O)_{6}]^{3 +}$ absorbs at $\overset{ν}{ˉ}_{1} = 16, 500 cm^{- 1}$ and $[Co (NH_{3})_{6}]^{3 +}$ at $\overset{ν}{ˉ}_{2} = 21, 000 cm^{- 1}$ . Which absorbs bluer light, which has the larger $Δ_{o}$ , and does this obey the spectrochemical series? Forecast: NH₃ vs H₂O — which is the stronger-field ligand? Predict which has the bigger $\overset{ν}{ˉ}$ .

Step 1. Recall $Δ_{o} = h c \overset{ν}{ˉ}$ , so larger $\overset{ν}{ˉ}$ ⇒ larger $Δ_{o}$ directly (no inversion, because wavenumber ∝ energy). Why this step? Comparing wavenumbers is comparing energies without conversion.

Step 2. $\overset{ν}{ˉ}_{2} = 21, 000 > \overset{ν}{ˉ}_{1} = 16, 500$ , so the ammine complex has the larger gap: $Δ_{o} (NH_{3}) > Δ_{o} (H_{2} O) .$ Why this step? Bigger wavenumber = higher-energy photon absorbed = wider $t_{2 g}$ – $e_{g}$ split.

Step 3. Higher energy = shorter wavelength = bluer. Convert to check: $λ_{2} = 1/21, 000 cm = 476 nm$ (blue) vs $λ_{1} = 1/16, 500 = 606 nm$ (orange). The ammine absorbs bluer. Why this step? Turns abstract wavenumbers into a wavelength you can name a colour for.

Step 4. Spectrochemical order: NH₃ sits above H₂O in the Spectrochemical Series, so NH₃ should give the larger $Δ_{o}$ — and it does. Why this step? The twist is checking the data against theory; they agree.

Verify: $1/21000 cm = 4.76 \times 1 0^{- 5} cm = 476 nm$ ✓; $1/16500 = 606 nm$ ✓. NH₃ stronger-field → bigger gap → bluer absorption, consistent. ✓ (Forecast: NH₃ bigger — correct.)

Recall

Recall Which cell does each ion hit?

$d^{0}$ Cr(VI) chromate ::: Case F — LMCT, allowed, intense $d^{10}$ Zn²⁺ aqua ::: Case D — no empty $e_{g}$ , colourless High-spin $d^{5}$ Mn²⁺ ::: Case E — Laporte + spin forbidden, pale Strong-field CN⁻ complex ::: Case A — large $Δ_{o}$ , absorbs bluer