Colour of complexes — d-d transitions: Worked examples

Everything here uses only two ideas you already earned: $Δ = \frac{h c}{λ} ⟺ λ = \frac{h c}{Δ}$ and the rule seen colour = complement of absorbed colour (see Complementary Colours & the Colour Wheel).

A reminder of the symbols, so nothing is unearned:

$h = 6.626 \times 1 0^{- 34} J s$ (Planck's constant — the "price tag" on one photon per unit frequency, from Planck's Equation E=hν).
$c = 3.0 \times 1 0^{8} m s^{- 1}$ (speed of light).
$λ$ (lambda) = wavelength of the absorbed light, in metres. $1 nm = 1 0^{- 9} m$ .
$Δ$ = crystal-field splitting energy (the gap an electron must leap), in joules per photon.
$N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ (Avogadro) — multiplies a per-photon energy up to a per-mole energy.

The scenario matrix

Every d-d colour question is really one of these ten cells. Each worked example below is tagged with its cell.

#	Cell (the "input class")	What's special	Example
A	$d^{0}$ — empty shell	no electron to jump	Ex. 1
B	$d^{10}$ — full shell	no empty slot to jump into	Ex. 1
C	$d^{1}$ — one clean jump	textbook baseline	Ex. 2
D	colour → energy (forward)	given $λ$ , find $Δ$	Ex. 2, 3
E	energy → colour (backward)	given $Δ$ , find $λ$ & seen colour	Ex. 4
F	weak vs strong ligand	same metal, $Δ$ shifts	Ex. 5
G	geometry change (oct ↔ tet)	$Δ_{t} = \frac{4}{9} Δ_{o}$	Ex. 6
H	real-world word problem	ruby / gem colour	Ex. 7
I	exam twist (sign/limit trap)	"bigger $Δ$ = longer $λ$ ?" & boundary of visible	Ex. 8, 9
J	high-spin vs low-spin ( $d^{4}$ – $d^{7}$ )	$Δ$ vs pairing energy branch	Ex. 10

Worked example Example 1 — Cells A & B: the two colourless extremes

Predict the colour of (i) $[Sc (H_{2} O)_{6}]^{3 +}$ and (ii) $[Zn (H_{2} O)_{6}]^{2 +}$ .

Forecast: Guess before reading — will either be coloured?

Count d-electrons. $Sc^{3 +}$ : scandium is $[Ar] 3 d^{1} 4 s^{2}$ ; removing 3 electrons empties everything → $d^{0}$ . $Zn^{2 +}$ : zinc is $[Ar] 3 d^{10} 4 s^{2}$ ; removing 2 electrons leaves $3 d^{10}$ → $d^{10}$ . Why this step? d-d colour needs an electron and a vacant d-level; the electron count decides both at once.
Check the jump for Sc³⁺ ( $d^{0}$ ). The lower $t_{2 g}$ set is empty — there is no electron sitting there to be kicked up. The splitting $Δ$ still exists, but nobody can use it. Why this step? This is the "zero-input" case A: gap present, occupant absent.
Check the jump for Zn²⁺ ( $d^{10}$ ). Both $t_{2 g}$ (6 electrons) and $e_{g}$ (4 electrons) are completely full. An electron could jump only if the upper set had a free slot; it doesn't. Why this step? Case B: occupant present, destination absent — Pauli exclusion blocks the move.
Conclude. Neither absorbs any visible photon → all white light transmits → both colourless.

Verify: Sanity check against the parent's rule "colour needs $d^{1}$ to $d^{9}$ ." Both $d^{0}$ and $d^{10}$ lie outside that range → colourless. Consistent. ✓

Worked example Example 2 — Cells C & D:

$[Ti (H_{2} O)_{6}]^{3 +}$ , colour → energy The purple titanium complex absorbs at $λ = 500 nm$ . Find $Δ_{o}$ in joules per photon and in kJ mol⁻¹.

Forecast: Purple is seen — so which colour is absorbed? (Use the colour-wheel inset above.)

Identify the transition. $Ti^{3 +}$ is $d^{1}$ : one electron in $t_{2 g}$ (the low set). Absorbing a photon lifts it $t_{2 g} \to e_{g}$ (low → high). Why this step? With a single electron there is exactly one possible d-d move, so exactly one absorption band — the cleanest case (cell C).
Convert wavelength to metres. $λ = 500 nm = 500 \times 1 0^{- 9} m = 5.00 \times 1 0^{- 7} m$ . Why this step? SI must be consistent or the joules come out wrong by powers of ten.
Apply $Δ_{o} = h c / λ$ (per photon). $Δ_{o} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{5.00 \times 1 0 ^{- 7}} = 3.98 \times 1 0^{- 19} J$ Why this step? This is the whole theory: a measured colour becomes a measured energy gap (cell D, forward direction).
Scale to a mole. $Δ_{o} \times N_{A} = (3.98 \times 1 0^{- 19}) (6.022 \times 1 0^{23}) = 2.40 \times 1 0^{5} J mol^{- 1} \approx 240 kJ mol^{- 1}$ . Why this step? Chemists compare gaps per mole; multiply by Avogadro's number.

Verify: $240 kJ mol^{- 1}$ is in the typical crystal-field range ( $100$ – $400 kJ mol^{- 1}$ ). And 500 nm is green-yellow; its complement on the wheel is purple — matches the observed colour. ✓

Worked example Example 3 — Cell D again, but in wavenumbers (exam-standard units)

A complex shows its absorption maximum at $\overset{ν}{ˉ} = 20, 000 cm^{- 1}$ . Find $λ$ in nm and $Δ_{o}$ in J.

Forecast: Wavenumber is "how many waves per cm." Is a bigger wavenumber a bigger or smaller wavelength?

Define the wavenumber. $\overset{ν}{ˉ} = 1/ λ$ (waves per unit length). So $λ = 1/ \overset{ν}{ˉ}$ . Why this step? Spectroscopists quote $\overset{ν}{ˉ}$ because it is directly proportional to energy ( $E = h c \overset{ν}{ˉ}$ ), so it's linear in $Δ$ — convenient.
Get $λ$ . $\overset{ν}{ˉ} = 20, 000 cm^{- 1} = 20, 000 \times 100 = 2.0 \times 1 0^{6} m^{- 1}$ . $λ = \frac{1}{2.0 \times 1 0 ^{6}} = 5.0 \times 1 0^{- 7} m = 500 nm$ Why this step? Convert cm⁻¹ → m⁻¹ (×100) before inverting, else the metres are off.
Get $Δ_{o} = h c \overset{ν}{ˉ}$ . $Δ_{o} = (6.626 \times 1 0^{- 34}) (3.0 \times 1 0^{8}) (2.0 \times 1 0^{6}) = 3.98 \times 1 0^{- 19} J$ Why this step? Same gap as Example 2 — this confirms $20, 000 cm^{- 1}$ and $500 nm$ describe the same photon.

Verify: $λ = 500 nm$ matches Example 2, and $Δ_{o} = 3.98 \times 1 0^{- 19} J$ matches too. Two different units, one physical gap. ✓

Worked example Example 4 — Cell E: energy → colour (the backward direction)

A complex has $Δ_{o} = 3.0 \times 1 0^{- 19} J$ . Which wavelength does it absorb, and what colour does it appear?

Forecast: Bigger or smaller gap than Ti³⁺'s $3.98 \times 1 0^{- 19} J$ ? Will $λ$ be longer or shorter?

Rearrange for wavelength. $λ = h c / Δ_{o}$ . Why this step? Now the energy is the known input and the colour is the unknown output — the reverse of Example 2 (cell E).
Plug in. $λ = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{3.0 \times 1 0 ^{- 19}} = 6.63 \times 1 0^{- 7} m = 663 nm$ Why this step? Direct substitution; keep SI.
Read the colour. $663 nm$ is red light absorbed. Its complement (across the wheel) is green. Why this step? We see white-minus-absorbed = complement.

Verify: The gap ( $3.0 \times 1 0^{- 19}$ ) is smaller than Ti³⁺'s ( $3.98 \times 1 0^{- 19}$ ), so $λ$ should be longer than 500 nm — and $663 > 500$ nm. The inverse law behaves correctly. ✓

Worked example Example 5 — Cell F: same metal, two ligands (spectrochemical shift)

$[Cu (H_{2} O)_{6}]^{2 +}$ absorbs at $\approx 800 nm$ ; on adding ammonia, $[Cu (NH_{3})_{4}]^{2 +}$ absorbs at $\approx 600 nm$ . Show the gap grows, and explain the colour change from pale blue to deep blue.

Forecast: Which is the stronger ligand, $H_{2} O$ or $NH_{3}$ ? (Check the spectrochemical-series reminder above — $NH_{3}$ sits to the right of $H_{2} O$ .)

Compute both gaps. Using $Δ = h c / λ$ : $Δ_{aqua} = \frac{h c}{800 \times 1 0 ^{- 9}} = 2.48 \times 1 0^{- 19} J, Δ_{ammine} = \frac{h c}{600 \times 1 0 ^{- 9}} = 3.31 \times 1 0^{- 19} J$ Why this step? Turning both colours into energies lets us compare ligand strength as a number.
Compare. $Δ_{ammine} > Δ_{aqua}$ — the gap grew because $NH_{3}$ sits higher (further right) in the spectrochemical series than $H_{2} O$ . Why this step? This is the series in action: stronger field ⇒ larger $Δ$ .
Track the colour. Aqua absorbs 800 nm (far red) → looks pale blue. Ammine absorbs 600 nm (orange) → complement is deep blue/violet, and more intense. Why this step? As absorbed $λ$ shifts red→orange, seen colour shifts oppositely and deepens.

Verify: Ratio $Δ_{ammine} / Δ_{aqua} = 800/600 = 1.33$ — the gap rose by exactly the inverse wavelength ratio, as $Δ \propto 1/ λ$ demands. ✓

Worked example Example 6 — Cell G: octahedral → tetrahedral geometry flip

An octahedral complex has $Δ_{o} = 3.0 \times 1 0^{- 19} J$ (absorbs 663 nm, from Ex. 4). Rebuild the same metal + ligands as a tetrahedral complex. What $Δ_{t}$ , what $λ$ , and does it still absorb visible light?

Forecast: Tetrahedral splitting is only $\frac{4}{9}$ of octahedral. Will $λ$ move toward red or violet?

The figure below is a wavelength axis with the visible band (380–750 nm) shaded. The blue dot marks where the octahedral complex absorbs (663 nm, inside the band); the green arrow then slides that dot rightward to the orange dot — the tetrahedral absorption at 1492 nm, which lands outside the shaded band in the infra-red. The picture makes concrete the rule "smaller gap → longer $λ$ → can leave the visible window."

Scale the gap. $Δ_{t} = \frac{4}{9} Δ_{o} = \frac{4}{9} (3.0 \times 1 0^{- 19}) = 1.33 \times 1 0^{- 19} J$ . Why this step? In tetrahedral fields only 4 ligands approach and none point straight at the d-orbitals (see Octahedral vs Tetrahedral Splitting), so the gap shrinks by the standard factor $\frac{4}{9}$ .
Find the new wavelength. $λ = h c / Δ_{t} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{1.33 \times 1 0 ^{- 19}} = 1.49 \times 1 0^{- 6} m = 1492 nm$ . Why this step? Smaller gap ⇒ longer wavelength; look at the green arrow in the figure sliding the dot rightward off the visible band.
Check the visible window. Visible light is ~380–750 nm. $1492 nm$ is deep infra-red — invisible to us. Why this step? If the absorbed band leaves the visible range, no visible colour is removed.

Verify: $1492/663 = 2.25 = 9/4$ exactly — dividing $Δ$ by $\frac{4}{9}$ multiplies $λ$ by $\frac{9}{4}$ , so the numbers close the loop. A very small tetrahedral gap can push absorption out of view (this is why some tetrahedral $d$ -systems look nearly colourless or only faintly tinted). ✓

Worked example Example 7 — Cell H: real-world word problem (why ruby is red)

A ruby is $Cr^{3 +}$ in an alumina lattice, an octahedral $d^{3}$ centre. It absorbs strongly in the violet (~410 nm) and the green-yellow (~560 nm), transmitting red. Estimate the two gap energies (kJ mol⁻¹) and explain the deep red.

Forecast: Two absorptions, both energies — does the deeper red come from absorbing more colours or fewer?

Gap for the violet band. $Δ_{1} = h c / λ_{1} = \frac{h c}{410 \times 1 0 ^{- 9}} = 4.85 \times 1 0^{- 19} J$ ; × $N_{A}$ → $292 kJ mol^{- 1}$ . Why this step? Each absorption band corresponds to one allowed $d^{3}$ transition; converting to energy pins the band.
Gap for the green band. $Δ_{2} = h c / λ_{2} = \frac{h c}{560 \times 1 0 ^{- 9}} = 3.55 \times 1 0^{- 19} J$ ; × $N_{A}$ → $214 kJ mol^{- 1}$ . Why this step? $d^{3}$ ions have several excited states, hence two visible bands — richer than the single $d^{1}$ band.
Combine the missing colours. Violet and green-yellow are removed from white light; the surviving transmitted mix is dominated by red → the ruby glows red. Why this step? Colour is white minus everything absorbed; removing two chunks leaves red.

Verify: $Δ_{1} > Δ_{2}$ because $410 nm < 560 nm$ ( $Δ \propto 1/ λ$ ) — consistent. Both gaps ( $292$ , $214 kJ mol^{- 1}$ ) sit in the normal $Cr^{3 +}$ range. ✓

Worked example Example 8 — Cell I: the sign trap

True or false: "Complex X has a larger $Δ_{o}$ than complex Y, so X absorbs the longer wavelength." Justify with numbers.

Forecast: Trust your gut, then test it.

Write the relation. $λ = h c / Δ_{o}$ . Since $h c$ is a positive constant, $λ$ is inversely proportional to $Δ_{o}$ . Why this step? The trap hides in the word "larger": bigger $Δ$ divides into $h c$ , giving a smaller $λ$ .
Numeric counter-check. Let $Δ_{X} = 4.0 \times 1 0^{- 19}$ J, $Δ_{Y} = 2.0 \times 1 0^{- 19}$ J. Then $λ_{X} = h c / Δ_{X} = 497 nm$ , $λ_{Y} = h c / Δ_{Y} = 994 nm$ . Why this step? Concrete numbers kill hand-waving: the larger-gap X absorbs the shorter wave.
Conclude. The statement is FALSE: larger $Δ$ ⇒ shorter $λ$ (more energetic photon).

Verify: $λ_{X} < λ_{Y}$ while $Δ_{X} > Δ_{Y}$ — the inequality flips, confirming inverse proportionality. This is exactly the parent note's steel-manned mistake. ✓

Worked example Example 9 — Cell I: the limiting / boundary case

A complex has $Δ_{o} = 5.2 \times 1 0^{- 19} J$ . Does it absorb inside the visible window (380–750 nm)? What if $Δ_{o}$ were only $2.0 \times 1 0^{- 19} J$ ?

Forecast: Larger gap → pushes absorption toward which edge, red or violet?

Test the large gap. $λ = h c / Δ_{o} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{5.2 \times 1 0 ^{- 19}} = 3.82 \times 1 0^{- 7} m = 382 nm$ . Why this step? We must locate $λ$ relative to the violet edge (380 nm) — the high-energy boundary of visibility.
Judge the large-gap case. $382 nm$ is just inside the violet edge → the complex absorbs faint violet and would look a pale yellow-green (the complement of violet). Why this step? Right at the boundary the absorbed band is barely visible, so the colour is weak — a genuine limiting case.
Test the small gap. $λ = h c / (2.0 \times 1 0^{- 19}) = 9.94 \times 1 0^{- 7} m = 994 nm$ → deep infra-red, outside the visible range → no visible colour removed → appears colourless (just like the tetrahedral case in Ex. 6). Why this step? This shows the other boundary: too small a gap leaves visible light untouched.
State the full range. As $Δ_{o}$ runs from small to large, the absorbed $λ$ slides from IR (>750 nm, colourless) through the whole visible band and out to the violet edge (<380 nm, colourless again). Colour only appears when $λ$ lands inside 380–750 nm. Why this step? Naming both edges completes the "every limit" requirement of the matrix.

Verify: As $Δ_{o}$ goes $2.0 \to 5.2 \times 1 0^{- 19}$ J, $λ$ goes $994 \to 382$ nm — a monotonic slide from IR through visible to the violet edge. Every band of "gap size" is now covered: too big → violet edge, mid → visible colour, too small → IR/colourless. ✓

Worked example Example 10 — Cell J: high-spin vs low-spin

$d^{6}$ ( $Fe^{2 +}$ ) $Fe^{2 +}$ is $d^{6}$ . Compare $[Fe (H_{2} O)_{6}]^{2 +}$ (weak field) with $[Fe (CN)_{6}]^{4 -}$ (strong field). Which is high-spin, which is low-spin, and why do their colours differ?

Forecast: $CN^{-}$ is at the far-right (strongest) of the series — will it force pairing or not?

Aqua case (weak field). $H_{2} O$ gives a small $Δ_{o}$ , so $Δ_{o} < P$ . The 6 electrons spread out: $t_{2 g}^{4} e_{g}^{2}$ → 4 unpaired, high-spin. Why this step? When the gap is cheaper to avoid than pairing is to pay, electrons occupy $e_{g}$ rather than double up.
Cyanide case (strong field). $CN^{-}$ gives a large $Δ_{o}$ , so $Δ_{o} > P$ . All 6 electrons crowd the lower set: $t_{2 g}^{6} e_{g}^{0}$ → 0 unpaired, low-spin. Why this step? Now paying $P$ is cheaper than jumping the huge gap, so electrons pair inside $t_{2 g}$ .
Link to colour. The strong-field cyanide has the larger $Δ_{o}$ → absorbs shorter $λ$ than the aqua complex ( $λ = h c /Δ$ ). So the two $d^{6}$ complexes, though the same ion, absorb different colours and look different — pale green vs. yellow, respectively. Why this step? Ligand strength sets both the spin state and (through $Δ$ ) the absorbed colour — one cause, two visible consequences.

Verify: Cross-check with magnetism: high-spin $d^{6}$ has 4 unpaired electrons ( $[Fe (H_{2} O)_{6}]^{2 +}$ is paramagnetic), low-spin $d^{6}$ has 0 ( $[Fe (CN)_{6}]^{4 -}$ is diamagnetic) — exactly what experiment reports. And $Δ_{CN} > Δ_{aqua}$ matches $CN^{-}$ being right of $H_{2} O$ in the series. ✓

Common mistake Edge case the calculation alone misses — forbidden transitions

The formula $Δ = h c / λ$ tells you which wavelength is absorbed, but not how strongly. Two selection rules make many "allowed-by-energy" d-d bands very weak:

Laporte rule: in a perfectly symmetric octahedron a $d \to d$ jump is "Laporte-forbidden" (no change in orbital-type symmetry), so the band is faint. Slight vibrations/distortions relax this, letting a little light through — which is why $[Ti (H_{2} O)_{6}]^{3 +}$ is only pale purple, not vivid.
Spin rule: a transition that would require flipping an electron's spin is "spin-forbidden" and fainter still — this is why $[Mn (H_{2} O)_{6}]^{2 +}$ ( $d^{5}$ , high-spin) is only very pale pink. Contrast: charge-transfer bands (see Charge-Transfer Spectra (why KMnO4 is intensely coloured)) are not d-d and not Laporte-forbidden, so they are hundreds of times more intense — that is why $KMnO_{4}$ is deep purple while a genuine d-d solution is pale. So the full picture is: energy sets the colour, selection rules set the intensity.

Recall Self-test — cover the answers

Sc³⁺ and Zn²⁺ are both colourless — same reason? ::: No: Sc³⁺ is $d^{0}$ (no electron), Zn²⁺ is $d^{10}$ (no vacancy). Different cells, same result. What are the $t_{2 g}$ and $e_{g}$ sets? ::: $t_{2 g}$ = lower three d-orbitals (point between ligands); $e_{g}$ = upper two (point at ligands). A d-d jump goes $t_{2 g} \to e_{g}$ . A complex absorbs $20, 000 cm^{- 1}$ . What is $λ$ ? ::: $500 nm$ (invert after converting cm⁻¹→m⁻¹). $Δ_{o} = 3.0 \times 1 0^{- 19}$ J → absorbed colour and seen colour? ::: Absorbs red (663 nm), looks green. Convert $Δ_{o}$ from octahedral to tetrahedral for the same ML set. ::: Multiply by $\frac{4}{9}$ ; $λ$ then × $\frac{9}{4}$ (longer, often IR). "Bigger gap absorbs longer wavelength" — true? ::: False; $λ = h c /Δ$ is inverse, bigger gap → shorter $λ$ . Why is $[Mn (H_{2} O)_{6}]^{2 +}$ only faintly pink? ::: Its d-d transitions are spin-forbidden (and Laporte-forbidden), so absorption is very weak. When is a $d^{6}$ ion low-spin? ::: When the field is strong enough that $Δ_{o} > P$ (pairing energy), e.g. with $CN^{-}$ ; then $t_{2 g}^{6} e_{g}^{0}$ , 0 unpaired.

Connections

3.3.05 Colour of complexes — d-d transitions (Hinglish) — the parent topic
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