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

We will use one worked system the whole way down: $[\text{Ti}({\text{H}}_2\text{O}{)}_6{]}^{3+}$, a $d^1$ ion — the simplest possible coloured complex (exactly one electron to jump).

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Step 1 — What "white light" actually is

WHY start here? Because "colour" only makes sense once we know that white = the full set. If you remove one colour from the set, the leftover mixture is a new colour. That subtraction is the entire story — so we must first see the full set to remove from.

PICTURE. The strip below is the visible spectrum. Short waves (tight wiggles) on the left are violet; long waves (loose wiggles) on the right are red. The number under each is its wavelength $\lambda$ — the distance between two wave crests, in nanometres ($1\text{ nm}=10^{-9}\text{ m}$).

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Step 2 — Each colour is a packet of energy (the photon)

WHY do we need this? An electron can only jump if it is handed exactly the right amount of energy. So we must know how much energy each colour of photon carries. That is the bridge from "colour" to "energy."

PICTURE. Same spectrum strip, now with an energy arrow: it grows as we move to short wavelength. The equation that ties them together:

$E=\frac{hc}{\lambda }$

• $E$ — the energy carried by one photon (in joules, J).
• $h$ — Planck's constant $=6.626\times 10^{-34}\text{ J s}$; the fixed "conversion rate" from wave-frequency to energy. It never changes.
• $c$ — the speed of light $=3.0\times 10^8{\text{ m s}}^{-1}$; also fixed.
• $\lambda$ — the wavelength from Step 1. It sits in the denominator, which is the whole point: bigger $\lambda$ ⇒ smaller $E$. Red light = low energy; violet = high energy.

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Step 3 — The d-orbitals split into two shelves

WHY does this matter for colour? Two shelves at different heights create a gap. A gap is a target energy: hand the electron a photon worth exactly ${\Delta }_o$ and it climbs from the low shelf to the high shelf. No gap, no jump, no colour.

PICTURE. Two horizontal shelves. The lower shelf ($t_{2g}$) holds our single $d^1$ electron (red dot). The upper shelf ($e_g$) is empty and waiting. The red double arrow between them is ${\Delta }_o$.

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Step 4 — The jump: a photon of energy exactly ${\Delta }_o$ is absorbed

WHY "exactly"? Energy is quantised — the electron can only sit on a shelf, never between. So only a photon carrying precisely ${\Delta }_o$ can complete the climb. Too little energy: can't reach. Too much: overshoots a non-existent level. This is why a complex absorbs a narrow slice, not everything.

PICTURE. The electron leaps from the lower to the upper shelf; the incoming red photon-wave is swallowed at the moment of the jump. Setting the photon energy (Step 2) equal to the gap (Step 3):

${\Delta }_o=\frac{hc}{{\lambda }_{\text{abs}}}$

• ${\Delta }_o$ — the gap the electron must clear (left side).
• $\frac{hc}{{\lambda }_{\text{abs}}}$ — the energy of the one photon able to do it.
• ${\lambda }_{\text{abs}}$ — the absorbed wavelength: the specific colour that goes missing. Solve for it and you have predicted which colour disappears.

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Step 5 — Rearrange to find the missing colour

WHY wavenumber? Because ${\Delta }_o=hc\bar{\nu }$: energy and $\bar{\nu }$ are locked together by the fixed factor $hc$. The constant hasn't vanished — you still multiply by $hc$ when you finally want joules — but reporting in $\bar{\nu }$ postpones that step, so ranking and comparing gaps is as simple as comparing wavenumbers, and inverting one gives $\lambda$ directly.

PICTURE. A dial: on one side the measured ${\Delta }_o=20000{\text{ cm}}^{-1}$; turning the crank $\lambda =1/\bar{\nu }$ lands the pointer at 500 nm — green light.

${\lambda }_{\text{abs}}=\frac{1}{\bar{\nu }}=\frac{1}{20000{\text{ cm}}^{-1}}=5.0\times 10^{-5}\text{ cm}=500\text{ nm}$

• $\bar{\nu }=20000{\text{ cm}}^{-1}$ — the measured gap, read off the spectrum.
• $1/\bar{\nu }$ — undoing the wavenumber definition to recover a length.
• Unit chain, spelled out: $\frac{1}{20000{\text{ cm}}^{-1}}=5\times 10^{-5}\text{ cm}$. Now convert cm→nm in two known hops: $1\text{ cm}=10^{-2}\text{ m}$ and $1\text{ m}=10^9\text{ nm}$, so $1\text{ cm}=10^7\text{ nm}$. Hence $5\times 10^{-5}\text{ cm}\times 10^7\frac{\text{nm}}{\text{cm}}=500\text{ nm}$.

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Step 6 — From absorbed colour to seen colour (the flip)

WHY the complement, not the absorbed colour? Because you never see the absorbed light — it's gone. You see everything else. On a colour wheel, removing one colour leaves you seeing the colour directly opposite it.

PICTURE. A colour wheel: absorbed colour (green, marked in red) on one side; the arrow points across to its opposite (purple) — the reported colour of $[\text{Ti}({\text{H}}_2\text{O}{)}_6{]}^{3+}$.

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Step 7 — Edge cases: when nothing happens (and when everything does)

First, one new symbol we will lean on:

• $d^0$ (Sc³⁺, Ti⁴⁺): the low shelf is empty. No electron to lift ⇒ no d–d absorption.
• $d^{10}$ (Zn²⁺, Cu⁺): the high shelf is full. No vacancy to land in ⇒ no d–d absorption.
• High-spin $d^5$ (${\text{Mn}}^{2+}$): every orbital holds one electron with parallel spin. Any jump would force a spin-flip, so the transition is doubly forbidden (Magnetic Properties of Complexes tracks these spins) ⇒ extremely pale ($\epsilon \approx 0.01$–$0.1$).
• Charge transfer (${\text{MnO}}_4^{-}$, $d^0$): no d–d possible, yet it is intensely coloured — because an electron leaps from ligand to metal (LMCT), a fully allowed jump with $\epsilon \sim 10^4$.

WHY show all four? So no reader ever meets a coloured (or colourless) complex the derivation didn't cover.

PICTURE. Four mini shelf-diagrams side by side: $d^0$ (empty low), $d^{10}$ (full high), $d^5$ (all half-filled, red "spin-clash" mark), and a CT arrow leaping the whole way from a separate ligand box.

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The one-picture summary

Everything in a single flow: white light → the one matching photon lifts the $t_{2g}\to e_g$ electron across ${\Delta }_o$ → that wavelength goes missing → the complement reaches your eye. The rendered flowchart below carries the whole story.

Recall Feynman retelling (hide and re-explain)

White light is a full box of coloured crayons. The metal ion has two shelves — a low one with a spare electron, a high empty one — separated by a gap called ${\Delta }_o$. A crayon (photon) only gets used if its energy exactly matches the gap; that one crayon is spent lifting the electron up, so it disappears from the box. Whatever crayons are left, mixed together, is the colour you see — the opposite of the one that vanished. If the low shelf is empty ($d^0$) or the high shelf is full ($d^{10}$), no electron can jump and the box stays complete — colourless. If the electron would have to awkwardly flip its spin ($d^5$), it barely jumps at all, giving a faint pastel. And if an electron takes a huge leap all the way from the ligand onto the metal (charge transfer), that jump is bold and allowed, giving a deep, vivid colour like purple permanganate.

Recall Numbers to keep

• ${\Delta }_o=20000{\text{ cm}}^{-1}$ absorbs at ::: $\lambda =1/20000\text{ cm}=500\text{ nm}$ (green) ⇒ looks purple.
• $E=hc/\lambda$: larger ${\Delta }_o$ ⇒ ::: shorter $\lambda$ (bluer absorption).
• Colourless d-counts ::: $d^0$ and $d^{10}$.
• Intense colour with $d^0$ means ::: charge transfer, not d–d.

See also: Electronic Spectra & Term Symbols, Oxidation States of Transition Metals.