Colour of complexes — d-d transitions: Visual walkthrough

Step 1 — Five identical shelves (the free ion)

WHAT. Picture a lonely transition-metal ion floating in space with nothing near it. It carries some electrons in its outermost d-orbitals. A d-orbital is just a region of space where a d-electron is likely to be found — think of it as a shelf an electron can sit on. There are exactly five of these shelves.

WHY. Before we can talk about a "gap", we need to know what things look like with no gap at all. When the ion is alone, all five shelves sit at the same height — the fancy word is degenerate (equal energy). That is our zero point.

PICTURE. Five flat lines at the same level. An electron on any of them costs the same energy.

Step 2 — The five shelves have two shapes

WHAT. The five d-orbitals are not all shaped alike. Two of them stick their lobes straight along the x, y, z axes: these are $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ . The other three point into the gaps between the axes (diagonally): these are $d_{x y}$ , $d_{y z}$ , $d_{z x}$ .

WHY. In Step 1 the shape did not matter (nothing was nearby to notice it). In the next step we put charges on the axes, so which direction a shelf points suddenly decides how hard it gets pushed. We must separate the two shape-families now, before the push arrives.

PICTURE. Left: an "axis-pointing" orbital aiming straight at where a ligand will sit. Right: a "between-the-axes" orbital aiming into empty space.

Step 3 — Six ligands arrive along the axes (octahedral)

WHAT. Bring six ligands up to the ion, one along $+ x$ , one along $- x$ , and likewise for $y$ and $z$ . This six-along-the-axes arrangement is called octahedral.

WHY. Every ligand carries negative charge; d-electrons are also negative. Like charges repel. So the ligands push up the energy of any electron whose shelf points towards them. The two axis-pointing orbitals ( $d_{x^{2} - y^{2}}$ , $d_{z^{2}}$ ) stare directly into the ligands and get shoved up hard. The three between-the-axes orbitals ( $d_{x y}$ , $d_{y z}$ , $d_{z x}$ ) dodge the ligands and are pushed up less.

PICTURE. The ion at the centre; six red ligand blobs on the axes; arrows showing the axis-orbitals catching the full repulsion and the diagonal ones slipping past.

Step 4 — The shelves split into two groups: $e_{g}$ (high) and $t_{2 g}$ (low)

WHAT. The five degenerate shelves separate into:

$e_{g}$ = ${d_{x^{2} - y^{2}}, d_{z^{2}}}$ — the upper pair (they faced the ligands).
$t_{2 g}$ = ${d_{x y}, d_{y z}, d_{z x}}$ — the lower trio (they dodged).

The vertical energy distance between these two groups is the star of the whole story:

$Δ_{o}$

Read this as: $Δ$ (Greek "delta", meaning a gap/difference), subscript $o$ for octahedral. It is measured in energy units (joules per photon, or kilojoules per mole).

WHY. This gap is the "shelf-jump height" an electron must be given to hop from the low trio to the high pair. Everything about the colour now hides inside the size of $Δ_{o}$ .

PICTURE. The energy-level diagram: the low $t_{2 g}$ trio, the high $e_{g}$ pair, and $Δ_{o}$ drawn as the arrow between them.

Step 5 — An electron jumps the gap by eating a photon

WHAT. Shine white light on the complex. Light is a stream of photons — little packets of energy. A photon of energy $E_{photon}$ can be absorbed only if that energy exactly matches the jump height:

$E_{photon} = Δ_{o}$

$E_{photon}$ — the energy carried by one packet of light.
$Δ_{o}$ — the gap we just built (the required jump height).
$=$ — the "exactly fits" condition: too small a photon can't lift the electron; too big and it isn't the resonant match either — only the exact-size packet is swallowed.

WHY. An electron cannot sit between shelves. To climb from $t_{2 g}$ to $e_{g}$ it must receive the whole $Δ_{o}$ in one bite. Light delivers energy only in whole photons, so we need one photon worth exactly $Δ_{o}$ .

PICTURE. An electron on the $t_{2 g}$ trio absorbs an incoming wavy photon and lands on the $e_{g}$ pair; the photon of that one colour vanishes from the beam.

Step 6 — Turn "energy" into "which colour" (Planck + the wave law)

WHAT. We know the energy the complex eats. We want the colour. Colour is set by wavelength $λ$ (the length of one ripple of the light wave). Two ready-made facts bridge energy and wavelength.

First, Planck's rule:

$E_{photon} = h ν$

$h$ — Planck's constant, $6.626 \times 1 0^{- 34} J\cdots$ , the fixed "exchange rate" between a wave's frequency and its energy.
$ν$ — ("nu") the frequency: how many ripples pass per second.
So a faster-wiggling wave carries proportionally more energy. Why this tool and not another? Planck's rule is the only law that converts a light-wave's rhythm into an energy — exactly the conversion we need.

Second, the wave relation every wave obeys:

$c = λ ν \Rightarrow ν = \frac{c}{λ}$

$c$ — the speed of light, $3 \times 1 0^{8} m/s$ .
$λ$ — the wavelength (metres), what our eye reads as colour.
This just says: speed = (length of one ripple) × (ripples per second). We solve it for $ν$ because we want to eliminate frequency in favour of wavelength.

WHY. We have $E$ in terms of $ν$ , but we think in wavelength/colour. Swapping $ν \to c / λ$ rewrites everything in the language of colour.

PICTURE. One wave drawn twice: a short-wavelength (blue, high-energy) ripple and a long-wavelength (red, low-energy) ripple, with $λ$ marked on each.

Step 7 — Assemble the master formula

WHAT. Chain the three equalities from Steps 5 and 6:

$Δ_{o} = E_{photon} = h ν = h \cdot \frac{c}{λ}$

which collapses to the boxed result:

$Δ_{o} = \frac{h c}{λ} ⟺ λ = \frac{h c}{Δ _{o}}$

Reading the second box term by term:

$λ$ on top of a fraction with $Δ_{o}$ on the bottom ⇒ $λ$ and $Δ_{o}$ are inversely related.
$h$ and $c$ are fixed constants — they set the scale but never change.
So: big gap $Δ_{o}$ ⇒ small $λ$ absorbed (energetic, blue-end light). Small gap ⇒ large $λ$ absorbed (weak, red-end light).

WHY. This single line is the whole theory: it turns a measured gap into a predicted absorbed colour, and vice-versa. This is exactly what the parent note boxed — now you have earned every symbol.

PICTURE. A see-saw: $Δ_{o}$ on one side, $λ$ on the other — push one up, the other drops, because $h c$ is a fixed weight in the middle.

Step 8 — From absorbed colour to seen colour (the complement)

WHAT. The complex removes the absorbed wavelength from white light. What passes through to your eye is "white minus the absorbed colour" — which the eye reads as the complementary colour (the opposite slice of the colour wheel).

WHY. We built what gets absorbed; but you never see the absorbed light — you see what survives. On the wheel, remove one wedge and the leftover mix balances out to the wedge directly opposite.

PICTURE. A colour wheel: absorbed green-yellow arrow pointing to its opposite, the purple you actually see — worked for $[T i (H_{2} O)_{6}]^{3 +}$ .

Step 9 — The degenerate cases: when NO colour appears

WHAT. The whole chain needs two things at Step 5: an electron on a lower shelf and an empty upper shelf to land on. Remove either and nothing is absorbed.

$d^{0}$ (e.g. $S c^{3 +}$ , $T i^{4 +}$ ): the shelves exist but no electron sits on them. Nobody to jump.
$d^{10}$ (e.g. $Z n^{2 +}$ , $C u^{+}$ ): both $t_{2 g}$ and $e_{g}$ are completely full. There is no empty seat in $e_{g}$ to jump into.

WHY. These are the boundary cases the formula cannot rescue: with no possible d-d jump, no visible photon is swallowed, white light passes untouched ⇒ colourless.

PICTURE. Three level-diagrams side by side: $d^{0}$ (empty), $d^{1}$ (jump possible → coloured), $d^{10}$ (full → blocked).

The one-picture summary

The single figure above compresses all nine steps: ligands push in → five shelves split into $t_{2 g}$ and $e_{g}$ with gap $Δ_{o}$ → one photon of exactly $Δ_{o}$ is eaten ( $λ = h c / Δ_{o}$ ) → that colour is subtracted from white → you see the complement.

Recall Feynman retelling (cover and narrate it yourself)

A metal ion has five little shelves for its electrons, all at the same height. Push six magnets (ligands) at it along the axes; the shelves that face the magnets get shoved higher, the ones that dodge stay low. Now there's a gap, $Δ_{o}$ . Shine sunlight — a bag of colour-kicks of every size. An electron grabs the one kick that exactly equals the gap and hops up; that colour vanishes from the light. Because a big gap needs a strong (short-wave) kick and a small gap needs a weak (long-wave) kick, the size of the gap picks which colour disappears. You never see the swallowed colour — you see the leftovers, which mix to the opposite colour on the wheel. And if the shelves are empty ( $d^{0}$ ) or totally full ( $d^{10}$ ), no electron can hop, no kick is grabbed, and the liquid stays clear.

Connections

3.3.05 Colour of complexes — d-d transitions (Hinglish)
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

Visual walkthrough — Colour of complexes — d-d transitions