Colour of complexes — d-d transitions — Chemistry notes

WHAT is happening?

Octahedral case:

$e_g$ set = $d_{x^2-y^2},\ d_{z^2}$ (point at ligands) → raised.
$t_{2g}$ set = $d_{xy},\ d_{yz},\ d_{zx}$ (point between ligands) → lowered.

HOW the size of $\Delta$ controls the colour

$\Delta$ depends on:

Ligand strength (spectrochemical series, weak→strong): $I^-<Br^-<Cl^-<F^-<OH^-<H_2O<NH_3<en<NO_2^-<CN^-<CO$ Stronger ligand ⇒ larger $\Delta$ ⇒ shorter $\lambda$ absorbed.
Metal oxidation state: higher charge pulls ligands closer ⇒ larger $\Delta$ .
Geometry: $\Delta_t=\tfrac{4}{9}\Delta_o$ (tetrahedral splitting is much smaller).
Position in transition series (3d < 4d < 5d for $\Delta$ ).

Figure — Colour of complexes — d-d transitions

Worked example Worked Example 1 —

$[Ti(H_2O)_6]^{3+}$ (the textbook classic) $Ti^{3+}$ is $d^1$ : a single electron sits in $t_{2g}$ .

Step: This one electron absorbs light and jumps $t_{2g}\to e_g$ . Why? It's the only possible d-d move.
The complex absorbs around $\lambda \approx 500\text{ nm}$ (green-yellow region).
Step: Removing green-yellow from white light leaves purple/violet — and indeed the solution is purple.
Estimate $\Delta_o$ : $\Delta_o=\dfrac{hc}{\lambda}=\dfrac{(6.626\times10^{-34})(3\times10^{8})}{500\times10^{-9}}\approx3.98\times10^{-19}\text{ J} \approx 240\ \text{kJ mol}^{-1}.$
Why this step? Plugging the observed absorbed wavelength back into $\Delta=hc/\lambda$ converts a colour into an energy, the whole point of the theory.

Common mistake Steel-manned misconceptions

"The complex shows the colour it absorbs." Why it feels right: a red rose looks red, so "thing = colour of light" seems natural. The fix: opaque objects show reflected colour; a solution shows transmitted = white minus absorbed = the complementary colour. $[Ti(H_2O)_6]^{3+}$ absorbs green-yellow but looks purple.

"Bigger $\Delta$ means longer wavelength absorbed." Why it feels right: "bigger" sounds like "bigger wavelength". The fix: $\lambda=hc/\Delta$ is inverse — bigger gap needs a more energetic = shorter wavelength photon.

"All transition metal ions are coloured." Why it feels right: most common ones ( $Cu^{2+},Fe^{3+},Cr^{3+}$ ) are. The fix: $d^0$ (Sc³⁺, Ti⁴⁺) and $d^{10}$ (Zn²⁺, Cu⁺) have no possible d-d transition ⇒ colourless.

Recall Feynman: explain to a 12-year-old

Imagine the metal atom has five identical shelves for its tiny electron-balls. When you push six magnets (ligands) around it, some shelves get raised a bit higher. Now an electron-ball can hop from a low shelf to a high shelf — but only if you give it exactly the right "kick". White sunlight is a bag of colour-kicks of all sizes. The ball grabs the one kick that fits the shelf-gap and uses it to jump; that colour disappears from the light. So when you look through the bottle, the rainbow is missing one colour — and the leftover mix is the colour you see! No electron to jump (full shelves) or no empty shelf? Then nothing is grabbed and the liquid stays clear.

Flashcards

Why are most transition metal complexes coloured?

A d-electron absorbs a visible photon and jumps between split d-orbital sets (d-d transition); the seen colour is the complement of the absorbed colour.

Formula linking splitting energy and absorbed wavelength?

\Delta_o = hc/\lambda

, so

\lambda = hc/\Delta_o

Larger

\Delta_o

means the complex absorbs which wavelength?

Shorter (more energetic) wavelength, because

\lambda \propto 1/\Delta_o

Why is

[Zn(H_2O)_6]^{2+}

colourless?

Zn^{2+}

d^{10}

; all d-orbitals are full, so no d-d transition is possible.

Why is

Sc^{3+}

colourless?

It is

d^0

; there is no d-electron available to undergo a transition.

What colour does

[Ti(H_2O)_6]^{3+}

appear and why?

Purple/violet; it absorbs green-yellow (~500 nm), and we see the complement.

Spectrochemical series (weak→strong, key members)?

I^-<Br^-<Cl^-<F^-<H_2O<NH_3<en<NO_2^-<CN^-<CO

Effect of stronger ligand on

\Delta

and observed colour?

Larger

\Delta

⇒ shorter

\lambda

absorbed ⇒ colour shifts (e.g. pale blue → deep blue on

H_2O→NH_3

Relation between

\Delta_t

and

\Delta_o

\Delta_t = \tfrac{4}{9}\,\Delta_o

(tetrahedral splitting is smaller).

Which two d-orbitals form

e_g

in an octahedral field?

d_{x^2-y^2}

and

d_{z^2}

(they point directly at the ligands).

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, transition metal complexes coloured kyun dikhte hain? Asli baat hai d-orbitals ki splitting. Free metal ion mein paanchon d-orbital same energy ke hote hain (degenerate). Jab ligands paas aate hain, unke electron pairs metal ke d-electrons ko repel karte hain. Jo orbitals seedha ligand ki taraf point karte hain ( $e_g$ ) woh upar chale jaate hain, aur jo beech mein hote hain ( $t_{2g}$ ) woh neeche reh jaate hain. In dono ke beech ka gap hi hai $\Delta_o$ — crystal field splitting energy.

Ab colour ka kheL: jab white light solution se guzarti hai, ek electron neeche wale set se upar wale set mein jump karta hai, lekin sirf tabhi jab photon ki energy exactly $\Delta_o$ ke barabar ho. Yani $\Delta_o = hc/\lambda$ . Jo colour absorb hua woh light se gayab ho gaya, aur humein dikhta hai uska complementary (opposite) colour. Isiliye $[Ti(H_2O)_6]^{3+}$ green-yellow absorb karta hai par humein purple dikhta hai. Yaad rakho: jo absorb hota hai woh dikhta nahi, uska ulta dikhta hai.

$\Delta$ ka size ligand pe depend karta hai — spectrochemical series mein jitna strong ligand ( $CN^-$ , $CO$ jaisa), utna bada $\Delta$ , utna chhota $\lambda$ absorbed. Isiliye $H_2O$ se $NH_3$ pe jaane se $Cu^{2+}$ ka rang pale blue se deep blue ho jaata hai. Aur ek important point: agar metal $d^0$ (jaise $Sc^{3+}$ ) ya $d^{10}$ (jaise $Zn^{2+}$ ) hai, toh jump karne ke liye ya toh electron nahi hai ya empty orbital nahi — isliye woh colourless hote hain. Bas yahi 80/20 funda exam ke liye kaafi hai!

Colour of complexes — d-d transitions

WHAT is happening?

WHY does this give colour?

HOW the size of $\Delta$ controls the colour

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — d-Block (Transition Metals) & f-Block

Connections

WHAT is happening?

WHY does this give colour?

HOW the size of Δ\DeltaΔ controls the colour

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — d-Block (Transition Metals) & f-Block

Connections

HOW the size of $\Delta$ controls the colour