Colour & Why Some Complexes Are Colourless

CHEBonding & Colour

Transition-metal complexes are coloured because of d–d transitions: an electron absorbs visible light to jump from $t_{2g}$ to $e_g$ , and the complement of the absorbed colour is seen. The flip side — why $Sc^{3+}$ , $Zn^{2+}$ and $d^{10}$ / $d^0$ species are colourless — is equally examinable.

d–d transition energy

\Delta_o = h\nu = \dfrac{hc}{\lambda_{abs}}

\lambda_{abs}

= wavelength absorbed; the observed colour is complementary to

\lambda_{abs}

Colour arises from a d–d electronic transition: an electron is promoted from $t_{2g}$ to $e_g$ absorbing light of energy $\Delta_o$ ; the observed colour is complementary to the absorbed wavelength.
A d–d transition needs partially filled $d$ -orbitals ( $d^1$ to $d^9$ ); $d^0$ and $d^{10}$ ions have no possible d–d jump and are colourless.
$[Ti(H_2O)_6]^{3+}$ ( $Ti^{3+}$ , $d^1$ ) is purple — its single electron absorbs in the visible (~ $500$ nm); $[Sc(H_2O)_6]^{3+}$ ( $Sc^{3+}$ , $d^0$ ) is colourless (no $d$ -electron to excite).
$[Zn(NH_3)_4]^{2+}$ is colourless because $Zn^{2+}$ is $d^{10}$ — the $d$ -levels are full, so no d–d transition is possible.
The size of $\Delta_o$ (hence the colour) depends on the ligand's field strength: a stronger-field ligand gives a larger $\Delta_o$ , absorbing shorter wavelengths.
$[Mn(H_2O)_6]^{2+}$ ( $d^5$ high spin) is very pale because its d–d transitions are spin-forbidden (they require a spin flip); strong-field $[Mn(CN)_6]^{4-}$ is low spin, has spin-allowed transitions and is deeply coloured.
Observed colour is the complement of the absorbed colour: e.g. a complex absorbing green/yellow appears red/violet; this is why $[Ti(H_2O)_6]^{3+}$ , which absorbs green-yellow, looks purple.
Changing the ligand changes $\Delta_o$ and therefore the colour for the same metal ion: aqua $[Cu(H_2O)_4]^{2+}$ is pale blue, whereas ammine $[Cu(NH_3)_4]^{2+}$ (stronger field, larger $\Delta_o$ ) is deep blue.
Other colourless/white examples: $Cu^+$ ( $d^{10}$ ), $Ag^+$ ( $d^{10}$ ) and $Ti^{4+}$ / $Sc^{3+}$ ( $d^0$ ) — all lack a d–d transition.
Intensity rules: d–d transitions are Laporte-forbidden, so colours are usually faint; they appear strong only when symmetry is lowered or when charge-transfer transitions occur, as in the intense purple of $MnO_4^-$ .
Charge-transfer (not d–d) gives very intense colours: $MnO_4^-$ ( $Mn$ is $d^0$ , so the colour is NOT d–d) is deep purple due to ligand-to-metal charge transfer; this distinguishes it from ordinary complex colour.
Heating or dehydration can change colour by altering the ligand field, e.g. blue hydrated $CuSO_4\cdot 5H_2O$ turns white anhydrous $CuSO_4$ on losing the aqua ligands.

Where the marks go

Saying the complex is the colour it absorbs — it shows the COMPLEMENTARY colour to the wavelength absorbed.
Calling a $d^{10}$ ion ( $Zn^{2+}$ , $Cu^+$ , $Ag^+$ ) coloured: with full $d$ -orbitals there is no d–d transition, so it is colourless.
Attributing the intense colour of $MnO_4^-$ or $Cr_2O_7^{2-}$ to d–d transitions — those metals are $d^0$ ; the colour is charge transfer, not d–d.
Forgetting that high-spin $d^5$ ( $Mn^{2+}$ , $Fe^{3+}$ ) complexes are only faintly coloured because their d–d transitions are spin-forbidden.
Ignoring the ligand's effect on colour: the same metal ion shows different colours with different ligands because $\Delta_o$ (and the absorbed wavelength) changes.

How the board asks it

Give reasons $d^0$ and $d^{10}$ ions have no d–d transition
Give reasons: $[Sc(H_2O)_6]^{3+}$ ( $d^0$ ) and $[Zn(NH_3)_4]^{2+}$ ( $d^{10}$ ) are colourless, whereas $[Ti(H_2O)_6]^{3+}$ ( $d^1$ ) is purple.
Predict the product $d$ -electron count and the d–d transition condition
Predict whether $[Cu(H_2O)_4]^{2+}$ ( $d^9$ ) and $[Ag(NH_3)_2]^{+}$ ( $d^{10}$ ) are coloured or colourless, justifying each answer from its $d$ -electron count.
Give reasonsligand field strength changing $\Delta_o$
Account for the fact that $[Cu(H_2O)_4]^{2+}$ is pale blue while $[Cu(NH_3)_4]^{2+}$ is deep blue, although both contain the same metal ion.
Distinguishd–d transition versus charge-transfer colour
Distinguish between the origin of the colour of $[Ti(H_2O)_6]^{3+}$ and that of $MnO_4^{-}$ , noting that $Mn$ in $MnO_4^{-}$ is $d^0$ .
Assertion–Reasoncomplementary colour and the d–d transition
Assertion: $[Ti(H_2O)_6]^{3+}$ appears purple. Reason: it absorbs visible light in the green–yellow region. State whether both are correct and whether the Reason explains the Assertion.
Numericalthe d–d transition energy $\Delta_o = hc/\lambda$
$[Ti(H_2O)_6]^{3+}$ absorbs visible light of wavelength $500\ \text{nm}$ . Calculate the crystal-field splitting energy $\Delta_o$ (in $\text{J}$ ). (Take $h = 6.6\times10^{-34}\ \text{J s}$ , $c = 3\times10^{8}\ \text{m s}^{-1}$ .)

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.