Colour of Transition Metal Ions

CHEColour & Magnetic Behaviour

Most aqueous transition-metal ions are coloured because partly filled $d$ -orbitals allow $d\text{-}d$ electronic transitions of visible light. Ions with $d^0$ or $d^{10}$ are colourless, while deep colours of $d^0$ oxoanions come instead from charge transfer.

Energy gap and absorbed wavelength

\Delta E = h\nu = \dfrac{hc}{\lambda}

\Delta E

is the

d

-orbital splitting; the ion absorbs at

\lambda

and shows the complementary colour.

Colour arises from $d\text{-}d$ transitions: in a ligand field the $d$ -orbitals split, and electrons absorb visible light to jump between them; the complementary colour is seen.
$Sc^{3+}$ ( $3d^0$ ) and $Zn^{2+}$ ( $3d^{10}$ ) are colourless because they have no partly filled $d$ -subshell, so no $d\text{-}d$ transition is possible.
$KMnO_4$ is intensely purple even though $Mn$ is $d^0$ in $MnO_4^-$ : the colour is from a charge-transfer (ligand-to-metal, $O \rightarrow Mn$ ) transition, not a $d\text{-}d$ one.
Charge-transfer bands are far more intense than $d\text{-}d$ bands, which are weak, so $MnO_4^-$ and $Cr_2O_7^{2-}$ are deeply coloured even at low concentration.
Number of unpaired $d$ electrons (increasing): $Zn^{2+}$ ( $d^{10}$ , $0$ ) $<$ $Ti^{3+}$ ( $d^1$ , $1$ ) $=$ $Cu^{2+}$ ( $d^9$ , $1$ ) $<$ $Fe^{2+}$ ( $d^6$ , $4$ ) $<$ $Mn^{2+}$ ( $d^5$ , $5$ ).
Colour also depends on the ligand and oxidation state: different ligands change the size of $d$ -orbital splitting and hence the wavelength absorbed.
The observed colour is complementary to the colour absorbed: e.g. an ion absorbing in the green-yellow region appears violet/red.
A larger $d$ -orbital splitting $\Delta$ means higher-energy (shorter-wavelength) light is absorbed, so colour shifts as the ligand-field strength changes.
Hydrated ions have characteristic colours, e.g. $[Cu(H_2O)_6]^{2+}$ blue, $[Fe(H_2O)_6]^{2+}$ pale green, $[Ti(H_2O)_6]^{3+}$ purple, $[Mn(H_2O)_6]^{2+}$ very pale pink.
$Mn^{2+}$ ( $d^5$ ) is only very faintly coloured because its $d\text{-}d$ transitions are spin-forbidden (no transition keeps all five spins parallel), making them extremely weak.
Changing the oxidation state changes the $d$ -electron count and the splitting, so the same metal can give different colours in different states (e.g. $Cr^{3+}$ green, $Cr_2O_7^{2-}$ orange).

Where the marks go

Saying $Sc^{3+}$ is colourless because of a full $d$ -subshell; it is $d^0$ (empty), while $Zn^{2+}$ is colourless because it is $d^{10}$ (full).
Reporting the absorbed colour as the observed colour; the ion shows the colour complementary to what it absorbs.
Explaining the deep colour of $MnO_4^-$ / $Cr_2O_7^{2-}$ by $d\text{-}d$ transitions; these are $d^0$ , so the colour is charge transfer.
Assuming every transition-metal ion is strongly coloured; $d^5$ ions like $Mn^{2+}$ are almost colourless because their transitions are spin-forbidden.
Confusing 'unpaired electrons' (magnetism) with 'partly filled $d$ -subshell' (colour); a $d^{10}$ ion has no unpaired electrons and no colour, but the requirement for colour is a partly filled $d$ -shell.

How the board asks it

Give reasons $d\text{-}d$ transitions in partly filled $d$ -orbitals
Account for the fact that aqueous solutions of most transition-metal salts are coloured, whereas $Sc^{3+}$ and $Zn^{2+}$ salts are colourless.
Give reasonsligand-to-metal charge transfer in $d^0$ oxoanions
Give reasons: $KMnO_4$ is intensely purple even though manganese is in the $d^0$ configuration in $MnO_4^-$ .
Predict the productcharacteristic colours of hydrated ions
Predict the colour of the hydrated ions $[Ti(H_2O)_6]^{3+}$ , $[Fe(H_2O)_6]^{2+}$ and $[Cu(H_2O)_6]^{2+}$ , and justify your answer in terms of $d\text{-}d$ transitions.
Distinguish $d^5$ spin-forbidden transitions
Explain why $Mn^{2+}$ ( $d^5$ ) is almost colourless in aqueous solution while $Cu^{2+}$ ( $d^9$ ) gives a distinct blue colour.
Assertion–Reasonoxidation state changing $d$ -electron count and splitting
Assertion: $Cr^{3+}$ is green whereas $Cr_2O_7^{2-}$ is orange. Reason: a change in oxidation state changes the $d$ -electron count and the magnitude of $d$ -orbital splitting. State whether both statements are correct and whether the reason correctly explains the assertion.

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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.