Q.$[\text{Sc}(\text{H}_2\text{O})_6]^{3+}$ is colourless, while $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ is purple. Account for this difference in terms of the $d$-electron configurations of $\text{Sc}^{3+}$ and $\text{Ti}^{3+}$ and the origin of colour in coordination compounds.

[!TLDR] $\text{Sc}^{3+}$ is $d^0$ (no electron to promote); $\text{Ti}^{3+}$ is $d^1$ (one electron undergoes a $d$-$d$ transition). [!ANSWER] $[\text{Sc}(\text{H}_2\text{O})_6]^{3+}$ is colourless because $\text{Sc}^{3+}$ has no d-electrons at all; $[\text{Ti}(\text{H}_2\text{O} …

Colour in a transition-metal complex arises from a $d$-$d$ transition: an electron in the lower-energy $t_{2g}$ orbital set absorbing a photon of visible light and being promoted to the higher-energy e_g} set, across the crystal field gap $\Delta_o$. For this to happen at all, the metal ion must have at least one d-electron available to promote. $\text{Sc}^{3+}$ has the electron configuration $[\text{Ar}]3d^0$ — it has lost all three of its valence electrons (two 4s, one 3d) to reach the $+3$ state, leaving it with zero d-electrons. With no electron to promote, no $d$-$d$ transition is possible, no visible light is absorbed, and $[\text{Sc}(\text{H}_2\text{O})_6]^{3+}$ is colourless. $\text{Ti}^{3+}$, immediately next to scandium, has the configuration $[\text{Ar}]3d^1$ — it retains one d-electron after losing its three valence electrons (two 4s, one 3d, leaving one 3d behind is incorrect phrasing; more precisely, Ti has configuration $[\text{Ar}]3d^24s^2$, and losing 3 electrons to form $\text{Ti}^{3+}$ leaves $[\text{Ar}]3d^1$). This single d-electron sits in the lower $t_{2g}$ set and readily absorbs a photon matching $\Delta_o$ (in the yellow-green region of the visible spectrum) to jump into the empty $e_g$ set. Because yellow-green light …

Determine the d-electron count of each metal ion; colour requires at least one electron in a lower orbital set AND at least one va …

Assuming all transition-metal complexes are automatically coloured simply because the metal is a 'transition metal' — the $d^0$ and …