Q.Explain why most transition metal compounds are coloured, and state, with reasons, why $\text{Sc}^{3+}$ and $\text{Zn}^{2+}$ are colourless.

When a transition metal ion is surrounded by ligands (such as water molecules in a simple aqueous solution), its five d orbitals -- degenerate in a bare, isolated ion -- are split by the surrounding ligands into two or more sets of slightly different energy. If the ion's d subshell is only partially filled (containing between one and nine d electrons), an electron in a lower-energy orbital of this split set can absorb a photon and be promoted to a higher-energy orbital of the same set -- a d-d transition. Because the energy gap involved usually corresponds to a photon in the visible part of the spectrum, the compound absorbs part of the visible spectrum and transmits (or reflects) the rest, which is what we perceive as its colour.

$\text{Sc}^{3+}$ is formed by scandium losing all three of its valence electrons ($4s^23d^1$), giving the configuration $[\text{Ar}]$ -- a $d^0$ ion. With no d electrons present at all, there is simply no electron available to undergo a d-d transition, so $\text{Sc}^{3+}$ compounds are colourless.

$\text{Zn}^{2+}$ is formed by zinc losing its two $4s$ electrons, leaving the $3d^{10}$ subshell completely untouched: $[\text{Ar}]3d^{10}$ -- a $d^{10}$ ion. Here every one of the five d orbitals already holds a full pair of electrons, so even though d electrons are present, there is no vacant orbital within the split d set for an electron to be excited into. A d-d transition is therefore equally impossible, and $\text{Zn}^{2+}$ compounds are also colourless. …

Explain the general d-d transition mechanism for colour, then work out the configuration of Sc3+ and Zn2+ explicitly to show why eac …

Stating 'transition metals are coloured' without explaining the mechanism; forgetting that BOTH an electron to excite AND a vacant orbital to receive …