The colour of coordination compounds and their ability to absorb visible light is primarily due to d-d transitions of electrons within the split d-orbitals. For this to occur, the central metal ion must have an incomplete d-subshell (unpaired electrons, configuration $d^1$ to $d^9$). Ions with $d^0$ (empty) or $d^{10}$ (fully filled) configurations do not undergo d-d transitions and are typically colourless.

Let's analyze the electronic configuration of the central metal ion in each option:

1. [Sc(H₂O)₃(NH₃)₃]³⁺: Sc is in the +3 oxidation state. Configuration of Sc (Z=21) is $[Ar]3d^1 4s^2$. $Sc^{3+}$ is $3d^0$. No d-electrons, so no colour.
2. [Ti(en)₂(NH₃)₂]⁴⁺: Ti is in the +4 oxidation state. Configuration of Ti (Z=22) is $[Ar]3d^2 4s^2$. $Ti^{4+}$ is $3d^0$. No d-electrons, so no colour.
3. [Zn(NH₃)₆]²⁺: Zn is in the +2 oxidation state. Configuration of Zn (Z=30) is $[Ar]3d^{10} 4s^2$. $Zn^{2+}$ is $3d^{10}$. Fully filled d-orbitals, so no d-d transitions possible. Colourless.
4. [Cr(NH₃)₆]³⁺: Cr is in the +3 oxidation state. Configuration of Cr (Z=24) is $[Ar]3d^5 4s^1$. $Cr^{3+}$ is $3d^3$. It has 3 unpaired electrons in the $t_{2g}$ level. d-d transitions are possible, so it absorbs visible light and appears coloured .