Which of the following pairs of ions is colourless? (a) \(\mathrm{T}_{1}^{3+}, \mathrm{Cu}^{2+}\) (b) \(\mathrm{Sc}^{3+}, \mathrm{Zn}^{2+}\) (c) \(\mathrm{Co}^{2+}, \mathrm{Fe}^{3+}\) (d) \(\mathrm{Ni}^{2+}, \mathrm{V}^{3+}\)

The vibrance of color in transition metal ions is intricately linked to 'd-orbital transitions'. These are essentially quantum jumps of electrons between different levels within the d-orbitals, that is, the subshell of an atom that can contain up to ten electrons. When light hits an ion, if the energy of the light matches the energy gap between the d-orbitals, the electron can absorb the light and jump to a higher energy state. Subsequently, as it returns to a lower energy state, it emits light, often in the visible spectrum, which gives the compound its characteristic color.

For instance, the lovely blue hue of copper sulfate is due to such transitions of electrons in the copper ion's d-orbitals. On the flip side, colorlessness in ions like \(\mathrm{Sc}^{3+}\) and \(\mathrm{Zn}^{2+}\), as seen in the exercise, indicates an absence of these transitions because the d-orbitals are either completely full or completely empty, leaving no room for electrons to jump to or from.

The presence of 'unpaired electrons' plays a dual role in an ion's properties, influencing both color and magnetism. Electrons have a property called spin, and they prefer to pair up with others that have an opposite spin. In the d-orbitals, if all electrons are paired, the substance typically lacks color because there's no mismatch in energy levels to absorb visible light. However, if there are unpaired electrons, they can absorb the right wavelengths of light, causing color.

To illustrate, \(\mathrm{Cu}^{2+}\) ions showcase color due to the presence of unpaired electrons that make d-d transitions possible. In contrast, \(\mathrm{Zn}^{2+}\) ions, which lack unpaired electrons, do not show the same phenomenon and therefore are colorless. Students should remember that the number of unpaired electrons can also affect the magnetic properties of an ion, making it paramagnetic (attracted to magnetic fields) or diamagnetic (not attracted to magnetic fields).

The 'electronic configuration' of an ion denotes the distribution of electrons among the various shells and subshells in an atom. This arrangement is key to understanding the chemical behaviors and properties of an element, including its color as previously discussed. Transition metals can have multiple oxidation states, leading to different configurations and the loss or gain of d-orbital electrons.

For example, \(\mathrm{Sc}^{3+}\) has a noble gas configuration of \(\mathrm{[Ar]}\), indicating that it has lost all its 3d electrons and is colorless. Conversely, \(\mathrm{Fe}^{3+}\) may have five 3d electrons which is why it can show color due to transitions among the d-orbitals at varying energy states. Students should be comfortable determining whether an ion’s d-orbitals are empty, partially filled with unpaired electrons, or completely filled, as these factors dictate the likelihood of observing color in a transition metal compound.