Explain why Sc 3+(aq) and Zn 2+(aq) are not coloured.

• Scandium (Sc) has an atomic number of 21. Its ground state electronic configuration is $[\text{Ar}] 3d^1 4s^2$.
• When Sc forms $\text{Sc}^{3+}$, it loses all three valence electrons (two from $4s$ and one from $3d$). $$\text{Sc}^{3+}: [\text{Ar}] 3d^0$$
• Zinc (Zn) has an atomic number of 30. Its ground state electronic configuration is $[\text{Ar}] 3d^{10} 4s^2$.
• When Zn forms $\text{Zn}^{2+}$, it loses its two $4s$ valence electrons. $$\text{Zn}^{2+}: [\text{Ar}] 3d^{10}$$

Transition metal ions are colored when they have partially filled d-orbitals, allowing for d-d electronic transitions. In these transitions, electrons absorb specific wavelengths of visible light to jump from a lower energy d-orbital to a higher energy d-orbital. The remaining unabsorbed light is transmitted, giving the ion its observed color.

• $\text{Sc}^{3+}$ has a $d^0$ configuration, meaning it has no d-electrons to undergo d-d transitions.
• $\text{Zn}^{2+}$ has a $d^{10}$ configuration, meaning its d-orbitals are completely filled. There are no vacant higher-energy d-orbitals for electrons to jump into, so d-d transitions cannot occur.

Since neither $\text{Sc}^{3+}$ nor $\text{Zn}^{2+}$ can undergo d-d electronic transitions, they do not absorb visible light and therefore appear colorless.

The final answer is $\boxed{\text{Sc}^{3+} \text{ has a } d^0 \text{ configuration and } \text{Zn}^{2+} \text{ has a } d^{10} \text{ configuration. Neither can undergo d-d electronic transitions, so they are colorless.}}$.