Among the given options, the pair of ions that is colorless is $(2)$ $\text{Sc}^{3+}$ and $\text{Zn}^{2+}$.
The color exhibited by transition metal ions in solution is typically due to the presence of unpaired electrons in their partially filled d orbitals. These unpaired electrons can absorb certain wavelengths of light, resulting in the absorption of specific colors and the appearance of color in the solution.
In option $(2)$, both $\text{Sc}^{3+}$ (scandium ion) and $\text{Zn}^{2+}$ (zinc ion) do not have any unpaired electrons in their d orbitals. Scandium (Sc) is a d-block element but has an electron configuration of $[Ar] 3d^0 4s^2$, where the d orbitals are empty. Zinc (Zn) is also a d-block element with an electron configuration of $[Ar] 3d^{10} 4s^2$, which means all the d orbitals are completely filled. Since both ions lack unpaired electrons, they do not absorb visible light and appear colorless in solution.
On the other hand, options $(1)$, $(3)$, and $(4)$ involve ions with unpaired electrons in their d orbitals. These ions can absorb specific wavelengths of light and exhibit colors in solution. $\text{Ti}^{3+}$ (titanium ion) and $\text{Cu}^{2+}$ (copper ion) are known to exhibit colors, $\text{Co}^{2+}$ (cobalt ion) and $\text{Fe}^{3+}$ (iron ion) also exhibit colors, and $\text{Ni}^{2+}$ (nickel ion) and $\text{V}^{3+}$ (vanadium ion) are typically colored as well.
Therefore, the correct answer is $(2)$ $\text{Sc}^{3+}$ and $\text{Zn}^{2+}$, as they form a pair of colorless ions. |
