Question

Which of the following ions is colourless in a solution?
A.${{\text{V}}^{{\text{3 + }}}}$
B.${\text{C}}{{\text{r}}^{{\text{3 + }}}}$
C.${\text{C}}{{\text{o}}^{{\text{2 + }}}}$
D.${\text{S}}{{\text{c}}^{{\text{3 + }}}}$

Answer 1

Hint: To solve this question, it is required to have knowledge about why certain compounds are coloured while other are not. Colour of d-block transition elements are mainly coloured due to d-d transitions which occur due to the splitting of d-orbital. Only the compound in which d-d transition is possible will show colour.

Complete step by step answer:
As we know that, in presence of ligands, the d-orbital of transition metals split up into ${{\text{t}}_{2g}}$ and ${{\text{e}}_{\text{g}}}$ orbitals. The elements which can show d-d transition of electrons between the two split orbitals can show colour.
In option A, vanadium with atomic number 23, in normal form has an electronic configuration of $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{3}}}{\text{4}}{{\text{s}}^{\text{2}}}$ . So, ${{\text{V}}^{{\text{3 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{2}}}{\text{4}}{{\text{s}}^{\text{0}}}$ which means that it has two unpaired electrons in its ${{\text{t}}_{2g}}$ orbital and can thus get excited into ${{\text{e}}_{\text{g}}}$ orbital and give a colour compound in a solution.
In option B, chromium with atomic number 24, in its normal form has an electronic configuration of $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^5}{\text{4}}{{\text{s}}^1}$ . So, ${\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{3}}}{\text{4}}{{\text{s}}^{\text{0}}}$ which means that it has three unpaired electrons in its ${{\text{t}}_{2g}}$ orbital which can show d-d transition. Thus, ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ will also form coloured compounds in a solution.
In option C, cobalt has an atomic number of 27. In its normal form, it has an electronic configuration of $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^7}{\text{4}}{{\text{s}}^{\text{2}}}$ . So, ${\text{C}}{{\text{o}}^{{\text{2 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{7}}}{\text{4}}{{\text{s}}^{\text{0}}}$ which means that it has three unpaired electrons, one of them is in ${{\text{t}}_{2g}}$ orbital and can show d-d transition. Thus, ${\text{C}}{{\text{o}}^{{\text{2 + }}}}$ will also form coloured compounds.
In option D, scandium has an atomic mass of 21. It has an electronic configuration of $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^1}{\text{4}}{{\text{s}}^2}$. So, ${\text{S}}{{\text{c}}^{{\text{3 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{0}}}{\text{4}}{{\text{s}}^{\text{0}}}$ which means that it has no electrons in its d-orbitals. So, d-d transition is not possible for ${\text{S}}{{\text{c}}^{{\text{3 + }}}}$ . Thus, it will form colourless compounds in a solution.
$\therefore $ The correct option is option D, i.e. ${\text{S}}{{\text{c}}^{{\text{3 + }}}}$ .

Note: Colour of a compound due to d-d transitions is only possible in a compound if it has unpaired electrons in the split orbital with lower energy. In case the ligand attached is also given, only the compounds which form high spin complexes will be able to shoe d-d transitions. This is because strong field ligands pair up the electrons in the lower split orbital and do not allow for d-d transitions due to its high splitting power.