Question

Which one of the following ions is the most stable in aqueous solution?
[Atomic Number; Ti = 22, V = 23, Cr = 24, Mn = 25]

$${\text{A}}{\text{. C}}{{\text{r}}^{3 + }} \\\ {\text{B}}{\text{. }}{{\text{V}}^{3 + }} \\\ {\text{C}}{\text{. T}}{{\text{i}}^{3 + }} \\\ {\text{D}}{\text{. M}}{{\text{n}}^{3 + }} \\\$$

Answer 1

Hint: Here, we will proceed by writing down the electronic configuration corresponding to all the three ions given in the options and then, we will see which of these generate fully filled or half filled ${t_{2g}}$ and ${e_g}$ orbitals.
The d orbital is basically split into different energy levels for their stability, reactivity and magnetic properties.

Complete answer:
It is important to note that d orbital is split into two energy levels i.e., ${t_{2g}}$ orbital (completely filled consisted of 6 electrons and half-filled consists of 3 electrons) and ${e_g}$ orbital (completely filled consisted of 4 electrons and half-filled consists of 2 electrons)
The atomic number of Ti is 22 and its electronic configuration is given below.
Ti = $\left[ {{\text{Ar}}} \right]3{d^2}4{s^2}$
When Ti exists in +3 oxidation state (i.e., it loses 3 electrons which includes 2 from the s orbital and 1 from the d orbital). The electron configuration of ${\text{T}}{{\text{i}}^{3 + }}$ is given below.
${\text{T}}{{\text{i}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^1}4{s^0}$
The one electron in the d orbital is actually placed in the ${t_{2g}}$ orbital and no electron in ${e_g}$ orbital. Here, the ${t_{2g}}$ orbital is not fully filled or half-filled. So, it does not generate the most stable configuration in aqueous solution (i.e., in the presence of ${{\text{H}}_2}{\text{O}}$ as the weak ligand).
The atomic number of V is 23 and its electronic configuration is given below.
V = $\left[ {{\text{Ar}}} \right]3{d^3}4{s^2}$
When V exists in +3 oxidation state (i.e., it loses 3 electrons which includes 2 from the s orbital and 1 from the d orbital). The electron configuration of ${{\text{V}}^{3 + }}$ is given below.
${{\text{V}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^2}4{s^0}$
The two electrons in the d orbital are actually placed in the ${t_{2g}}$ orbital and no electron in ${e_g}$ orbital. Here, the ${t_{2g}}$ orbital is not fully filled or half-filled. So, it does not generate the most stable configuration in aqueous solution (i.e., in the presence of ${{\text{H}}_2}{\text{O}}$ as the weak ligand).
The atomic number of Cr is 24 and its electronic configuration is given below.
Cr = $\left[ {{\text{Ar}}} \right]3{d^4}4{s^2}$
When Cr exists in +3 oxidation state (i.e., it loses 3 electrons which includes two from the s orbital and 1 from the d orbital). The electron configuration of ${\text{C}}{{\text{r}}^{3 + }}$ is given below.
${\text{C}}{{\text{r}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^3}4{s^0}$
The three electrons in the d orbital are actually placed in the ${t_{2g}}$ orbital and no electron in ${e_g}$ orbital. Here, the ${t_{2g}}$ orbital is half-filled and ${e_g}$ orbital is completely empty. So, it generates the most stable configuration in aqueous solution (i.e., in the presence of ${{\text{H}}_2}{\text{O}}$ as the weak ligand).
The atomic number of Mn is 25 and its electronic configuration is given below.
Mn = $\left[ {{\text{Ar}}} \right]3{d^5}4{s^2}$
When, Mn exists in +3 oxidation state (i.e., it loses 3 electrons which includes two from the s orbital and 1 from the d orbital). The electron configuration of ${\text{M}}{{\text{n}}^{3 + }}$ is given below.
${\text{M}}{{\text{n}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^4}4{s^0}$
Out of the four electrons present in the d orbital, there are three electrons placed in ${t_{2g}}$ orbital and one electron in ${e_g}$ orbital. Here, the ${t_{2g}}$ orbital is half-filled but along with that there is ${e_g}$ orbital (which is neither fully filled or half-filled), resulting in the decrease of stability of the ion. So, it does not generate the most stable configuration in aqueous solution (i.e., in the presence of ${{\text{H}}_2}{\text{O}}$ as the weak ligand).
Therefore, ${\text{C}}{{\text{r}}^{3 + }}$ ion is most stable in aqueous solution.

Hence, option C is correct.

Note: Breaking of degeneracies of the electron orbital states mostly d or f orbitals because of the static electric field which is generated by the surrounding charge distribution is referred to as the crystal field theory (CFT). The two energy levels i.e., ${t_{2g}}$ and ${e_g}$ are filled in such a way that firstly till three electrons enter in ${t_{2g}}$ because of the crystal field splitting energy.