Which one of these tripositive ions is the most stable in aqueous sol…

The correct answer is Option (1) → $Cr^{3+}$

Electronic configuration for each ion:

$\text{Cr (Z = 24): }[\text{Ar}]\,3d^5\,4s^1$

$\textbf{Cr}^{3+}:[\text{Ar}]\,3d^3$

$\text{Ti (Z = 22): }[\text{Ar}]\,3d^2\,4s^2$

$\textbf{Ti}^{3+}:[\text{Ar}]\,3d^1$

$\text{V (Z = 23): }[\text{Ar}]\,3d^3\,4s^2$

$\textbf{V}^{3+}:[\text{Ar}]\,3d^2$

$\text{Mn (Z = 25): }[\text{Ar}]\,3d^5\,4s^2$

$\textbf{Mn}^{3+}:[\text{Ar}]\,3d^4$

In an aqueous solution, water molecules surround the metal ion, forming an octahedral complex.

According to Crystal Field Theory (CFT), this splits the d-orbitals into two sets:

$t_{2g}$ (lower energy, 3 orbitals) and $e_{g}$ (higher energy, 2 orbitals)

Electrons fill the lower energy $t_{2g}$ set first:

$\text{Ti}^{3+}(d^1):t_{2g}^1$

$\text{V}^{3+}(d^2):t_{2g}^2$

$\text{Cr}^{3+}(d^3):t_{2g}^3$

$\text{Mn}^{3+}(d^4):t_{2g}^3\,e_{g}^1$

The $\text{Cr}^{3+}$ ion has a $t_{2g}^3$ configuration, which is a perfectly half-filled $t_{2g}$ subshell.

This gives extra stability. Hence, $Cr^{3+}$ is the most stable ion among the given choices.