To determine the electronic configuration of ions, electrons are removed first from the outermost shell (4s) and then from the penultimate shell (3d) .

1. Titanium ($Z=22$): Ground state is $[Ar]3d^2 4s^2$. For $Ti^{2+}$, remove 2 electrons from 4s $\rightarrow [Ar]3d^2$.
2. Vanadium ($Z=23$): Ground state is $[Ar]3d^3 4s^2$. For $V^{3+}$, remove 2 electrons from 4s and 1 from 3d $\rightarrow [Ar]3d^2$.
3. Chromium ($Z=24$): Ground state is $[Ar]3d^5 4s^1$. For $Cr^{4+}$, remove 1 electron from 4s and 3 from 3d $\rightarrow [Ar]3d^2$.
4. Manganese ($Z=25$): Ground state is $[Ar]3d^5 4s^2$. For $Mn^{5+}$, remove 2 electrons from 4s and 3 from 3d $\rightarrow [Ar]3d^2$.

Thus, the set $Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$ all share the $3d^2$ configuration.