Why does Fe(II) have electrons in the 3d subshell?
You might want to compare iron(II) not with neutral iron, but with neutral chromium.
Both iron(II) and neutral chromium have 24 electrons, but there the similarity ends:
$\ce{Fe^{2+}}:  [\text{Ar}]3d^6$
$\ce{Cr^{0}}:  [\text{Ar}]3d^54s^1$
Think of the Bohr model.  According to this model the energy level of electrons depends only on the shell number $n$, so we would expect 24 electrons to follow the $[\text{Ar}]3d^6$ configuration.  In real life that happens exactly only for single-electron atoms where there are only electron-nucleus interactions.  When there are electron-electron interactions they could fill the shells not in order, like the chromium atom described above.
But in a multielectron atom, if you add more nuclear charge you make the electron-nucleus interaction stronger, and the Bohr-model configuration becomes more favorable.  One might suppose that dozens of added protons might be needed to get chromium's $[\text{Ar}]3d^54s^1$ to the Bohr-predicted $[\text{Ar}]3d^6$.  Instead, the energies of the two configurations are so closely spaced that two extra protons, converting neutral chromium to ferrous iron, is enough.