Hello
Single electrons in a whole orbital tend to be thrown out to give a more stable electronic configuration; that is why potassium with electron configuration [Ar]4s
1 throws one electron to reach the most stable configuration, [Ar].
Titanium(III) chloride is a classical d
1 ion, thus very eager to give away its single (and somewhat troublesome) electron, to reach titanium(IV) ion, again [Ar].
So why is argon so stable, thus not reactive at all? Its basic electronic shells are COMPLETELY filled, with no half-filled orbitals. If you manage to bombard with an electron, you'll end up argon again, after a system interconversion, Ar^-\longrightarrow Ar+e^-.
Let me show the other alternative, namely, pulling out an electron; in this case, a cation like Ar
+ will be formed, but this process is extremely energy consuming, so the negligible amount of ionization energy is very eagerly satisfied by the system, to give finally the most stable form, Ar: Ar^++e^-\longrightarrow Ar
You are okay to think that 3d orbital can be filled, and if you manage to do this (as this will be VERY hard, due to 3d orbitals of extremely higher level than that of 4s and 3p), with enough number of electrons, the super anion (probably be 7-)
might be somewhat stable, but another problem occurs here; argon is argon, with its small radius will be very unstable with this number of extraordinary electrons; an unusual value of \displaystyle \frac {charge}{radius} will not possibly cause the anion to be stable. You must keep this anion at extremely low temperatures to avoid its decomposition (here, ionization energies will be very low to give away these extra electrons, finally meeting at the usual place, [Ar]

)
Hope this helps.