As we know, The complete electron configuration for titanium is as follows:
1s2 2s2 2p6 3s2 3p6 3d2 4s2 or [Ar]3d2 4s2
Also, for transition metals, and in particular those in columns 3 through 7, the effective number of valence electrons is equal to the number of electrons that occur after a noble-gas-like core. Also, the 3d and 4s electrons are so close in energy that they behave as if they are in the same energy level (more-or-less ). Consequently, titanium has 4 valence electrons.
TiCl3 is an odd electron specie. So, we'll firstly calculate the steric number:
S.N. = (4 + 3)/2 i.e. 3.5
Now, since Cl is a more electronegative element, it will decrease the P.E. of Ti when approaching towards it. Thus, the energy gap within the 3d & 4d will further decrease and 3d orbitals will participate in hybridisation. So, the hybridisation of Ti in TiCl4 will be sd3 (first excitation of 4s electron; four unpaired electron distributed in 3 d-orbital and 1 s-orbital) and electron (which is localised at central atom) will be present in a hybridised ordbital.
This is what I could think of right now. I would have to confirm with my teacher, for more.