Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)? In most of the cases I have learnted, halogens act as a weak field ligand. However, I recently learned somewhere on the web that the strength of a ligand also depends upon the size of the atom. A classic example is tetrachloridoplatinate(II): $\ce{[Pt(Cl)4]^2-}$. 

Why is it so, and how can I predict the hybridization correctly in such exceptions, if I come across them for the first time?
I would like to add an answer stating a few exceptions which would come handy to you in future: 


$\ce{Ni^{2+}}$ has a $3d^8$ configuration. In this case strong field ligands ($\ce{C}$ or $\ce{N}$ donors) will make the complex square planar. In all other cases, the complex will be tetrahedral.
In the case of $4d^8$ and $5d^8$ configuration and any ligand the hybridisation will be that of a square planar complex.
With $\ce{Co^{3+}}$, nearly all ligands behave as strong field ligands ($dsp^2$ hybridisation). Notable exceptions to this rule being $\ce{[CoF6]^{3-}}$ and $\ce{[Co(H2O)3F3]}$.
In case of bulky ligands (for instance $\ce{PPh3}$) present with $3d^8$ configuration, the ligands cannot get adjusted in the same plane leading to a tetrahedral shape ($sp^3$ configuration).