Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)?
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).