The colour of transition metal complexes is generally due to d-d electronic transitions. For this to occur, the metal ion must possess an incomplete d-subshell ($d^1$ to $d^9$) containing unpaired electrons. Ions with empty ($d^0$) or fully filled ($d^{10}$) d-orbitals do not exhibit d-d transitions and are typically colourless.

Let's analyze the oxidation state and electronic configuration of the metal in each species:

1. $TiF_6^{2-}$: Titanium (Z=22, $[Ar]3d^2 4s^2$). Oxidation state: $x + 6(-1) = -2 \Rightarrow x = +4$. Configuration of $Ti^{4+}$ is $[Ar]3d^0$. Since it has no d-electrons ($d^0$), it is colourless .
2. $Cu_2Cl_2$: Copper (Z=29, $[Ar]3d^{10} 4s^1$). This is copper(I) chloride. Oxidation state: $2x + 2(-1) = 0 \Rightarrow x = +1$. Configuration of $Cu^+$ is $[Ar]3d^{10}$. Since the d-subshell is fully filled ($d^{10}$), d-d transitions are not possible, making it colourless .
3. $CoF_6^{3-}$: Cobalt (Z=27). Oxidation state is +3. Configuration of $Co^{3+}$ is $3d^6$. It has unpaired electrons and is coloured.
4. $NiCl_4^{2-}$: Nickel (Z=28). Oxidation state is +2. Configuration of $Ni^{2+}$ is $3d^8$. It has unpaired electrons and is coloured.

Therefore, the colourless species are $TiF_6^{2-}$ and $Cu_2Cl_2$.