Identify the complex having the highest number of unpaired electrons from following.

In $[\text{Co(NH}_3)_6]^{3+}$, Co is in $+3$ oxidation state with $3d^6$ configuration. Since NH$_3$ acts as a strong field ligand for Co$^{3+}$, pairing of electrons takes place resulting in $t_{2g}^6 e_g^0$ configuration. Number of unpaired electrons $= 0$.

In $[\text{CoF}_6]^{3-}$, Co is in $+3$ oxidation state with $3d^6$ configuration. F$^-$ is a weak field ligand, so no pairing occurs. The configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons $= 4$.

In $[\text{NiCl}_4]^{2-}$, Ni is in $+2$ oxidation state with $3d^8$ configuration. Cl$^-$ is a weak field ligand, forming a tetrahedral complex with $e^4 t_2^4$ configuration. Number of unpaired electrons $= 2$.

In $[\text{Ni(CN)}_4]^{2-}$, Ni is in $+2$ oxidation state with $3d^8$ configuration. CN$^-$ is a strong field ligand, forming a square planar complex where all electrons are paired. Number of unpaired electrons $= 0$.

Thus, $[\text{CoF}_6]^{3-}$ has the highest number of unpaired electrons.

Answer: $[\text{CoF}_6]^{3-}$