Which among the following is diamagnetic?

In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, the oxidation state of Ni is $+2$. The electronic configuration of $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}] 3d^8$. Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals. The configuration becomes $t_{2g}^6 e_g^2$ (or $d_{xy}^2 d_{yz}^2 d_{zx}^2 d_{z^2}^2 d_{x^2-y^2}^0$), resulting in zero unpaired electrons, making it diamagnetic.

In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$, Fe is in $+3$ state ($3d^5$). With a strong field ligand, it has one unpaired electron, making it paramagnetic.

In $\left[\mathrm{Co}(\mathrm{F})_6\right]^{3-}$, Co is in $+3$ state ($3d^6$). F is a weak field ligand, resulting in four unpaired electrons, making it paramagnetic.

In $\left[\mathrm{NiCl}_4\right]^{2-}$, Ni is in $+2$ state ($3d^8$). Cl is a weak field ligand, resulting in two unpaired electrons, making it paramagnetic.

Answer: $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$