The diamagnetic species is

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(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:$1$. In $[Ni(CN)_

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$[Ni(CN)_4]^{2-}$

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(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons. Thus,it is diamagnetic.$2$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ $(3d^8)$ is bonded to $Cl^-$ (a weak field ligand),which does not cause pairing,resulting in two unpaired electrons. Thus,it is paramagnetic.$3$. In $[CoCl_4]^{2-}$,$Co^{2+}$ $(3d^7)$ is bonded to $Cl^-$ (a weak field ligand),resulting in three unpaired electrons. Thus,it is paramagnetic.$4$. In $[CoF_6]^{2-}$,$Co^{4+}$ $(3d^5)$ is bonded to $F^-$ (a weak field ligand),resulting in five unpaired electrons. Thus,it is paramagnetic.Therefore,the correct option is $A$.

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