Which of the following is paramagnetic?

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(D) To determine if a complex is paramagnetic,we check for the presence of unpaired electrons in the central metal ion.$1$. In $[Fe(CN)_6]^{4-}$,$Fe$

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$[CoF_6]^{3-}$

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(D) To determine if a complex is paramagnetic,we check for the presence of unpaired electrons in the central metal ion.$1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons (diamagnetic).$2$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).$3$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).$4$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,so no pairing occurs. The configuration is $t_{2g}^4 e_g^2$,which has $4$ unpaired electrons,making it paramagnetic.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

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