Amongst $Ni (CO)_{4}, {[Ni (CN)_{4}]}^{2 -}, {NiCl}_{4}^{2 -}$ :

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${NiCl}_{4}^{2 -}$ and ${[Ni (CN)_{4}]}^{2 -}$ are diamagnetic and $Ni (CO)_{4}$ is paramagnetic.

$Ni (CO)_{4}$ is diamagnetic and, ${NiCl}_{4}^{2 -}$ - and ${[Ni (CN)_{4}]}^{2 -}$ are paramagnetic

$Ni (CO)_{4}$ and ${NiCl}_{4}^{2 -}$ are diamagnetic and ${[Ni (CN)_{4}]}^{2 -}$ is paramagnetic.

$Ni (CO)_{4}$ and ${[Ni (CN)_{4}]}^{2 -}$ are diamagnetic and ${NiCl}_{4}^{2 -}$ is paramagnetic.

answer is C.

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Detailed Solution

Since there are no unpaired electrons in $[Ni (CO)_{4}]$ having configuration $Ni = [Ar] 3 d^{8} 4 s^{2}$ the complex is diamagnetic and have zero magnetic moment of spin.

Since there are two unpaired electrons in ${NiCl}_{4}^{2 -}$ having configuration ${Ni}^{2 +} = [Ar] 3 d^{8}$ the complex is paramagnetic and have some magnetic moment of spin as,