Coordination Compounds Part 4

Concept:Magnetic behaviour depends on the number of unpaired electrons. Complexes with all electrons paired are diamagnetic, those with unpaired electrons are paramagnetic. Ligand field strength influences pairing in $d^8$ configuration.Explanation:1) $\mathrm{Ni}(\mathrm{CO})_4$: CO is neutral ligand. Oxidation state of Ni is $0$. Electronic configuration is $3d^{10}4s^0$. All electrons are paired. Hence it is diamagnetic.2) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$: Oxidation state of Ni is $+2$ ( $x + 4(-1) = -2$ ). Ni$^{2+}$ has $3d^8$ configuration. CN$^{-}$ is a strong field ligand, it forces pairing of electrons. The complex becomes square planar (inner orbital) with all electrons paired. Hence it is diamagnetic.3) $\left[\mathrm{NiCl}_4\right]^{2-}$: Oxidation state of Ni is $+2$ again, Ni$^{2+}$ is $3d^8$. Cl$^{-}$ is a weak field ligand, does not cause pairing. The complex adopts tetrahedral geometry (outer orbital) with 2 unpaired electrons. Hence it is paramagnetic.Thus $\mathrm{Ni}(\mathrm{CO})_4$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are diamagnetic, while $\left[\mathrm{NiCl}_4\right]^{2-}$ is paramagnetic.Shortcut:CO and CN$^{-}$ are strong field ligands – they always cause pairing. Cl$^{-}$ is weak field. For Ni$^{2+}$ ($d^8$): strong field $\rightarrow$ diamagnetic, weak field $\rightarrow$ paramagnetic. Ni(0) with $d^{10}$ is always diamagnetic.Answer:Option D: $\mathrm{Ni}(\mathrm{CO})_4$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are diamagnetic and $\left[\mathrm{NiCl}_4\right]^{2-}$ is paramagnetic.