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Q.
The number of diamagnetic species among the following is
______$Ni ( CO )_4, [NiCl _4]^{2-}, [CoF _6]^{3-}, [Cu(NH _3)_4]^{2+}, [Ni(CN)_4]^{2-}$
NTA AbhyasNTA Abhyas 2022
Solution:
$\left[\right.Ni\left(\right. CO \left.\right)_{4}\left]\right.:$ In this complex nickel is in $0$ oxidation state and valence electronic configuration is $4s^{2}3d^{8}$ . In presence of strong field ligand $CO$ , two electrons present in $4s$ orbital get shifted to $3d$ orbital. Due to this, d orbital is completely filled in complex and all electrons are paired due to which it is diamagnetic.
$\left[\right. NiCl_{4} \left]\right.^{2 -}$ : In this complex nickel is in $+2$ oxidation state and outermost electronic configuration is $3d^{8}$ and since $Cl$ is weak field ligand it can't pair electrons in orbital of nickel due to which it is paramagnetic with two unpaired electrons.
$\left[\right. CoF_{6} \left]\right.^{3 -}$ : In this complex cobalt is in $+3$ oxidation state and outermost electronic configuration is $3d^{6}$ and since fluoride is weak field ligand due to which it can't pair the electrons present in orbital of cobalt and complex is paramagnetic with four unpaired electrons.
$\left[\right.Cu\left(\right. \left(NH\right)_{3} \left.\right)_{4}\left(\left]\right.\right)^{2 +}$ : In this complex copper is in $+2$ oxidation state and configuration is $3d^{9}$ and these electrons can't be paired anyhow therefore complex is paramagnetic with one unpaired electron.
$\left[ Ni ( CN ]^{2-}:\right.$ : In this complex nickel is in $+2$ oxidation state and outermost electronic configuration is $3d^{8}$ and since $CN$ is strong field ligand it will pair electrons in orbital of nickel due to which it is diamagnetic with two unpaired electrons.
Thus, $\left[\right.Ni\left(\right. CO \left.\right)_{4}\left]\right.\text{and}\left[\right.Ni\left(\right. CN \left.\right)_{4}\left(\left]\right.\right)^{2 -}$ are diamagnetic.