Question
$[NIC14]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why?
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Answer
Though both $\left[ NICl _4\right]^{2-}$ and $\left[ Ni ( CO )_4\right]$ are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. $CN ^{-}$is a weak field ligand and it does not cause the pairing of unpaired $3d$ electrons. Hence, $\left[ NICl _4\right]^{2-}$ is paramagnetic.
But $CO$ is a strong field ligand. Therefore, it causes the pairing of unpaired $3d$ electrons. Also, it causes the $4s$ electrons to shift to the $3d$ orbital, thereby giving rise to $sp³$ hybridization. Since no unpaired electrons are present in this case, $[Ni(CO)_4]$ is diamagnetic.
But $CO$ is a strong field ligand. Therefore, it causes the pairing of unpaired $3d$ electrons. Also, it causes the $4s$ electrons to shift to the $3d$ orbital, thereby giving rise to $sp³$ hybridization. Since no unpaired electrons are present in this case, $[Ni(CO)_4]$ is diamagnetic.
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Column II (Element)
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Radioactive lanthanoid
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Lanthanoid which has $4f^7$ electronic configuration in +3 oxidation state
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(v)
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Lanthanoid which has $4f^{14}$ electronic configuration in +3 oxidation state
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$(e)$
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$Gd$
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|
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$(f)$
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$Dy$
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a.
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Calcium
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Iron
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5.
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Magnesium
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