Question
$\ce{[NIC14]^{2-}}$ is paramagnetic while $\ce{[Ni(CO)_4]}$ is diamagnetic though both are tetrahedral. Why?
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Answer
Though both $\ce{[ NICl _4]^{2-}}$ and $\ce{[ Ni ( CO )_4]}$ are tetrahedral, their magnetic characters are different.
This is due to a difference in the nature of ligands. $\ce{CN ^{-}}$ is a weak field ligand and it does not cause the pairing of unpaired $3d$ electrons.
Hence, $\ce{[ NICl _4]^{2-}}$ 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 $sp^3$ hybridization.
Since no unpaired electrons are present in this case, $\ce{[Ni(CO)_4]}$ is diamagnetic.
This is due to a difference in the nature of ligands. $\ce{CN ^{-}}$ is a weak field ligand and it does not cause the pairing of unpaired $3d$ electrons.
Hence, $\ce{[ NICl _4]^{2-}}$ 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 $sp^3$ hybridization.
Since no unpaired electrons are present in this case, $\ce{[Ni(CO)_4]}$ is diamagnetic.
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