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 sp³hybridization. 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 sp³hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.
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