Both [Ni(CO)_4] and [Ni(CN)_4]^{2-} are diama | vedclass

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(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pa

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$sp^3, dsp^2$

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(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,leading to $sp^3$ hybridization and tetrahedral geometry.In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The electronic configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the $3d$ electrons to pair up,resulting in $dsp^2$ hybridization and square planar geometry.

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes,respectively,are

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