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

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

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$sp^3, dsp^2$

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(B) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state,so the electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes the pairing of $4s$ electrons into the $3d$ subshell,resulting in a $3d^{10}$ configuration. This leaves one $4s$ and three $4p$ orbitals vacant,leading to $sp^3$ hybridization.In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state,so the electronic configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the pairing of electrons in the $3d$ subshell,leaving one $3d$ orbital vacant. This vacant $3d$ orbital,along with one $4s$ and two $4p$ orbitals,undergoes hybridization to form $dsp^2$ hybridization.

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

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