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

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(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong fiel

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$sp^3, dsp^2$

Vedclass · Answer

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10}$ configuration. The hybridisation is $sp^3$ (tetrahedral geometry).In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the two unpaired electrons in the $3d$ orbitals to pair up. This creates one empty $3d$ orbital,which is used for $dsp^2$ hybridisation (square planar geometry).Therefore,the hybridisation types are $sp^3$ and $dsp^2$ respectively.

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

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