The hybridization states of the central metal | vedclass

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(D) $1$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. $CO$ is a strong field ligand,causing pairing of $3d$ electrons. The hybridization is $sp^3$.

Vedclass · Accepted Answer

$sp^3, dsp^2, sp^3$

Vedclass · Answer

(D) $1$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. $CO$ is a strong field ligand,causing pairing of $3d$ electrons. The hybridization is $sp^3$.$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons. The hybridization is $dsp^2$.$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,not causing pairing of $3d$ electrons. The hybridization is $sp^3$.Thus,the hybridization states are $sp^3, dsp^2, sp^3$ respectively.

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic

The hybridization states of the central metal atom in the complexes $[Ni(CO)_4]$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$ are respectively:

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