Among [Ni(CO)_4],[Ni(CN)_4]^{2-},and [NiCl_4] | vedclass

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(D) $1$. For $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$,res

Vedclass · Accepted Answer

$sp^3, dsp^2, sp^3$ (Atomic number of $Ni = 28$)

Vedclass · Answer

(D) $1$. For $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$,resulting in $sp^3$ hybridization.$2$. For $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons,resulting in $dsp^2$ hybridization.$3$. For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $sp^3$ hybridization.Therefore,the hybridization states are $sp^3, dsp^2, sp^3$ respectively. The correct option is $D$.

Among $[Ni(CO)_4]$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$ species,the hybridization states at the $Ni$ atom are,respectively:

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