Hybridization in [Ni(CO)_4] and [Ni(CN)_4]^{2 | 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

Vedclass · Accepted Answer

$sp^3$ and $dsp^2$

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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 field ligand,it causes pairing of electrons,resulting in $3d^{10} 4s^0 4p^0$. Thus,the hybridization is $sp^3$.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,forcing the two unpaired electrons in the $3d$ orbitals to pair up,creating one empty $3d$ orbital. This leads to $dsp^2$ hybridization.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

Hybridization in $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are respectively . . . . . . .

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