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

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Question

(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

$sp^3, dsp^2$

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.

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

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

Similar Questions

Which of the following statements is incorrect?

The correct order of $[FeF_6]^{3-}, [CoF_6]^{3-}, [Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ complex species based on the number of unpaired electrons present is $:$

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.
$A. sp^3$ $(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$ $(ii). [Ni(CO)_4]$
$C. sp^3d^2$ $(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$ $(iv). [CoF_6]^{3-}$
$(v). [Fe(CO)_5]$

For which of the following $d^n$ configurations of octahedral complexes,can they $NOT$ exist in both high spin and low spin forms?
$(I)$ $d^3$,$(II)$ $d^5$,$(III)$ $d^6$,$(IV)$ $d^8$

Which one of the following has a square planar geometry?

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