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

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.

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

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

Identify the coordination complexes in which the central metal ion has $d^4$ configuration.
$(A)$ $[FeO_4]^{2-}$
$(B)$ $[Mn(CN)_6]^{3-}$
$(C)$ $[Fe(CN)_6]^{3-}$
$(D)$ $[Cr(H_2O)_6]^{2+}$
$(E)$ $[NiF_6]^{2-}$
Choose the correct answer from the options given below:

What is the shape of $Fe(CO)_5$?

The hybridization of $[Ni(CO)_4]$ is:

The structure of iron pentacarbonyl is:

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