The incorrect statement about Ni(CO)_4 is | vedclass

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(D) $1$. $Ni(CO)_4$ is a tetrahedral complex where $Ni$ is in $0$ oxidation state.$2$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In th

Vedclass · Accepted Answer

The complex cannot act as oxidizing or reducing agent according to sidwick $EAN$ rule

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(D) $1$. $Ni(CO)_4$ is a tetrahedral complex where $Ni$ is in $0$ oxidation state.$2$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization. It is diamagnetic.$3$. Due to back-bonding from $Ni$ to $CO$,the $C-O$ bond order decreases compared to free $CO$ molecule,while the $Ni-C$ bond order increases (becomes greater than $1$).$4$. According to the $EAN$ rule,$Ni(CO)_4$ has $28 + 4 	imes 2 = 36$ electrons,which is the atomic number of $Kr$. This makes it stable,but it can still act as a reducing agent in certain chemical reactions. Thus,the statement that it cannot act as an oxidizing or reducing agent is incorrect.

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 incorrect statement about $Ni(CO)_4$ is

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