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.

Column-$I$ Complex	Column-$II$ Structure / Geometry / Property
$A$. $[Ni(CN)_4]^{2-}$	$I$. Tetrahedral / Paramagnetic
$B$. $[Ni(CO)_4]$	$II$. Tetrahedral / Diamagnetic
$C$. $[NiCl_4]^{2-}$	$III$. Square planar / Diamagnetic

The incorrect statement about $Ni(CO)_4$ is

Similar Questions

Match the following:

Column-$I$ Complex Column-$II$ Structure / Geometry / Property

$A$. $[Ni(CN)_4]^{2-}$ $I$. Tetrahedral / Paramagnetic

$B$. $[Ni(CO)_4]$ $II$. Tetrahedral / Diamagnetic

$C$. $[NiCl_4]^{2-}$ $III$. Square planar / Diamagnetic

The correct match is:

Square-planar geometry is shown by

For the complexes showing the square pyramidal structure,the $d-$orbital involved in the hybridisation is

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
$(i)$ $[Fe(CN)_6]^{4-}$
$(ii)$ $[FeF_6]^{3-}$
$(iii)$ $[Co(C_2O_4)_3]^{3-}$
$(iv)$ $[CoF_6]^{3-}$

The hybridisation and geometry of $[Ni(CN)_4]^{2-}$ are:

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