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

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Question

(A) In the complex $[Ni(CO)_4]$,the central metal atom is Nickel $(Ni)$.The atomic number of $Ni$ is $28$. Its ground state electronic configuration i

Vedclass · Accepted Answer

$sp^3$

Vedclass · Answer

(A) In the complex $[Ni(CO)_4]$,the central metal atom is Nickel $(Ni)$.The atomic number of $Ni$ is $28$. Its ground state electronic configuration is $[Ar] 3d^8 4s^2$.In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$ because $CO$ is a neutral ligand.$CO$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbital. The two electrons from the $4s$ orbital shift into the $3d$ orbital,making it fully filled $(3d^{10})$.This leaves one $4s$ orbital and three $4p$ orbitals vacant,which undergo $sp^3$ hybridization to accommodate the four electron pairs donated by four $CO$ ligands.Therefore,the hybridization of $[Ni(CO)_4]$ is $sp^3$.

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

Similar Questions

The magnetic moment of $[MnCl_{4}]^{2-}$ is $5.92 \ BM$. Explain the reason for this value.

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