The hybridization of $Ni {(CO)}_{4}$ is

$d s p^{2}$

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$s p^{2}$

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$s p^{3} d$

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$s p^{3}$

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Solution

The correct option is D
$s p^{3}$

Explanation for correct option
(D) $s p^{3}$
$Ni$ is in zero oxidation state in $Ni {(CO)}_{4}$ .
So the electronic configuration of Ni is $3 d^{8} 4 s^{2}$ .
As $CO$ is a strong ligand, it pushes all the electrons in the $3 d$ orbital as shown in the given diagram:-
Therefore the hybridisation of $Ni {(CO)}_{4}$ is $s p^{3}$ and it has tetrahedral geometry.
Explanation for incorrect option
The hybridization of $Ni {(CO)}_{4}$ is $s p^{3}$ . Hence, options (A), (B), and (C) are incorrect.
Hence, option (D) is correct. The hybridization of $Ni {(CO)}_{4}$ is $s p^{3}$ .

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