Question

$Ni{\left( {CO} \right)_4}$ is:
(A) Tetrahedral and paramagnetic
(B) Square planar and diamagnetic
(C) Tetrahedral and diamagnetic
(D) Square planar and paramagnetic

Answer 1

The molecular geometry of a compound shows its $3 - D$ arrangement of an atom within a molecule. It depends upon the hybridization, whereas magnetic behavior of an atom can either be paramagnetic or diamagnetic based upon the pairing of electrons in the molecule.

Complete step by step answer:
$Ni{\left( {CO} \right)_4}$ is nickel tetracarbonyl. It is a colourless liquid, high in toxicity.
The atomic number of Nickel ($Ni$) is $28$ and its electronic configuration is $3{d^8}4{s^2}$
In ground state,

In an excited state, when $CO$ ligand approaches it.

In an excited state, all the ten electrons are shifted into the $3d -$ orbital and get paired up.
The $4s$ and three $4p$ orbitals are empty, so they undergo $s{p^3}$ hybridization. Upon hybridization, the form bonds with $CO$ ligands and give rise to $Ni{\left( {CO} \right)_4}$

As all the electrons in $Ni{\left( {CO} \right)_4}$ are paired. Thus, the geometry of the molecule depends upon the hybridization. So, the geometry of $Ni{\left( {CO} \right)_4}$ will be tetrahedral due to $s{p^3}$ hybridization.

Hence, $Ni{\left( {CO} \right)_4}$ is tetrahedral and diamagnetic in nature.
Hence, the correct option is C, tetrahedral and diamagnetic.

Additional Information:
Ludwig Mond synthesized $Ni{\left( {CO} \right)_4}$ for the first time in $1890$ the performed direct reaction of $CO$ with $Ni$ to obtain $Ni{\left( {CO} \right)_4}$.
Now-a-days it is prepared in laboratories by carbonylation of bis(cyclooctadiene) nickel ($O$).
The vapours of $Ni{\left( {CO} \right)_4}$ are able to auto ignite and decompose in air quickly.

Note:
Oxidation state of nickel in $Ni{\left( {CO} \right)_4}$ can be found mathematically as:
We know the oxidation state of $CO$ is neutral, that is zero.
Let oxidation state of $Ni = x$
Therefore, $x + \left( {4 \times 0} \right) = 0$
$x = 0$
Hence, oxidation state of nickel is zero in $Ni{\left( {CO} \right)_4}$