The problem involves understanding the magnetic properties of coordination complexes. Let's analyze both the complexes given in the question.
Complex: \([\text{Ni(CO)}_4]\)
Nickel (Ni) in this complex is in the zero oxidation state. Therefore, the electronic configuration of Ni is \([Ar]3d^{8}4s^{2}\).
Carbon monoxide (CO) is a strong field ligand, which causes pairing of electrons in the 3d orbitals of nickel. As a result, all the electrons in \([\text{Ni(CO)}_4]\) become paired due to the strong field nature of CO.
Because all the electrons are paired, \([\text{Ni(CO)}_4]\) is diamagnetic.
Complex: \([\text{NiCl}_4]^{2-}\)
Here, Nickel (Ni) is in the +2 oxidation state, giving it the electronic configuration of \([Ar]3d^{8}\).
Chloride (Cl-) is a weak field ligand and does not cause pairing of electrons in the 3d orbitals of nickel.
Due to the presence of unpaired electrons in the 3d orbitals of Ni2+ in \([\text{NiCl}_4]^{2-}\), this complex is paramagnetic.
Thus, the correct answer is that \([\text{Ni(CO)}_4]\) is diamagnetic, and \([\text{NiCl}_4]^{2-}\) is paramagnetic.
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Approach Solution -2
To determine the magnetic properties of the complexes \(\text{[Ni(CO)}_4\text{]}\) and \(\text{[NiCl}_4\text{]}^{2-}\), we need to consider the following:
\(\text{[Ni(CO)}_4\text{]}\):
Nickel in its zero oxidation state is given in this complex. The electronic configuration of Ni is \(\text{[Ar]} \, 3d^{8} \, 4s^{2}\).
CO is a strong field ligand and causes the pairing of electrons. The electrons from the 3d orbitals pair up and thus all the 3d orbitals get filled with paired electrons.
After pairing, the 3d orbitals are completely filled, leading to no unpaired electrons in the Ni 3d subshell.
Therefore, \(\text{[Ni(CO)}_4\text{]}\) is diamagnetic.
\(\text{[NiCl}_4\text{]}^{2-}\):
Nickel in this complex is in a +2 oxidation state. Thus, the electronic configuration is \(\text{[Ar]} \, 3d^{8}\).
Cl is a weak field ligand and doesn't cause pairing of electrons.
As a result, Ni retains unpaired electrons in its d orbitals.
This implies that \(\text{[NiCl}_4\text{]}^{2-}\) is paramagnetic due to the presence of unpaired electrons.
The correct property for the given options is:
\(\text{[Ni(CO)}_4\text{]}\) is diamagnetic because it has no unpaired electrons.
\(\text{[NiCl}_4\text{]}^{2-}\) is paramagnetic due to the presence of unpaired electrons.
Therefore, the correct option is: \( \text{[Ni(CO)}_4\text{]} \) diamagnetic, \( \text{[NiCl}_4\text{]}^{2-} \) paramagnetic.
