Select the option with correct property -
- \( \text{[Ni(CO)}_4\text{]} \) and \( \text{[NiCl}_4\text{]}^{2-} \) both diamagnetic
- \( \text{[Ni(CO)}_4\text{]} \) and \( \text{[NiCl}_4\text{]}^{2-} \) both paramagnetic
- \( \text{[NiCl}_4\text{]}^{2-} \) diamagnetic, \( \text{[Ni(CO)}_4\text{]} \) paramagnetic
- \( \text{[Ni(CO)}_4\text{]} \) diamagnetic, \( \text{[NiCl}_4\text{]}^{2-} \) paramagnetic
The Correct Option is D
Approach Solution - 1
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.
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.
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