Question

Select the option with correct property -

• A. \( \text{[Ni(CO)}_4\text{]} \) and \( \text{[NiCl}_4\text{]}^{2-} \) both diamagnetic
• B. \( \text{[Ni(CO)}_4\text{]} \) and \( \text{[NiCl}_4\text{]}^{2-} \) both paramagnetic
• C. \( \text{[NiCl}_4\text{]}^{2-} \) diamagnetic, \( \text{[Ni(CO)}_4\text{]} \) paramagnetic
• D. \( \text{[Ni(CO)}_4\text{]} \) diamagnetic, \( \text{[NiCl}_4\text{]}^{2-} \) paramagnetic

Answer 1

The Correct Option is D

The problem involves understanding the magnetic properties of coordination complexes. Let's analyze both the complexes given in the question.

1. 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.
2. 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.

Answer 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:

1. \(\text{[Ni(CO)}_4\text{]}\):
   • 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.
2. \(\text{[NiCl}_4\text{]}^{2-}\):
   • 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.