Question

Select the option with correct property -

• A. $[Ni(CO)_4]$ and $[NiCl_4]^{2-}$ both diamagnetic
• B. $[Ni(CO)_4]$ and $[NiCl_4]^{2-}$ both paramagnetic
• C. $[NiCl_4]^{2-}$ diamagnetic, $[Ni(CO)_4]$ paramagnetic
• D. $[Ni(CO)_4]$ diamagnetic, $[NiCl_4]^{2-}$ paramagnetic

Answer 1

The Correct Option is D

To determine the properties of the complexes \([Ni(CO)_4]\) and \([NiCl_4]^{2-}\), we need to analyze their electronic configurations and types of bonding.

Electronic Configuration of Nickel

The electronic configuration of a Nickel (Ni) atom in its elemental form (atomic number 28) is:

\([Ar] \, 3d^8 \, 4s^2\)

In the formation of complexes, the oxidation state and the nature of ligands influence the electronic configuration.

Analysis of \([Ni(CO)_4]\)

\([Ni(CO)_4]\) is a complex where Ni is in the zero oxidation state. Carbon monoxide (CO) is a strong field ligand and causes the pairing of electrons in the 3d subshell.

1. CO, being a strong field ligand, leads to pairing of electrons in the 3d orbital.
2. Resulting electron pairing: \(3d^{10} \, 4s^0 \, 4p^0\)
3. With all electrons paired, the complex is diamagnetic.

Analysis of \([NiCl_4]^{2-}\)

\([NiCl_4]^{2-}\) is a complex where Ni is in the +2 oxidation state.

1. Chloride ion (Cl^-) is a weak field ligand.
2. It does not cause pairing. Thus, Ni²⁺ configuration: \(3d^8\)
3. With two unpaired electrons, the complex is paramagnetic.

Conclusion

Based on the analysis, the correct property is:

• \([Ni(CO)_4]\) is diamagnetic.
• \([NiCl_4]^{2-}\) is paramagnetic.

Therefore, the correct option is: \([Ni(CO)_4]\) diamagnetic, \([NiCl_4]^{2-}\) paramagnetic.

Answer 2

In \([\text{Ni}(\text{CO})_4]\), nickel is in the zero oxidation state and has a \(3d^{10}\) electron configuration, leading to a fully paired electron configuration, making it diamagnetic.

In contrast, in \([\text{NiCl}_4]^{2-}\), nickel is in the \(+2\) oxidation state with a \(3d^8\) electron configuration. The chloride ligands are weak-field ligands and do not pair up the electrons, resulting in unpaired electrons and a paramagnetic nature.

Thus, the correct answer is:

\[ [\text{Ni}(\text{CO})_4] \text{ is diamagnetic and } [\text{NiCl}_4]^{2-} \text{ is paramagnetic.} \]