Select the option with correct property -
- $[Ni(CO)_4]$ and $[NiCl_4]^{2-}$ both diamagnetic
- $[Ni(CO)_4]$ and $[NiCl_4]^{2-}$ both paramagnetic
- $[NiCl_4]^{2-}$ diamagnetic, $[Ni(CO)_4]$ paramagnetic
- $[Ni(CO)_4]$ diamagnetic, $[NiCl_4]^{2-}$ paramagnetic
The Correct Option is D
Approach Solution - 1
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.
- CO, being a strong field ligand, leads to pairing of electrons in the 3d orbital.
- Resulting electron pairing: \(3d^{10} \, 4s^0 \, 4p^0\)
- 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.
- Chloride ion (Cl-) is a weak field ligand.
- It does not cause pairing. Thus, Ni2+ configuration: \(3d^8\)
- 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.
Approach Solution -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:
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