From the magnetic behaviour of \([NiCl_4]^{2-}\) (paramagnetic) and [Ni\((CO)_4\)] (diamagnetic), choose the correct geometry and oxidation state.
From the magnetic behaviour of \([NiCl_4]^{2-}\) (paramagnetic) and [Ni\((CO)_4\)] (diamagnetic), choose the correct geometry and oxidation state.
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\([NiCl_4]^2-\) : Ni²⁺, square planar [Ni\((CO)_4\)] : Ni(0), square planar
\([NiCl_4]^2-\) : Ni²⁺, tetrahedral [Ni\((CO)_4\)] : Ni(0), tetrahedral
\([NiCl_4]^2-\): Ni²⁺, tetrahedral [Ni\((CO)_4\)] : \(Ni^{2+}\), square planar
\([NiCl_4]^2-\) : Ni(0), tetrahedral [Ni\((CO)_4\)] : Ni(0), square planar
The Correct Option is B
Solution and Explanation
The solution involves determining the geometry and oxidation state of nickel in two complexes: \([NiCl_4]^{2-}\) and \([Ni(CO)_4]\).
1. Analyze \([NiCl_4]^{2-}\):
Nickel in \([NiCl_4]^{2-}\) is in the +2 oxidation state (Ni²⁺). The chloride ions, Cl⁻, are weak field ligands and do not cause pairing of electrons in the d-orbitals of nickel. Thus, the configuration remains high-spin, resulting in unpaired electrons, making it paramagnetic. The paramagnetic property suggests a tetrahedral geometry, as this structure does not allow for complete electron pairing like in square planar complexes.
2. Check \([Ni(CO)_4]\):
In \([Ni(CO)_4]\), nickel remains in the zero oxidation state (Ni(0)). Carbon monoxide, CO, is a strong field ligand, leading to the pairing of electrons in the d-orbitals, thus making the complex diamagnetic. The electron pairing results in no unpaired electrons. In this case, since the complex is diamagnetic and the hybridization involved is sp³, the geometry is tetrahedral.
Conclusion:
| Complex | Oxidation State | Geometry |
|---|---|---|
| \([NiCl_4]^{2-}\) | Ni²⁺ | Tetrahedral |
| \([Ni(CO)_4]\) | Ni(0) | Tetrahedral |
Thus, the correct configuration for the complexes according to the magnetic properties is: \([NiCl_4]^{2-}\): Ni²⁺, tetrahedral; \([Ni(CO)_4]\): Ni(0), tetrahedral.
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