The magnetic moment of coordination compounds can be measured by the magnetic susceptibility experiments. The results can be used to obtain information about the number of unpaired electrons and hence structures adopted by metal complexes. A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with up to three electrons in the d orbitals, like Ti³⁺(d¹); V³⁺(d²); Cr³⁺(d³); two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund’s rule). Thus, for d⁴ (Cr²⁺, Mn³⁺), d⁵ (Mn²⁺, Fe³⁺), d⁶ (Fe²⁺, Co³⁺) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d⁶ ions. However, with species containing d⁴ and d⁵ ions there are complications. [Mn(CN)₆]^3– has magnetic moment of two unpaired electrons while [MnCl₆]^3– has a paramagnetic moment of four unpaired electrons. [Fe(CN)₆]^3– has magnetic moment of a single unpaired electron while [FeF₆]^3– has a paramagnetic moment of five unpaired electrons. [CoF₆]^3– is paramagnetic with four unpaired electrons while [Co(C₂O₄)₃]^3– is diamagnetic. This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. [Mn(CN)₆]^3–, [Fe(CN)₆]^3– and [Co(C₂O₄)₃]^3– are inner orbital complexes involving d²sp³ hybridization, the former two complexes are paramagnetic and the latter diamagnetic. On the other hand, [MnCl₆]^3–, [FeF₆]^3– and [CoF₆]^3– are outer orbital complexes involving sp³d² hybridization and are paramagnetic corresponding to four, five and four unpaired electrons.

Which one of the following is paramagnetic in nature?

[NiCl₄]^2-

The correct answer is option 3. \([NiCl_4]^{2-}\).

1. \(Ni(CO)_4\)

Oxidation State: Ni is in the zero oxidation state.

Electronic Configuration: For \(Ni\), it’s \([Ar] 3d^8 4s^2\).

Ligands and Hybridization: \(CO\) is a strong field ligand, causing the formation of \(sp^3\) hybrid orbitals in a tetrahedral geometry. Strong field ligands lead to pairing of the 3d electrons.

Unpaired Electrons: All the electrons in the d-orbitals are paired due to the strong field effect of \(CO\).

Magnetic Property: Since all electrons are paired, \(Ni(CO)_4\) is diamagnetic.

2. \([Ni(CN)_4]^{2-}\)

Oxidation State: \(Ni\) is in the +2 oxidation state.

Electronic Configuration: For \(Ni^{2+}\), it’s \([Ar] 3d^8\).

Ligands and Hybridization: \(CN^-\) is a strong field ligand, leading to \(dsp^2\) hybridization in a square planar geometry. Strong field ligands cause pairing of the \(3d\) electrons.

Unpaired Electrons: All the d-electrons are paired because of the strong field effect.

Magnetic Property: Since all electrons are paired, \([Ni(CN)_4]^{2-}\) is diamagnetic.

3. \([NiCl_4]^{2-}\)

Oxidation State: Ni is in the +2 oxidation state.

Electronic Configuration: For \(Ni^{2+}\), it’s \([Ar] 3d^8\).

Ligands and Hybridization: \(Cl^-\) is a weak field ligand, which typically leads to \(sp^3\) hybridization in a tetrahedral geometry. Weak field ligands do not cause significant pairing of the \(3d\) electrons.

Unpaired Electrons: In tetrahedral complexes with weak field ligands, d-electrons are not fully paired, resulting in unpaired electrons.

Magnetic Property: Since there are unpaired electrons, \([NiCl_4]^{2-}\) is paramagnetic.

4. \([Fe(CN)_]^{4-}\)

Oxidation State: \(Fe\) is in the \(+2\) oxidation state.

Electronic Configuration: For \(Fe^{2+}\), it’s \([Ar] 3d^6\).

Ligands and Hybridization: \(CN^-\) is a strong field ligand, leading to \(d^2sp^3\) hybridization in an octahedral geometry. Strong field ligands cause pairing of the \(3d\) electrons.

Unpaired Electrons: All the d-electrons are paired because of the strong field effect.

Magnetic Property: Since all electrons are paired, \([Fe(CN)_6]^{4-}\) is diamagnetic.

Summary:

Paramagnetic: A substance is paramagnetic if it has unpaired electrons.

Diamagnetic: A substance is diamagnetic if all electrons are paired.

Among the given options, only \([NiCl_4]^{2-}\) has unpaired electrons and thus exhibits paramagnetic behavior.