How many unpaired electrons are found in NiBr₄^2- (tetrahedral)?

Question

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Nickel in its +2 oxidation state (Ni^2+) has the electronic configuration [Ar]3d^8. In a tetrahedral complex, the crystal field splitting does not lead to the pairing of all electrons due to the lower splitting energy compared to octahedral complexes. Thus, Ni^2+ in a tetrahedral complex would have 2 unpaired electrons, as the 3d orbitals would be filled with 8 electrons, with the last two occupying separate orbitals according to Hund's rule.

How Many Unpaired Electrons Are Found in NiBr42- (Tetrahedral)

How Many Unpaired Electrons Are Found in NiBr₄^2- (Tetrahedral)