nickel does not form low spin octahedral complexes

Nickel usually forms octahedral complexes and exits in +2 oxidation state. The electronic configuration of Ni is [Ar] 3d8 4s2. So the electronic configuration of Ni+2 ion will be [Ar] 3d8. We know that in the presence of ligands, the 5 degenerate d-orbitals split into 2 set of orbitals - the t2g set and the eg set. Strong ligands cause pairing of electrons and result in low spin complexes. On the other hand, weak ligands do not cause the pairing of electrons and result in high spin complexes. As there are 8 electrons in d-orbitals of Ni+2 ion, therefore for both strong field and weak field ligands, the electronic configuration will be (t2g)2(eg)2. Hence, Ni2+ ion will always contain 2 unpaired electrons in the eg set and will therefore form high spin complexes, irrespective of the strength of ligand. 

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 Nickel does form square planare complexes, but it also forms lots of octahedralcomplexes.* (It also tends to form tetrahedral complexes with weak-field ligands, e.g., NiCl₄²⁻.) 

Both [Ni(NH₃)₆]²⁺ and [Ni(H₂O)₆]²⁺ are octahedral complexes. Most octahedral complexes of Ni are Ni(II) complexes, and for a d⁸ complex there is no distinction between 'high-spin' and 'low-spin'. For example, [Ni(H₂O)₆]²⁺, has a (t_2g)⁶(e_g)² d-electron configuration. Even though the ligand field of the water ligands is weak, you still get the t_2g orbitals completely filled. Likewise, it doesn't matter how big the ligand field splitting is ([Ni(NH₃)₆]²⁺ has a significantly larger splitting), if the complex is octahedral, you can't get more than 6 electrons into t_2g – all octahedral N
i(II) complexes are (t_2g)⁶(e_g)².

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