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VBT for [Ni(NH₃)₆]²⁺: geometry and magnetism

Short answer

[Ni(NH3)6]2+ forms by sp3d2 hybridization using outer 4d orbitals, resulting in an octahedral geometry. It is paramagnetic with two unpaired 3d electrons and a magnetic moment of 2.83 BM.

Explain the formation of [Ni(NH₃)₆]²⁺ on the basis of valence bond approach and predict its geometry and magnetic behavior.

Ni (Z = 28); Ni²⁺ is 3d⁸. NH₃ is a ligand with a lone pair on N, and coordination number is 6.

Formation: the 3d electrons of Ni²⁺ stay unpaired (even with pairing, d⁸ can free only one 3d orbital, not the two needed for d²sp³). The six vacant orbitals used are one 4s, three 4p and two 4d, giving sp³d² hybridisation.

Ni²⁺ 3d: [↑↓][↑↓][↑↓][↑ ][↑ ]   4s [ ]  4p [ ][ ][ ]  4d [ ][ ]
                                 \______ sp³d² ______/
                                  six NH₃ lone pairs →
  • Geometry: octahedral (outer orbital complex).
  • Magnetism: two unpaired 3d electrons, so it is paramagnetic, μ = √(2×4) = 2.83 BM.

✓ Answer checked by Claude Opus 5.5 (high effort)

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