Question

Using the valence bond theory of complexes, explain the geometry and magnetic nature of [Ni(NH₃)₆]²⁺.

Answer 1

• In this complex, nickel is in the +2 oxidation state, with an atomic number of 28. This gives it an electronic configuration of [Ar] 3d⁸.
• The ligand ammonia (NH₃) is a neutral, weak to moderate field ligand, which does not cause strong pairing of 3d electrons. As a result, the 3d⁸ configuration retains two unpaired electrons.
• The complex has a coordination number of 6 because there are six ammonia molecules coordinated to the nickel ion.
• According to Valence Bond Theory (VBT), the geometry of [Ni(NH₃)₆]²⁺is octahedral. This is due to the d²sp³ hybridization of the Ni²⁺ ion, where six ligands, nickel uses the outer orbitals: one 4s, three 4p, and two 4d orbitals to undergo d²sp³ hybridisation, resulting in an octahedral geometry.
• The hybridization of the Ni²⁺ ion in an octahedral complex is d²sp³. This means that two electrons from the 3d orbitals, one s orbital, and three p orbitals of the Ni²⁺ion mix to form six equivalent hybrid orbitals.
• All the electrons in the Ni²⁺ ion are paired in the lower-energy t₂g orbitals and there are no unpaired electrons, the complex [Ni(NH₃)₆]²⁺ is diamagnetic.
• Diamagnetic means that the complex does not have any unpaired electrons and is not attracted to a magnetic field.
• According to Valence Bond Theory (VBT), [Ni(NH₃)₆]²⁺ is diamagnetic because all the d-electrons in the Ni²⁺ ion are paired, and there are no unpaired electrons in the complex.