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Structure and magnetism of [NiCl₄]²⁻ and [Ni(CN)₄]²⁻

Short answer

[NiCl4]2- has tetrahedral geometry with sp3 hybridization and two unpaired electrons (paramagnetic). [Ni(CN)4]2- is square planar with dsp2 hybridization and all electrons paired (diamagnetic). The difference arises from the weak field Cl- versus strong field CN- ligands.

How would you account for the difference in structures and magnetic properties between [NiCl₄]²⁻ and [Ni(CN)₄]²⁻?

Ni²⁺ has the configuration 3d⁸. The difference comes from the ligand strength.

[NiCl₄]²⁻

Cl⁻ is a weak-field ligand, so the 3d electrons are not paired. Ni²⁺ uses one 4s and three 4p orbitals (sp³ hybridisation).

3d: [↑↓][↑↓][↑↓][↑ ][↑ ]   4s 4p: [ ][ ][ ][ ] sp³
  • Geometry: tetrahedral.
  • Two unpaired electrons: paramagnetic, μ = √8 = 2.83 BM.

[Ni(CN)₄]²⁻

CN⁻ is a strong-field ligand, so the two unpaired 3d electrons pair up and one 3d orbital is vacated. dsp² hybridisation (one 3d, one 4s, two 4p).

3d: [↑↓][↑↓][↑↓][↑↓][  ]   dsp² (4 empty orbitals)
  • Geometry: square planar.
  • No unpaired electrons: diamagnetic.

Strong-field CN⁻ gives a square planar diamagnetic complex; weak-field Cl⁻ gives a tetrahedral paramagnetic one.

✓ Answer checked by Claude Opus 5.5 (high effort)

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