JEE Main 2023 · 10 April · morning · Single correct · +4 / −1
The octahedral diamagnetic low spin complex among the following is :
- (a) [Co(NH₃)₆]³⁺
- (b) [NiCl₄]²⁻
- (c) [CoCl₆]³⁻
- (d) [CoF₆]³⁻
Answer and solution
Answer (a) [Co(NH₃)₆]³⁺
Solution
Chemistry ma’am · 3 steps
All three cobalt ions are d⁶, and only strong-field NH₃ pairs all six electrons.
- Find the d count. In all three cobalt complexes cobalt is Co³⁺, which is 3d⁶. In [NiCl₄]²⁻ nickel is Ni²⁺, which is 3d⁸, and it has only four ligands, so it is not octahedral.
- Ligand strength decides. In an octahedral field a strong-field ligand makes Δₒ larger than the pairing energy, so all six electrons pair up in t_(2g). A weak-field ligand such as F⁻ or Cl⁻ gives a small Δₒ, so electrons spread into e_g and stay unpaired.
Complex Geometry Ligand field Configuration Unpaired [Co(NH₃)₆]³⁺ octahedral NH₃, strong for Co³⁺ t_(2g)⁶ e_g⁰, low spin 0, diamagnetic [NiCl₄]²⁻ tetrahedral Cl⁻, weak e⁴ t₂⁴ 2, paramagnetic [CoCl₆]³⁻ octahedral Cl⁻, weak t_(2g)⁴ e_g², high spin 4, paramagnetic [CoF₆]³⁻ octahedral F⁻, weak t_(2g)⁴ e_g², high spin 4, paramagnetic - So. Only [Co(NH₃)₆]³⁺ is octahedral, low spin and diamagnetic, which is option (a).
The idea to keep. For a d⁶ octahedral ion, the ligand decides everything. A strong field gives t_(2g)⁶, low spin and diamagnetic. A weak field gives t_(2g)⁴ e_g² with 4 unpaired electrons.
Answer from the NTA final answer key. Solution by PrepWiser.