[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
[Cr(NH3)6]3+ is paramagnetic: In this complex, Cr is in +3 oxidation state with 3d³ electronic configuration. NH₃ is a strong ligand that causes pairing of electrons and forms d²sp³ hybridisation (inner orbital complex). However, the three 3d electrons remain unpaired, making the complex paramagnetic.
[Ni(CN)4]2- is diamagnetic: In this complex, Ni is in +2 oxidation state with 3d⁸ configuration. CN⁻ is a strong ligand that causes dsp² hybridisation forming a square planar complex. All the electrons pair up in the d orbitals, leaving no unpaired electron, making the complex diamagnetic.
Explanation
The context explains that [Ni(CN)4]2- involves dsp² hybridisation with Ni²+ having 3d⁸ configuration, and the absence of unpaired electrons makes it diamagnetic. For [Cr(NH3)6]3+, the context shows NH₃ acts as a strong ligand causing d²sp³ hybridisation (inner orbital complex). With Cr³+ having 3d³ configuration, three unpaired electrons remain, making it paramagnetic.
Solution Steps
Step 1: Identify oxidation state and electronic configuration - Cr³+ has 3d³, Ni²+ has 3d⁸
Step 2: Identify ligand strength - both NH₃ and CN⁻ are strong ligands
Step 3: Determine hybridisation - d²sp³ for [Cr(NH3)6]3+, dsp² for [Ni(CN)4]2-
Step 4: Count unpaired electrons - 3 unpaired in Cr complex (paramagnetic), 0 unpaired in Ni complex (diamagnetic)