Q.Based on VB theory explain why [Cr(NH₃)₆]³⁺ is paramagnetic, while [Ni(CN)₄]²⁻ is diamagnetic.
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Start your 14-day free trial to unlock the full solution →Step 1. Cr in [Cr(NH₃)₆]³⁺ is Cr³⁺, outer configuration 3d³. With only 3 electrons available for the 3 t2g orbitals, Hund's rule places one electron in each t2g orbital, unpaired -- there is no possibility of pairing since there aren't enough electrons to force it, regardless of whether NH₃ is a strong-field ligand.
Step 2. The coordination number is 6, hybridisation is d²sp³ (inner orbital, since only 3 of the 5 d orbitals are needed by the metal's own electrons, leaving 2 free (n-1)d orbitals for hybridisation), giving an octahedral geometry with 3 unpaired electrons -- [Cr(NH₃)₆]³⁺ is paramagnetic.
Step 3. Ni in [Ni(CN)₄]²⁻ is Ni²⁺, outer configuration 3d⁸. …
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