The complex exhibits paramagnetism with 3 unpaired electrons. What is its hybridisation state accord — Coordination Compounds Chemistry Question

💡 Solution & Explanation

$Cr^{3+}$ has a $3d^3$ configuration. According to Hund's rule, the 3 electrons occupy three 3d orbitals singly, leaving two inner 3d orbitals completely empty. Thus, it seamlessly undergoes inner orbital $d^2sp^3$ hybridisation regardless of ligand strength.