According to Valence Bond Theory (VBT), what is the specific hybridization and geometry of the ion i — Coordination Compounds Chemistry Question

💡 Solution & Explanation

$Cu^{2+}$ has a $3d^9$ electronic configuration. The complex $[Cu(NH_3)_4]^{2+}$ is experimentally known to be square planar. In VBT, this is explained by the promotion of the single unpaired electron from the $3d$ orbital to a higher energy $4p$ orbital, vacating one $3d$ orbital. This allows for $dsp^2$ hybridization (using the $d_{x^2-y^2}$ orbital), resulting in a square planar geometry and leaving $1$ unpaired electron, making it paramagnetic.