Assign the hybridization, shape and magnetic moment of [Cu(CN)4]: — Coordination Compounds Chemistry Question

💡 Solution & Explanation

# Solution: $K_3[Cu(CN)_4]$ Analysis **Step 1: Identify the metal ion and electron configuration** - Cu in this complex is Cu⁺ (since $K_3$ provides +3 charge, and $[Cu(CN)_4]^{3-}$ makes it neutral) - Cu⁺: $[Ar]3d^{10}4s^0$ (d¹⁰ configuration) **Step 2: Determine hybridization** - CN⁻ is a strong field ligand → pairs electrons in d-orbitals - 4 CN⁻ ligands around Cu⁺ require **sp³ hybridization** - (Uses one s and three p orbitals for bonding) **Step 3: Determine shape** - sp³ hybridization with 4 ligands → **tetrahedral geometry** **Step 4: Determine magnetic moment** - Cu⁺ has d¹⁰ configuration (completely filled d-orbitals) - All electrons are **paired** → **diamagnetic** - Magnetic moment μ = **0 B.M.** (Bohr magnetons) $$\boxed{\text{Hybridization: sp}^3 \text{ | Shape: Tetrahedral | Magnetic moment: 0 B.M.}}$$ **Why this is correct:** The d¹⁰ configuration ensures all d-electrons are paired (no unpaired electrons). Despite the tetrahedral geometry typically suggesting sp³, the key here is recognizing Cu⁺ is diamagnetic due to having no unpaired electrons, which uniquely identifies option B.