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

💡 Solution & Explanation

# Analysis of $K_3[Cu(CN)_4]$ **Step 1: Identify the central metal and its oxidation state** - Cu in $K_3[Cu(CN)_4]$ has oxidation state +1 (since $3(+1) + x + 4(-1) = 0$ gives $x = +1$) - Electronic configuration: Cu⁺ is $[Ar]3d^{10}$ (d¹⁰ configuration) **Step 2: Analyze the ligand environment** - CN⁻ is a strong field ligand - 4 CN⁻ ligands coordinate to Cu⁺ → tetrahedral geometry **Step 3: Determine hybridization** - With 4 ligands in tetrahedral arrangement: $sp^3$ hybridization **Step 4: Determine shape** - 4 coordinate covalent bonds, no lone pairs on Cu⁺ - **Tetrahedral shape** **Step 5: Calculate magnetic moment** - Cu⁺ has $d^{10}$ configuration: all d-orbitals are fully filled - Number of unpaired electrons = 0 - Magnetic moment: $$\mu = \sqrt{n(n+2)} \text{ B.M.} = \sqrt{0(0+2)} = 0 \text{ B.M.}$$ - The complex is **diamagnetic** **Answer: Hybridization = $sp^3$, Shape = Tetrahedral, Magnetic moment = 0 B.M. (diamagnetic)**