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

💡 Solution & Explanation

# Solution: Hybridization, Shape, and Magnetic Moment of $K_3[Cu(CN)_4]$ **Step 1: Identify the central metal ion** Cu is in the +3 oxidation state: $[Cu(CN)_4]^{3-}$ Electronic configuration of Cu³⁺: $[Ar]3d^8$ (8 d-electrons) **Step 2: Analyze the ligand** $CN^-$ is a strong field ligand (σ-donor with π-acceptor capability) **Step 3: Determine electron pairing in $d^8$ configuration** With a strong field ligand causing maximum crystal field splitting: $$\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow\downarrow \quad \text{(—)}$$ All 8 electrons pair up completely → **0 unpaired electrons** **Step 4: Assign hybridization** - 4 ligands (CN⁻) require 4 hybrid orbitals - Square planar geometry with $d^8$ configuration uses $dsp^2$ hybridization - Orbitals involved: one $d$, one $s$, two $p$ → **$dsp^2$** **Step 5: Determine shape** 4 coordination number with $dsp^2$ hybridization → **Square planar geometry** **Step 6: Calculate magnetic moment** $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$ where $n$ = number of unpaired electrons = 0 $$\mu = \sqrt{0(0+2)} = \mathbf{0 \text{ B.M.}} \text{ (diamagnetic)}$$ --- **Answer B contains:** $dsp^2$ hybridization, **square planar** shape, and **0 B.M.** (diamagnetic character)