[Cr()6] (atomic number of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d — Coordination Compounds Chemistry Question

💡 Solution & Explanation

# Solution: Electron Distribution in [Cr(H₂O)₆]Cl₃ **Step 1: Determine the oxidation state of Cr** In $[Cr(H_2O)_6]Cl_3$: - $H_2O$ is a neutral ligand - 3 $Cl^-$ ions provide charge balance - Therefore: Cr oxidation state = +3 **Step 2: Write electron configuration of Cr³⁺** Cr (Z = 24): $[Ar]3d^5 4s^1$ Cr³⁺: Remove 3 electrons (one $4s^1$ and two $3d$ electrons) $$[Cr]^{3+} = [Ar]3d^3$$ **Step 3: Analyze the ligand and crystal field** $H_2O$ is a weak field ligand → causes **high-spin** configuration The three $3d$ electrons occupy separate orbitals with parallel spins: $$\uparrow \quad \uparrow \quad \uparrow \quad \_ \quad \_$$ **Step 4: Calculate magnetic moment** Number of unpaired electrons: $n = 3$ $$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ B.M.}$$ This matches the given value of 3.83 B.M. (within experimental error). **Answer A is correct** because it shows the $3d^3$ configuration with three unpaired electrons in separate orbitals, consistent with water being a weak-field ligand and the observed magnetic moment.