[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: Finding 3d Electron Configuration 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 balance the charge - Therefore, Cr is in +3 oxidation state: $Cr^{3+}$ **Step 2: Write the electron configuration of Cr³⁺** Neutral Cr (Z = 24): $[Ar] 3d^5 4s^1$ $Cr^{3+}$ loses 3 electrons (one 4s and two 3d): $$Cr^{3+}: [Ar] 3d^3$$ **Step 3: Calculate magnetic moment from electron configuration** For $3d^3$ configuration, all three electrons occupy separate orbitals with parallel spins (following Hund's rule), giving: - Number of unpaired electrons: $n = 3$ Magnetic moment formula: $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$ $$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ B.M.} \approx 3.83 \text{ B.M.} \checkmark$$ **Step 4: Verify the electron distribution** The 3d³ configuration in $[Cr(H_2O)_6]Cl³$ with all unpaired electrons: $$\boxed{t_{2g}^3 \text{ (high spin configuration)}}$$ This matches the observed magnetic moment of 3.83 B.M., confirming **Option A** represents the correct 3d electron distribution: **three unpaired electrons in a high-spin arrangement**.