[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: 3d Electron Distribution in [Cr(H₂O)₆]Cl₃ **Step 1: Determine Chromium's Oxidation State** In $[Cr(H_2O)_6]Cl_3$: - $H_2O$ is a neutral ligand (no charge) - Overall charge is +3 - Therefore, Cr has oxidation state **+3** **Step 2: Find Electron Configuration of Cr³⁺** Cr (Z = 24): $[Ar]3d^5 4s^1$ Cr³⁺ loses 3 electrons (one from 4s and two from 3d): $$Cr^{3+}: [Ar]3d^3$$ **Step 3: Use Magnetic Moment to Determine Spin State** Given: $\mu = 3.83$ B.M. The formula for magnetic moment is: $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$ where $n$ = number of unpaired electrons $$3.83 = \sqrt{n(n+2)}$$ $$14.67 \approx n(n+2)$$ $$n = 3$$ **Step 4: Determine Electron Distribution** With 3 unpaired electrons in a $3d^3$ system: - Water is a **weak field ligand** (doesn't cause pairing) - The 3 electrons occupy **three different d orbitals with parallel spins** (Hund's rule) **Distribution:** $t_{2g}^1 e_g^1 t_{2g}^1$ or simply **three unpaired electrons in separate d orbitals** The correct answer (A) shows three singly occupied 3d orbitals with parallel spins (↑ ↑ ↑), confirming the high-spin octahedral complex configuration.