[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 **Step 1: Determine the oxidation state of Cr** In $[Cr(H_2O)_6]Cl_3$: - Each $Cl^-$ contributes –1, so 3 $Cl^-$ = –3 total - Each $H_2O$ is neutral - Therefore: Cr oxidation state = +3 **Step 2: Write the electron configuration of Cr³⁺** Cr (Z = 24): $[Ar] 3d^5 4s^1$ Removing 3 electrons (one from 4s and two from 3d): $$Cr^{3+}: [Ar] 3d^3$$ **Step 3: Calculate unpaired electrons from magnetic moment** Given: $\mu = 3.83$ B.M. $$\mu = \sqrt{n(n+2)}$$ where $n$ = number of unpaired electrons $$3.83 = \sqrt{n(n+2)}$$ $$14.67 \approx n(n+2)$$ $$n = 3$$ **Step 4: Determine the 3d electron distribution** With 3 electrons in 3d orbitals and 3 unpaired electrons required: Following **Hund's rule** (maximum spin multiplicity), the distribution is: $$\boxed{t_{2g}^0 \, e_g^0 \text{ or } 3d: \, \uparrow \quad \uparrow \quad \uparrow \quad \_ \quad \_}$$ That is, **three unpaired electrons in three different d orbitals** with parallel spins. This confirms **High Spin (HS)** configuration of $d^3$, which gives exactly 3 unpaired electrons matching the observed magnetic moment. **Answer A is correct** because it shows the three 3d electrons distributed singly across three orbitals (high-spin configuration), consistent with the magnetic moment measurement.