In the ground state of atomic Fe (Z = 26), the spin-only magnetic moment is ................... . × — d and f Block Elements Chemistry Question

💡 Solution & Explanation

**Step 1: Write the electron configuration of Fe (Z = 26)** Fe: [Ar] 3d⁶ 4s² For magnetic moment calculations, only unpaired d-electrons contribute significantly (4s electrons pair up). **Step 2: Determine unpaired electrons in 3d⁶** The 3d⁶ configuration with maximum spin multiplicity (Hund's rule): ↑ ↑ ↑ ↑ ↑ ↑ 3d orbitals: all five d orbitals get one electron each, plus one more Number of unpaired electrons: **n = 4** **Step 3: Apply the spin-only magnetic moment formula** $$\mu_s = \sqrt{n(n+2)} \text{ BM}$$ where n = number of unpaired electrons **Step 4: Calculate** $$\mu_s = \sqrt{4(4+2)}$$ $$\mu_s = \sqrt{4 × 6}$$ $$\mu_s = \sqrt{24}$$ $$\mu_s = 4.899 \text{ BM}$$ **Step 5: Convert to × 10 BM** $$\mu_s = 4.899 × 10 \text{ BM} ≈ 49 × 10^{-1} \text{ BM}$$ When expressed as "................... × 10 BM": 4.899 × 10 = 48.99 ≈ **49.00** Therefore, the answer is 49.00.