In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ____________ × 10⁻¹ BM. (Round off to the Nearest Integer). [Given : √3 = 1.73, √2 = 1.41 ]

Solution

Chemistry ma’am · 4 steps

Neutral Fe is 3d⁶ 4s²: the 4s pair adds nothing, so count only the unpaired 3d electrons.

1. Ground-state configuration. For Z = 26, the configuration is [Ar] 3d⁶ 4s². The two 4s electrons share one orbital, so they are paired.
2. Unpaired electrons. By Hund's rule, the six 3d electrons first go singly into all five d orbitals. The sixth one then pairs up, which leaves n = 4 unpaired electrons.

   μ = √n(n+2) = √(4 × 6) = √24 = 2√2 √3 BM

3. Put in numbers. μ = 2 × 1.41 × 1.73 = 4.88 BM, which is 48.8 × 10⁻¹ BM.
4. So. To the nearest integer, the spin-only moment is 49 × 10⁻¹ BM.

The common slip: 59. Taking 5 unpaired electrons, as in Fe³⁺ (3d⁵), gives √35 = 5.92 BM and so 59. Neutral Fe is 3d⁶, which has only 4 unpaired.

The idea to keep. For spin-only moments, first write the correct configuration of the exact species asked for (atom or ion), then count its unpaired electrons and use √n(n+2).

Answer from the NTA final answer key. Solution by PrepWiser.