For the octahedral complex [Fe(H₂O)₆]²⁺ (high‑spin), calculate the number of unpaired electrons and the spin‑only magnetic moment (in Bohr magnetons).

medium 3 marks Class 12 · Chemistry · d- and f-Block Elements

Answer

4 unpaired electrons; μ_eff ≈ 4.90 BM

The correct answer is 4 unpaired electrons; μ_eff ≈ 4.90 BM — Thus the complex has 4 unpaired electrons and μ_eff ≈ 4.90 BM.

Step-by-step solution

Fe²⁺ has the electron configuration d⁶.
In a weak‑field (high‑spin) octahedral field, the d‑orbitals split into t₂g (lower) and e_g (higher) with electrons filling according to Hund's rule.
High‑spin d⁶ distribution: t₂g³ e_g³, giving four unpaired electrons (one in each of the three t₂g orbitals and one in each of the two e_g orbitals, but two electrons pair in one t₂g).
Use the spin‑only formula μ_eff = √[n(n+2)], where n = number of unpaired electrons.
Substitute n = 4: μ_eff = √[4(4+2)] = √24 ≈ 4.90 Bohr magnetons.
Thus the complex has 4 unpaired electrons and μ_eff ≈ 4.90 BM.

Common mistake: Using n = 6 (total d electrons) in the magnetic moment formula instead of the number of unpaired electrons.

Concept tested: Crystal field theory and magnetic moment of transition metal complexes