JEE Main 2023 · 11 April · evening · Single correct · +4 / −1

The magnetic moment is measured in Bohr Magneton (BM).
Spin only magnetic moment of Fe in [Fe(H₂ O)₆]³⁺ and [Fe(CN)₆]³⁻ complexes respectively is :

  1. (a) 3.87 B.M. and 1.732 B.M.
  2. (b) 5.92 B.M. and 1.732 B.M.
  3. (c) 6.92 B.M. in both
  4. (d) 4.89 B.M. and 6.92 B.M.
Answer and solution

Answer (b) 5.92 B.M. and 1.732 B.M.

Solution

Chemistry ma’am · 3 steps

Both are Fe³⁺, d⁵. The ligand decides how many electrons stay unpaired.

  1. Oxidation state. In both complexes iron is Fe³⁺. Fe is [Ar] 3d⁶ 4s², so Fe³⁺ is 3d⁵.
  2. Field strength. H₂O is a weak-field ligand, so the splitting Δₒ is small and the five electrons spread out as t_(2g)³ e_g² (high spin). CN⁻ is a strong-field ligand, so the electrons pair up in the lower level as t_(2g)⁵ e_g⁰ (low spin).

    μ = √n(n+2) BM

    ComplexConfigurationUnpaired nμ (BM)
    [Fe(H₂O)₆]³⁺t_(2g)³ e_g²5√35 = 5.92
    [Fe(CN)₆]³⁻t_(2g)⁵ e_g⁰1√3 = 1.732
  3. So. The spin-only moments are 5.92 BM and 1.732 BM, which is option (b).

The idea to keep. For the same metal ion, first fix the d-electron count. Then let the ligand decide: a weak field spreads the electrons out, a strong field pairs them. Count the unpaired ones and use √n(n+2).

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

Study Coordination Compounds

Asked about 250 times in JEE Main since 2020. This question is from 2023.