The spin-only magnetic moment of a transition metal ion is given by μ = √(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons. The Fe^3+ ion has the configuration 3d^5, and in a weak octahedral field the ligands cause little splitting, so all five d-electrons remain unpaired in a high-spin arrangement, giving n = 5. Substituting, μ = √(5(5+2)) = √(5 × 7) = √35 = 5.92 Bohr magnetons. This relatively high value reflects the maximum number of unpaired electrons possible for a d^5 ion. The option 1.73 BM corresponds to one unpaired electron. The option 3.87 BM corresponds to three unpaired electrons. The option 2.83 BM corresponds to two unpaired electrons, none of which match a high-spin d^5 ion. This calculation combining configuration and crystal field reasoning is a frequent JEE Advanced task. Carefully relating the data to the governing principle ensures the reasoning remains valid even when the numbers or species in the question are changed. Such questions reward conceptual clarity, since a student who truly grasps magnetic moment can solve many superficially different variants with the same approach. Plausibility check: substituting n = 5 reproduces √35 = 5.92 Bohr magnetons, the largest spin-only value for a first-series ion, confirming the answer.