The spin-only magnetic moment of a transition metal ion is calculated using the formula μ = √(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons. To find n, we set √(n(n+2)) = 3.87 and square both sides to get n(n+2) = 14.98, approximately 15. Testing n = 3 gives 3 × 5 = 15, which matches, so the ion has three unpaired electrons. This corresponds, for example, to a d^3 or d^7 configuration such as Cr^3+ or Co^2+ depending on the ion. The option of two unpaired electrons gives √8 ≈ 2.83 magnetons. The option of four gives √24 ≈ 4.90 magnetons. The option of five gives √35 ≈ 5.92 magnetons, none of which match 3.87. This spin-only calculation is a frequent JEE Advanced numerical from NCERT. Understanding magnetic moment in this way ties directly into the wider study of d- and f-block elements, where the same reasoning recurs across many problems. Working through the logic step by step, rather than memorising the result, makes it clear why unpaired electrons governs the behaviour seen here. Plausibility check: substituting n = 3 back into the formula reproduces √15 = 3.87 magnetons exactly, confirming three unpaired electrons.