Magnetic Moment

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A tetrahedral nickel complex has two unpaired electrons; using the spin-only formula, what is its approximate magnetic moment in Bohr magnetons?

RankGuru · Accepted Answer

The magnetic behaviour of many transition-metal complexes is estimated from the spin-only formula, μ = √(n(n+1)) Bohr magnetons, where n is the number of unpaired electrons. This relation ignores orbital contributions, which is a good approximation for first-row transition metals. For a tetrahedral nickel(II) species the d^8 ion typically keeps two unpaired electrons, so n = 2. Substituting gives μ = √(2 × 3) = √6 ≈ 2.83 BM. The value 1.73 BM corresponds to a single unpaired electron, 3.87 BM corresponds to three unpaired electrons, and 0 BM would indicate a diamagnetic species with all electrons paired, none of which match n = 2. Tetrahedral nickel(II) complexes are commonly paramagnetic, in line with NCERT's treatment of crystal field effects on geometry and spin. Plausibility check: the computed 2.83 BM lies between the one-electron and three-electron values, exactly where a two-electron paramagnet should fall, confirming the result is reasonable.

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