How to Calculate Spin-Only Magnetic Moment | Chemistry

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Formula Basics
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Formula Basics

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    Introduces spin-only magnetic moment formula μ = √n(n+2).

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    Defines n as the number of unpaired electrons in the ion.

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    Demonstrates calculation with Mn²⁺ yielding 5.92 Bohr magnetons.

Understanding the electronic configuration of transition metal elements (d-block) and how they form cations by losing outer-shell electrons first.
Familiarity with Hund's Rule of Maximum Multiplicity and the Aufbau Principle to correctly populate d-orbitals with electrons.
The ability to identify and count 'unpaired electrons' in a given orbital diagram.
Basic knowledge of magnetism, specifically the qualitative difference between paramagnetic (attracted to magnetic fields) and diamagnetic (repelled) substances.
Introduction to Crystal Field Theory (CFT) and how ligand field strength (spectrochemical series) influences high-spin versus low-spin electronic configurations.
Understanding deviations from the spin-only formula due to orbital angular momentum contribution (spin-orbit coupling).
Using experimental magnetic susceptibility measurements to deduce the coordination geometry (e.g., tetrahedral vs. square planar) of coordination compounds.
Exploring cooperative magnetic phenomena in solid-state chemistry, such as ferromagnetism, antiferromagnetism, and ferrimagnetism.
3.6K views70likes7:26@EasyChemistrybyPraveenOriginal Release: 2020-10-02

The spin only magnetic moment (μ) of a compound or ion is calculated using the formula μ = √[n(n+2)] Bohr magnetons, where n represents the number of unpaired electrons. To calculate: (1) determine the electronic configuration of the ion, (2) count the unpaired electrons (n), and (3) substitute into the formula. For example, Mn²⁺ (3d⁵) has n=5, giving μ = √35 ≈ 5.92 BM, while Fe²⁺ (3d⁶) has n=4, giving μ = √24 ≈ 4.92 BM. Common values to remember: n=1 gives 1.73 BM, n=2 gives 2.83 BM, n=3 gives 3.87 BM, n=4 gives 4.90 BM, and n=5 gives 5.92 BM.