The Spin-Only Magnetic Moment Formula

μ = √(n(n+2)) BM and the Bohr magneton as a unit

Lesson 2704 of 4,500 · Coordination Chemistry and CFT

Learning objectives

Introduction

Counting unpaired electrons predicts whether a complex is paramagnetic, but a numerical magnetic moment needs another step. For many first-row complexes, a useful first approximation is μ so=√[n(n+2)] Bohr magnetons, where n is the number of unpaired electrons. The square root matters: four unpaired electrons do not produce “4 BM,” and a measured value need not match the approximation exactly.

Core explanation

The Bohr magneton, written μ B or BM, is a unit of magnetic moment, equal to eħ/(2m e) in magnitude. It is about 9.274×10⁻²⁴ joules per tesla. The spin-only expression estimates the effective moment of a paramagnetic centre when orbital angular momentum contributes little and spins combine in the usual high-spin way. It does not count how many moles of material are present; a molar susceptibility experiment is used to infer a per-centre effective moment.

For n unpaired electrons with parallel spin one-half, the total spin quantum number is S=n/2. The quantum-mechanical magnitude associated with spin is μ≈g√[S(S+1)]μ B, where the electron spin g factor is close to 2. Substitute S=n/2: μ≈2√[(n/2)(n/2+1)]μ B=√[n(n+2)]μ B. This derivation explains the +2 term and why simply multiplying n by one electron’s moment is wrong.

If n=0, the spin-only value is zero. The material still has diamagnetic response, so “zero spin-only moment” does not mean no magnetic physics. For n=1, μ so=√3≈1.73 BM. For n=2, √8≈2.83 BM. These are effective moments, not the spin projection along an arbitrarily chosen axis. Quantum vector magnitude and measured thermal response differ from a picture of rigid arrows that always align fully with a field.

The input n must come from an electron configuration. Octahedral high-spin d⁶ has t₂g⁴e g² and n=4, giving √24≈4.90 BM. Low-spin d⁶ has t₂g⁶ and n=0, giving zero. A wrong oxidation-state assignment leads to a wrong n and therefore a wrong moment even if the formula is applied flawlessly.

The formula has limitations. Orbital angular momentum may not be completely quenched by the ligand field, especially in some Co²⁺ complexes and configurations with orbital degeneracy. Spin–orbit coupling changes the measured effective moment. Magnetic exchange between metal centres, temperature-dependent spin crossover and experimental background corrections can also alter a reported value. When measured and spin-only moments differ, investigate the chemistry rather than forcing n to a non-integer value.

Units deserve attention. BM and μ B denote the same unit in most course notation; some authors write only the numerical value followed by μ B. A magnetic moment of 4.90 BM is not a 4.90 tesla magnetic field. A tesla measures field strength, while joules per tesla measures magnetic moment. The formula predicts a per-centre magnitude in Bohr magnetons.

Step-by-step reasoning

Find formal d count, geometry and spin state. Draw the orbital filling and count n arrows without partners. Compute n(n+2), take its square root and append BM. If comparing to measurement, state that the result is spin-only and consider whether orbital or collective effects could explain any deviation.

Visual explanation

Draw n separate spin-up arrows, then a bracket indicating total S=n/2. Beneath write μ so=2√[S(S+1)]μ B and substitute S=n/2 to obtain √[n(n+2)]μ B. Keep a separate small note for n=0: no spin moment, yet a diamagnetic response may remain.

Real-world analogy

Adding small directional vectors is not the same as adding their numerical lengths. Quantum spins combine according to angular-momentum rules, so the final magnetic magnitude contains a square root. The formula is a calibrated vector-style rule, not an arithmetic count of tiny bar magnets.

Real-world example

Magnetic susceptibility of a high-spin Fe(II) salt can be converted into an effective moment. A value near the four-unpaired spin-only estimate of 4.90 BM supports high-spin d⁶; a nearly diamagnetic Fe(II) complex supports low-spin d⁶, provided coupling and other effects are ruled out.

Why?

Why does one unpaired electron give about 1.73 rather than 1 BM? The effective quantum spin magnitude is proportional to √[S(S+1)], and with S=1/2 and g≈2 the result is √3 μ B, not exactly one μ B.

Common misconception

“The magnetic moment equals the number of unpaired electrons in Bohr magnetons.” The spin-only relationship is nonlinear. For n=3 it predicts √15≈3.87 BM, not 3 BM.

Worked example

For octahedral high-spin Co²⁺, the formal count is d⁷ and the filling t₂g⁵e g² has three unpaired electrons. Spin-only μ so=√[3(3+2)]=√15≈3.87 BM. A measured Co²⁺ moment may be higher because orbital contributions can be significant; the difference does not imply a fractional unpaired-electron count.

Quick check

1. What is spin-only moment for n=2? Answer: √[2(2+2)]=√8≈2.83 BM. 2. What does BM measure? Answer: Magnetic moment, not magnetic field strength or ligand-field splitting.

Exam focus

Show n from a justified orbital diagram, substitute it explicitly into √[n(n+2)], round sensibly and state BM. Do not present the result as an exact experimental prediction.

Advanced insight

The spin-only formula assumes an effective g factor near 2 and negligible orbital contribution. More complete treatments use spin–orbit coupling and the symmetry of electronic states to calculate anisotropic g values or moments that vary with temperature.

Summary

For n unpaired electrons, μ so=√[n(n+2)] BM follows from S=n/2 and the quantum spin-moment expression. It is a useful estimate whose limits must be checked against orbital and collective magnetic effects.

Practice questions

1. Derive the formula for n=4 and give the value. Answer: S=4/2=2, so μ≈2√[2(2+1)]μ B=√24μ B≈4.90 BM. Equivalently, use √[4(4+2)]. 2. A low-spin d⁶ complex has what spin-only moment? Answer: t₂g⁶ has n=0, so μ so=0 BM. It can still exhibit weak diamagnetism. 3. Why is 4.90 BM not a precise requirement for every high-spin d⁶ compound? Answer: The formula neglects orbital contributions, spin–orbit coupling, magnetic interactions and temperature effects that can shift measured effective moments.