Spin-Only Magnetic Moment Formula

A chemist weighs a sample of green nickel(II) sulfate solution on a Gouy balance. The sample is pulled into the magnetic field — the salt is paramagnetic. From the strength of that pull, a single number emerges. That number is a magnetic moment near 2.83 Bohr magnetons. Remarkably, that number alone tells the chemist something important. Nickel(II) in this compound has exactly two unpaired electrons. This is known without ever needing a spectrometer or a diffractometer. This is the everyday magic of the spin-only formula. It is one of the cheapest and most diagnostic tools in the coordination chemist's kit. The formula matters because magnetism is a direct, quantitative fingerprint of electron configuration. First, a metal ion's $d^n$ configuration and its high-spin or low-spin state must be known — the subject of High-Spin vs Low-Spin Configurations. Once these are known, the number of unpaired electrons follows immediately. This comes from filling the $d$ orbitals according to Hund's rule within the crystal-field splitting pattern. The spin-only formula then converts that unpaired-electron count into a predicted magnetic moment. That moment can be measured in the laboratory and compared directly against theory. Agreement (or disagreement) becomes evidence for oxidation state, spin state, and even geometry. The big picture is that magnetism sits at the intersection of quantum mechanics and practical bench chemistry. Each unpaired electron carries an intrinsic spin magnetic moment. In the absence of orbital motion contributing extra magnetism, these spins add according to the total spin quantum number $S$ . The resulting paramagnetic moment follows a strikingly simple square-root law. For the first-row transition metals, orbital angular momentum is often "quenched" by the ligand field. There, this simple law works with startling accuracy. It correctly predicts moments to within a few percent for many octahedral and tetrahedral complexes. This article builds the spin-only formula from first principles. It shows exactly how to count unpaired electrons for any $d^n$ ion in any common geometry. It works through three complete examples spanning routine identification. These include an applied puzzle in bioinorganic chemistry, and a subtler case. In that subtler case, the simple formula visibly strains at its limits. It also flags precisely where orbital angular momentum reasserts itself. This sets up the natural next step into Orbital Contributions to Magnetic Moment and the practical business of Measuring Magnetic Moments.

Key takeaways

Every unpaired electron has spin $s = \tfrac12$ ; total spin quantum number is $S = n/2$ for $n$ unpaired electrons.
Spin-only formula: $\mu_{\text{s.o.}} = \sqrt{n(n+2)}\ \mu_B$ , derived from $\mu = g\sqrt{S(S+1)}\,\mu_B$ with $g \approx 2$ .
Works best for first-row (3d) transition metal ions where orbital angular momentum is largely quenched by the ligand field.
Count $n$ by filling $d$ -orbitals per Hund's rule within the crystal-field splitting diagram appropriate to the geometry and spin state.
Measured moments for 3d ions with $n = 1$ –5 unpaired electrons typically lie within ~0.5–1 $\mu_B$ of the spin-only prediction.
Systematic deviations (especially for $t_{2g}$ -heavy configurations like $d^6$ , $d^7$ octahedral, or 4d/5d ions) signal real orbital contributions.
Key idea: magnetic moment is a direct, measurable proxy for the number of unpaired $d$ electrons, via $\mu = \sqrt{n(n+2)}$ .

Where this lives today

Magnetic moment measurements remain a fast, cheap, non-destructive structural probe used across chemistry and biochemistry. In bioinorganic chemistry, the spin state of iron in heme proteins (deoxyhemoglobin vs oxyhemoglobin vs cyanomethemoglobin) is routinely rationalized using exactly the high-spin/low-spin $d^6$ analysis of Worked Example 2, connecting oxygen-binding cooperativity to a measurable change in unpaired electron count. In catalysis research, magnetic susceptibility measurements help identify the oxidation and spin state of reactive metal intermediates (e.g., high-valent iron-oxo species implicated in C–H activation) when crystallography is difficult on short-lived species. In materials science, spin-crossover complexes — molecules that switch between high-spin and low-spin states in response to temperature, pressure, or light — are being explored as molecular switches and memory elements; their defining signature is precisely a temperature-dependent jump in $\mu_{\text{eff}}$ between the two spin-only-predicted values, monitored by SQUID magnetometry (a modern, far more sensitive descendant of the Gouy balance, covered under Measuring Magnetic Moments). In industrial quality control, magnetic susceptibility balances remain used to check oxidation-state purity of transition-metal reagents and catalysts, since a stray amount of the wrong oxidation state or spin state shows up immediately as an anomalous moment. And pedagogically, the spin-only formula remains the first quantitative bridge every student crosses between electron configuration diagrams and a real, measurable physical quantity — the launching point toward the richer physics of Orbital Contributions to Magnetic Moment where the $T$ -term exceptions of the table above get their full quantum mechanical treatment.

:::quiz[K₃[Co(CN)₆] is diamagnetic. What does this tell you about the Co³⁺ ion inside it?]

It must be tetrahedral
It is low-spin d⁶ with all six d electrons paired ✓
It has five unpaired electrons
Cobalt must be in the +2 oxidation state

Ask the tutor: If a measured moment lies between two spin-only predictions (say 4.3 μB, between the n=3 value 3.87 and the n=4 value 4.90), what does that tell us about the sample?

The physical picture

An unpaired electron in an atom or ion is a tiny magnet. Quantum mechanically, this magnetism comes from two sources: the electron's intrinsic spin, and its orbital motion around the nucleus. In many first-row transition-metal complexes, the surrounding ligand field distorts the $d$ orbitals so effectively that the orbital motion contributes almost nothing to the observable moment — it is said to be quenched. What remains is a "spin-only" magnet, whose strength depends purely on how many electrons are unpaired. Each electron carries spin angular momentum characterized by quantum number $s = \tfrac12$ . When electrons are unpaired, their spins align (Hund's rule keeps them parallel in different orbitals to minimize electron-electron repulsion), and the total spin quantum number for $n$ unpaired electrons is simply

S = \frac{n}{2}

This total spin generates a magnetic moment whose magnitude, expressed in Bohr magnetons ( $\mu_B = e\hbar/2m_e$ , the natural unit of atomic-scale magnetism), follows the general Landé formula for pure spin angular momentum. Because the electron's spin $g$ -factor is very close to 2 (2.0023, from quantum electrodynamics, but universally rounded to exactly 2 for this formula), the moment simplifies to a function of $n$ alone. This is the origin of the compact, memorable expression $\mu_{\text{s.o.}} = \sqrt{n(n+2)}$ . The crucial chemistry lies not in the formula itself but in correctly finding $n$ . This requires knowing the metal's oxidation state (hence its $d^n$ electron count), the geometry of the complex (octahedral, tetrahedral, square planar, etc.), and — for octahedral and square-planar $d^4$ – $d^7$ ions — whether the complex is high-spin or low-spin, which depends on the relative size of the crystal-field splitting energy $\Delta_o$ compared to the pairing energy $P$ .

Figure

Filling d-orbitals: high-spin vs low-spin d⁶ in an octahedral field

The figure makes the essential point of the whole topic: the electron count $n$ is not fixed by the metal alone. It is fixed by the metal and the field it sits in. Once $n$ is correctly read off the orbital diagram, the spin-only formula converts it into a number you can check against a balance reading.

Figure

The square-root law μ = √n(n+2) for n = 0 to 5

Notice the curve rises steeply at first, then flattens — the moment is not linear in $n$ . Going from 0 to 1 unpaired electron jumps the moment by 1.73 $\mu_B$ , but going from 4 to 5 adds only about 1.02 $\mu_B$ . This nonlinearity is itself diagnostic: it means moderately different unpaired-electron counts produce well-separated, easily distinguished experimental values.

The core ideas

The general quantum-mechanical expression for the magnetic moment of an ion with total spin angular momentum $S$ and total orbital angular momentum $L$ combined into total angular momentum $J$ is the full Landé (or "spin-and-orbit") formula. When orbital contributions are negligible — the working assumption for most first-row $M^{2+}/M^{3+}$ octahedral and tetrahedral complexes — this collapses to the spin-only form:

\mu_{\text{s.o.}} = g_e\sqrt{S(S+1)}\ \mu_B \approx 2\sqrt{S(S+1)}\ \mu_B

where:

$\mu_{\text{s.o.}}$ is the spin-only magnetic moment, in Bohr magnetons ( $\mu_B$ )
$g_e \approx 2.0023$ is the free-electron spin $g$ -factor, conventionally rounded to 2
$S$ is the total spin quantum number of the ion, $S = n/2$ for $n$ unpaired electrons (assuming maximum multiplicity, i.e. Hund's-rule ground state)
$\mu_B = e\hbar/2m_e = 9.274\times10^{-24}\ \text{J T}^{-1}$ is the Bohr magneton, the fundamental unit of atomic magnetic moment

Substituting $S = n/2$ turns this into the working formula every coordination chemist memorizes:

\mu_{\text{s.o.}} = \sqrt{n(n+2)}\ \mu_B

where:

$n$ is the number of unpaired electrons in the ion's ground-state electron configuration
the expression is dimensionless inside the square root; the unit $\mu_B$ is appended by convention

Reading: this formula says the paramagnetic pull on a sample depends only on a head-count of unpaired spins — not on which orbitals they occupy, not on the metal's identity, not on the ligands present (except insofar as ligands determine spin state). It is a beautifully "dumb" formula in the best sense: all the hard chemistry is front-loaded into determining $n$ , and the arithmetic afterward is trivial. A companion relationship worth keeping close is the direct link between $S$ and the total spin degeneracy, since some texts quote moments via the multiplicity $2S+1$ :

\mu_{\text{s.o.}} = \sqrt{(2S+1)^2 - 1}\ \mu_B

where:

$2S+1$ is the spin multiplicity (the superscript label in term symbols, e.g. the "4" in $^4A_{2g}$ )

This is algebraically identical to $\sqrt{n(n+2)}$ once $S = n/2$ is substituted, and is useful when the ground term symbol is already known from a Tanabe–Sugano-style analysis rather than a simple orbital-filling diagram.

Derivation AFrom spin angular momentum to the square-root law

Goal: show explicitly how $\mu_{\text{s.o.}} = \sqrt{n(n+2)}$ follows from the quantum mechanics of spin angular momentum, with no unjustified steps.

Step 1 — magnetic moment of a single angular momentum. For any angular momentum quantum number $j$ (spin, orbital, or total), quantum mechanics gives the magnitude of the associated magnetic moment as

\mu = g\sqrt{j(j+1)}\ \mu_B

This is the general " $\sqrt{j(j+1)}$ " law that appears throughout atomic physics: the magnitude of an angular momentum vector of quantum number $j$ is $\hbar\sqrt{j(j+1)}$ , and multiplying by the appropriate gyromagnetic ratio ( $g\mu_B/\hbar$ ) converts this into a magnetic moment.

Step 2 — apply it to pure spin, $j = S$ . For an ion whose magnetism comes entirely from unpaired electron spins (orbital motion quenched), set $j = S$ and $g = g_e \approx 2$ :

\mu_{\text{s.o.}} = 2\sqrt{S(S+1)}\ \mu_B

This is exact in the pure-spin, $g=2$ approximation; no orbital term has been added.

Step 3 — total spin from unpaired electron count. Each unpaired electron has $s = \tfrac12$ ; unpaired electrons couple their spins parallel in the ground state (Hund's rule maximizes total spin to minimize repulsion energy). For $n$ unpaired electrons, all aligned,

S = n\times\frac12 = \frac{n}{2}

Step 4 — substitute and simplify. Insert $S = n/2$ into the Step 2 result:

\mu_{\text{s.o.}} = 2\sqrt{\frac{n}{2}\left(\frac{n}{2}+1\right)}\ \mu_B

Expand the product inside the square root:

\frac{n}{2}\left(\frac{n}{2}+1\right) = \frac{n}{2}\cdot\frac{n+2}{2} = \frac{n(n+2)}{4}

\mu_{\text{s.o.}} = 2\sqrt{\frac{n(n+2)}{4}}\ \mu_B = 2\cdot\frac{\sqrt{n(n+2)}}{2}\ \mu_B

The factor of 2 cancels the $\sqrt{4}=2$ in the denominator exactly:

\mu_{\text{s.o.}} = \sqrt{n(n+2)}\ \mu_B

Limiting case check. Setting $n=0$ (a closed-shell, diamagnetic ion, e.g. low-spin $d^6$ Fe(II) or $d^{10}$ Zn(II)) gives $\mu_{\text{s.o.}} = \sqrt{0\times 2} = 0$ , correctly predicting no paramagnetic moment — consistent with experiment, since such ions are indeed measured as diamagnetic (weakly repelled by a field, moment effectively zero on this scale).

Go deeper · Derivation B · 2 minCounting n correctly: filling d-orbitals under a ligand field

Derivation BCounting n correctly: filling d-orbitals under a ligand field

Goal: derive $n$ systematically for any $d^x$ configuration in octahedral and tetrahedral fields, including the high-spin/low-spin branch point, so that $\mu_{\text{s.o.}}$ can be predicted without guesswork.

Step 1 — establish the splitting pattern for the geometry. In an octahedral field, the five degenerate free-ion $d$ orbitals split into a lower triplet $t_{2g}$ and an upper doublet $e_g$ , separated by $\Delta_o$ . In a tetrahedral field the ordering inverts and the splitting $\Delta_t$ is smaller (roughly $\Delta_t \approx \tfrac{4}{9}\Delta_o$ for comparable metal-ligand distances), so tetrahedral complexes are essentially always high-spin.

Step 2 — place electrons according to the Aufbau/Hund procedure, but competing against pairing energy $P$ . For $d^1$ – $d^3$ and $d^8$ – $d^{10}$ octahedral configurations there is only one sensible filling (electrons enter $t_{2g}$ then $e_g$ singly before pairing, or fully occupy both sets at $d^{10}$ ); no high-spin/low-spin ambiguity arises. The ambiguity is confined to $d^4$ – $d^7$ : here the ion "chooses" between

High-spin: electrons keep spreading across all five orbitals before any pairing, costing $\Delta_o$ to promote an electron from $t_{2g}$ to $e_g$ rather than pair it.
Low-spin: electrons pair up within $t_{2g}$ first, costing pairing energy $P$ but avoiding the $\Delta_o$ promotion.

The ion adopts whichever configuration has lower total energy: if $\Delta_o > P$ , low-spin is favored; if $\Delta_o < P$ , high-spin is favored. Weak-field ligands (e.g. halide, $\text{H}_2\text{O}$ ) and 3d metals favor high-spin; strong-field ligands (e.g. CN⁻, CO) and heavier 4d/5d metals favor low-spin.

Step 3 — tabulate $n$ for octahedral $d^4$ – $d^7$ , both spin states. Working through Hund's rule filling explicitly:

$d^x$	High-spin filling	$n$ (HS)	Low-spin filling	$n$ (LS)
$d^4$	$t_{2g}^3 e_g^1$	4	$t_{2g}^4 e_g^0$	2
$d^5$	$t_{2g}^3 e_g^2$	5	$t_{2g}^5 e_g^0$	1
$d^6$	$t_{2g}^4 e_g^2$	4	$t_{2g}^6 e_g^0$	0
$d^7$	$t_{2g}^5 e_g^2$	3	$t_{2g}^6 e_g^1$	1

Each entry follows by placing one electron in each of the five orbitals before any pairing (high-spin) versus filling $t_{2g}$ completely before touching $e_g$ (low-spin), then counting how many orbitals hold a lone, unpaired electron.

Step 4 — convert $n$ to $\mu_{\text{s.o.}}$ using the Step-1-derivation result. Apply $\mu_{\text{s.o.}} = \sqrt{n(n+2)}$ to each row. For example, high-spin $d^6$ ( $n=4$ ):

\mu_{\text{s.o.}} = \sqrt{4(4+2)} = \sqrt{24} = 4.90\ \mu_B

while low-spin $d^6$ ( $n=0$ ) gives $\mu_{\text{s.o.}} = 0$ — a difference of nearly 5 $\mu_B$ between the two spin states of the same ion, which is exactly why magnetic measurements are such a powerful spin-state diagnostic.

Limiting case check. For tetrahedral complexes, $\Delta_t$ is small enough (never observed to exceed typical pairing energies for 3d ions) that only the high-spin filling ever applies; setting the "choice" in Step 2 to always favor high-spin recovers the familiar rule that virtually all tetrahedral 3d complexes are high-spin, with $n$ found by the same one-electron-per-orbital-first logic applied to a lower $e$ pair and upper $t_2$ triplet.

Worked examples

Example 1 — Foundational: high-spin octahedral $\text{[Ni(H}_2\text{O)}_6]^{2+}$ . Nickel(II) is $d^8$ (Ni: $[\text{Ar}]3d^8 4s^2$ ; remove two electrons from $4s$ first, leaving $3d^8$ ). For $d^8$ octahedral complexes there is no high-spin/low-spin ambiguity: $t_{2g}$ must fill completely (6 electrons, all paired) before $e_g$ receives its two electrons, and by Hund's rule those two $e_g$ electrons occupy the two different $e_g$ orbitals singly rather than pairing:

t_{2g}^6\, e_g^2, \quad \text{with both } e_g \text{ electrons unpaired}

So $n = 2$ . Applying the formula:

\mu_{\text{s.o.}} = \sqrt{2(2+2)} = \sqrt{8} = 2.83\ \mu_B

This matches textbook experimental values for octahedral Ni(II) complexes (typically 2.9–3.4 $\mu_B$ — close to, if slightly above, spin-only, an early hint of a small orbital contribution addressed in Orbital Contributions to Magnetic Moment). As a limiting-case sanity check: if Ni(II) were (hypothetically) square planar and low-spin with all electrons paired ( $n=0$ ), the predicted moment would collapse to zero, matching the well-known diamagnetism of square-planar $\text{[Ni(CN)}_4]^{2-}$ — a useful internal consistency check on the method, not just the number.

Example 2 — Applied: distinguishing spin states of iron in a heme-like model complex. Deoxyhemoglobin contains high-spin Fe(II) ( $d^6$ ) in a five-coordinate, weak-field-like pocket, while the oxygenated or cyanide-bound forms push iron toward low-spin. Suppose a bioinorganic chemist synthesizes a six-coordinate iron(II) model complex $\text{[Fe(L)}_6]^{2+}$ and measures $\mu_{\text{eff}} = 5.1\ \mu_B$ at room temperature. Which spin state is present? From the derivation table above, octahedral Fe(II) ( $d^6$ ) has two possibilities:

\text{high-spin: } n=4,\quad \mu_{\text{s.o.}} = \sqrt{4(6)} = \sqrt{24} = 4.90\ \mu_B

\text{low-spin: } n=0,\quad \mu_{\text{s.o.}} = 0\ \mu_B

The measured 5.1 $\mu_B$ sits close to the high-spin prediction (4.90 $\mu_B$ ) and is utterly incompatible with the low-spin prediction of zero. The chemist concludes the complex is high-spin Fe(II), consistent with weak-field donor atoms such as carboxylate-O or imidazole-N (rather than strong-field CO or CN⁻) occupying the coordination sites. The small excess above 4.90 $\mu_B$ (about 0.2 $\mu_B$ ) is a typical, modest orbital contribution rather than evidence of a mixed-spin or intermediate-spin state — a distinction made rigorous by comparing against literature ranges (Fe(II) high-spin octahedral: typically 5.1–5.7 $\mu_B$ ).

Example 3 — Exam/competition level: a trick with oxidation state AND geometry, $\text{K}_3[\text{CoF}_6]$ vs $\text{K}_3[\text{Co(CN)}_6]$ . A common exam trap gives two cobalt salts and asks the student to predict and explain contrasting magnetic behavior, forcing correct reasoning about oxidation state, $d$ -count, and field strength together — not just the arithmetic. Step A — oxidation state and $d$ -count. In $\text{K}_3[\text{CoF}_6]$ , potassium is $+1$ (three of them, $+3$ total) and fluoride is $-1$ (six of them, $-6$ total), so cobalt must be $+3$ to balance: $3(+1) + \text{Co} + 6(-1) = 0 \Rightarrow \text{Co} = +3$ . The same charge-balance applies to $\text{K}_3[\text{Co(CN)}_6]$ since cyanide is also $-1$ : cobalt is again $+3$ . Co(III) is $d^6$ in both salts ( $\text{Co}: [\text{Ar}]3d^7 4s^2 \to \text{Co}^{3+}: 3d^6$ ) — the trap is that both compounds contain the identical $d^6$ ion, so any difference in magnetism must come entirely from ligand field strength, not oxidation state. Step B — field strength via the spectrochemical series. Fluoride is a weak-field ligand (low on the spectrochemical series), so $\text{[CoF}_6]^{3-}$ is high-spin: $\Delta_o < P$ . Cyanide is one of the strongest-field ligands known, so $\text{[Co(CN)}_6]^{3-}$ is low-spin: $\Delta_o > P$ . Step C — apply the $d^6$ table.

\text{[CoF}_6]^{3-}\ (\text{high-spin}): n=4,\quad \mu_{\text{s.o.}} = \sqrt{24} = 4.90\ \mu_B

\text{[Co(CN)}_6]^{3-}\ (\text{low-spin}): n=0,\quad \mu_{\text{s.o.}} = 0\ \mu_B

Step D — the exam-style conclusion. $\text{K}_3[\text{CoF}_6]$ is paramagnetic (measured moments cluster near 4.5–5.2 $\mu_B$ in real salts) while $\text{K}_3[\text{Co(CN)}_6]$ is diamagnetic — despite both containing "the same" Co³⁺ $d^6$ ion. As a limiting-case check on the reasoning: if fluoride were, counterfactually, as strong-field as cyanide, both salts would be low-spin and both would be diamagnetic; the entire observed contrast is attributable only to $\Delta_o$ vs $P$ , confirming that ligand identity — not metal oxidation state — is the variable under test in this problem.

Limiting cases, scope & common traps

Situation	What spin-only predicts	What actually happens	Why
First-row (3d) $M^{2+}$ / $M^{3+}$ , $A$ or $E$ ground terms (e.g. $d^3, d^5$ high-spin, $d^8$ )	$\mu_{\text{s.o.}}$	Very close match (within ~0.2–0.4 $\mu_B$ )	Orbital angular momentum genuinely quenched by the field
First-row ions with $T$ ground terms (e.g. octahedral high-spin $d^1,d^2,d^6,d^7$ ; tetrahedral $d^3,d^8$ )	$\mu_{\text{s.o.}}$	Systematically higher (often by 0.5–1 $\mu_B$ )	Residual orbital angular momentum not fully quenched — see Orbital Contributions to Magnetic Moment
4d and 5d transition metals	$\mu_{\text{s.o.}}$	Often poorly matched	Strong spin–orbit coupling scrambles $S$ and $L$ into $J$ ; spin-only assumption breaks down
Lanthanides ( $4f$ ions)	$\mu_{\text{s.o.}}$	Usually badly wrong	Orbital contribution is large and not quenched; must use full Landé $g\sqrt{J(J+1)}$ formula instead
Antiferromagnetically coupled polynuclear clusters	$\mu_{\text{s.o.}}$ per metal center	Reduced total moment	Spins on adjacent centers couple and partially cancel; single-ion formula does not apply directly
$n=0$ (diamagnetic closed shell)	$\mu = 0$	Small residual diamagnetic susceptibility observed, always negative and tiny	Diamagnetism from paired electrons is universal but far weaker than any paramagnetic signal; the formula correctly predicts zero paramagnetic moment

Common trapMisconception: "high spin-only moment always means more unpaired electrons on that specific atom, full stop"

A measured moment close to a spin-only value is strong evidence for a particular $n$ , but the mapping from $n$ back to a unique electron configuration is not always one-to-one. Different $d^x$ configurations in different spin states can give numerically close (though rarely identical) spin-only predictions — for instance, high-spin $d^7$ ( $n=3$ , $\mu = 3.87\ \mu_B$ ) and high-spin $d^3$ ( $n=3$ , $\mu = 3.87\ \mu_B$ ) are indistinguishable by the spin-only value alone. Oxidation-state and stoichiometric context (as in Worked Example 3) must always accompany the magnetic measurement; magnetism confirms a hypothesis, it does not generate one from nothing.

The chemist behind it

The spin-only formula is not the work of a single named discoverer in the way a named reaction is; it emerges directly from the vector model of atomic angular momentum developed in the 1920s. But its systematic application to transition-metal coordination chemistry — turning bench magnetism into a routine structural tool — owes an enormous debt to Linus Pauling and, especially, to the ligand-field theorists who followed, most notably Leslie Orgel and John Van Vleck, whose theory of electric and magnetic susceptibilities (1932) put the quantum mechanics of paramagnetism in transition-metal ions on rigorous footing.

Origins · 1 minJohn Hasbrouck Van Vleck and how this was discovered

John Hasbrouck Van Vleck

1899–1980 · American theoretical physicist, pioneer of the quantum theory of magnetism in solids and complexes

Van Vleck built the theoretical bridge between atomic quantum mechanics and the bulk magnetic susceptibilities chemists actually measure on a balance. His 1932 monograph "The Theory of Electric and Magnetic Susceptibilities" derived the temperature dependence of paramagnetic susceptibility for ions in a crystal field, explaining both why the spin-only approximation works so well for first-row transition-metal ions and precisely why it fails for lanthanides and heavier transition metals with strong spin–orbit coupling. He shared the 1977 Nobel Prize in Physics for related work on the electronic structure of magnetic and disordered systems.

History

- 1900s — early classical theories of paramagnetism (Curie, Langevin) establish that

\chi \propto 1/T

for many paramagnetic salts.

- 1926–1928 — quantum mechanical vector model of angular momentum (

\sqrt{j(j+1)}

law) established by Heisenberg, Dirac, and others, giving the theoretical basis for magnetic moment formulas.

- 1929 — Hans Bethe introduces crystal-field theory, explaining orbital splitting of

d

orbitals in a ligand environment.

- 1932 — Van Vleck publishes the full quantum theory of magnetic susceptibility, formally justifying the spin-only approximation for quenched-orbital ions.

- 1950s–1960s — Gouy and Faraday balance measurements become routine in inorganic laboratories; spin-only moments become a standard tool for assigning oxidation states and spin states of coordination compounds.

Why is the g-factor taken as exactly 2 in the spin-only formula?

The true free-electron spin $g$ -factor is 2.0023, a small correction arising from quantum electrodynamic effects (the electron's magnetic moment is not perfectly classical). For essentially all coordination-chemistry purposes this 0.1% correction is far smaller than other sources of deviation — orbital contributions, temperature-independent paramagnetism, or measurement uncertainty — so it is universally rounded to 2. Using 2.0023 instead of 2 changes a predicted moment of, say, 4.90 $\mu_B$ to 4.906 $\mu_B$ , a difference no bench Gouy balance could resolve, so the simplification introduces no meaningful error in practice.

Does the spin-only formula apply to organic radicals or main-group paramagnetic species?

Yes, with the same logic: any species with unpaired electrons and negligible orbital angular momentum contribution follows $\mu = \sqrt{n(n+2)}$ . A simple organic radical with one unpaired electron ( $n=1$ ) is predicted to have $\mu = 1.73\ \mu_B$ , and this is indeed observed for many stable radicals such as TEMPO. The formula is fundamentally about spin angular momentum counting, not about $d$ -orbitals specifically; transition metals are simply the setting where it is taught first because crystal-field splitting gives such rich, tunable examples of different $n$ for the "same" ion.

How is the magnetic moment actually measured in practice?

Historically, the Gouy balance measures the force exerted on a sample suspended partly in and partly out of a magnetic field gradient; a paramagnetic sample is pulled into the field (weight increases), while a diamagnetic sample is pushed out (weight decreases slightly). The Evans NMR method, using the field-dependent shift of a reference proton signal, is a common modern solution-phase alternative. Today, SQUID (superconducting quantum interference device) magnetometers give the most precise and temperature-resolved measurements, essential for detecting subtle spin-crossover transitions — the full technique is treated under Measuring Magnetic Moments.

Why do t2g-heavy configurations like high-spin d⁶ or d⁷ octahedral complexes deviate more from the spin-only value than d³, d⁵, or d⁸?

Configurations whose ground term is an "A" or "E" term (like $^4A_{2g}$ for $d^3$ or $^6A_{1g}$ for high-spin $d^5$ ) have their orbital angular momentum genuinely quenched — there is no way to redistribute electrons among degenerate orbitals without changing energy, so no first-order orbital circulation survives. Configurations with a "T" ground term (like $^4T_{1g}$ for high-spin $d^7$ ) retain orbitals related by symmetry operations that permit electrons to "circulate" between them at no energy cost, generating a genuine residual orbital magnetic moment that adds to the spin-only value. This distinction is the entire subject of orbital contribution corrections.

Can two different d-electron configurations give exactly the same spin-only moment?

Yes — the formula depends only on $n$ , so any two configurations sharing the same unpaired-electron count give numerically identical spin-only predictions, regardless of which specific $d^x$ ion or spin state produced that count. High-spin $d^3$ , high-spin $d^7$ , and low-spin $d^9$ -type situations (hypothetically) could all show $n=3$ and hence $\mu = 3.87\ \mu_B$ . This is precisely why magnetic moment measurements are used to support a structural hypothesis derived from stoichiometry and the spectrochemical series, rather than to determine oxidation state from magnetism alone with no other information.

Is the spin-only formula ever exactly correct, or always an approximation?

It is an approximation, though an excellent one for many first-row ions. It assumes $g=2$ exactly (a ~0.1% approximation), assumes zero orbital angular momentum contribution (good for A/E ground terms, worse for T ground terms), and neglects small temperature-independent paramagnetism (TIP) contributions from field-induced mixing with excited states. For high-spin $d^3$ , $d^5$ , and $d^8$ octahedral ions the formula typically lands within 0.1–0.3 $\mu_B$ of experiment; for $d^1,d^2,d^6,d^7$ configurations deviations of 0.5–1.0 $\mu_B$ above the spin-only prediction are common and expected.

Why does tetrahedral geometry almost always give high-spin complexes?

The tetrahedral crystal-field splitting $\Delta_t$ is intrinsically smaller than the octahedral splitting $\Delta_o$ for comparable metal–ligand distances and identical ligands — approximately $\Delta_t \approx \frac{4}{9}\Delta_o$ , a consequence of the geometric overlap factors between $d$ orbitals and four versus six ligand positions. Because $\Delta_t$ rarely exceeds typical pairing energies $P$ for any common ligand set, the energetic incentive to pair electrons within the lower set is essentially never large enough to overcome Hund's-rule spin-maximization, so virtually every tetrahedral 3d complex observed in practice is high-spin.

Spin-Only Magnetic Moment Formula

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