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Spin-Only Magnetic Moment Calculator

Chemistry

Predict the spin-only moment μ_so = √(n(n+2)) μB for a complex, a metal ion, a dⁿ count or a number of unpaired electrons, then compare a measured μ_eff with the possible spin states. Turn raw susceptibility, Evans NMR or balance data into μ_eff, and see why lanthanide moments follow J instead.

Read as [Fe(CN)₆]³⁻

Presets

Problems (example values)

Default: CN 6 → octahedral
[Fe(CN)₆]³⁻Fe(III)d⁵octahedrallow spinbasis: observed

1 unpaired electron · S = 1/2 (doublet) · μ_so = √(1 × 3) = 1.73 μB · paramagnetic

μ_so = 1.732 μB with g = 2 and S = n/2.

Moment scale

Spin-only lines for n = 0–7. Candidate states are bold (filled dot = predicted); shaded bands are typical observed ranges.

n = 011.73LS23455.92HS67typical 2.0–2.5typical 5.7–6.0

Occupancy

Levels not to scale — see the Crystal Field Stabilization Energy Calculator for energies.

High spin

t₂geg

5 unpaired · S = 5/2 · μ_so = 5.92 μB

Low spin (predicted)

t₂geg

1 unpaired · S = 1/2 · μ_so = 1.73 μB

Unpaired electrons are drawn in the primary colour, paired ones (↑↓) muted.

d-electron count

Fe is in group 8 → 8 − 3 = d⁵

Geometry

CN 6 → octahedral

Configuration

t₂g⁵ eg⁰

1 unpaired electron, S = 1/2 (doublet)

Ground term and expected μ_eff

²T₂g: first-order orbital contribution → above μ_so

A T ground term keeps a first-order orbital contribution, and above half filling L and S add, so μ_eff is above μ_so.

Typical observed

2.0–2.5 μB at room temperature

Predicted at 298 K

χ_M = 1.259 × 10⁻³ cm³/mol (1.582 × 10⁻⁸ m³/mol SI)

χ_M·T = 0.3751 cm³ K mol⁻¹

Candidate spin states

StateUnpairedSμ_so (μB)Ground termExpected μ_effTypical

High spin

55/25.92⁶A₁g≈ equal to μ_so5.7–6.0 μB

Low spin

predicted
11/21.73²T₂gabove μ_so2.0–2.5 μB

Same d⁵ in other geometries

Why [NiCl₄]²⁻ is paramagnetic but [Ni(CN)₄]²⁻ is diamagnetic: the geometry and the field strength decide how many electrons stay unpaired.

ArrangementUnpairedμ_so (μB)
Free ion55.92
Octahedral, high spin55.92
Octahedral, low spin11.73
Tetrahedral55.92
Square planar11.73

Tetrahedral is taken as high spin and square planar as low spin, as almost all real examples are.

Worked steps

[Fe(CN)6]³⁻: charge −3; 6 × CN⁻ (−1) → Fe oxidation state = −3 − (−6) = +3 Fe: group 8 → d electrons = 8 − 3 = 5 (d⁵) CN 6 → octahedral; [Fe(CN)6]³⁻ is observed low spin Fill: t₂g⁵ eg⁰ → 1 unpaired electron, S = 1/2 μso = √(n(n + 2)) = √(1 × 3) = 1.73 μB Ground term ²T₂g: first-order orbital contribution → μeff usually above μso (typical 2.0–2.5 μB)

Related: the Crystal Field Stabilization Energy Calculator explains why a complex is high or low spin (Δ vs P), the Electron Configuration Generator covers free atoms, and the Quantum Numbers Calculator lists n, l, m_l and m_s per electron.

About This Tool

Spin-Only Magnetic Moment Calculator – Unpaired Electrons, High vs Low Spin and μ_eff

Every electron carries a spin, and a spin is a tiny magnet. Paired electrons cancel, so the magnetism of a transition-metal compound comes from its unpaired electrons. This spin-only magnetic moment calculator predicts the moment from a complex, a metal ion, a dⁿ count or a plain number of unpaired electrons. It also works backwards from a measured effective magnetic moment μ_eff, or from the susceptibility data behind it, to the number of unpaired electrons and the most likely spin state.

The spin-only formula

Moments are measured in Bohr magnetons (μB = 9.274 × 10⁻²⁴ J/T). With n unpaired electrons the total spin is S = n/2, and with the free-electron g = 2 the moment is:

μso = g√(S(S + 1)) μB = √(n(n + 2)) μB n = 1: 1.73 n = 2: 2.83 n = 3: 3.87 n = 4: 4.90 n = 5: 5.92 μB

Going the other way, n = √(1 + μ²) − 1 gives a continuous estimate. The calculator reports the spin-only value nearest the measurement instead of rounding that estimate, because the two can disagree.

Magnetism as a structural probe

A d⁵ Fe(III) ion can be high spin with five unpaired electrons (5.92 μB) or low spin with one (1.73 μB). A single magnetic measurement tells them apart, so [Fe(H₂O)₆]³⁺ and [Fe(CN)₆]³⁻ are easy to distinguish. The same trick separates geometries: tetrahedral [NiCl₄]²⁻ keeps two unpaired electrons and is paramagnetic, while square-planar [Ni(CN)₄]²⁻ pairs all eight d electrons and is diamagnetic. The Crystal Field Stabilization Energy Calculator explains why: it compares the splitting Δ with the pairing energy P.

Orbital contributions: above or below spin-only?

Orbiting electrons also produce a moment, but the ligand field quenches most of it. How much survives depends on the ground term. A T term (such as ²T₂g for low-spin d⁵ or ⁴T₁g for high-spin d⁷) keeps a first-order orbital contribution. An A₂ or E term has none and only mixes in an excited T term through spin–orbit coupling, a second-order effect:

μeff ≈ μso (1 − αλ / Δ) α = 4 for A₂, 2 for E

The spin–orbit constant λ is positive below half filling and negative above it, so d³ Cr(III) reads slightly below 3.87 μB while d⁸ Ni(II) reads above 2.83 μB, around 3.2. High-spin d⁵ (⁶A₁g) has no orbital moment at all and sits almost exactly on 5.92 μB.

Measuring the moment

Chemists measure the molar susceptibility χ_M and convert it with μeff = 2.828 √(χM·T) (cgs units). A Gouy balance or a magnetic susceptibility balance weighs the pull on a solid sample. The Evans NMR method measures how far a paramagnetic solute shifts a reference signal: for a superconducting magnet χ_M = 3Δf / (4π f c′). Either way, first subtract the diamagnetic correction from the core electrons, roughly −M/2 × 10⁻⁶ cm³/mol. Measuring at several temperatures tests the Curie law (χT constant) or fits the Curie–Weiss form χ = C/(T − θ). Molar masses come from the Molar Mass Calculator.

Lanthanides: when J takes over

The 4f orbitals are buried under filled 5s and 5p shells, so ligands barely touch them and the orbital moment survives. The moment then follows the total angular momentum: μJ = gJ √(J(J + 1)), with the Landé factor g_J from Hund's-rules L, S and J. Dy³⁺ and Ho³⁺ reach about 10.6 μB, far above their spin-only values. Gd³⁺ (4f⁷, L = 0) is the one ion where spin-only works. Sm³⁺ and Eu³⁺ read higher than μ_J because their excited J levels lie close enough to be populated at room temperature.

Limits of the spin-only model
The formula assumes g = 2, isolated metal centres and a fully quenched orbital moment. Orbital contributions, spin crossover, magnetic coupling between neighbouring ions and paramagnetic impurities all move real values. Treat the nearest spin-only count as evidence, not proof.

Frequently Asked Questions

Is the Spin-Only Magnetic Moment Calculator free?

Yes, Spin-Only Magnetic Moment Calculator is totally free :)

Can I use the Spin-Only Magnetic Moment Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Spin-Only Magnetic Moment Calculator?

Yes, any data related to Spin-Only Magnetic Moment Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

What is the spin-only magnetic moment?

It is the magnetic moment an ion would have from electron spin alone: μ_so = √(n(n+2)) μB, where n is the number of unpaired electrons and the Bohr magneton μB = 9.274 × 10⁻²⁴ J/T. For n = 1 to 5 it gives 1.73, 2.83, 3.87, 4.90 and 5.92 μB.

How do you find the number of unpaired electrons from μ_eff?

Invert the formula, n = √(1 + μ²) − 1, then pick the whole number whose spin-only value is nearest the measurement rather than rounding n. Remember the parity: an odd d-electron count can only have an odd number of unpaired electrons, and orbital contributions can push μ_eff past the next spin-only line.

How does the magnetic moment tell high-spin from low-spin complexes?

Compare the measurement with the spin-only values the d-count allows. [Fe(CN)₆]³⁻ is d⁵, so it can be high spin (5 unpaired, 5.92 μB) or low spin (1 unpaired, 1.73 μB); a measured 2.3 μB is far closer to 1.73, so it is low spin. The extra 0.57 μB is the orbital contribution expected from its ²T₂g ground term.

Why do measured moments differ from the spin-only value?

The ligand field quenches most orbital angular momentum, but not all. T ground terms keep a first-order orbital contribution, while A and E terms only gain a small second-order one through spin–orbit coupling. Above half filling both push μ_eff up, which is why octahedral Ni(II) reads about 3.2 μB instead of 2.83 and high-spin Co(II) about 4.8 μB instead of 3.87.

How do you get μ_eff from magnetic susceptibility?

Use μ_eff = 2.828 √(χ_M·T) with χ_M in cm³/mol (cgs) after subtracting the diamagnetic correction χ_dia, which is negative and roughly −M/2 × 10⁻⁶ cm³/mol. In the Evans NMR method χ_M = 3Δf / (4π f c′) for a superconducting magnet, where Δf is the shift of the reference peak and c′ the concentration in mol/cm³.

Why doesn't the spin-only formula work for lanthanides?

The 4f electrons are shielded by the 5s and 5p shells, so the ligand field barely quenches their orbital motion and the moment follows the total angular momentum J: μ_J = g_J √(J(J+1)). Dy³⁺ and Ho³⁺ reach about 10.6 μB, far above spin-only. Gd³⁺ (4f⁷, L = 0) is the exception where spin-only works.