Spin-Only Magnetic Moment — Counting Unpaired Electrons in Complexes

By Aniket Bhardwaj · 14 September 2026 · Advanced Chemistry

A magnetic measurement is one of the few experiments that tells you, almost directly, how the d electrons are arranged inside a coordination complex. Measure the magnetic moment, put it through one small formula, and you get the number of unpaired electrons — and from that you can decide whether a complex is high spin or low spin, octahedral or square planar. This calculation appears in IIT-JAM, GATE and CSIR-NET inorganic sections almost every year, usually as a short numerical or as a structure-deduction question.

The spin-only formula

μ_s.o. = √[n(n + 2)] BM (equivalently μ = 2√[S(S + 1)] BM, with S = n/2)

What the symbols mean:

μ_s.o. — the spin-only magnetic moment, in Bohr magnetons (BM). 1 BM = 9.274 × 10⁻²⁴ J T⁻¹.
n — the number of unpaired electrons. Not the total number of d electrons.
S — total spin quantum number, equal to n × ½.

The name says exactly what the approximation is: only electron spin is counted. Any contribution from orbital motion of the electrons is ignored. For first-row transition metals in most octahedral and tetrahedral environments this is a good approximation, because the ligand field largely "quenches" the orbital contribution.

The five numbers worth memorising

You should be able to write these without a calculator, because most exam questions use one of them:

Unpaired electrons, n	n(n + 2)	μ_s.o. / BM
0	0	0 (diamagnetic)
1	3	√3 = 1.73
2	8	√8 = 2.83
3	15	√15 = 3.87
4	24	√24 = 4.90
5	35	√35 = 5.92

Step 1 — get the d-electron count right

Everything downstream depends on this. The rule is: find the oxidation state of the metal, subtract that many electrons from the neutral atom, and remove the 4s electrons before the 3d electrons.

Fe³⁺: neutral Fe is [Ar] 3d⁶ 4s² (26 electrons). Removing three electrons — the two 4s first, then one 3d — leaves [Ar] 3d⁵. So Fe³⁺ is d⁵.
Ni²⁺: Ni is [Ar] 3d⁸ 4s²; remove the two 4s electrons → [Ar] 3d⁸. So Ni²⁺ is d⁸.
Co³⁺: Co is [Ar] 3d⁷ 4s²; remove three → [Ar] 3d⁶. So Co³⁺ is d⁶.

Step 2 — high spin or low spin?

In an octahedral field the d orbitals split into a lower t_2g set (three orbitals) and an upper e_g set (two orbitals), separated by Δ_o. Which arrangement wins depends on the size of Δ_o against the pairing energy P:

Δ_o < P (weak-field ligands: I⁻, Br⁻, Cl⁻, F⁻, H₂O) → electrons spread out → high spin, maximum n.
Δ_o > P (strong-field ligands: CN⁻, CO, NO₂⁻, PPh₃) → electrons pair in t_2g first → low spin, minimum n.

The two possibilities differ only for d⁴ to d⁷. For d¹, d², d³, d⁸, d⁹ and d¹⁰ there is only one octahedral arrangement, so the magnetic moment cannot distinguish ligand strength there.

dⁿ	High spin: n unpaired	μ / BM	Low spin: n unpaired	μ / BM
d¹	1	1.73	1	1.73
d²	2	2.83	2	2.83
d³	3	3.87	3	3.87
d⁴	4	4.90	2	2.83
d⁵	5	5.92	1	1.73
d⁶	4	4.90	0	0
d⁷	3	3.87	1	1.73
d⁸	2	2.83	2	2.83
d⁹	1	1.73	1	1.73
d¹⁰	0	0	0	0

Worked example 1 — the same metal ion, two answers

[Fe(H₂O)₆]³⁺
Fe³⁺ is d⁵. H₂O is a weak-field ligand, so the complex is high spin: t_2g³ e_g², all five electrons unpaired, n = 5.
μ = √[5 × (5 + 2)] = √35 = 5.92 BM.

[Fe(CN)₆]³⁻
Fe³⁺ is still d⁵, but CN⁻ is a strong-field ligand, so the complex is low spin: t_2g⁵ e_g⁰, and only one electron is unpaired, n = 1.
μ = √[1 × (1 + 2)] = √3 = 1.73 BM.

Same metal, same oxidation state, same geometry — and the measured moment differs by a factor of more than three. That difference is the whole experimental case for ligand field strength.

Worked example 2 — geometry from magnetism (a classic)

Two nickel(II) complexes, both d⁸:

[NiCl₄]²⁻ is tetrahedral. In a tetrahedral field the splitting is small (Δ_t ≈ 4/9 Δ_o), so it is always high spin: e⁴ t₂⁴ with n = 2.
μ = √[2 × 4] = √8 = 2.83 BM → paramagnetic.

[Ni(CN)₄]²⁻ is square planar. Here the d_x²−y² orbital is pushed far above the other four, so all eight electrons occupy the lower four orbitals in pairs: n = 0.
μ = 0 BM → diamagnetic.

So a single magnetic measurement decides the geometry. This is exactly how the square planar structure of d⁸ cyanide complexes is argued in textbooks, and it is a standard JAM/NET question format.

Worked example 3 — working backwards from a measured value

Problem: a manganese complex has a measured moment of 3.88 BM. How many unpaired electrons does it have?

Square both sides: n(n + 2) = (3.88)² = 15.0544
Rearrange: n² + 2n − 15.0544 = 0
Discriminant: b² − 4ac = 4 + 4 × 15.0544 = 4 + 60.2176 = 64.2176; √64.2176 = 8.0136
n = (−2 + 8.0136) ÷ 2 = 6.0136 ÷ 2 = 3.007

n must be a whole number, so n = 3 (predicted μ = 3.87 BM, agreeing with the measurement to 0.01 BM). For manganese that points to Mn(IV), d³, or to a low-spin d³-like arrangement — not to Mn(II), which as high-spin d⁵ would give 5.92 BM.

When the spin-only value is not enough

The spin-only formula deliberately throws away the orbital contribution, and there are cases where that contribution refuses to disappear:

Octahedral Co²⁺ (high-spin d⁷) has a T ground term. Its measured moments sit well above the spin-only 3.87 BM, commonly in the 4.3–5.2 BM range. Reporting "3.87 BM, so n = 3" is right about n but should carry the caution that experiment reads higher.
Lanthanides (4f) are the big exception. The 4f orbitals are buried under the 5s and 5p shells, so the ligand field barely touches them and orbital angular momentum survives. For these, use the Landé expression μ = g√[J(J + 1)] BM with g = 1 + [J(J+1) + S(S+1) − L(L+1)] ÷ [2J(J+1)], not the spin-only formula. Gd³⁺ (f⁷, L = 0) is the one case where the two agree, because with L = 0 there is no orbital contribution to begin with.
The full first-row expression that keeps both contributions is μ_S+L = √[4S(S + 1) + L(L + 1)] BM. Exams usually ask for the spin-only value, but read the question — if it says "including orbital contribution", it wants this one.

Common mistakes that cost marks

Putting total d electrons into the formula. For high-spin d⁶ the answer is √24 = 4.90 BM (n = 4), not √48. Only unpaired electrons count.
Getting the d count wrong by removing 3d before 4s. Fe³⁺ is 3d⁵, not 3d³4s². For cations the 4s electrons always go first.
Applying high-spin/low-spin logic to d¹–d³, d⁸–d¹⁰. There is only one octahedral arrangement for those, so a moment of 3.87 BM for a d³ complex tells you nothing about the ligand.
Forgetting that square-planar d⁸ is diamagnetic. Students often answer 2.83 BM for [Ni(CN)₄]²⁻ because they assume tetrahedral by habit.
Using the spin-only formula on a lanthanide. For Dy³⁺ or Ho³⁺ the spin-only answer is far below the measured value.
Assuming tetrahedral complexes can be low spin. Δ_t is roughly 4/9 of Δ_o for the same metal and ligands, so it almost never beats the pairing energy — treat tetrahedral first-row complexes as high spin.

Where this appears in exams

Exam	Typical use
IIT-JAM	Direct μ calculation from a given complex; matching μ values to complexes
GATE (Chemistry)	Numerical answer type: μ from d count, or n from a measured μ
CSIR-NET	Deducing geometry and spin state together with CFSE and spectral data
CUET-PG / M.Sc. entrance	High-spin vs low-spin comparison for the same metal ion

Always check the current official syllabus and notification for the paper you are sitting — topic lists are revised from time to time.

Get the d-electron count right first. The Electron Configuration calculator writes out the ground-state configuration for an element or ion, so you can confirm that Fe³⁺ really is 3d⁵ before you ever touch the spin-only formula. The square root itself is one keystroke on the Scientific Calculator.

Open the Electron Configuration Calculator →

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