Magnetic properties — paramagnetism via spin-only formula μ = √(n(n+2)) BM: Question bank

Before we start, three words you must not confuse:

Paired electrons — two electrons sharing one orbital, spinning opposite ways, their tiny magnets cancel.
Unpaired electron — one electron alone in an orbital, its magnet has no partner, so it counts in $n$ .
High-spin / low-spin — two different ways the same d-count arranges itself depending on ligand strength (see Crystal Field Theory).

Spot the error

Fe^{3 +}

[Ar] 3 d^{3} 4 s^{2}

, so it has 3 unpaired d-electrons."

Wrong configuration. In ions the 4s empties before 3d, so

Fe^{3 +} = 3 d^{5}

(not

3 d^{3} 4 s^{2}

). It has 5 unpaired electrons — see Electronic Configuration of Ions.

μ = n (n + 2)

, and

n

is the total number of d-electrons."

n

is the number of unpaired electrons only, never the total d-count. For

d^{10}

(ten electrons)

n = 0

, so

μ = 0

, not

120

"A complex measures

μ = 1.73

BM, so

n^{2} + 2 n - 3 = 0

gives

n = 1

n = - 3

; the answer is both."

A negative electron count is physically impossible. Discard

n = - 3

; only

n = 1

is a real answer.

"Since the formula outputs BM, I should square the final answer to get the units right."

Dimensional confusion.

n

is a pure count;

n (n + 2)

is already in BM by the model's definition. Squaring the final

μ

is meaningless.

"Because the orbital term is dropped, we use

μ = 4 S (S + 1) + L (L + 1)

with

L = 0

The logic is right but check the wording — with

L = 0

that expression collapses to

4 S (S + 1)

, which simplifies (via

S = n /2

) to

n (n + 2)

. Writing the full formula and claiming it's spin-only is the slip; you must actually set

L = 0

Cr^{2 +}

(

d^{4}

) is diamagnetic because 4 is an even number."

Evenness of the electron count says nothing about pairing. High-spin

d^{4}

t_{2 g}^{3} e_{g}^{1}

= 4 unpaired, so

μ = 24 = 4.90

BM — strongly paramagnetic.

"CN⁻ is a weak-field ligand, so

Fe^{2 +}

with cyanide is high-spin."

CN⁻ is a strong-field ligand. It forces low-spin (

t_{2 g}^{6}

), giving

n = 0

. Reversing the field strength flips the whole answer — check the spectrochemical series in Crystal Field Theory.

Why questions

Why do we count unpaired electrons rather than all electrons?

Paired electrons spin in opposite directions, so their magnetic moments cancel exactly. Only a lone electron leaves an uncancelled magnet, and it is those uncancelled magnets that make the atom feel an external field.

Why does the orbital contribution "disappear" in most 3d complexes?

The ligands lock the d-orbitals into fixed spatial directions, so the electron cannot freely circulate around the nucleus. Circulation is what gives orbital angular momentum, so quenching it sends

L \to 0

and leaves only spin.

Why does

μ

reach its maximum exactly at the half-filled

d^{5}

ion?

Hund's rule fills each of the five d-orbitals singly before any pairs, so

d^{5}

gives the largest possible number of singles (5). Any further electron must double up in an already-occupied orbital, reducing

n

Why must we use the ion's configuration, not the neutral atom's?

The magnetism belongs to the ion that actually exists in the compound. Since ionisation removes 4s before 3d, the ion's d-count (and its unpaired-electron count) differs from what a neutral-atom configuration would suggest.

Why can the same metal ion show two different magnetic moments?

Strong-field ligands split the d-orbitals far enough apart that pairing electrons costs less energy than promoting them, giving low-spin (fewer singles). Weak-field ligands leave the split small, so electrons stay unpaired (high-spin). Same ion, different

n

Why is measuring

μ

a useful experimental probe?

It counts unpaired electrons without seeing them, so from a single number you can infer oxidation state, d-configuration, and whether a complex is high- or low-spin — which in turn reveals ligand field strength.

Why does a

d^{0}

ion still count as "diamagnetic" and not "non-magnetic"?

Every filled electron pair produces a tiny repulsion from a field (diamagnetism is universal). With no unpaired electrons the paramagnetic attraction vanishes, leaving only this weak repulsion — hence diamagnetic,

n = 0

Edge cases

d^{5}

always the maximum-

μ

configuration?

Only in the high-spin (weak-field) case. Low-spin

d^{5}

t_{2 g}^{5}

with just 1 unpaired electron (

μ = 3 \approx 1.73

), so strong ligands demolish the

d^{5}

peak entirely.

What is

μ

for a substance with all electrons paired?

Exactly

0

BM. With

n = 0

0 (0 + 2) = 0

— the defining case of a diamagnetic species.

Can

μ

be negative or fractional in this model?

It can be irrational (e.g.

3

15

) but never negative —

n

is a non-negative integer count. A "fractional

n

" from back-solving means your measured

μ

carries orbital contribution the spin-only model ignores.

Does the spin-only formula work for the f-block?

Generally no. In lanthanides the 4f electrons are buried and shielded, so the ligand field does not quench orbital motion;

L

survives and you need the full

4 S (S + 1) + L (L + 1)

(indeed the total-

J

form). See Lanthanide Magnetism.

If a

d^{6}

ion is diamagnetic, what does that tell you about its ligands?

They are strong-field. Only low-spin

d^{6}

(

t_{2 g}^{6} e_{g}^{0}

) has

n = 0

; high-spin

d^{6}

keeps 4 unpaired. So diamagnetism pins the ligands as strong-field pairing agents.

Two different ions both measure

μ \approx 3.87

BM — are they the same element?

Not necessarily.

μ \approx 3.87

means

n = 3

, and any ion with three unpaired electrons matches — e.g.

Cr^{3 +}

(

d^{3}

) and

Co^{2 +}

(

d^{7}

, high-spin).

μ

counts singles, it does not identify the element.

Recall One-line safety net

Before every magnetism problem, run this checklist ::: (1) build the ion config, 4s out before 3d; (2) apply Hund / check field strength for high- vs low-spin; (3) count unpaired electrons only; (4) $μ = n (n + 2)$ — do not square the result.

Connections

Crystal Field Theory — decides high-spin vs low-spin, i.e. how many singles remain.
Hund's Rule & Electron Configuration — why $d^{5}$ maximises unpaired electrons.
Electronic Configuration of Ions — the 4s-before-3d removal rule several traps rely on.
Colour in Transition Metal Complexes — same d-electrons, a different observable.
Bohr Magneton — the unit every $μ$ is quoted in.
Lanthanide Magnetism — where spin-only breaks and orbital motion returns.

Question bank — Magnetic properties — paramagnetism via spin-only formula μ = √(n(n+2)) BM

True or false — justify