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

Before anything, one reminder of the two symbols we will use over and over:

A few examples reach into the world of an ion sitting inside a complex. Two ideas from Crystal Field Theory are needed there, so we pin them down now before any symbol is used:

Definition Complex-ion terms (used in Cells E and F′)

== $Δ$ (crystal-field splitting)== = the energy gap the surrounding ligands open up between the two groups of d-orbitals. A big $Δ$ = "strong field"; a small $Δ$ = "weak field". Picture two shelves; $Δ$ is the height between them.
== $t_{2 g}$ == = the lower shelf — the group of three d-orbitals that end up lower in energy inside an octahedral complex.
== $e_{g}$ == = the upper shelf — the group of two d-orbitals that end up higher (octahedral).
Geometry matters: the shape of the complex changes which shelf is lower and how big $Δ$ is. In a tetrahedral complex the split is smaller (about $\frac{4}{9}$ of the octahedral one) and the two shelves swap labels to $e$ (lower, two orbitals) and $t_{2}$ (upper, three orbitals). Because that split is small, tetrahedral complexes are almost always high-spin.

So " $t_{2 g}^{4} e_{g}^{2}$ " just reads: 4 electrons on the lower shelf, 2 on the upper shelf. Which shelf an electron chooses is a tug-of-war between climbing (costs $Δ$ ) and pairing up (costs pairing energy).

The scenario matrix

Every question this topic can throw is ONE of these cells. The last column names the worked example that kills it.

Cell	Case class	The twist it tests	Killed by
A	$n = 0$ (degenerate)	$d^{0}$ or $d^{10}$ → diamagnetic, $μ = 0$	Ex 1
B	Forward, plain ion	remove 4s before 3d, then count	Ex 2
C	The peak, $d^{5}$	maximum $n = 5$ , half-filled	Ex 3
D	Backward from data	given $μ$ , solve for $n$ (reject $n < 0$ )	Ex 4
E	Same ion, two answers	high-spin vs low-spin ( $d^{6}$ ) — ligand field flips $n$	Ex 5
F	Descending side of curve	$d^{6}$ – $d^{9}$ : $μ$ falls as pairing resumes	Ex 6
F′	Non-octahedral geometry	tetrahedral $\Rightarrow$ small $Δ$ $\Rightarrow$ forced high-spin	Ex 7
G	Real-world word problem	balance-tube weighing → identify the ion	Ex 8
H	Exam twist / trap	orbital NOT quenched (f-block); why spin-only fails	Ex 9

The curve of $μ$ against d-electron count is the map of all these cells at once. In the figure below (Figure s01, filename dd-chemistry-3.3.04-d3-s01.png), each labelled dot is tagged with the Cell it belongs to, so you can read off exactly which worked example lives at which point:

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

Look at the amber peak in Figure s01: it sits at $d^{5}$ (Cell C). Everything to its left (Cells A, B) is climbing; everything to its right (Cell F) is falling as electrons pair back up. Cell E is the vertical amber double-arrow at $d^{6}$ — the jump between the high-spin dot ( $μ = 4.90$ ) and the low-spin dot ( $μ = 0$ ). The $d^{0} / d^{10}$ endpoints sitting on the axis are Cell A. Cell F′ (tetrahedral) lives on the upper high-spin branch, since tetrahedral complexes cannot reach the low-spin dots.

Worked examples

Cell A — the degenerate $n = 0$ case

Worked example 1. Is

$Zn^{2 +}$ magnetic? Find $μ$ for the zinc(II) ion, $Zn^{2 +}$ .

Forecast: it has ten d-electrons — surely a monster magnet? Guess before reading.

Write the ion config. Zn is $[Ar] 3 d^{10} 4 s^{2}$ . Remove two electrons — 4s empties first → $Zn^{2 +} = [Ar] 3 d^{10}$ . Why this step? Magnetism is decided by the ion, and 4s leaves before 3d (see Electronic Configuration of Ions).
Count unpaired electrons. All five d-orbitals hold 2 electrons each → every electron has a partner → $n = 0$ . Why this step? Paired electrons spin opposite ways; their tiny magnets cancel exactly. "Pairs cancel, singles sing" — and here there are no singles.
Apply the formula. $μ = 0 (0 + 2) = 0 = 0$ BM. Why this step? $n = 0$ is not an error case; the formula happily returns $0$ , the signature of a diamagnetic species.

Verify: ten electrons but zero lonely ones. A full $d^{10}$ shell is spherically balanced — no net magnet. So the "ten electrons" instinct is exactly the mistake the formula guards against. $μ = 0$ ✓. Same story for $Sc^{3 +}$ ( $d^{0}$ ): no electrons at all, $n = 0$ , $μ = 0$ .

Cell B — forward, plain ion

Worked example 2. Moment of

$Cr^{3 +}$ Find $μ$ for chromium(III).

Forecast: three lonely electrons — where does that land on the curve?

Ion config. Cr is $[Ar] 3 d^{5} 4 s^{1}$ . Remove 3 electrons: the lone 4s first, then two 3d → $Cr^{3 +} = [Ar] 3 d^{3}$ . Why this step? Always strip 4s before 3d for the ion.
Distribute by Hund's rule. Three d-electrons spread into three separate orbitals, all spins parallel → $n = 3$ . Why this step? Hund's Rule & Electron Configuration says fill singly before pairing, which maximises $n$ in a free ion or weak field.
Formula. $μ = 3 (3 + 2) = 15 \approx 3.87$ BM. Why this step? Straight substitution once $n$ is fixed.

Verify: back-solve $n = - 1 + 1 + 3.8 7^{2} = - 1 + 1 + 14.98 = - 1 + 3.998 \approx 3$ ✓. Units: $n$ is a pure count, output is BM by definition. $Cr^{3 +}$ is a textbook stable $d^{3}$ ion — this 3.87 BM is worth memorising.

Cell C — the peak, $d^{5}$

Worked example 3. The strongest first-row moment:

$Mn^{2 +}$ Find $μ$ for manganese(II) and explain why nothing in the first row beats it.

Forecast: guess the number of unpaired electrons and whether any $d^{n}$ can exceed it.

Ion config. Mn is $[Ar] 3 d^{5} 4 s^{2}$ . Remove two 4s → $Mn^{2 +} = [Ar] 3 d^{5}$ . Why this step? 4s before 3d, again.
Half-filled shell. Five orbitals, five electrons, one each, all parallel → $n = 5$ — the maximum possible for five orbitals. Why this step? You cannot have a sixth unpaired electron; the sixth would have to pair up. So $d^{5}$ high-spin is the ceiling.
Formula. $μ = 5 (5 + 2) = 35 \approx 5.92$ BM. Why this step? We use $μ = n (n + 2)$ because the count $n = 5$ is now fixed and this formula is exactly the tool that converts an unpaired-electron count into a moment in BM — one substitution finishes the job.

Verify: try $d^{6}$ next door — its 6th electron must pair, dropping to 4 unpaired, $μ = 24 = 4.90 < 5.92$ . So the curve genuinely peaks here (the amber dot in Figure s01). $Fe^{3 +}$ is also $d^{5}$ → same 5.92 BM.

Cell D — backward from measured data

Worked example 4. Given the moment, name the count

A complex is measured at $μ = 4.90$ BM. How many unpaired electrons does the metal ion have?

Forecast: is 4.90 closer to 4 or 5 unpaired? Guess.

Square both sides. $μ^{2} = n (n + 2)$ → $4.9 0^{2} = 24.01 \approx 24 = n^{2} + 2 n$ . Why this step? The square-root is undone by squaring, turning the problem into a plain quadratic.
Rearrange to zero. $n^{2} + 2 n - 24 = 0$ . Why this step? Standard form so we can factor or use the reverse formula.
Solve. $(n + 6) (n - 4) = 0$ → $n = 4$ or $n = - 6$ . Why this step? Two algebraic roots always appear; set each bracket to zero: $n + 6 = 0 \Rightarrow n = - 6$ , and $n - 4 = 0 \Rightarrow n = 4$ .
Reject the impossible root. A count of electrons cannot be negative → discard $n = - 6$ → $n = 4$ . Why this step? This is the corner people miss: the math offers a negative root ( $- 6$ ), physics forbids it.

Verify: forward-check $4 (6) = 24 = 4.90$ ✓. Equivalently the reverse formula: $n = - 1 + 1 + 24 = - 1 + 5 = 4$ ✓. Four unpaired → e.g. high-spin $Fe^{2 +}$ or $Cr^{2 +}$ ( $d^{4}$ ).

Cell E — same ion, two possible answers

Worked example 5.

$Fe^{2 +}$ ( $d^{6}$ ): weak field vs strong field Give both possible $μ$ values for iron(II) and say what decides which one you see.

Forecast: can one and the same ion give two different magnetic readings? Guess before deciding.

Ion config. Fe is $[Ar] 3 d^{6} 4 s^{2}$ ; remove the two 4s → $Fe^{2 +} = [Ar] 3 d^{6}$ . Why this step? For the ion we strip 4s before 3d (the same rule as every forward example) — so both 4s electrons go and the six d-electrons remain to decide the magnetism.
Weak (high-spin) field, e.g. $[Fe(H_{2} O)_{6}]^{2 +}$ : small orbital split ( $Δ$ small), so electrons prefer separate orbitals → $t_{2 g}^{4} e_{g}^{2}$ → 4 unpaired. Why this step? When the field is weak (small $Δ$ , the shelf-gap defined above), climbing to the upper $e_{g}$ shelf is cheaper than pairing, so electrons stay unpaired — see Crystal Field Theory.
$μ_{high} = 4 (4 + 2) = 24 \approx 4.90$ BM. Why this step? Once the high-spin count $n = 4$ is fixed, $μ = n (n + 2)$ is the tool that turns that count into a moment — we simply substitute $n = 4$ .
Strong (low-spin) field, e.g. $[Fe(CN)_{6}]^{4 -}$ : large split ( $Δ$ big) forces all six into the lower $t_{2 g}$ set → $t_{2 g}^{6} e_{g}^{0}$ → 0 unpaired. Why this step? A big $Δ$ makes the energy gap so large that pairing up on the lower shelf is cheaper than climbing to $e_{g}$ .
$μ_{low} = 0 (2) = 0$ BM (diamagnetic). Why this step? With zero unpaired electrons every micro-magnet is cancelled by its partner, so the formula returns $μ = 0 (0 + 2) = 0$ — the same $n = 0$ logic as Cell A, reached here by strong-field pairing.

Verify: $24 = 4.90$ ✓ and $0 = 0$ ✓. Same electron count (6), two unpaired counts (4 vs 0) → magnetism is a spy on ligand strength. This is why measuring $μ$ tells a chemist whether a ligand is weak or strong.

Cell F — the descending arm of the curve

Worked example 6.

$Cu^{2 +}$ ( $d^{9}$ ) — the far right of the curve Find $μ$ for copper(II) and locate it on Figure s01.

Forecast: with 9 d-electrons, is $μ$ large or small? Guess.

Ion config. Cu is $[Ar] 3 d^{10} 4 s^{1}$ ; remove 4s then one 3d → $Cu^{2 +} = [Ar] 3 d^{9}$ . Why this step? Same ion rule throughout — the 4s electron leaves before any 3d electron, so the lone 4s goes first and then one 3d to reach the $+ 2$ charge (see Electronic Configuration of Ions).
Count singles. Nine electrons in five orbitals: four orbitals are full (8 electrons paired) and one holds a single electron → $n = 1$ . Why this step? Past $d^{5}$ , each new electron must double up, removing a lonely electron. So $n$ falls: $d^{6} : 4, d^{7} : 3, d^{8} : 2, d^{9} : 1$ .
Formula. $μ = 1 (1 + 2) = 3 \approx 1.73$ BM. Why this step? With the count pinned at $n = 1$ , we apply $μ = n (n + 2)$ because it is exactly the rule that turns one unpaired electron into its moment in BM — a single substitution.

Verify: $3 = 1.73$ ✓. On the curve this is the near-right dot, symmetric with $d^{1}$ ( $Ti^{3 +}$ ) which also has $n = 1$ → the curve is a hill, not a ramp. $Cu^{2 +}$ paramagnetic but only weakly.

Cell F′ — non-octahedral geometry forces high-spin

Worked example 7. Tetrahedral

$[CoCl_{4}]^{2 -}$ ( $d^{7}$ ) Find $μ$ for the cobalt(II) ion in the tetrahedral complex $[CoCl_{4}]^{2 -}$ , and say why you don't even need to ask "is it low-spin?"

Forecast: a $d^{7}$ ion could in principle be low-spin (fewer unpaired). Does the tetrahedral shape allow that? Guess.

Ion config. Co is $[Ar] 3 d^{7} 4 s^{2}$ ; strip the two 4s → $Co^{2 +} = [Ar] 3 d^{7}$ . Why this step? Same ion rule — 4s before 3d.
Invoke the geometry rule. The complex is tetrahedral, so its split is small — about $Δ_{t} \approx \frac{4}{9} Δ_{o}$ (smaller than octahedral). A small $Δ$ can never beat the pairing energy → tetrahedral complexes are always high-spin. Why this step? Geometry, not just ligand identity, sets the size of $Δ$ . Because $Δ_{t}$ is tiny, electrons always prefer to spread out and stay unpaired — the "low-spin" option is off the table (see Crystal Field Theory).
Fill high-spin and count. Seven electrons in five orbitals, spread as far as possible: five orbitals get one each (5 electrons), the last two electrons pair up in two of them → $5 - 2 = 3$ unpaired → $n = 3$ . Why this step? High-spin filling maximises unpaired electrons; the two "extra" electrons past the half-filled set are the only forced pairs.
Formula. $μ = 3 (3 + 2) = 15 \approx 3.87$ BM. Why this step? Count fixed at $n = 3$ → one substitution into $μ = n (n + 2)$ .

Verify: measured $[CoCl_{4}]^{2 -}$ moments run a little above 3.87 BM (some orbital contribution survives in tetrahedral $Co^{2 +}$ ), but the spin-only value $15 = 3.87$ ✓ is the correct baseline. Key lesson: geometry can lock in the high-spin count — you never have to worry about a low-spin tetrahedral $d^{7}$ .

Cell G — real-world word problem

Worked example 8. The magnetic-balance detective

A salt of an unknown $+ 2$ transition-metal ion is placed in a Gouy/magnetic balance. Its measured spin-only moment is $μ = 3.87$ BM, and independent analysis says the metal is a first-row transition element in the $+ 2$ state with a $3 d^{n}$ configuration where $n$ -electrons $= 7$ . Identify the ion and confirm consistency.

Forecast: which $+ 2$ ion has $μ = 3.87$ BM AND seven d-electrons? Guess the element.

From $μ$ , get unpaired count. $3.8 7^{2} = 14.98 \approx 15 = n (n + 2)$ → $n^{2} + 2 n - 15 = 0$ → $(n + 5) (n - 3) = 0$ → $n = 3$ (reject $- 5$ ). Why this step? Convert the measurement into a physical count first.
Match to a $3 d^{7}$ ion. For $d^{7}$ high-spin: fill five orbitals singly (5 electrons), then pair two → $3$ unpaired. So $n = 3$ agrees with $d^{7}$ . ✓ Why this step? The two clues (moment and electron count) must point at the same ion, or the sample is misidentified.
Name the ion. A first-row $M^{2 +}$ that is $3 d^{7}$ : cobalt is $[Ar] 3 d^{7} 4 s^{2}$ , so $Co^{2 +} = [Ar] 3 d^{7}$ . Why this step? $d^{7}$ + $M^{2 +}$ + first row uniquely fixes cobalt.

Verify: forward-check $Co^{2 +}$ , $d^{7}$ , high-spin → 3 unpaired → $μ = 15 = 3.87$ BM ✓. Both clues consistent → the salt is a cobalt(II) salt.

Cell H — the exam trap: when spin-only fails

Worked example 9. Why

$Eu^{3 +}$ breaks spin-only (and $Gd^{3 +}$ doesn't) A lanthanide problem gives $Eu^{3 +}$ ( $4 f^{6}$ ) and $Gd^{3 +}$ ( $4 f^{7}$ ). Their observed moments are ~3.4 BM and ~7.9 BM. Explain why the plain spin-only formula gets one badly wrong and the other right.

Forecast: is the trap "wrong $n$ ", or "wrong formula entirely"? Guess.

Recall the assumption behind spin-only. $μ = n (n + 2)$ is the special case of the full $μ = 4 S (S + 1) + L (L + 1)$ with the orbital part $L \to 0$ . Why this step? You can't judge when a formula fails without knowing what it threw away.
Ask: is orbital motion quenched? In 3d ions the ligand field locks the orbitals so $L \to 0$ — spin-only works. But 4f orbitals are buried deep inside the atom, shielded from ligands, so they keep circulating → $L \neq = 0$ . Why this step? The whole justification for dropping $L$ evaporates for f-block (see Lanthanide Magnetism).
Consequence for $Eu^{3 +}$ . Because its buried 4f electrons keep circulating, its $L \neq = 0$ ; the deleted $L (L + 1)$ term genuinely mattered. Spin-only for its 6 unpaired says $6 \cdot 8 = 48 = 6.93$ BM, but experiment gives ~3.4 BM — a huge miss. Why this step? This shows the failure is a wrong-formula trap, not a wrong- $n$ trap: no re-counting of $n$ can rescue $6.93$ down to $3.4$ ; only restoring the $L$ term (or using the total- $J$ formula) works.
Why $Gd^{3 +}$ is the lucky exception. $Gd^{3 +}$ is $4 f^{7}$ — a half-filled f-shell, so like a half-filled d-shell it has $L = 0$ by symmetry. With $L = 0$ the full formula collapses back to spin-only: $7 (7 + 2) = 63 = 7.94$ BM, matching the observed ~7.9 BM. Why this step? It pinpoints exactly when spin-only survives in the f-block — only when $L$ happens to vanish (half-filled $4 f^{7}$ ).

Verify (numeric sanity): $Gd^{3 +}$ spin-only $63 = 7.94$ BM ≈ observed 7.9 BM ✓ (because $L = 0$ ); $Eu^{3 +}$ spin-only $48 = 6.93$ BM is ~3.5 BM away from the observed ~3.4 BM ✗. Lesson: spin-only is a 3d/4d tool; in the f-block it works only by accident (half-filled), and you should reach for the full $4 S (S + 1) + L (L + 1)$ .

Recall Self-test (cover the answers)

Cell A: $μ$ of $Sc^{3 +}$ ? ::: $0$ BM (diamagnetic, $d^{0}$ ). Cell B: $μ$ of $Cr^{3 +}$ ? ::: $15 = 3.87$ BM ( $n = 3$ ). Cell C: $μ$ of $Mn^{2 +}$ ? ::: $35 = 5.92$ BM (peak, $n = 5$ ). Cell D: $μ = 4.90$ BM → $n$ ? ::: $n = 4$ (reject $n = - 6$ ). Cell E: two $μ$ of $Fe^{2 +}$ ? ::: high-spin $4.90$ BM, low-spin $0$ BM. Cell F: $μ$ of $Cu^{2 +}$ ? ::: $3 = 1.73$ BM ( $n = 1$ ). Cell F′: $μ$ of tetrahedral $[CoCl_{4}]^{2 -}$ ? ::: $15 = 3.87$ BM — forced high-spin ( $n = 3$ ). Cell G: $μ = 3.87$ , $d^{7}$ $M^{2 +}$ ? ::: $Co^{2 +}$ . Cell H: why spin-only fails for $Eu^{3 +}$ ? ::: 4f shielded → $L$ not quenched → need $4 S (S + 1) + L (L + 1)$ ; $Gd^{3 +}$ works only because $4 f^{7}$ has $L = 0$ .

Connections

Parent: Magnetic properties (Hinglish)
Crystal Field Theory — decides high-spin vs low-spin (Cells E, F′); defines $Δ$ , $t_{2 g}$ , $e_{g}$ , and geometry effects.
Hund's Rule & Electron Configuration — fixes $n$ in each example.
Electronic Configuration of Ions — the 4s-before-3d rule used in every forward case.
Bohr Magneton — the unit of every $μ$ here.
Colour in Transition Metal Complexes — same d-electrons, sister observable.
Lanthanide Magnetism — why Cell H needs the full formula.

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