Magnetic moments of complexes: Worked examples — Chemistry deep dive

Before anything else, let's be crystal clear about the words and symbols we keep using.

The scenario matrix

Every question this topic throws is one cell of this grid. Each example below is tagged with its cell letter. (The last column of the ambiguous row explains why the field is powerless there.)

Cell	Case class	What makes it distinct	Example
A	$n = 0$ , diamagnetic	zero unpaired → $μ = 0$ ; the "degenerate" input	Ex 1
B	High-spin, weak-field	maximum unpaired for that $d$ -count	Ex 2
C	Low-spin, strong-field	same ion, pairing forced, small $μ$	Ex 3
D	Ambiguous $d$ -count where HS = LS	$d^{1}, d^{2}, d^{3}, d^{8}, d^{9}, d^{10}$ : electrons fill without ever being forced to choose pairing, so $Δ_{o}$ can't change $n$	Ex 4
E	Tetrahedral (always high-spin)	small $Δ_{t}$ → never low-spin	Ex 5
F	Square-planar $d^{8}$	geometry deduced from $μ = 0$	Ex 6
G	Reverse problem (given $μ$ , find $n$ )	invert the formula	Ex 7
H	Non-integer measured $μ$ (orbital contribution / exam twist)	when spin-only fails	Ex 8
I	Real-world word problem	translate lab words into $n$	Ex 9
J	Limiting/max case ( $n = 5$ , half-filled)	the biggest first-row $μ$	Ex 2 doubles here

Prerequisites feeding into this: Crystal Field Theory (sets $Δ_{o}$ ), Spectrochemical Series (ranks ligands), Electronic Configuration of d-block ions (gives the $d$ -count), Hund's Rule and Pairing Energy (spin state), Square Planar vs Tetrahedral Geometry (Cells E, F).

Here is the master map every example follows:

Intuition What the master map shows (full description)

The figure has two parts. The top row is four boxes joined left-to-right by black arrows — the only four moves you ever make on any question: (1) black box "Ion d-count" — strip the neutral atom's electrons to get how many $d$ -electrons the ion has; (2) black box "Field strong/weak" — read the ligand off the spectrochemical series to decide high vs low spin; (3) the red box "Count unpaired n" — the single pivotal number, drawn red because everything hinges on it; (4) black box " $μ = n (n + 2)$ " — the formula that turns $n$ into a moment. The bottom panel is the graph of that formula: horizontal axis is $n$ (unpaired electrons, $0$ to $5$ ), vertical axis is $μ$ in $μ_{B}$ ; the red curve rises from $0$ and bends over, and six black dots mark the only legal whole-number answers ( $n = 0, 1, 2, 3, 4, 5$ , i.e. $μ = 0, 1.73, 2.83, 3.87, 4.90, 5.92$ ). The curve is visibly not a straight line — that curvature is precisely why $μ \neq = n$ .

Worked examples

Cell A — the zero case, $[Zn(H_{2} O)_{6}]^{2 +}$

Worked example Cell A:

$n = 0$ diamagnetic (degenerate input) Statement. Find $μ$ for $[Zn(H_{2} O)_{6}]^{2 +}$ . Forecast: guess now — will this ion be attracted to a magnet or not?

Step 1 — oxidation state & $d$ -count. Zn is $[Ar] 3 d^{10} 4 s^{2}$ . Remove the two $4 s$ electrons first → $Zn^{2 +}$ is $3 d^{10}$ . Why this step? The formula counts electrons of the ion, not the atom; and we always strip $4 s$ before $3 d$ .

Step 2 — count unpaired. Ten electrons in five $d$ -boxes means every box is $↑↓$ . So $n = 0$ . Why this step? $n$ is the whole input to the formula.

Step 3 — apply formula. $μ = 0 (0 + 2) = 0 = 0 μ_{B}$ . Why this step? Plugging $n = 0$ tests the degenerate end of the formula.

Verify: $0 \cdot 2 = 0$ . ✓ All paired → diamagnetic → weakly repelled, not attracted. Matches the forecast that filled $d^{10}$ ions are never magnetic.

Cell B (and J) — high-spin maximum, $[Mn(H_{2} O)_{6}]^{2 +}$

Worked example Cell B/J: weak field, high spin, the

$n = 5$ ceiling Statement. Find $μ$ for $[Mn(H_{2} O)_{6}]^{2 +}$ . Forecast: how many boxes can you fill with lonely arrows before pairing starts?

Step 1 — $d$ -count. Mn is $[Ar] 3 d^{5} 4 s^{2}$ ; remove two $4 s$ → $Mn^{2 +}$ is $3 d^{5}$ . Why this step? $d^{5}$ is the special half-filled count; getting it right is the input to everything after.

Step 2 — spin state. $H_{2} O$ is a weak-field ligand (Spectrochemical Series) → small $Δ_{o}$ → Hund's rule wins → each of the five $d$ -orbitals gets one arrow before any pairing. $n = 5$ . Why this step? Weak field means spreading out costs less than pairing, so this step fixes how many arrows stay lonely.

Step 3 — apply formula. $μ = 5 (5 + 2) = 35 = 5.92 μ_{B}$ . Why this step? $n$ is now known; the formula is the only remaining move — it converts "5 lonely arrows" into a measurable magnet strength.

Verify: $5 \times 7 = 35$ , $35 \approx 5.916$ . ✓ This is the largest spin-only moment possible for a first-row ion (you can't have more than 5 unpaired in 5 boxes) — the limiting case J.

Cell C — low-spin, $[Mn(CN)_{6}]^{4 -}$

Worked example Cell C: same ion, strong field, spin squeezed down

Statement. Find $μ$ for $[Mn(CN)_{6}]^{4 -}$ . Forecast: same $d^{5}$ ion as Ex 2 — will $μ$ be the same? Guess before reading.

Step 1 — $d$ -count. Charge bookkeeping: six $CN^{-}$ give $- 6$ ; overall $- 4$ ; so Mn is $+ 2$ → $3 d^{5}$ . (Same $d$ -count as Ex 2.) Why this step? To prove that $d$ -count alone does not fix $μ$ .

Step 2 — spin state. $CN^{-}$ is a strong-field ligand → large $Δ_{o}$ → pairing in the lower $t_{2 g}$ set is cheaper than climbing to $e_{g}$ . Five electrons fill $t_{2 g}$ as $(↑↓) (↑↓) (↑)$ . So $n = 1$ . Why this step? Strong field flips the Hund-vs-pairing competition, changing how many arrows stay lonely.

Step 3 — apply formula. $μ = 1 (1 + 2) = 3 = 1.73 μ_{B}$ . Why this step? With $n = 1$ established, the formula converts it to the final measurable moment.

Verify: $1 \times 3 = 3$ , $3 \approx 1.732$ . ✓ Same ion as Ex 2 but $5.92 \to 1.73$ — the ligand, not the metal, decides.

Cell D — the field can't change anything, $[Cr(H_{2} O)_{6}]^{3 +}$ vs $[Cr(CN)_{6}]^{3 -}$

Worked example Cell D:

$d^{3}$ — high spin = low spin Statement. Show that $Cr^{3 +}$ ( $d^{3}$ ) gives the same $μ$ with weak or strong ligands. Forecast: three electrons, three low $t_{2 g}$ boxes — is there ever a reason to pair?

Step 1 — $d$ -count. Cr is $[Ar] 3 d^{5} 4 s^{1}$ ; remove one $4 s$ and two $3 d$ → $Cr^{3 +}$ is $3 d^{3}$ . Why this step? $d^{3}$ is one of the counts where field strength is irrelevant, and we must confirm the count to claim that.

Step 2 — spin state. The three lowest $t_{2 g}$ orbitals each take one electron — no orbital is forced to double up regardless of $Δ_{o}$ . So $n = 3$ either way. Why this step? Pairing only becomes a choice once electrons exceed the number of low orbitals. With 3 electrons and 3 $t_{2 g}$ boxes, there's no choice. (Same logic: $d^{1}, d^{2}, d^{8}, d^{9}, d^{10}$ .)

Step 3 — apply formula. $μ = 3 (3 + 2) = 15 = 3.87 μ_{B}$ for both complexes. Why this step? Both complexes share $n = 3$ , so the formula must (and does) return one identical answer — that is the whole point.

Verify: $3 \times 5 = 15$ , $15 \approx 3.873$ . ✓ Field strength changed nothing — a common exam trap defused.

Cell E — tetrahedral is always high-spin, $[CoCl_{4}]^{2 -}$

Worked example Cell E: tetrahedral geometry

Statement. Find $μ$ for $[CoCl_{4}]^{2 -}$ . Forecast: tetrahedral splitting $Δ_{t}$ is small — will pairing ever happen?

Step 1 — $d$ -count. Four $Cl^{-}$ give $- 4$ ; overall $- 2$ ; so Co is $+ 2$ → $3 d^{7}$ . Why this step? The $d$ -count is the raw material; every later move depends on knowing it is 7.

Step 2 — geometry & splitting. Tetrahedral fields split the $d$ -orbitals the other way up (lower $e$ , upper $t_{2}$ ) and, crucially, $Δ_{t} \approx \frac{4}{9} Δ_{o}$ — always small (recall the splitting definitions above). So tetrahedral complexes are essentially always high-spin (Square Planar vs Tetrahedral Geometry). Why this step? Small $Δ_{t}$ means pairing energy always beats the gap → Hund's rule wins → this fixes the spin state as high before we count.

Step 3 — fill and count. Seven electrons, high-spin, over five orbitals: first give each of the five orbitals one arrow (uses 5 electrons → five lonely arrows). Then the remaining two electrons go in, each one pairing up one already-singly-occupied orbital — so exactly two orbitals become $↑↓$ . Occupancies: $2, 2, 1, 1, 1$ . Three orbitals still hold one lonely arrow, so $n = 3$ . Why this step? Carefully placing electron 6 and electron 7 (two electrons total, not four) is what turns " $d^{7}$ high-spin" into the concrete number $n = 3$ .

Step 4 — apply formula. $μ = 3 (3 + 2) = 15 = 3.87 μ_{B}$ . Why this step? With $n = 3$ fixed, the formula delivers the final magnet strength — the only quantity the question actually asked for.

Verify: $15 \approx 3.87$ . ✓ (Experimentally $[CoCl_{4}]^{2 -}$ runs a bit higher, ~4.4–4.8, because tetrahedral Co(II) has an unquenched orbital contribution — a real hint toward Cell H below.)

Cell F — geometry deduced from $μ$ , $[Ni(CN)_{4}]^{2 -}$

Worked example Cell F:

$μ = 0$ proves square planar Statement. $[Ni(CN)_{4}]^{2 -}$ is measured diamagnetic. Deduce its geometry. Forecast: two four-coordinate shapes exist — which one hides all its electrons?

Step 1 — $d$ -count. Four $CN^{-}$ give $- 4$ ; overall $- 2$ ; Ni is $+ 2$ → $3 d^{8}$ . Why this step? We can't compare candidate geometries until we know how many $d$ -electrons must be housed.

Step 2 — two candidate shapes. A four-coordinate $d^{8}$ can be tetrahedral (high-spin, $n = 2$ , paramagnetic) or square planar (the four lower orbitals fill as pairs, leaving the high $d_{x^{2} - y^{2}}$ empty, $n = 0$ , diamagnetic). See the level diagram below. Why this step? Each shape predicts a different $n$ , so the two shapes make competing, testable magnetic predictions.

Step 3 — match to measurement. Measured $μ = 0$ ⇒ $n = 0$ ⇒ square planar. Strong-field $CN^{-}$ makes this favourable. Why this step? This is the key inference: the measured moment acts as a fingerprint that selects one geometry and rejects the other — magnetism becomes structural evidence.

Step 4 — the formula check. $μ = 0 (0 + 2) = 0$ . ✓ Why this step? Feeding the deduced $n = 0$ back through the formula confirms it reproduces the observed zero — a consistency check.

Verify: tetrahedral $d^{8}$ would give $n = 2$ , $μ = 8 = 2.83 μ_{B} \neq = 0$ . The observed zero rules it out. Magnetism became a structural proof.

Intuition What the level diagram shows (full description)

The figure is an energy-level diagram: the vertical axis is orbital energy (increasing upward, marked with an " $E$ " arrow and tick marks — higher line means costlier to occupy), and there is no horizontal axis (horizontal position just separates the two geometries). On the left, labelled "Square planar," five horizontal energy levels are drawn at increasing heights; the lowest four each carry a full pair $↑↓$ (black arrows), and the topmost level — the $d_{x^{2} - y^{2}}$ , drawn and labelled in red — sits noticeably higher and is empty. Because that top orbital is pushed so high (the four in-plane ligands raise it), leaving it empty and pairing below costs less energy, so every electron has a partner: $n = 0$ , diamagnetic. On the right, labelled "Tetrahedral," the levels split into a lower pair (the $e$ set) and an upper trio (the $t_{2}$ set) separated by only a small gap $Δ_{t}$ ; filling eight electrons high-spin fills the $e$ set with two pairs and puts $(↑↓) (↑) (↑)$ in the $t_{2}$ set, leaving two lonely red arrows — $n = 2$ , paramagnetic. The small vertical gap on the right versus the large jump to the red level on the left is exactly why one filling is all-paired and the other is not.

Cell G — the reverse problem, given $μ$

Worked example Cell G: invert the formula

Statement. A complex shows $μ = 4.90 μ_{B}$ . Find $n$ and suggest an ion. Forecast: which whole number of arrows gives about 4.9?

Step 1 — set up the inversion. $n (n + 2) = 4.90 \Rightarrow n (n + 2) = 4.9 0^{2} = 24.01 \approx 24$ . Why this step? Squaring both sides undoes the square root — the only way to release $n$ from inside it, which is what a "reverse" problem demands.

Step 2 — solve the quadratic in words. We need two numbers, $n$ and $n + 2$ , multiplying to 24. Test: $4 \times 6 = 24$ . So $n = 4$ . Why this step? $n$ must be a whole number (you can't have half a lonely electron in the spin-only model), so we hunt for the integer that fits rather than solving formally.

Step 3 — identify a candidate ion. Four unpaired electrons come from a high-spin $d^{4}$ ( $Cr^{2 +}$ , $Mn^{3 +}$ ) or a high-spin $d^{6}$ ( $Fe^{2 +}$ with a weak-field ligand). Any of these is a valid answer. Why this step? Converting $n$ back into a real ion is what makes the abstract number chemically meaningful and answers the "suggest an ion" half of the question.

Step 4 — check against the formula. Put $n = 4$ back in: $μ = 4 (4 + 2) = 24 = 4.90 μ_{B}$ . Why this step? A reverse problem is only finished when the recovered $n$ reproduces the measured $μ$ — this closes the loop.

Verify: $4 \times 6 = 24 = 4.899 \approx 4.90$ . ✓ The inversion is self-consistent.

Cell H — when spin-only fails (the twist)

Worked example Cell H: non-integer measured

$μ$ , orbital contribution Statement. An octahedral $Co^{2 +}$ aqua complex is measured at $μ \approx 4.8 μ_{B}$ , but the spin-only formula predicts less. Explain the discrepancy and give the correction. Forecast: $Co^{2 +}$ is $d^{7}$ high-spin with $n = 3$ ; what does spin-only say — and does the measurement agree?

Step 1 — spin-only prediction. $d^{7}$ high-spin → $n = 3$ → $μ_{so} = 3 (3 + 2) = 15 = 3.87 μ_{B}$ . Why this step? We need the baseline the pure spin model gives so we can see whether reality overshoots it.

Step 2 — spot the gap. Measured $4.8 > 3.87$ . The extra magnetism is orbital angular momentum that is not fully quenched for this configuration (a partly-filled $t_{2 g}$ set can still circulate around the metal and add its own magnetism). Why this step? The mismatch is the whole point of the cell: a measurement above spin-only is a signal, not an error — it tells us the "frozen-orbital" assumption has broken.

Step 3 — the fuller formula. When orbital motion matters, use $μ = 4 S (S + 1) + L (L + 1) μ_{B},$ where $S = n /2$ is the total spin quantum number (add $\frac{1}{2}$ per unpaired electron) and $L$ is the total orbital quantum number. To get $L$ : give each $d$ -orbital a magnetic quantum number $m_{ℓ} \in {- 2, - 1, 0, + 1, + 2}$ , fill the electrons by Hund's rule, and add up the occupied $m_{ℓ}$ values; $L$ is the magnitude of that sum ( $0, 1, 2, 3, 4, \dots$ ). A half-filled or completely filled shell sums to $0$ , giving $L = 0$ — the quenched case where spin-only is already exact. Why this step? Naming and pinning down $L$ is what lets the reader actually use this formula instead of leaving the new symbol floating.

Step 4 — sanity of the correction. With $n = 3$ : $S = \frac{3}{2}$ , so the spin part is $4 S (S + 1) = 4 \cdot \frac{3}{2} \cdot \frac{5}{2} = 15$ , reproducing $15 = 3.87$ when $L = 0$ . Any $L > 0$ adds $L (L + 1)$ inside the root and pushes $μ$ up toward the observed $4.8$ . Why this step? It shows the new formula collapses back to spin-only in the quenched limit and explains the direction of the correction.

Verify: spin-only $15 = 3.87 < 4.8$ measured, and adding a positive $L (L + 1)$ raises the prediction — consistent. ✓ A non-round measured number is a clue (unquenched orbital motion), not an arithmetic slip.

Cell I — real-world word problem

Worked example Cell I: from lab words to

$n$ Statement. A student weighs a chromium salt on a balance placed between the poles of a magnet. Switching the magnet on makes the sample appear heavier, by an amount consistent with $μ \approx 3.9 μ_{B}$ . Find the chromium oxidation state and spin state. Forecast: "heavier when the field is on" — attracted or repelled? What does that already tell you before any arithmetic?

Step 1 — translate the observation. Appearing heavier means the sample is pulled into the field ⇒ paramagnetic ⇒ at least one unpaired electron. Why this step? Paramagnets are attracted; diamagnets are pushed out (would read lighter). Reading the physical sign correctly is what turns a lab observation into a statement about $n$ .

Step 2 — get $n$ from $μ$ . Invert the formula: $n (n + 2) = 3.9 \Rightarrow n (n + 2) = 3. 9^{2} \approx 15.2 \approx 15 \Rightarrow n = 3$ . Why this step? This is the same "square and split" inversion as Cell G — converting the measured magnet strength into the concrete number of lonely electrons.

Step 3 — identify the ion. Three unpaired electrons for chromium point to $Cr^{3 +}$ , which is $d^{3}$ with exactly $n = 3$ — and from Cell D that value is field-independent, so the spin state is unambiguous. Why this step? Mapping $n$ back to a real ion answers the actual question (oxidation state $+ 3$ , spin state fixed) instead of stopping at a bare number.

Step 4 — check. Put $n = 3$ back in: $μ = 3 (3 + 2) = 15 = 3.87 \approx 3.9 μ_{B}$ . Why this step? Confirms the deduced ion reproduces the measured tug.

Verify: $15 = 3.873 \approx 3.9$ . ✓ A word problem reduced to the same one formula and the same inversion.

Recall Self-test the whole matrix

$[Zn(H_{2} O)_{6}]^{2 +}$ : $d$ -count, $n$ , $μ$ ? ::: $d^{10}$ , $n = 0$ , $μ = 0$ (diamagnetic) $[Mn(CN)_{6}]^{4 -}$ vs $[Mn(H_{2} O)_{6}]^{2 +}$ : both $d^{5}$ , why different $μ$ ? ::: CN⁻ strong field → low-spin $n = 1$ (1.73); H₂O weak field → high-spin $n = 5$ (5.92) Why is $d^{3}$ field-independent? ::: 3 electrons fill the 3 $t_{2 g}$ orbitals singly; no orbital is ever forced to pair, so $n = 3$ for any ligand What is $μ_{B}$ and its unit? ::: The Bohr magneton, the electron-sized yardstick for magnetic moment; unit J T⁻¹ ( $\approx 9.274 \times 1 0^{- 24}$ ) What are $Δ_{o}$ and $Δ_{t}$ , and how do they relate? ::: The $d$ -orbital splitting gaps in octahedral and tetrahedral fields; $Δ_{t} \approx \frac{4}{9} Δ_{o}$ , so tetrahedral gaps are small → high spin Why are tetrahedral complexes essentially always high-spin? ::: $Δ_{t} \approx \frac{4}{9} Δ_{o}$ is too small to beat the pairing energy Measured $μ = 4.90$ : find $n$ ::: $n (n + 2) = 24 \Rightarrow n = 4$ How do you find $L$ in the fuller formula? ::: Sum the $m_{ℓ}$ values of the occupied $d$ -orbitals; $L$ is the magnitude of that sum (0 for a half- or fully-filled shell) A sample gets heavier in a magnetic field — para or dia? ::: Paramagnetic (attracted), so it has unpaired electrons Non-integer $μ$ above the spin-only value signals what? ::: Unquenched orbital angular momentum; use $4 S (S + 1) + L (L + 1)$

Connections

Parent: Magnetic moments of complexes
Crystal Field Theory — the $Δ_{o}$ behind every high/low-spin decision here.
Spectrochemical Series — tells you weak (Cell B) vs strong (Cell C) field.
Electronic Configuration of d-block ions — Step 1 of every example.
Hund's Rule and Pairing Energy — the competition in Cells B, C, D, E.
Square Planar vs Tetrahedral Geometry — Cells E and F.
Color of Coordination Compounds — same $Δ_{o}$ that sets spin also sets colour.