Visual walkthrough — Magnetic moments of complexes

We assume nothing beyond "a moving charge makes a magnetic field." Everything else — spin, the quantum number $S$, even the units — is built in front of you.

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Step 1 — A spinning charge is a tiny magnet

WHAT. Picture a single electron as a small ball carrying negative electric charge, and imagine it spinning about its own axis.

WHY this picture. The only fact we import from physics is: a charge that goes in a loop makes a magnetic field. A spinning charged ball is charge going round and round — a current loop — so it must produce a magnetic field. A magnet with a north and south end is exactly what we call a magnetic moment, written $\vec{\mu }$.

A crucial sign. The electron's charge is negative. For a current loop the moment is $\vec{\mu }=I\vec{A}$ (current $\times$ enclosed area). Because the charge is negative, the "current" runs opposite to the way a positive charge would circulate — so $\vec{\mu }$ ends up pointing opposite to the spin arrow. Keep this in mind: the magnet points against the spin.

PICTURE. Blue arrows show charge circulating; the green arrow is the spin $\vec{S}$; the yellow arrow $\vec{\mu }$ points the other way because the charge is negative.

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Step 2 — Two paired electrons cancel; a lone one survives

WHAT. Electrons come with two possible spin directions, drawn as ↑ and ↓. When two electrons share one orbital they must spin opposite ways.

WHY this matters. Two opposite tiny magnets sit north-to-south and their fields cancel — the pair contributes zero net moment. So only the electrons without a partner — the unpaired electrons — leave any magnetism behind. This is why the whole problem reduces to counting unpaired electrons, a number we call $n$.

PICTURE. Left: a paired box, two opposite yellow arrows summing to nothing. Right: a lone ↑ electron, one surviving yellow arrow (drawn down, since $\vec{\mu }$ opposes the up-spin).

Recall Why does pairing force opposite spins?

Pauli exclusion ::: two electrons in the same orbital must differ in spin, so one is ↑ and the other ↓ — their moments oppose and cancel.

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Step 3 — Add up the lonely spins into one number $S$

WHAT. Give each unpaired electron a spin value of $+\frac{1}{2}$ and add them. With $n$ unpaired electrons all pointing the same way:

$S=\underset{\text{one lonely spin}}{\underbrace{\frac{1}{2}}}\times \underset{\text{how many}}{\underbrace{n}}=\frac{n}{2}$

Here $S$ (the total spin quantum number) is our bookkeeping total, $n$ is the count of unpaired electrons, and $\frac{1}{2}$ is what each contributes.

WHY add them. All unpaired electrons in a subshell align the same direction (this is Hund's rule — see Hund's Rule and Pairing Energy), so their spins simply pile up. $S$ is the single number that summarises "how much spin the whole atom has."

PICTURE. Three parallel ↑ arrows stacking into one tall arrow labelled $S=\frac{3}{2}$.

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Step 4 — The quantum surprise: length is $\sqrt{S(S+1)}$, not $S$

WHAT. We need the length of the spin arrow $\vec{S}$, because the magnet's strength will be proportional to it. First, a unit issue: raw angular momentum is measured in a fixed natural chunk called $\hslash$ ("h-bar") — think of it simply as "one quantum unit of spin," the way a metre is one unit of length. So we write the length as (a pure number) $\times \hslash$.

In everyday geometry a "size-$S$" arrow has length $S$. In quantum mechanics it does not — its true length is $\sqrt{S(S+1)}$ of these $\hslash$ units.

WHY this tool and not plain $S$. A spin arrow can never point exactly along the measuring (up) axis; it is forced to tilt slightly. Its tallest possible shadow on that axis is $S$, but the arrow itself is longer. By Pythagoras, (true length)${}^2$ = (shadow)${}^2$ + (sideways wobble)${}^2$, and quantum mechanics fixes the wobble so that the total comes out to $\sqrt{S(S+1)}$. We use the square root because we want the true magnitude of a vector that is never allowed to lie flat.

$|\vec{S}|=\underset{\text{pure number}}{\underbrace{\sqrt{S(S+1)}}}\underset{\text{unit of spin}}{\underbrace{\hslash }}\text{not just }S\hslash$

PICTURE. A right triangle: the vertical side is the shadow $S$, and the hypotenuse (the real arrow) is longer — labelled $\sqrt{S(S+1)}\hslash$. The gap is the quantum "you can't align perfectly" tilt.

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Step 5 — Turn spin length into magnetic strength (where $\hslash$ vanishes and ${\mu }_B$ is born)

WHAT. Now convert the spin vector $\vec{S}$ into the magnetic moment $\vec{\mu }$. From Step 1, a moving charge circulating makes $\vec{\mu }=I\vec{A}$; working that out for a charge $e$ of mass $m_e$ whirling with angular momentum $\vec{S}$ gives a moment proportional to $\vec{S}$:

$\vec{\mu }=-g\frac{e}{2m_e}\vec{S}$

Read it term by term: $\vec{S}$ is the spin (Step 4); $\frac{e}{2m_e}$ is what the loop-current calculation spits out (charge over twice the mass); the minus sign is the negative-charge flip from Step 1 (so $\vec{\mu }$ opposes $\vec{S}$); and $g\approx 2$ is a measured "extra doubling" special to the electron's spin.

WHY a natural unit appears. The clump of fundamental constants $\frac{e\hslash }{2m_e}$ turns up every single time. Rather than drag it around, we name it once:

$\boxed{{\mu }_B\equiv \frac{e\hslash }{2m_e}}(\text{the }\text{Bohr magneton})$

${\mu }_B$ is literally "the magnetic moment of one unit ($\hslash$) of angular momentum" — it is derived from $e$, $\hslash$, $m_e$, not pulled from thin air.

WHERE $\hslash$ goes. Take the length of $\vec{\mu }$ using $|\vec{S}|=\sqrt{S(S+1)}\hslash$ from Step 4:

$|\vec{\mu }|=g\frac{e}{2m_e}\underset{=|\vec{S}|}{\underbrace{\sqrt{S(S+1)}\hslash }}=g\underset{={\mu }_B}{\underbrace{\frac{e\hslash }{2m_e}}}\sqrt{S(S+1)}=g\sqrt{S(S+1)}{\mu }_B$

The $\hslash$ did not disappear — it merged into ${\mu }_B$. That is exactly why the moment comes out in units of ${\mu }_B$: the unit of spin ($\hslash$) got baked into the unit of magnetism (${\mu }_B$).

PICTURE. The spin arrow $\vec{S}$ (length $\sqrt{S(S+1)}\hslash$) → multiply by $\frac{e}{2m_e}$ (which eats the $\hslash$ and makes ${\mu }_B$) and by $g$ → the red $\vec{\mu }$ arrow, pointing the opposite way.

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Step 6 — Substitute $S=n/2$ and watch it simplify

WHAT. Put Step 3 ($S=\frac{n}{2}$) into Step 5, with $g=2$, and simplify carefully:

${\mu }_{\text{so}}=2\sqrt{\frac{n}{2}\left(\frac{n}{2}+1\right)}{\mu }_B$

Rewrite the inside over a common denominator — $\frac{n}{2}+1=\frac{n+2}{2}$:

$=2\sqrt{\frac{n}{2}\cdot \frac{n+2}{2}}{\mu }_B=2\sqrt{\frac{n(n+2)}{4}}{\mu }_B$

The $\sqrt{\frac{1}{4}}=\frac{1}{2}$ pulls out of the root and cancels the leading $2$:

$=2\cdot \frac{1}{2}\sqrt{n(n+2)}{\mu }_B=\sqrt{n(n+2)}{\mu }_B$

Each move: common denominator (so the two fractions multiply cleanly), pull the $\frac{1}{4}$ out as $\frac{1}{2}$, then the $2\cdot \frac{1}{2}=1$ erases the constant. The $g=2$ was engineered to disappear here.

PICTURE. A cancellation ladder: the $2$ out front and the $\frac{1}{2}$ from the root annihilate, leaving the clean boxed result.

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Step 7 — Every case, plotted: $n=0,1,2,\dots$

WHAT. Feed in each possible $n$ and read the moment. This is the full range — nothing is left out.

$n$	$n(n+2)$	${\mu }_{\text{so}}$ (${\mu }_B$)	meaning
0	0	$0$	diamagnetic (all paired)
1	3	$1.73$	e.g. low-spin $d^5$
2	8	$2.83$
3	15	$3.87$	e.g. ${\text{Cr}}^{3+}$, $d^3$
4	24	$4.90$
5	35	$5.92$	high-spin $d^5$

WHY plot the curve. Notice $\mu$ is not equal to $n$ — the gap grows then shrinks. Drawing $\sqrt{n(n+2)}$ against the straight line $y=n$ makes the curved relationship unmistakable, and kills the classic mistake "$\mu =n$."

Degenerate case $n=0$. Then $n(n+2)=0$, so $\mu =0$: no lonely electrons, no magnet — the substance is repelled, not attracted. This is exactly how $[{\text{Ni(CN)}}_4{]}^{2-}$ is diagnosed square planar (see Square Planar vs Tetrahedral Geometry).

PICTURE. Blue dots on the curve $\sqrt{n(n+2)}$ versus a yellow dashed line $y=n$; the vertical gaps are annotated.

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The one-picture summary

Everything above, compressed into a single left-to-right flow: lonely spins → total $S=\frac{n}{2}$ → tilted arrow of length $\sqrt{S(S+1)}\hslash$ → times $g\frac{e}{2m_e}$ (which makes ${\mu }_B$ and flips the sign) → clean $\sqrt{n(n+2)}{\mu }_B$.

Recall Feynman retelling of the whole walkthrough

A spinning charged ball is a tiny magnet — but because the electron's charge is negative, its magnet points opposite to the way it spins. Inside an atom, electrons spin like that. When two share a room they spin opposite ways and cancel — so only the lonely electrons make magnetism. Count them: call it $n$. Each lonely spin is "half a unit," so the total spin is $S=n/2$. A spin arrow in quantum-land can never stand perfectly straight — it always leans — so its real length isn't $S$, it's the hypotenuse $\sqrt{S(S+1)}$, measured in the natural spin-unit $\hslash$. To turn spin into magnetism you multiply by $e/2m_e$; that combination times $\hslash$ is exactly the Bohr magneton ${\mu }_B$, our unit of magnetism — so $\hslash$ doesn't vanish, it becomes ${\mu }_B$. The electron is doubly magnetic ($g=2$). Plug $S=n/2$ into $2\sqrt{S(S+1)}$, tidy up, and a $2$ cancels a hidden $\frac{1}{2}$, leaving the beautifully simple $\sqrt{n(n+2)}{\mu }_B$. The whole trick is a hypotenuse and a cancellation.

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Connections

• Parent: Magnetic Moments of Complexes — the summary this page expands.
• Hund's Rule and Pairing Energy — why unpaired spins align (Step 3).
• Crystal Field Theory · Spectrochemical Series — decide $n$ via high/low spin.
• Electronic Configuration of d-block ions — gives the $d$-count that feeds $n$.
• Square Planar vs Tetrahedral Geometry — the $n=0$ diamagnetic case as proof.
• Color of Coordination Compounds — same ${\Delta }_o$ that sets spin also sets colour.