Magnetic moments of complexes

WHAT are we measuring?

WHY does it matter? For a transition-metal complex, $\mu$ is dominated by the number of unpaired $d$ -electrons, $n$ . Measure $\mu$ → deduce $n$ → deduce geometry, spin state, and oxidation state.

HOW the formula arises (derivation from scratch)

An electron has two sources of magnetism:

Spin angular momentum (it acts like a spinning charge).
Orbital angular momentum (it circulates around the nucleus).

Spin contribution. Quantum mechanics gives the magnetic moment of a particle with total spin quantum number $S$ as

$\mu_S = g\sqrt{S(S+1)}\;\mu_B$

where $g \approx 2$ for the electron.

Why $\sqrt{S(S+1)}$ and not just $S$ ? Because in quantum mechanics the magnitude of an angular momentum vector of quantum number $S$ is $\sqrt{S(S+1)}\,\hbar$ , not $S\hbar$ . The magnetic moment is proportional to this magnitude.

Now relate $S$ to the number of unpaired electrons $n$ . Each unpaired electron contributes spin $\tfrac12$ , all aligned, so

$S = \frac{n}{2}.$

Substitute (with $g=2$ ):

$\mu_S = 2\sqrt{\frac{n}{2}\left(\frac{n}{2}+1\right)} = 2\sqrt{\frac{n}{2}\cdot\frac{n+2}{2}} = 2\cdot\frac12\sqrt{n(n+2)}$

$\boxed{\;\mu_S = \sqrt{n(n+2)}\;\;\mu_B\;}$

Quick reference table

$n$	$\mu_{\text{so}} = \sqrt{n(n+2)}$	$\mu_B$
0	$\sqrt{0}$	0 (diamagnetic)
1	$\sqrt{3}$	1.73
2	$\sqrt{8}$	2.83
3	$\sqrt{15}$	3.87
4	$\sqrt{24}$	4.90
5	$\sqrt{35}$	5.92

WHY high-spin vs low-spin changes everything

The same metal ion can give different $n$ depending on the ligand field:

Strong field ligand (e.g. $\text{CN}^-$ , CO) → large $\Delta_o$ → electrons pair up in the lower $t_{2g}$ → low spin, fewer unpaired electrons → small $\mu$ .
Weak field ligand (e.g. $\text{H}_2\text{O}$ , $\text{F}^-$ ) → small $\Delta_o$ → electrons spread out (Hund) → high spin, more unpaired electrons → large $\mu$ .

So measuring $\mu$ tells you the spin state directly.

Worked examples

Common mistakes (steel-manned)

Spin-only magnetic moment formula

\mu = \sqrt{n(n+2)}\ \mu_B

, where

n

= number of unpaired electrons

Why

\sqrt{n(n+2)}

and not

n

It comes from

g\sqrt{S(S+1)}

with

S=n/2

and

g=2

; angular-momentum magnitude is

\sqrt{S(S+1)}\hbar

, not

S\hbar

\mu

for

n=4

unpaired electrons

\sqrt{24}=4.90\ \mu_B

Unit of magnetic moment

Bohr magneton,

\mu_B

High-spin vs low-spin: which has larger

\mu

High-spin (more unpaired electrons, weak field ligand)

[\text{Fe(CN)}_6]^{3-}

[\text{Fe(H}_2\text{O})_6]^{3+}

d^5

low-spin (

n=1

, 1.73) vs high-spin (

n=5

, 5.92)

Why is spin-only formula good for 3d metals

Orbital contribution is largely quenched by the ligand field

When does spin-only formula fail

For lanthanides (4f) and heavy metals where orbital angular momentum is not quenched

[\text{Ni(CN)}_4]^{2-}

magnetism

d^8

square planar,

n=0

, diamagnetic

A complex shows

\mu=2.83\ \mu_B

. Find

n

n(n+2)=8 \Rightarrow n=2

Recall Feynman: explain it to a 12-year-old

A spinning top with electric charge acts like a tiny magnet. Inside a metal atom, electrons spin like tops. If two electrons sit together they spin opposite ways and their magnetism cancels — boring. But a lonely electron has no partner to cancel it, so it stays a little magnet. The more lonely electrons an atom has, the stronger a magnet it becomes. We pull the substance toward a big magnet and measure how strongly it's tugged: a strong tug means lots of lonely electrons. The formula $\sqrt{n(n+2)}$ just turns "number of lonely electrons" into "strength of the magnet."

Connections

Crystal Field Theory — sets $\Delta_o$ , which decides high-spin vs low-spin.
Spectrochemical Series — ranks ligand field strength.
Electronic Configuration of d-block ions — gives the $d$ -count.
Square Planar vs Tetrahedral Geometry — diamagnetism as structural proof.
Hund's Rule and Pairing Energy — competition behind spin state.
Color of Coordination Compounds — same $\Delta_o$ controls both color and magnetism.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, har electron ek chhota magnet jaisa hota hai kyunki wo spin karta hai aur charge bhi rakhta hai. Jab do electron ek saath pair ban jaate hain, unka magnetism cancel ho jaata hai. Lekin jo electron akela (unpaired) hota hai, wo apna magnetism dikhata hai. Isliye complex ka magnetic moment basically ye batata hai ki kitne unpaired electrons hain — yahi hai poora khel.

Formula yaad rakho: $\mu = \sqrt{n(n+2)}\ \mu_B$ , jahan $n$ unpaired electrons ki ginti hai. Ye seedha $n$ nahi hai — quantum mechanics se aata hai ( $g\sqrt{S(S+1)}$ wala derivation), isliye thoda sa tedha relation hai. $n=1$ par $1.73$ , $n=5$ par $5.92$ . Plug karo aur answer mil jaata hai.

Sabse important baat: same metal ion bhi alag-alag magnetism de sakta hai ligand ke according. Strong field ligand (jaise $CN^-$ ) electrons ko paas-paas pair kara deta hai — low spin, kam unpaired, chhota $\mu$ . Weak field ligand (jaise $H_2O$ ) electrons ko spread hone deta hai — high spin, zyada unpaired, bada $\mu$ . Isi liye $[\text{Fe(CN)}_6]^{3-}$ ka $\mu=1.73$ hai but $[\text{Fe(H}_2\text{O})_6]^{3+}$ ka $5.92$ — dono mein Fe $^{3+}$ , $d^5$ hai!

Ek caution: ye spin-only formula sirf 3d (first row) metals ke liye accha chalta hai, kyunki wahan orbital contribution "quench" ho jaata hai. Lanthanides (4f) ke liye ye fail karta hai — wahan orbital angular momentum bhi count karna padta hai. Exam mein hamesha pehle ion ka $d$ -count nikaalo, fir ligand strength dekho, fir $n$ ginke formula lagao.

WHAT are we measuring?

HOW the formula arises (derivation from scratch)

Quick reference table

WHY high-spin vs low-spin changes everything

Worked examples

Common mistakes (steel-manned)

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Coordination Chemistry

Connections