Magnetic moments of complexes: Question bank — Chemistry deep dive

Before the traps, this page builds every symbol it will test you on — so no reveal ever uses a term you have not yet met.

The toolkit — every symbol earned first

Look at the picture below: the left panel shows the two ways an electron makes magnetism (spin and orbital circulation), and the ruler on the right marks off magnetism in units of $μ_{B}$ .

True or false — justify

A single number is a lie detector for a lot of chemistry. Test whether you know why each statement holds or breaks.

A diamagnetic complex must have zero unpaired electrons

True — diamagnetism means every electron is paired, so

n = 0

and

μ_{so} = 0 = 0

; there is no lone spin left to be attracted.

Two complexes with the same

μ

must have the same number of unpaired electrons

True for the spin-only regime —

μ

is a one-to-one function of

n

there, so equal

μ

forces equal

n

(though the metals and geometries can still differ).

A larger

Δ_{o}

always gives a larger magnetic moment

False — larger

Δ_{o}

favours pairing (low spin), which reduces the number of unpaired electrons and hence lowers

μ

Every

d^{5}

complex is high-spin with

μ = 5.92 μ_{B}

False —

d^{5}

is high-spin (

n = 5

) only with a weak field; a strong field (

CN^{-}

) forces low-spin

t_{2 g}^{5}

, giving

n = 1

and

μ = 1.73

The spin-only formula works equally well for

Nd^{3 +}

4 f

ion)

False — for

4 f

ions the orbital angular momentum is not quenched, so spin-only fails; you need the full

g_{J} J (J + 1)

expression with total angular momentum

J

d^{10}

ion is always diamagnetic regardless of ligand or geometry

True — all ten

d

-orbitals are full, every electron is paired, so

n = 0

no matter the field strength or shape.

Doubling

n

doubles

μ

False — the relationship

n (n + 2)

is curved, not linear;

n = 1

gives

1.73

but

n = 2

gives

2.83

, not

3.46

A paramagnetic reading proves the metal is a transition metal

False — any species with unpaired electrons is paramagnetic (e.g.

O_{2}

, free radicals); paramagnetism alone says "unpaired electron present," not "transition metal."

Measured

μ

is the same at any temperature

False — by Curie's law the susceptibility scales as

1/ T

, so a paramagnet appears weaker when hotter; quoted

μ

values are the (nearly temperature-independent) effective moments extracted at a stated temperature.

Spot the error

Each line contains one flawed statement. Name the specific mistake, not just "it's wrong."

Fe^{3 +}

3 d^{5}

because we remove three

3 d

electrons from

3 d^{6} 4 s^{2}

Wrong removal order — you strip the two

4 s

electrons first, then only one

3 d

; the result

3 d^{5}

is right but the reasoning is not.

μ = 2.83 μ_{B}

means

n = 2.83

unpaired electrons."

n

must be a whole number;

2.83 = 8 = 2 (2 + 2)

, so this is

n = 2

, and

μ

is never equal to

n

itself.

[Ni(CN)_{4}]^{2 -}

is tetrahedral, so with

d^{8}

it has 2 unpaired electrons."

Strong-field

CN^{-}

makes this

d^{8}

complex square planar, not tetrahedral; square-planar

d^{8}

is diamagnetic (

n = 0

). See Square Planar vs Tetrahedral Geometry.

"Orbital angular momentum always adds to the spin moment, so real

μ

is always above spin-only."

3 d

complexes the orbital contribution is largely quenched by the ligand field, so measured values sit close to spin-only, not systematically above it.

CN^{-}

is a weak-field ligand, so

[Fe(CN)_{6}]^{3 -}

is high-spin."

CN^{-}

is a strong-field ligand near the top of the Spectrochemical Series; the complex is low-spin with

n = 1

"Since

μ_{S} = g S (S + 1)

uses

S

, the formula

n (n + 2)

needs

g \neq = 2

The

n (n + 2)

form comes from setting

g = 2

and

S = n /2

; it already bakes in

g = 2

, so nothing extra is needed.

"All

d^{5}

ions are paramagnetic, so

μ

can never be zero for

d^{5}

True that no

d^{5}

arrangement gives

n = 0

(odd electron count guarantees at least one unpaired), but the reasoning "

d^{5}

so paramagnetic" should cite the odd electron count, which is the real guarantee.

"A dimer of two

Cu^{2 +}

(

d^{9}

, one unpaired each) must show

μ

for two unpaired electrons."

If the two spins couple antiferromagnetically through a bridge, they cancel and

μ

drops toward zero; magnetic exchange coupling can hide unpaired electrons.

Cover everything and answer: "Why does measuring how hard a substance is pulled into a magnet tell a chemist the oxidation state and geometry of a metal?" Because the pull measures unpaired electrons ( $n$ ); $n$ pins down the $d$ -count and spin state, and those together fix the oxidation state and (via $Δ_{o}$ vs $P$ ) the geometry — as long as no exchange coupling or unquenched orbital moment is confusing the reading.

Connections

Crystal Field Theory — the $Δ_{o}$ splitting behind every high/low-spin trap here.
Spectrochemical Series — tells you which ligand gives large or small $Δ_{o}$ .
Electronic Configuration of d-block ions — the source of every $d$ -count.
Square Planar vs Tetrahedral Geometry — the geometry traps in the edge cases.
Hund's Rule and Pairing Energy — the $Δ_{o}$ vs $P$ tug-of-war deciding $n$ .
Color of Coordination Compounds — same $Δ_{o}$ , two observables.

Question bank — Magnetic moments of complexes

The toolkit — every symbol earned first

True or false — justify

Spot the error

Why questions

Edge cases

Connections