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

WHAT are we measuring?

WHY transition metals matter: they have partially filled d-orbitals, so many of their ions carry unpaired electrons → most are coloured and paramagnetic.

HOW the formula arises (derivation from scratch)

An electron has two sources of magnetism:

Orbital angular momentum (it circulates around the nucleus).
Spin angular momentum (intrinsic).

The full moment is: $\mu_{S+L} = \sqrt{4S(S+1) + L(L+1)}\ \ \mu_B$

Step 1 — spin quantum number for $n$ unpaired electrons. Each unpaired electron contributes spin $+\tfrac12$ . With all spins aligned (Hund's rule), $S = \frac{n}{2}.$ Why this step? Total spin = sum of individual $+\tfrac12$ values; $n$ electrons give $S=n/2$ .

Step 2 — plug into the spin formula $\mu = \sqrt{4S(S+1)}$ : $\mu = \sqrt{4\cdot\frac{n}{2}\left(\frac{n}{2}+1\right)}$ Why? Setting $L=0$ leaves only the $4S(S+1)$ term under the root.

Step 3 — simplify the algebra: $\mu = \sqrt{4\cdot\frac{n}{2}\cdot\frac{n+2}{2}} = \sqrt{\frac{4n(n+2)}{4}} = \sqrt{n(n+2)}$

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

Reference table (memorise the shape, not the digits)

$n$	$\mu=\sqrt{n(n+2)}$	Example ion (config)
0	0.00 BM (diamagnetic)	$\text{Sc}^{3+}$ ( $d^0$ ), $\text{Zn}^{2+}$ ( $d^{10}$ )
1	$\sqrt3\approx1.73$	$\text{Ti}^{3+}$ ( $d^1$ ), $\text{Cu}^{2+}$ ( $d^9$ )
2	$\sqrt8\approx2.83$	$\text{V}^{3+}$ ( $d^2$ ), $\text{Ni}^{2+}$ ( $d^8$ )
3	$\sqrt{15}\approx3.87$	$\text{Cr}^{3+}$ ( $d^3$ ), $\text{Co}^{2+}$
4	$\sqrt{24}\approx4.90$	$\text{Cr}^{2+}$ ( $d^4$ ), $\text{Mn}^{3+}$
5	$\sqrt{35}\approx5.92$	$\text{Mn}^{2+}$ ( $d^5$ ), $\text{Fe}^{3+}$

Worked examples

Common mistakes (Steel-man + fix)

Flashcards

Spin-only magnetic moment formula?

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

BM, where

n

= number of unpaired electrons.

What does paramagnetism require?

At least one unpaired electron (

n\ge1

Why is orbital contribution often ignored in 3d complexes?

It is "quenched" by the ligand field, so

L\to0

and only spin survives.

Magnetic moment of

\text{Mn}^{2+}

(

d^5

\sqrt{35}=5.92

BM.

Magnetic moment of

\text{Sc}^{3+}

\text{Zn}^{2+}

0 BM — diamagnetic (

d^0

d^{10}

A complex has

\mu=2.83

BM. How many unpaired electrons?

n=2

(since

\sqrt{8}=2.83

Derive

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

from

\sqrt{4S(S+1)}

Put

S=n/2

\sqrt{4\cdot\frac n2(\frac n2+1)}=\sqrt{n(n+2)}

Which electrons leave first when forming a transition-metal ion?

The 4s electrons (before 3d).

Why does

\mu

peak at

d^5

d^5

has the maximum 5 unpaired electrons (half-filled, Hund); beyond it electrons start pairing.

Same

\text{Fe}^{2+}

, why two different

\mu

values?

High-spin (weak field) = 4 unpaired = 4.90 BM; low-spin (strong field) = 0 unpaired = 0 BM.

Recall Feynman: explain to a 12-year-old

Imagine each electron is a tiny spinning top that acts like a baby magnet. When two electrons share a room (an orbital) they spin opposite ways, so their magnets cancel — boring, no magnetism. But if an electron sits alone in its room, its little magnet has no partner to cancel it, so the whole atom can be pulled toward a big magnet. To guess how strong the pull is, just count the lonely electrons ( $n$ ) and do $\sqrt{n(n+2)}$ . More lonely electrons = stronger magnet.

Connections

Crystal Field Theory — explains high-spin vs low-spin → changes $n$ .
Hund's Rule & Electron Configuration — sets how many electrons stay unpaired.
Colour in Transition Metal Complexes — same d-electrons, different observable (d–d transitions).
Electronic Configuration of Ions — 4s-before-3d removal rule.
Bohr Magneton — the unit $\mu_B=\frac{eh}{4\pi m_e}$ .
Lanthanide Magnetism — f-block keeps orbital contribution, needs $\sqrt{4S(S+1)+L(L+1)}$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, magnetism ka asli funda simple hai: har electron ek chhota magnet hai kyunki woh spin karta hai. Jab do electron ek orbital mein paired hote hain, unka spin opposite hota hai, toh dono ke magnets cancel ho jaate hain. Lekin jab koi electron akela (unpaired) baitha hota hai, uska magnet cancel nahi hota — isliye poora atom bahar ke magnet ki taraf khinchta hai. Isi attraction ko paramagnetism kehte hain. Agar saare paired hain toh substance diamagnetic (halka repel) hota hai.

Magnet ki strength naapne ke liye hum spin-only formula use karte hain: $\mu=\sqrt{n(n+2)}$ BM, jahan $n$ = unpaired electrons ki ginti. Yeh formula bade waale $\sqrt{4S(S+1)}$ se aata hai, bas $S=n/2$ rakh do aur simplify karo. Transition metals mein orbital ka contribution mostly "quench" ho jaata hai ligand field ki wajah se, isliye sirf spin count karna kaafi hota hai.

Important trick: ion banate waqt pehle 4s electron nikalte hain, phir 3d. Jaise $\text{Fe}^{3+}$ = $3d^5$ , toh 5 unpaired, $\mu=\sqrt{35}=5.92$ BM. Aur $\text{Zn}^{2+}$ ( $d^{10}$ ) ya $\text{Sc}^{3+}$ ( $d^0$ ) ka $\mu=0$ — diamagnetic. Yaad rakho: sirf akele electron count hote hain, total nahi.

Yeh cheez exam mein bahut kaam aati hai aur deep concept bhi clear karti hai — same ion ke do alag $\mu$ ho sakte hain (high-spin vs low-spin) jo batata hai ki ligand strong hai ya weak. Toh magnetism actually ek "detector" hai jo andar ki electron arrangement bata deta hai bina dekhe!

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

WHAT are we measuring?

HOW the formula arises (derivation from scratch)

Reference table (memorise the shape, not the digits)

Worked examples

Common mistakes (Steel-man + fix)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — d-Block (Transition Metals) & f-Block

Connections