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

3.3.4 · Chemistry › d-Block (Transition Metals) & f-Block

HUM KYA measure kar rahe hain?

Transition metals kyun important hain: unke paas partially filled d-orbitals hote hain, isliye unke bahut se ions mein unpaired electrons hote hain → zyaatar coloured aur paramagnetic hote hain.

FORMULA kaise aata hai (scratch se derivation)

Ek electron mein magnetism ke do sources hain:

Orbital angular momentum (yeh nucleus ke around circulate karta hai).
Spin angular momentum (intrinsic).

Full moment yeh hai: $μ_{S + L} = 4 S (S + 1) + L (L + 1) μ_{B}$

Step 1 — $n$ unpaired electrons ke liye spin quantum number. Har unpaired electron spin $+ \frac{1}{2}$ contribute karta hai. Jab saare spins aligned hain (Hund's rule), $S = \frac{n}{2} .$ Yeh step kyun? Total spin = individual $+ \frac{1}{2}$ values ka sum; $n$ electrons dete hain $S = n /2$ .

Step 2 — spin formula mein plug karo $μ = 4 S (S + 1)$ : $μ = 4 \cdot \frac{n}{2} (\frac{n}{2} + 1)$ Kyun? $L = 0$ set karne se root ke andar sirf $4 S (S + 1)$ term bachta hai.

Step 3 — algebra simplify karo: $μ = 4 \cdot \frac{n}{2} \cdot \frac{n + 2}{2} = \frac{4 n ( n + 2 )}{4} = n (n + 2)$

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

Reference table (shape yaad karo, exact digits nahi)

$n$	$μ = n (n + 2)$	Example ion (config)
0	0.00 BM (diamagnetic)	$Sc^{3 +}$ ( $d^{0}$ ), $Zn^{2 +}$ ( $d^{10}$ )
1	$3 \approx 1.73$	$Ti^{3 +}$ ( $d^{1}$ ), $Cu^{2 +}$ ( $d^{9}$ )
2	$8 \approx 2.83$	$V^{3 +}$ ( $d^{2}$ ), $Ni^{2 +}$ ( $d^{8}$ )
3	$15 \approx 3.87$	$Cr^{3 +}$ ( $d^{3}$ ), $Co^{2 +}$
4	$24 \approx 4.90$	$Cr^{2 +}$ ( $d^{4}$ ), $Mn^{3 +}$
5	$35 \approx 5.92$	$Mn^{2 +}$ ( $d^{5}$ ), $Fe^{3 +}$

Worked examples

Common mistakes (Steel-man + fix)

Flashcards

Spin-only magnetic moment formula kya hai?

μ = n (n + 2)

BM, jahan

n

= unpaired electrons ki number.

Paramagnetism ke liye kya chahiye?

Kam se kam ek unpaired electron (

n \geq 1

3d complexes mein orbital contribution kyun often ignore hoti hai?

Yeh ligand field dwara "quenched" ho jaati hai, isliye

L \to 0

aur sirf spin bachta hai.

Mn^{2 +}

(

d^{5}

) ka magnetic moment kya hai?

35 = 5.92

BM.

Sc^{3 +}

Zn^{2 +}

ka magnetic moment kya hai?

0 BM — diamagnetic (

d^{0}

d^{10}

Ek complex ka

μ = 2.83

BM hai. Kitne unpaired electrons hain?

n = 2

(kyunki

8 = 2.83

4 S (S + 1)

μ = n (n + 2)

derive karo.

S = n /2

rakho:

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

Transition-metal ion banate waqt pehle kaun se electrons jaate hain?

4s electrons (3d se pehle).

μ

d^{5}

par peak kyun karta hai?

d^{5}

mein maximum 5 unpaired electrons hote hain (half-filled, Hund); uske baad electrons pairing shuru karte hain.

Same

Fe^{2 +}

, do alag

μ

values kyun?

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

Recall Feynman: 12-saal ke bache ko samjhao

Socho har electron ek tiny spinning top hai jo ek baby magnet ki tarah kaam karta hai. Jab do electrons ek room (orbital) share karte hain to woh opposite taraf spin karte hain, isliye unke magnets cancel ho jaate hain — boring, koi magnetism nahi. Lekin agar ek electron akele apne room mein baitha hai, to uske chhote magnet ka koi partner nahi jo use cancel kare, isliye poora atom ek bade magnet ki taraf pull ho sakta hai. Pull kitna strong hai yeh guess karne ke liye, bas lonely electrons count karo ( $n$ ) aur $n (n + 2)$ karo. Zyada lonely electrons = zyada strong magnet.

Connections

Crystal Field Theory — high-spin vs low-spin explain karta hai → $n$ change hota hai.
Hund's Rule & Electron Configuration — set karta hai ki kitne electrons unpaired rehte hain.
Colour in Transition Metal Complexes — same d-electrons, alag observable (d–d transitions).
Electronic Configuration of Ions — 4s-before-3d removal rule.
Bohr Magneton — the unit $μ_{B} = \frac{e h}{4 π m _{e}}$ .
Lanthanide Magnetism — f-block orbital contribution rakhta hai, $4 S (S + 1) + L (L + 1)$ chahiye.