Unpaired electrons and magnetic moment

1. Magnetic properties and catalytic activity

2. Same metal, two chlorides

FeCl2 FeCl X_{2}

and

FeCl3 FeCl X_{3}

both contain iron. Are they equally magnetic?

Pick the right answer

Predict: which has the larger magnetic moment?

A. They are equal: both contain iron, $3d⁶ 3 d^{6}$
B. $FeCl2 FeCl X_{2}$ : iron keeps more electrons
C. $FeCl3 FeCl X_{3}$

Explanation

FeCl3 FeCl X_{3}

has the larger magnetic moment.

Fe³⁺ Fe X^{3 +}

is

3d⁵ 3 d^{5}

with 5 unpaired electrons, while

Fe²⁺ Fe X^{2 +}

is

3d⁶ 3 d^{6}

with 4, and the magnetic moment increases with the number of unpaired electrons.

3. Count on the ion

Compound → Oxidation state of the metal
Oxidation state of the metal → Ion's 3d configuration (4s lost first)
Ion's 3d configuration (4s lost first) → 3d boxes by Hund's rule
3d boxes by Hund's rule → Count unpaired electrons
Count unpaired electrons → Classify and order

In a compound the metal is an ion. So count the unpaired electrons of the ion, not of the atom.

4. Worked: FeCl₃

Book, Q7.9: iron(III) chloride is paramagnetic. Open each step and read its reason.

Step 1: oxidation state

Cl is −1. Three Cl give −3, and the compound is neutral, so Fe is +3.

Step 2: the ion's configuration

Fe is

[Ar]\ 4s²\ 3d⁶

. It loses the two 4s electrons first, then one 3d:

Fe³⁺ Fe X^{3 +}

is

[Ar]\ 3d⁵

.

Step 3: boxes and count

↑↑↑↑↑

3d

Fe³⁺ Fe X^{3 +}

:

3d⁵ 3 d^{5}

One electron in each box: 5 unpaired, so

FeCl3 FeCl X_{3}

is paramagnetic.

5. Your part: Cr₂O₃

A compound from the book's exercise. Steps 1 and 2 are done. You finish steps 3 and 4.

O is −2. Three O give −6, shared by two Cr, so Cr is +3.
Cr is $[Ar]\ 4s¹\ 3d⁵$ . It loses 4s¹, then two 3d: $Cr³⁺ Cr X^{3 +}$ is $[Ar]\ 3d³$ .

Fill in the blanks

Cr³⁺ Cr X^{3 +}

has _____ unpaired electrons, so

Cr2O3 Cr X_{2} O X_{3}

is _____.

Explanation

Cr³⁺ Cr X^{3 +}

:

3d³ 3 d^{3}

=

[][][][\ ][\ ]

. 3 unpaired electrons, so paramagnetic.

Need a hint?

Draw five boxes for 3d.
Place the ion's 3d electrons one per box, before any pairing.
Any unpaired electron makes the substance paramagnetic.

6. One metal, four ions

Vanadium forms

VO VO

,

V2O3 V X_{2} O X_{3}

,

VO2 VO X_{2}

and

V2O5 V X_{2} O X_{5}

. Move through its oxidation states.

oxidation state of vanadium 2

↑↑↑

3d

VO: V²⁺ is [Ar] 3d³

7. Para or dia?

Sort into groups

Sort the compounds. Find each metal's oxidation state first.

$TiCl4 TiCl X_{4}$ $CuCl CuCl$ $Sc2O3 Sc X_{2} O X_{3}$ $CuSO4 CuSO X_{4}$ NiO

8. Ascending order

Put in order

Arrange in ascending order of magnetic moment (smallest first).

$TiCl3 TiCl X_{3}$
$NiSO4 NiSO X_{4}$
$CoSO4 CoSO X_{4}$
$MnCl2 MnCl X_{2}$

Need a hint?

For each compound, find the metal's oxidation state from the other part's charge.
Remove the 4s electrons first, then 3d electrons for higher states.
Fill the 3d boxes singly first; order by the number of boxes with one electron.

9. Summary

Oxidation state → the ion (4s lost first) → 3d boxes by Hund's rule → count unpaired electrons.
No unpaired electrons ( $3d⁰ 3 d^{0}$ , $3d¹⁰ 3 d^{10}$ ): diamagnetic. Any unpaired: paramagnetic.
More unpaired electrons, larger magnetic moment.

Back to the hook:

FeCl3 FeCl X_{3}

(5 unpaired) beats

FeCl2 FeCl X_{2}

(4). Can you explain why the iron atom gives the wrong count?

Reveal

In

FeCl3 FeCl X_{3}

iron is the ion

Fe³⁺ Fe X^{3 +}

, not the atom. It has lost the two 4s electrons and one 3d electron, so its 3d is

3d⁵ 3 d^{5}

(5 unpaired), not the atom's

3d⁶ 3 d^{6}

(4 unpaired).

↑↓↑↑↑↑

3d

Fe atom:

3d⁶ 3 d^{6}

, 4 unpaired