# Unpaired electrons and magnetic moment

> From a compound's formula to its magnetic moment


## Objectives

- Find the metal's oxidation state in a compound and write its ion's 3d configuration (4s lost first).
- Count unpaired electrons by Hund's rule and classify the compound as paramagnetic or diamagnetic.
- Arrange compounds in ascending order of magnetic moment.

## 1. Magnetic properties and catalytic activity




## 2. Same metal, two chlorides

Predict: which has the larger magnetic moment?

- $\ce{FeCl3}$ ✓
- $\ce{FeCl2}$: iron keeps more electrons
- They are equal: both contain iron, $3d^6$

$\ce{FeCl3}$ has the larger magnetic moment. $\ce{Fe^{3+}}$ is $3d^5$ with 5 unpaired electrons, while $\ce{Fe^{2+}}$ is $3d^6$ with 4, and the magnetic moment increases with the number of unpaired electrons.

[ar-EG] الـ $\ce{FeCl3}$ هو اللي الـ magnetic moment بتاعه أكبر. الـ $\ce{Fe^{3+}}$ هو $3d^5$ وفيه 5 unpaired electrons، إنما الـ $\ce{Fe^{2+}}$ هو $3d^6$ وفيه 4، والـ magnetic moment بيزيد بزيادة عدد الـ unpaired electrons.
## 3. Count on the ion

In a compound the metal is an **ion**. So count the unpaired electrons of the ion, not of the atom. Pair the electrons first: $d^6$ is three pairs, so 0 unpaired.


## 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 $[\text{Ar}]\ 4s^2\ 3d^6$. It loses the two 4s electrons first, then one 3d: $\ce{Fe^3+}$ is $[\text{Ar}]\ 3d^5$. Step 3: boxes and count $\ce{Fe^3+}$: $3d^5$ One electron in each box: **5 unpaired**, so $\ce{FeCl3}$ is **paramagnetic**. For $\ce{FeCl3}$, use the iron atom: $3d^6$, 4 unpaired.


## 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 $[\text{Ar}]\ 4s^1\ 3d^5$. It loses 4s¹, then two 3d: $\ce{Cr^3+}$ is $[\text{Ar}]\ 3d^3$. $\ce{Cr^3+}$ has ___ unpaired electrons, so $\ce{Cr2O3}$ is ___.


## 6. One metal, four ions

Vanadium forms $\ce{VO}$, $\ce{V2O3}$, $\ce{VO2}$ and $\ce{V2O5}$. Move through its oxidation states. oxidation state of vanadium VO: V²⁺ is [Ar] 3d³ V₂O₃: V³⁺ is [Ar] 3d² VO₂: V⁴⁺ is [Ar] 3d¹ V₂O₅: V⁵⁺ is [Ar] 3d⁰ Every transition-metal compound is paramagnetic.


## 7. Para or dia?

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


## 8. Ascending order

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


## 9. Summary

Oxidation state → the ion (4s lost first) → 3d boxes by Hund's rule → count unpaired electrons. No unpaired electrons ($3d^0$, $3d^{10}$): diamagnetic. Any unpaired: paramagnetic. More unpaired electrons, larger magnetic moment. Arrange by magnetic moment: TiO₂, Cr₂O₃, FeCl₃ (book p.14) $\ce{TiO2}$ ($\ce{Ti^4+}$, $3d^0$, 0) < $\ce{Cr2O3}$ ($\ce{Cr^3+}$, $3d^3$, 3) < $\ce{FeCl3}$ ($\ce{Fe^3+}$, $3d^5$, 5). Write the ion, its 3d configuration and its unpaired electrons for each. Back to the hook: $\ce{FeCl3}$ (5 unpaired) beats $\ce{FeCl2}$ (4). Can you explain why the iron atom gives the wrong count? In $\ce{FeCl3}$ iron is the ion $\ce{Fe^3+}$, not the atom. It has lost the two 4s electrons and one 3d electron, so its 3d is $3d^5$ (5 unpaired), not the atom's $3d^6$ (4 unpaired). Fe atom: $3d^6$, 4 unpaired $\ce{Fe^3+}$: $3d^5$, 5 unpaired



Chapter: Transition Elements (https://yaani.ai/en/learn/chemistry/chem-s3-en.ch1)
Source: https://yaani.ai/en/lesson/chem-s3-en.lesson-te.magnetic-moment-from-config