---
title: "Paramagnetism of transition metals"
subject: "Engineering Chemistry"
chapter: 5
key: "semester_2/engineering_chemistry:tm-paramagnetism"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tm-paramagnetism
site: BCT Engineering
---

# Paramagnetism of transition metals

Engineering Chemistry · Chapter 5: 3-d Transition Elements and their Applications

- Asked in: 2073 Magh, 2.5 marks; 2071 Bhadra, 2.5 marks; 2070 Magh, 2.5 marks (asked 3 times)
- Video: [Magnetic properties of transition metals | The d-block elements | Chemistry | Khan Academy](https://www.youtube.com/watch?v=hPohHCeBBeA) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tm-paramagnetism
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-5

## Question

Why do the transition metals show paramagnetism?

## Short answer

Paramagnetism is due to unpaired electrons with net spin magnetic moment. Transition metal ions have partly filled d orbitals, so most contain unpaired electrons and are paramagnetic. The spin-only magnetic moment is μ = √(n(n+2)) BM, where n is the number of unpaired electrons.

## Answer

**Paramagnetism** is the property of being weakly attracted by a magnetic field. It is due to the presence of **unpaired electrons**: each unpaired electron has a net spin magnetic moment (and orbital contribution) that aligns with the applied field. Substances with all electrons paired are diamagnetic (repelled weakly).

Transition metal ions have partly filled d orbitals, so most have unpaired electrons and are paramagnetic. The more unpaired electrons, the greater the paramagnetism (spin-only value shown below).

Spin-only magnetic moment: μ = √(n(n+2)) BM, where n = number of unpaired electrons.

| Ion (d-configuration) | Unpaired e⁻ (n) | μ (BM) |
|---|---|---|
| Sc²⁺/Ti³⁺ (d¹) | 1 | 1.73 |
| Ti²⁺ (d²) | 2 | 2.83 |
| V²⁺ (d³) | 3 | 3.87 |
| Cr²⁺ (d⁴) | 4 | 4.90 |
| Mn²⁺ (d⁵) | 5 | 5.92 |
| Fe²⁺ (d⁶) | 4 | 4.90 |
| Co²⁺ (d⁷) | 3 | 3.87 |
| Ni²⁺ (d⁸) | 2 | 2.83 |
| Cu²⁺ (d⁹) | 1 | 1.73 |
| Zn²⁺ (d¹⁰) | 0 | 0 (diamagnetic) |

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Paramagnetism is due to
   A. Unpaired electrons
   B. Paired electrons
   C. Neutrons
   D. Filled shells only
   Answer: A. Unpaired spins have a net moment that aligns with the field. (Key point: Cause)

2. Substances with all electrons paired are
   A. Strongly paramagnetic
   B. Diamagnetic (weakly repelled)
   C. Ferromagnetic
   D. Superconducting
   Answer: B. No unpaired electrons means no net moment. (Key point: Diamagnetism)

3. What is the spin-only magnetic moment?
   A. n(n+2) BM
   B. sqrt(n) BM
   C. sqrt(n(n+2)) BM
   D. 2n BM
   Answer: C. n is the number of unpaired electrons. (Key point: Spin-only formula)

4. Mn2+ (d5) has how many unpaired electrons and approx. moment?
   A. 3 and 3.87 BM
   B. 4 and 4.90 BM
   C. 1 and 1.73 BM
   D. 5 and 5.92 BM
   Answer: D. sqrt(5 x 7) = 5.92 BM. (Key point: Calculation)

## Related in Engineering Chemistry

- [Transition elements; configuration; paramagnetism](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atm-config-paramagnetism) — Engineering Chemistry, chapter 5
- [Zn and Cd not typical transition metals](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Azn-cd-not-typical) — Engineering Chemistry, chapter 5
- [d-block elements; 3d configuration; non-typical metals](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Adblock-config-not-typical) — Engineering Chemistry, chapter 5