---
title: "Unpaired electrons and paramagnetism; [NiCl₄]²⁻, [Co(NH₃)₆]³⁺"
subject: "Engineering Chemistry"
chapter: 6
key: "semester_2/engineering_chemistry:paramagnetism-unpaired-ni-co"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/paramagnetism-unpaired-ni-co
site: BCT Engineering
---

# Unpaired electrons and paramagnetism; [NiCl₄]²⁻, [Co(NH₃)₆]³⁺

Engineering Chemistry · Chapter 6: Coordination Complexes

- Asked in: 2068 Magh, 2+3 marks
- Video: [[co(NH3)6]3- || VBT](https://www.youtube.com/watch?v=OYVzh4ugRoI) — ASN CHEMISTRY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/paramagnetism-unpaired-ni-co
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-6

## Question

Why does the presence of unpaired electron make a substance paramagnetic in nature? Show geometry and magnetic properties of [NiCl₄]²⁻ and [Co(NH₃)₆]³⁺.

## Short answer

Unpaired electrons produce a net magnetic moment because opposite spins do not cancel. [NiCl₄]²⁻ is tetrahedral (sp³), 2 unpaired electrons, paramagnetic. [Co(NH₃)₆]³⁺ is octahedral (d²sp³), no unpaired electrons, diamagnetic.

## Answer

#### Unpaired electrons and paramagnetism
An electron has spin and so behaves like a tiny magnet. In a pair, the two opposite spins cancel each other's magnetic effect. An unpaired electron has no partner to cancel it, so the atom has a net magnetic moment and is attracted into an external magnetic field. Such a substance is paramagnetic. Magnetic moment μ = √(n(n+2)) BM, where *n* = number of unpaired electrons.

#### [NiCl₄]²⁻
Ni²⁺ is 3d⁸; Cl⁻ is a weak-field ligand, so no pairing. sp³ hybridisation.
```text
3d: [↑↓][↑↓][↑↓][↑ ][↑ ]  4s 4p → sp³
```
Geometry: **tetrahedral**. Two unpaired electrons, so **paramagnetic** (μ = 2.83 BM).

#### [Co(NH₃)₆]³⁺
Co³⁺ is 3d⁶; NH₃ is strong-field and forces pairing, so d²sp³.
```text
3d: [↑↓][↑↓][↑↓][  ][  ]  4s 4p → d²sp³
```
Geometry: **octahedral**. No unpaired electron, so **diamagnetic**.

## Check yourself

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

1. Why does an unpaired electron make a substance paramagnetic?
   A. Paired spins attract each other
   B. Unpaired electrons are charged
   C. Its spin has no partner to cancel it, giving a net magnetic moment attracted into a magnetic field
   D. Unpaired electrons repel a magnetic field
   Answer: C. In a pair, opposite spins cancel; an unpaired one leaves a net moment. (Key point: Why paramagnetic)

2. Which formula gives the magnetic moment?
   A. μ = sqrt(n(n+2)) BM, n = unpaired electrons
   B. μ = n(n+2) BM
   C. μ = sqrt(n) BM
   D. μ = n^2 BM
   Answer: A. The spin-only formula uses the number of unpaired electrons n. (Key point: Magnetic moment formula)

3. What are the geometry and magnetism of [NiCl4]2-?
   A. Square planar, diamagnetic
   B. Octahedral, paramagnetic
   C. Tetrahedral, diamagnetic
   D. Tetrahedral (sp3), paramagnetic with 2 unpaired electrons
   Answer: D. Cl- is weak-field so no pairing; Ni2+ 3d8 keeps 2 unpaired electrons, μ = 2.83 BM. (Key point: [NiCl4]2-)

4. What are the hybridisation and magnetism of [Co(NH3)6]3+?
   A. sp3d2, octahedral, paramagnetic
   B. d2sp3, octahedral, diamagnetic
   C. sp3, tetrahedral, diamagnetic
   D. d2sp3, octahedral, paramagnetic
   Answer: B. NH3 is strong-field, so Co3+ 3d6 pairs up in three d orbitals, leaving none unpaired. (Key point: [Co(NH3)6]3+)

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