---
title: "Structure and magnetism of [NiCl₄]²⁻ and [Ni(CN)₄]²⁻"
subject: "Engineering Chemistry"
chapter: 6
key: "semester_2/engineering_chemistry:nicl4-nicn4"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/nicl4-nicn4
site: BCT Engineering
---

# Structure and magnetism of [NiCl₄]²⁻ and [Ni(CN)₄]²⁻

Engineering Chemistry · Chapter 6: Coordination Complexes

- Asked in: 2072 Asoj, 2.5+2.5 marks; 2069 Poush, 2.5+2.5 marks (asked 2 times)
- Video: [Ni(CO)4, [Ni(CN)4]2-, [NiCl4]2-Structure-Hybridization-diamagnetic-VBT-IUPAC Name - IIT JEE NEET](https://www.youtube.com/watch?v=r_C4yyTUSjM) — AdiChemistry Aditya Vardhan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/nicl4-nicn4
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-6

## Question

How would you account for the difference in structures and magnetic properties between [NiCl₄]²⁻ and [Ni(CN)₄]²⁻?

## Short answer

[NiCl4]2- has tetrahedral geometry with sp3 hybridization and two unpaired electrons (paramagnetic). [Ni(CN)4]2- is square planar with dsp2 hybridization and all electrons paired (diamagnetic). The difference arises from the weak field Cl- versus strong field CN- ligands.

## Answer

Ni²⁺ has the configuration 3d⁸. The difference comes from the ligand strength.

#### [NiCl₄]²⁻
Cl⁻ is a weak-field ligand, so the 3d electrons are not paired. Ni²⁺ uses one 4s and three 4p orbitals (sp³ hybridisation).
```text
3d: [↑↓][↑↓][↑↓][↑ ][↑ ]   4s 4p: [ ][ ][ ][ ] sp³
```
- Geometry: **tetrahedral**.
- Two unpaired electrons: **paramagnetic**, μ = √8 = 2.83 BM.

#### [Ni(CN)₄]²⁻
CN⁻ is a strong-field ligand, so the two unpaired 3d electrons pair up and one 3d orbital is vacated. dsp² hybridisation (one 3d, one 4s, two 4p).
```text
3d: [↑↓][↑↓][↑↓][↑↓][  ]   dsp² (4 empty orbitals) 
```
- Geometry: **square planar**.
- No unpaired electrons: **diamagnetic**.

Strong-field CN⁻ gives a square planar diamagnetic complex; weak-field Cl⁻ gives a tetrahedral paramagnetic one.

## Check yourself

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

1. What is the d configuration of Ni2+?
   A. 3d10
   B. 3d8
   C. 3d6
   D. 3d9
   Answer: B. Ni2+ is 3d8. (Key point: Configuration)

2. [NiCl4]2- is tetrahedral and paramagnetic because Cl- is
   A. strong-field, pairing electrons
   B. a neutral ligand giving dsp2
   C. weak-field, so 3d electrons stay unpaired with sp3
   D. a pi-acceptor
   Answer: C. Two unpaired electrons remain; mu = 2.83 BM. (Key point: Cl- ligand)

3. [Ni(CN)4]2- is square planar and diamagnetic because CN- is
   A. weak-field with sp3
   B. weak-field with dsp2
   C. strong-field with sp3d2
   D. strong-field, pairing 3d electrons with dsp2
   Answer: D. One 3d orbital is vacated for dsp2. (Key point: CN- ligand)

## Related in Engineering Chemistry

- [VBT: magnetism of [Fe(CN)₆]⁴⁻/³⁻ and [Ni(CO)₄]](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-facts-fecn6-nico4) — Engineering Chemistry, chapter 6
- [VBT for [Ni(CN)₄]²⁻ and its magnetism](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-nicn4) — Engineering Chemistry, chapter 6
- [VBT: [Ni(CN)₄]²⁻ and [Ni(CO)₄]](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-nicn4-nico4) — Engineering Chemistry, chapter 6
- [VBT for [Fe(CN)₆]⁴⁻: geometry, magnetism](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-fecn6-4minus) — Engineering Chemistry, chapter 6, asked 2×
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- [VBT formation of [Ni(CO)₄] and [Fe(CN)₆]⁴⁻](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-nico4-fecn6-4) — Engineering Chemistry, chapter 6

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