---
title: "VBT: magnetism of [Fe(CN)₆]⁴⁻/³⁻ and [Ni(CO)₄]"
subject: "Engineering Chemistry"
chapter: 6
key: "semester_2/engineering_chemistry:vbt-facts-fecn6-nico4"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/vbt-facts-fecn6-nico4
site: BCT Engineering
---

# VBT: magnetism of [Fe(CN)₆]⁴⁻/³⁻ and [Ni(CO)₄]

Engineering Chemistry · Chapter 6: Coordination Complexes

- Asked in: 2067 Chaitra, 5 marks
- Video: [[Ni(CO)4] is tetrahedral and [Ni(CN)4]2-  is Square Planar ?Why coordinate compound VBT](https://www.youtube.com/watch?v=La_CXLjcsaw) — Vicky Science Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/vbt-facts-fecn6-nico4
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-6

## Question

How does the valence bond theory account for the following facts?
a) [Fe(CN)₆]⁴⁻ ion is diamagnetic but [Fe(CN)₆]³⁻ is paramagnetic
b) [Ni(CO)₄]⁰ is diamagnetic and tetrahedral.

## Short answer

[Fe(CN)₆]⁴⁻ has Fe²⁺ (d⁶), CN⁻ causes pairing → diamagnetic. [Fe(CN)₆]³⁻ has Fe³⁺ (d⁵) with one unpaired electron → paramagnetic. [Ni(CO)₄] has Ni(0) 3d¹⁰, sp³ hybridization giving tetrahedral shape, all electrons paired → diamagnetic.

## Answer

#### a) [Fe(CN)₆]⁴⁻ diamagnetic but [Fe(CN)₆]³⁻ paramagnetic
- [Fe(CN)₆]⁴⁻ contains Fe²⁺ (3d⁶). CN⁻ is strong-field, so all six electrons pair in three 3d orbitals; d²sp³ hybridisation. No unpaired electron: **diamagnetic**.
- [Fe(CN)₆]³⁻ contains Fe³⁺ (3d⁵). The electrons pair up as far as possible in the three lower orbitals, but five electrons cannot all pair. One unpaired electron remains: **paramagnetic** (1.73 BM), also d²sp³.
```text
Fe²⁺: [↑↓][↑↓][↑↓][  ][  ]     Fe³⁺: [↑↓][↑↓][↑ ][  ][  ]
```

#### b) [Ni(CO)₄]⁰ diamagnetic and tetrahedral
Ni(0) is 3d⁸ 4s². The strong-field ligand CO causes the 4s electrons to shift into 3d, giving 3d¹⁰ (all paired). The empty 4s and three 4p orbitals form sp³ hybrids, giving a **tetrahedral** shape. No unpaired electron, hence **diamagnetic**.
```text
3d: [↑↓][↑↓][↑↓][↑↓][↑↓]   sp³ (4s + 3×4p)
```

## Check yourself

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

1. Why is [Fe(CN)6]4- diamagnetic?
   A. Fe2+ has 5 unpaired electrons
   B. Fe2+ (3d6) pairs all six electrons in three 3d orbitals with strong CN-
   C. CN- is weak-field
   D. Fe2+ has 3d10
   Answer: B. d2sp3 with all electrons paired means no unpaired electron. (Key point: Fe2+ diamagnetic)

2. Why is [Fe(CN)6]3- paramagnetic?
   A. Fe3+ has 3d6
   B. It uses sp3d2
   C. CN- is weak-field here
   D. Fe3+ (3d5) cannot pair all five electrons in three orbitals, leaving one unpaired (1.73 BM)
   Answer: D. Five electrons in three lower orbitals leave one unpaired. (Key point: Fe3+ paramagnetic)

3. What gives [Ni(CO)4] its diamagnetism?
   A. 4s electrons shift to 3d giving fully paired 3d10
   B. Ni has 3d8 with two unpaired
   C. CO is weak-field
   D. Ni is Ni2+
   Answer: A. 3d10 has no unpaired electron. (Key point: Ni(CO)4 electrons)

4. Why is [Ni(CO)4] tetrahedral?
   A. d2sp3 hybridisation
   B. dsp2 hybridisation
   C. Empty 4s and three 4p form sp3 hybrids
   D. sp3d2 hybridisation
   Answer: C. sp3 hybridisation gives a tetrahedral arrangement. (Key point: Ni(CO)4 geometry)

## Related in Engineering Chemistry

- [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
- [VBT for [Fe(CN)₆]³⁻ and [FeF₆]³⁻ magnetism](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-fecn6-fef6) — 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 [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 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×
- [Tetrahedral and square planar complexes; [Ni(NH₃)₄]²⁺ VBT](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atetra-square-vbt-ni-nh3) — Engineering Chemistry, chapter 6