---
title: "VBT for [Fe(CN)₆]³⁻ inner/outer orbital"
subject: "Engineering Chemistry"
chapter: 6
key: "semester_2/engineering_chemistry:vbt-fecn6"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/vbt-fecn6
site: BCT Engineering
---

# VBT for [Fe(CN)₆]³⁻ inner/outer orbital

Engineering Chemistry · Chapter 6: Coordination Complexes

- Asked in: 2074 Bhadra, 3+2 marks
- Video: [Inner and outer-orbital octahedral complexes](https://www.youtube.com/watch?v=hhTWbZL5WmQ) — Chem-Matics (Chemistry+ Mathematics)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/vbt-fecn6
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-6

## Question

How does valence bond theory explain the formation of [Fe(CN)₆]³⁻? Also explain whether this complex is inner or outer orbital and why?

## Short answer

CN- pairs four 3d5 electrons, leaving one unpaired and vacating two 3d orbitals for d2sp3 hybridization. The complex is inner orbital, octahedral, and paramagnetic with one unpaired electron (1.73 BM).

## Answer

Fe (Z = 26); in [Fe(CN)₆]³⁻ the oxidation state of Fe is +3, so Fe³⁺ is 3d⁵.

**Formation:** CN⁻ is a strong-field ligand, so the five 3d electrons are forced to pair up in two orbitals (with one electron left unpaired), vacating two 3d orbitals. Two 3d, one 4s and three 4p orbitals form six **d²sp³** hybrid orbitals, which accept six lone pairs from CN⁻.
```text
Fe³⁺ 3d: [↑↓][↑↓][↑ ][  ][  ]  4s [ ]  4p [ ][ ][ ]
                   \_______ d²sp³ _______/
```
- **Geometry:** octahedral.
- **Inner or outer orbital:** it is an **inner orbital complex**, because the inner (n−1)d orbitals take part in hybridisation, and low spin because electrons are paired.
- **Magnetism:** one unpaired electron, so **paramagnetic**, μ = √3 = 1.73 BM.

## Check yourself

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

1. What is the d configuration of Fe3+ (Z = 26)?
   A. 3d5
   B. 3d6
   C. 3d3
   D. 3d10
   Answer: A. The correct option is: 3d5. This follows from the answer on fe3+ configuration. (Key point: Fe3+ configuration)

2. Why is [Fe(CN)6]3- an inner orbital complex?
   A. outer nd orbitals are used
   B. inner (n-1)d orbitals take part in hybridisation
   C. no hybridisation
   D. 4p only
   Answer: B. The correct option is: inner (n-1)d orbitals take part in hybridisation. This follows from the answer on inner orbital. (Key point: Inner orbital)

3. How many unpaired electrons does [Fe(CN)6]3- have?
   A. 0
   B. 3
   C. 1
   D. 5
   Answer: C. Paramagnetic. (Key point: Unpaired)

4. What is the spin-only magnetic moment of [Fe(CN)6]3-?
   A. 0
   B. 5.92 BM
   C. 2.83 BM
   D. 1.73 BM
   Answer: D. The correct option is: 1.73 BM. This follows from the answer on moment. (Key point: Moment)

## Related in Engineering Chemistry

- [Inner/outer orbital complexes; VBT of [Fe(CN)₆]³⁻](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ainner-outer-vbt-fecn6) — Engineering Chemistry, chapter 6
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