Question Restatement and Final Answer

Ferrous sulfate (FeSO₄) reacts with potassium hexacyanochromate(III) to give a
brick red coordination complex containing Fe and Cr linked through cyanide ligands in a
Prussian-blue–type lattice. In this brick red complex, iron exists as Fe(III) in a low-spin d⁵
configuration with exactly 1 unpaired electron.

Therefore, the number of unpaired electrons on Fe in the red complex is
1.

Stepwise Conceptual Explanation

Oxidation State and Nature of the Complex

Potassium hexacyanochromate(III) contains the anion [Cr(CN)₆]³⁻, where chromium is in the
+3 oxidation state with a d³ configuration in an octahedral cyanide field. Cyanide is a strong-field,
π-acceptor ligand that usually produces low-spin complexes in the first-row transition series.

When an aqueous Fe(II) solution (from ferrous sulfate) is added to this hexacyanochromate(III), a mixed Fe–Cr
cyanide network complex is formed, which is reported in inorganic and coordination-chemistry question banks as a
brick-red compound that later turns dark green on heating.

Even though different sources focus on identifying the exact stoichiometry of the green product (such as
Fe₄[Cr(CN)₆]₃ or related frameworks), the key point for this question is the
electronic situation of Fe in the initial brick red complex.

Oxidation State and Electron Configuration of Fe

Starting species:

Ferrous sulfate provides Fe(II), i.e. Fe²⁺, which in the gas phase has the configuration
[Ar] 3d⁶.

Redox change:

In cyanide-bridged Prussian-blue–type systems, Fe(II) often gets oxidized to Fe(III) in the final solid-state
complex while the polynuclear cyanide framework forms. This is consistent with analogous iron hexacyanide
chemistry, where Fe(III) cyanide frameworks are low-spin in the presence of CN⁻.

Electronic configuration of Fe³⁺:

Atomic Fe: [Ar] 3d⁶ 4s²
Fe³⁺: [Ar] 3d⁵

Because cyanide is a strong-field ligand, Fe(III) in an octahedral CN⁻ field tends to form a low-spin d⁵ complex.

Crystal Field Splitting and Unpaired Electrons

For an octahedral complex with a strong-field ligand like CN⁻, the five d-orbitals split into:

• t_2g (lower energy, set of three orbitals)
• e_g (higher energy, set of two orbitals)

In a low-spin d⁵ system:

• All five electrons fill the lower-energy t_2g set first, with pairing enforced by strong-field splitting.
• Population: t_2g⁵ e_g⁰.

This means there are 2 paired electron sets (4 electrons) and 1 unpaired electron in the t_2g set.

Thus, the number of unpaired electrons = 1 for low-spin Fe³⁺ in a cyanide octahedral environment.
This matches the expected behaviour of Fe(III) in hexacyanide-type complexes.

Discussion of Typical Options and Why 1 is Correct

In multiple-choice format, the options for “number of unpaired electrons on Fe” are usually:

• 0 unpaired electrons
• 1 unpaired electron
• 3 unpaired electrons
• 5 unpaired electrons

Here is how each possibility compares to the actual situation:

Option: 0 Unpaired Electrons

A value of 0 unpaired electrons would correspond to a low-spin d⁶ metal centre in an octahedral
strong-field complex (configuration t_2g⁶ e_g⁰), as seen in
[Fe(CN)₆]⁴⁻ with Fe(II) and CN⁻.

However, in the brick red Fe–Cr cyanide complex obtained from FeSO₄ + K₃[Cr(CN)₆],
Fe is effectively present as Fe(III) rather than Fe(II), giving a d⁵ rather than d⁶ configuration.

Therefore, 0 unpaired electrons is incorrect for this system.

Option: 3 Unpaired Electrons

Three unpaired electrons typically correspond to:

• High-spin d⁵: t_2g³ e_g² (octahedral, weak-field ligands), or
• Low-spin d⁷: t_2g⁶ e_g¹.

With CN⁻ as a strong-field ligand, Fe(III) does not remain high-spin; instead it becomes low-spin,
reducing the number of unpaired electrons.

Hence, 3 unpaired electrons does not match the cyanide environment of Fe in this brick red complex.

Option: 5 Unpaired Electrons

Five unpaired electrons correspond to high-spin d⁵ in an octahedral complex with weak-field ligands
(e.g. water or halides like Cl⁻), with configuration t_2g³ e_g².

CN⁻ is strong-field and significantly increases crystal field splitting, favouring pairing and a low-spin arrangement.

Thus, Fe(III) in a cyanide octahedral field does not keep all 5 electrons unpaired; this makes 5 unpaired electrons incorrect.

Option: 1 Unpaired Electron (Correct)

Low-spin Fe(III) (d⁵) in an octahedral strong-field complex such as a cyanide framework has
t_2g⁵ e_g⁰, resulting in 1 unpaired electron.

This matches:

• The oxidation state situation (Fe(III))
• The ligand field strength (CN⁻, strong-field)
• The behaviour of related hexacyanoferrate and hexacyanochromate frameworks

Therefore, 1 unpaired electron is the only option consistent with coordination chemistry principles and
the ligand field of the brick red complex.

Key Takeaways for Exams

• In cyanide complexes, always check if the metal is in a situation where low-spin is expected; CN⁻ is a strong-field ligand.
• Fe(III), d⁵, in a strong-field octahedral environment (like CN⁻) gives a low-spin d⁵ with 1 unpaired electron.
• For questions involving coloured Fe–CN or Fe–Cr–CN solids (such as Prussian-blue–type or mixed cyanide frameworks), identify:
  • Metal oxidation state
  • d-electron count
  • Ligand field strength
  • Then deduce spin state and number of unpaired electrons