Why the Same Iron Ion Can Have One or Five Unpaired Electrons

The same iron(III) ion is strongly magnetic in one complex and only weakly magnetic in another, and changing a single ligand can turn a solution from green to violet. Two bonding models explain these observations: valence bond theory, built on hybridisation, and crystal field theory, built on how ligands split the energies of d orbitals.

This lesson covers bonding by valence bond theory and crystal field theory, how crystal field splitting explains colour and magnetism, and how to calculate crystal field stabilisation energy.

Valence bond theory treats the metal as using hybrid orbitals to accept lone pairs from ligands, which predicts shape and magnetism, while crystal field theory treats ligands as point charges that split the metal's d orbitals into sets of different energy.

Valence bond theory:

- The metal ion supplies empty orbitals that hybridise to match the coordination number, and each ligand donates a lone pair into one of them
- $\mathrm{s}{\mathrm{p}}^3$ gives tetrahedral, $\mathrm{ds}{\mathrm{p}}^2$ square planar, and ${\mathrm{d}}^2\mathrm{s}{\mathrm{p}}^3$ or $\mathrm{s}{\mathrm{p}}^3{\mathrm{d}}^2$ octahedral complexes

Worked example 1 — cobalt(III). $\mathrm{C}{\mathrm{o}}^{3+}$ is $3d^6$. In $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$, ammonia pairs the d electrons into three 3d orbitals, leaving two empty for ${\mathrm{d}}^2\mathrm{s}{\mathrm{p}}^3$ hybridisation: an inner orbital, diamagnetic complex. In $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$, fluoride does not force pairing, so outer 4d orbitals are used for $\mathrm{s}{\mathrm{p}}^3{\mathrm{d}}^2$: an outer orbital complex with four unpaired electrons.

Worked example 2 — nickel(II). $\mathrm{N}{\mathrm{i}}^{2+}$ is $3d^8$. Cyanide pairs its electrons, freeing one 3d orbital, so $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ is $\mathrm{ds}{\mathrm{p}}^2$, square planar and diamagnetic; $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ is $\mathrm{s}{\mathrm{p}}^3$, tetrahedral, with two unpaired electrons.

Limitations. Valence bond theory cannot explain colour, gives no energies, and cannot predict inner or outer orbital bonding without magnetic data.

Crystal field theory:

- Ligands act as point charges or dipoles that repel the metal's d electrons
- In an octahedral field, the $e_g$ orbitals, which point at the ligands, rise in energy, while the $t_{2g}$ orbitals, which lie between them, fall; the gap is ${\Delta }_o$
- In a tetrahedral field, the splitting is reversed and smaller: ${\Delta }_t\approx \frac{4}{9}{\Delta }_o$
- The spectrochemical series ranks ligands by the splitting they cause: ${\mathrm{I}}^{-}<\mathrm{B}{\mathrm{r}}^{-}<\mathrm{C}{\mathrm{l}}^{-}<{\mathrm{F}}^{-}<\mathrm{O}{\mathrm{H}}^{-}<{\mathrm{H}}_2\mathrm{O}<\mathrm{N}{\mathrm{H}}_3<\mathrm{en}<\mathrm{C}{\mathrm{N}}^{-}<\mathrm{CO}$

An everyday example. The Hutti gold mines in Karnataka extract gold with cyanide, which binds gold as the very stable complex $[\mathrm{Au}(\mathrm{CN}{)}_2{]}^{-}$ and so pulls it out of the crushed ore.

The substance. The same metal ion can form inner or outer orbital complexes — the ligand, not the metal, decides whether d electrons are forced to pair.

Crystal field theory explains colour as light absorbed when a d electron jumps across the splitting gap, and magnetism as depending on whether that gap is large enough to force electrons to pair — giving low-spin complexes with strong-field ligands and high-spin complexes with weak-field ligands.

Colour:

- A d electron absorbs a photon whose energy matches ${\Delta }_o$ and jumps from $t_{2g}$ to $e_g$
- The colour seen is complementary to the light absorbed, so $[\mathrm{Ti}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$, which absorbs blue-green light, looks violet
- A stronger-field ligand increases ${\Delta }_o$ and shifts absorption to shorter wavelengths: green $[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ turns blue and then violet as ethane-1,2-diamine replaces the water

Magnetism — high spin and low spin:

- If ${\Delta }_o$ is smaller than the pairing energy P, electrons enter $e_g$ before pairing, giving a high-spin complex with more unpaired electrons
- If ${\Delta }_o$ is larger than P, electrons pair in $t_{2g}$ first, giving a low-spin complex

**Worked example — iron(III), $3d^5$:**

- $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$: weak-field fluoride gives high spin, $t_{2g}^3{e}_g^2$, with 5 unpaired electrons and $\mu =\sqrt{35}=5.92$ BM
- $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$: strong-field cyanide gives low spin, $t_{2g}^5$, with 1 unpaired electron and $\mu =\sqrt{3}=1.73$ BM

An everyday example. The blue copper sulphate in Bordeaux mixture, sprayed on grape vines around Nashik, owes its colour to a d-d transition in hydrated copper(II) ions.

The substance. Tetrahedral complexes are almost always high spin — their splitting, only about four-ninths of the octahedral value, is rarely large enough to force pairing.