Magnetic Properties of Coordination Compounds

The magnetic moment of a coordination compound can be measured experimentally through magnetic susceptibility measurements. Since magnetic moment depends directly on the number of unpaired electrons present, these measurements provide a window into the electronic structure — and hence the probable geometry and bonding — of a metal complex.

Where the simple picture works

For metal ions carrying up to three electrons in the $d$ orbitals — such as $\text{Ti}^{3+}$ ($d^1$), $\text{V}^{3+}$ ($d^2$), or $\text{Cr}^{3+}$ ($d^3$) — Hund's rule allows these electrons to occupy three separate $d$ orbitals singly, leaving two $d$ orbitals empty and available for octahedral hybridisation with the $4s$ and $4p$ orbitals. Because no rearrangement of electrons is forced, the free ion and its coordination entities show essentially the same magnetic behaviour.

Where complications appear

Once more than three $3d$ electrons are present, a vacant pair of $d$ orbitals is not automatically available — Hund's rule keeps electrons unpaired across all five $d$ orbitals as far as possible, so freeing two orbitals for hybridisation requires forcing some electrons to pair up. This is exactly the situation for:

• $d^4$ ions, e.g. $\text{Cr}^{2+}$, $\text{Mn}^{3+}$
• $d^5$ ions, e.g. $\text{Mn}^{2+}$, $\text{Fe}^{3+}$
• $d^6$ ions, e.g. $\text{Fe}^{2+}$, $\text{Co}^{3+}$

For these, forming the vacant pair of $d$ orbitals needed for octahedral ($d^2sp^3$) hybridisation means pairing up $3d$ electrons — leaving, respectively, two, one, and zero unpaired electrons if maximum pairing is assumed.

The experimental anomaly

Magnetic data agree with this "maximum spin pairing" picture in a great many cases — particularly for $d^6$ coordination entities. But for $d^4$ and $d^5$ species, real measurements reveal a genuine split in behaviour depending on the ligand:

• $[\text{Mn(CN)}_6]^{3-}$ has a magnetic moment corresponding to two unpaired electrons, while $[\text{MnCl}_6]^{3-}$ is paramagnetic with four unpaired electrons.
• $[\text{Fe(CN)}_6]^{3-}$ has a magnetic moment corresponding to a single unpaired electron, while $[\text{FeF}_6]^{3-}$ has a paramagnetic moment of five unpaired electrons.
• $[\text{CoF}_6]^{3-}$ is paramagnetic with four unpaired electrons, while $[\text{Co(C}_2\text{O}_4)_3]^{3-}$ is diamagnetic.

How Valence Bond Theory rationalises this

This split is explained within VBT by recognising two distinct hybridisation routes for octahedral entities, exactly as introduced for cobalt in the previous section:

• $[\text{Mn(CN)}_6]^{3-}$, $[\text{Fe(CN)}_6]^{3-}$, and $[\text{Co(C}_2\text{O}_4)_3]^{3-}$ are inner orbital complexes, using $d^2sp^3$ hybridisation (the strong-field ligand forces electron pairing within the $3d$ set, freeing orbitals for hybridisation). The first two of these remain paramagnetic (with fewer unpaired electrons than the free ion), while the third is fully diamagnetic. …