Magnetic properties

Diamagnetism vs paramagnetism. A substance is diamagnetic if every electron in it is paired; it is weakly repelled by an external magnetic field, because paired electron spins cancel each other's tiny magnetic effect. A substance is paramagnetic if it has one or more unpaired electrons; it is weakly attracted into an external field, because each unpaired electron behaves like a miniature bar magnet (its spin generates a magnetic moment) and these tiny magnets can partially align with the applied field.

Why transition metals are usually paramagnetic. Most transition-metal ions have partially filled d-orbitals, and a partially filled shell almost always means some electrons sit alone in an orbital rather than paired up. As you move across a d-block series the number of unpaired electrons — and hence the paramagnetism — rises, peaks around the half-filled $d^5$ configuration, and then falls again as pairing resumes on the way to a fully filled, diamagnetic $d^{10}$.

From configuration to magnetic moment. To find how many electrons are unpaired: write the ion's electronic configuration (for a cation, remove $ns$ electrons before $(n-1)d$ electrons), then distribute the d-electrons across the five d-orbitals using Hund's rule — one electron goes into each orbital before any orbital receives a second, paired electron. Counting the singly occupied orbitals gives $n$. The spin-only magnetic moment is then

where: $\mu$ is the magnetic moment, expressed in Bohr magnetons (BM) — the natural unit of magnetic moment for an electron — and $n$ is the number of unpaired electrons in the ion. The formula is called "spin-only" because it accounts only for the magnetism due to electron spin and deliberately ignores any contribution from orbital motion, an approximation that works well for most 3d ions at this level.

The goal is to show exactly where $\mu = \sqrt{n(n+2)}$ comes from, so it is a remembered relationship rather than a magic number, and to see why it depends only on the count $n$ of unpaired electrons and nothing else about the ion.

Step 1. Each unpaired electron carries two independent sources of magnetism at the quantum level: its spin and its orbital motion around the nucleus. For first-row (3d) transition-metal ions the orbital contribution is largely "quenched" — suppressed by the surrounding electric field of neighbouring atoms or ions — so to a good approximation only the spin part survives. This is the physical justification for calling the result "spin-only."

Step 2. Quantum theory gives the spin-only magnetic moment of a system with total spin quantum number $S$ as $\mu = 2\sqrt{S(S+1)}$ BM, where the factor $2$ is the electron's spin g-factor (close to $2.00$).

Step 3. Each unpaired electron contributes a spin of $\tfrac{1}{2}$, and for $n$ unpaired electrons all aligned (as Hund's rule demands within a given set of orbitals), the total spin is $S = \tfrac{n}{2}$.

Step 4. Substitute $S = \tfrac{n}{2}$ into Step 2: $\mu = 2\sqrt{\tfrac{n}{2}\left(\tfrac{n}{2}+1\right)} = 2\sqrt{\dfrac{n(n+2)}{4}} = \sqrt{n(n+2)}$ BM. The factor of $2$ and the $\tfrac{1}{4}$ inside the root cancel neatly, leaving the compact formula used throughout this topic — it depends on nothing but the number of unpaired electrons.

Problem: Calculate the number of unpaired electrons and the spin-only magnetic moment of $\text{Fe}^{3+}$ ($Z_{\text{Fe}} = 26$), and compare it with $\text{Mn}^{2+}$.

Step 1. Fe atom: $[\text{Ar}]3d^6 4s^2$. To form $\text{Fe}^{3+}$, remove 3 electrons — first the two $4s$ electrons, then one $3d$ electron — giving $[\text{Ar}]3d^5$.

Step 2. Distribute the 5 d-electrons among the five d-orbitals by Hund's rule: since there are exactly 5 orbitals and 5 electrons, each orbital receives one electron before any pairing — configuration $\uparrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow$, all unpaired, so $n = 5$.

Step 3. Magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\ \text{BM}$. This is the maximum spin-only value possible for a 3d ion, because $d^5$ is the half-filled configuration with the most unpaired electrons a d-subshell can hold.

Step 4. $\text{Mn}^{2+}$: Mn atom is $[\text{Ar}]3d^5 4s^2$; removing the two $4s$ electrons to form $\text{Mn}^{2+}$ also gives $[\text{Ar}]3d^5$ — the same $n = 5$ and the same $\mu \approx 5.92\ \text{BM}$ as $\text{Fe}^{3+}$, because both ions land on the identical, especially stable half-filled $d^5$ configuration. Answer: both $\text{Fe}^{3+}$ and $\text{Mn}^{2+}$ have $n = 5$ unpaired electrons and $\mu \approx 5.92\ \text{BM}$, the largest spin-only moment among common 3d ions.

Competitive stretch

For $\text{Cu}^{2+}$: Cu atom is the exceptional $[\text{Ar}]3d^{10}4s^1$; removing the $4s^1$ electron and one $3d$ electron gives $[\text{Ar}]3d^9$, with four orbitals paired and one orbital holding a lone electron, so $n = 1$ and $\mu = \sqrt{1\times 3} = \sqrt{3} \approx 1.73\ \text{BM}$ — weakly paramagnetic, matching the gentle pull seen for $\text{CuSO}_4$ solution described in the opening. By contrast $\text{Zn}^{2+}$ is $[\text{Ar}]3d^{10}$ with every d-orbital fully paired, $n = 0$, $\mu = 0$: diamagnetic, weakly repelled.

• Forgetting to remove the $ns$ electrons before $(n-1)d$ electrons when writing a transition-metal cation's configuration — this is the single most common source of wrong unpaired-electron counts.
• Applying Hund's rule incorrectly by pairing electrons in one orbital before every orbital in the set has at least one electron.
• Assuming paramagnetism rises monotonically with atomic number across the series — it actually peaks near the half-filled $d^5$ configuration (e.g. $\text{Mn}^{2+}$, $\text{Fe}^{3+}$) and then falls again toward $d^{10}$.
• Confusing diamagnetic (all paired, weakly repelled) with paramagnetic (unpaired electrons present, weakly attracted) — and note ferromagnetism (as in metallic iron) is a distinct, much stronger effect, not described by the spin-only formula.
• Treating the spin-only formula as exact for every case: it neglects orbital contribution, which is a fair approximation for most 3d ions but not the full physical picture used in advanced magnetochemistry.

Ask the tutor: want to practise the configuration-to-moment method on $\text{Cr}^{3+}$, $\text{Co}^{2+}$ or $\text{Ni}^{2+}$?