Magnetic properties of complexes — diamagnetic vs paramagnetic prediction

Question

How do you predict whether a coordination compound is paramagnetic or diamagnetic? What determines the number of unpaired electrons — and how do strong-field and weak-field ligands change the magnetic behaviour?

(JEE Main, JEE Advanced, CBSE 12 — magnetic properties via CFT are tested every year)

Solution — Step by Step

Paramagnetic substances have unpaired electrons — they are attracted into a magnetic field. Diamagnetic substances have all electrons paired — they are slightly repelled.

The number of unpaired electrons ( $n$ ) determines the magnetic moment:

\mu = \sqrt{n(n+2)} \text{ BM (Bohr Magnetons)}

To find $n$ , we need to know the d-electron configuration in the complex — and this depends on whether the ligands produce a strong field or weak field.

In an octahedral complex, d-orbitals split into $t_{2g}$ (lower, 3 orbitals) and $e_g$ (higher, 2 orbitals).

Weak-field ligands (I-, Br-, Cl-, F-, H2O): Small $\Delta_o$ . Electrons follow Hund’s rule and fill all orbitals singly before pairing. This gives the maximum number of unpaired electrons (high-spin complex).

Strong-field ligands (CN-, CO, NH3, en): Large $\Delta_o$ . Electrons fill $t_{2g}$ completely before going to $e_g$ , because pairing energy is less than $\Delta_o$ . This gives the minimum number of unpaired electrons (low-spin complex).

Fe2+ (d6) with different ligands:

With H2O (weak field): $t_{2g}^4 e_g^2$ → 4 unpaired electrons → paramagnetic, $\mu = \sqrt{4 \times 6} = 4.9$ BM

With CN- (strong field): $t_{2g}^6 e_g^0$ → 0 unpaired electrons → diamagnetic, $\mu = 0$

Co3+ (d6): $[\text{CoF}_6]^{3-}$ (weak field): 4 unpaired electrons — paramagnetic $[\text{Co(NH}_3\text{)}_6]^{3+}$ (strong field): 0 unpaired electrons — diamagnetic

Same metal, same oxidation state — completely different magnetic properties due to ligand field strength.

graph TD
    A["Coordination Complex"] --> B["Find metal d-electron count"]
    B --> C{"Ligand type?"}
    C -->|"Strong field: CN, CO, NH3"| D["Low spin - max pairing"]
    C -->|"Weak field: Cl, Br, H2O"| E["High spin - max unpaired"]
    D --> F["Count unpaired electrons"]
    E --> F
    F --> G{"n = 0?"}
    G -->|"Yes"| H["Diamagnetic"]
    G -->|"No"| I["Paramagnetic"]
    I --> J["mu = sqrt of n times n+2 BM"]

Why This Works

The magnetic behaviour is entirely controlled by the competition between two energies:

Crystal field splitting energy ( $\Delta_o$ ): the energy cost of placing an electron in the higher $e_g$ orbital
Pairing energy ( $P$ ): the energy cost of putting two electrons in the same orbital

If $\Delta_o > P$ (strong-field ligand): electrons pair up in $t_{2g}$ rather than go to $e_g$ → low spin → fewer unpaired electrons.

If $\Delta_o < P$ (weak-field ligand): electrons go to $e_g$ to avoid pairing → high spin → more unpaired electrons.

For d4, d5, d6, and d7 configurations, the spin state depends on the ligand. For d1, d2, d3, d8, d9, d10, the number of unpaired electrons is the same regardless of field strength.

Alternative Method

For JEE, remember which d-electron counts are affected by ligand field strength: only $d^4, d^5, d^6, d^7$ in octahedral complexes. For all others ( $d^1, d^2, d^3, d^8, d^9, d^{10}$ ), the number of unpaired electrons is the same whether the field is strong or weak. This eliminates many MCQ options instantly.

Also, the magnetic moment formula $\mu = \sqrt{n(n+2)}$ gives: $n=1 \rightarrow 1.73$ , $n=2 \rightarrow 2.83$ , $n=3 \rightarrow 3.87$ , $n=4 \rightarrow 4.90$ , $n=5 \rightarrow 5.92$ BM. Memorise these values.

Common Mistake

The biggest error: assuming all complexes are high-spin or all are low-spin. The spin state depends on the specific ligand. $[\text{Fe(H}_2\text{O)}_6]^{2+}$ is high-spin (4 unpaired), but $[\text{Fe(CN)}_6]^{4-}$ is low-spin (0 unpaired) — same Fe2+ ion, completely different magnetic behaviour. Always check the ligand position in the spectrochemical series before deciding.

Also, students often apply the spin-pairing analysis to tetrahedral complexes. In tetrahedral geometry, $\Delta_t$ is always small (about $\frac{4}{9}\Delta_o$ ), so tetrahedral complexes are almost always high-spin. Low-spin tetrahedral complexes are extremely rare. Do not apply the strong-field/weak-field analysis to tetrahedral cases.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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