Magnetic properties of complexes — diamagnetic vs paramagnetic prediction

Question

How do we predict whether a coordination complex is diamagnetic or paramagnetic? How does Crystal Field Theory help us determine the number of unpaired electrons and calculate the magnetic moment?

(JEE Main, JEE Advanced, CBSE 12 — magnetic moment calculation and high-spin vs low-spin prediction is a JEE Advanced favourite)

Solution — Step by Step

Diamagnetic: All electrons are paired. The substance is weakly repelled by a magnetic field.
Paramagnetic: One or more unpaired electrons exist. The substance is attracted to a magnetic field.

The number of unpaired electrons ( $n$ ) determines the strength of paramagnetism. The spin-only magnetic moment is:

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

$n$	$\mu$ (BM)
0	0 (diamagnetic)
1	1.73
2	2.83
3	3.87
4	4.90
5	5.92

For octahedral complexes, the filling depends on the relative magnitude of $\Delta_o$ (crystal field splitting energy) and $P$ (pairing energy):

If $\Delta_o \lt P$ : electrons prefer to go to the higher $e_g$ orbitals rather than pair up. High-spin complex. More unpaired electrons.
If $\Delta_o \gt P$ : electrons prefer to pair up in $t_{2g}$ rather than go to $e_g$ . Low-spin complex. Fewer unpaired electrons.

Weak field ligands ( $Cl^-$ , $F^-$ , $H_2O$ ) give small $\Delta_o$ and typically high-spin complexes. Strong field ligands ( $CN^-$ , $CO$ , $NH_3$ , $NO_2^-$ ) give large $\Delta_o$ and typically low-spin complexes.

$Fe^{3+}$ has $d^5$ configuration.

$[FeF_6]^{3-}$ with $F^-$ (weak field):

High spin: $t_{2g}^3 e_g^2$ — all 5 electrons unpaired
$\mu = \sqrt{5 \times 7} = 5.92$ BM (strongly paramagnetic)

$[Fe(CN)_6]^{3-}$ with $CN^-$ (strong field):

Low spin: $t_{2g}^5 e_g^0$ — only 1 unpaired electron
$\mu = \sqrt{1 \times 3} = 1.73$ BM (weakly paramagnetic)

Same metal, same oxidation state — completely different magnetic behaviour because of the ligand.

Tetrahedral splitting ( $\Delta_t$ ) is only about $\frac{4}{9}$ of octahedral splitting ( $\Delta_o$ ). This means $\Delta_t$ is almost never large enough to force pairing. So tetrahedral complexes are almost always high-spin.

Square planar complexes (typically $d^8$ with strong field ligands) have a large splitting that often forces all electrons to pair, making them diamagnetic. Example: $[Ni(CN)_4]^{2-}$ is diamagnetic (all electrons paired in the 4 lower orbitals).

flowchart TD
    A["Given complex"] --> B["Find metal ion d-electron count"]
    B --> C["Identify ligand strength"]
    C --> D{"Strong or weak field?"}
    D -->|"Strong field (CN⁻, CO, NH₃)"| E["Low spin: fill t₂g first, pair before going to eg"]
    D -->|"Weak field (Cl⁻, F⁻, H₂O)"| F["High spin: fill all orbitals singly first"]
    E --> G["Count unpaired electrons (n)"]
    F --> G
    G --> H["μ = √(n(n+2)) BM"]
    H --> I{"n = 0?"}
    I -->|"Yes"| J["Diamagnetic"]
    I -->|"No"| K["Paramagnetic"]

Why This Works

Magnetism in complexes is purely an electron spin phenomenon. Each unpaired electron has a magnetic moment from its spin. When all electrons are paired, their spins cancel out (diamagnetic). The crystal field splitting determines HOW the electrons fill the d-orbitals, which directly determines the number of unpaired electrons.

The spectrochemical series is the practical tool — it tells us which ligands produce enough splitting to overcome the pairing energy penalty. This is why the same metal ion shows different magnetic behaviour with different ligands.

Common Mistake

Students forget to determine the oxidation state of the metal FIRST before counting d-electrons. In $[Co(NH_3)_6]^{3+}$ , cobalt is Co(III) which is $d^6$ , not $d^7$ (which would be Co(II)). With strong-field $NH_3$ , this gives a low-spin $t_{2g}^6 e_g^0$ configuration — all electrons paired, diamagnetic. If you mistakenly use $d^7$ , you get 1 unpaired electron and the wrong magnetic moment. JEE Advanced penalises this heavily.

Only $d^4$ , $d^5$ , $d^6$ , and $d^7$ configurations show different behaviour in high-spin vs low-spin. For $d^1$ , $d^2$ , $d^3$ (always high-spin filling of $t_{2g}$ ) and $d^8$ , $d^9$ , $d^{10}$ (no choice), the number of unpaired electrons is the same regardless of field strength.

Question

Solution — Step by Step

Why This Works

Common Mistake

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