d and f Block Elements: Edge Cases and Subtle Traps (5)

Question

Calculate the spin-only magnetic moment of $\text{Fe}^{3+}$ and $\text{Mn}^{2+}$ ions in their high-spin state. Comment on which is more paramagnetic.

Solution — Step by Step

Why This Works

The spin-only formula assumes orbital angular momentum is “quenched” — true for first-row transition metals in most coordination environments, where the crystal field splits orbitals such that orbital motion is suppressed.

Both $\text{Fe}^{3+}$ and $\text{Mn}^{2+}$ are isoelectronic (both have $24$ electrons) and have the half-filled $d^5$ configuration, giving identical magnetic moments in spin-only treatment.

Alternative Method

Direct memorisation: for $d^5$ high-spin, $\mu = 5.92$ BM. JEE often gives multiple ions; for each, just read off the $d^n$ count and look up the value.

Common Mistake

The five classic d-block traps:

1. Counting all $d$ electrons as unpaired: only the truly unpaired ones contribute. For low-spin $d^6$ (like $[\text{Fe(CN)}_6]^{4-}$), all six pair up → $n = 0 \to \mu = 0$ BM (diamagnetic).

2. Removing $3d$ electrons before $4s$: when forming cations, $4s$ electrons are removed first, then $3d$. So $\text{Fe} \to \text{Fe}^{2+}$ loses $4s^2$, giving $3d^6$, not $3d^4 4s^2$.

3. Mixing high-spin and low-spin: depends on the ligand strength. Strong-field ligands (CN$^-$, CO) cause low-spin; weak-field (H$_2$O, F$^-$) cause high-spin. Default to high-spin if not specified.

4. Forgetting that lanthanides have orbital contribution: for f-block, the spin-only formula doesn’t work — must use $\mu = g\sqrt{J(J+1)}$ with $J$ the total angular momentum.

5. Confusing the units (BM vs J/T): 1 Bohr Magneton = $9.274 \times 10^{-24}$ J/T. JEE/NEET answers are conventionally in BM.

The fix: write the configuration carefully, identify $n$, apply the formula. For lanthanides, switch to the full Landé formula.