d and f Block Elements: Step-by-Step Worked Examples (2)

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Tags D And F Block Elements

Question

Calculate the spin-only magnetic moment (μs\mu_s) of Fe3+\text{Fe}^{3+} and Fe2+\text{Fe}^{2+} ions and explain which is more paramagnetic.

Solution — Step by Step

Iron atomic number = 26. Configuration: [Ar] 3d64s2[Ar]\, 3d^6 4s^2.

  • Fe3+\text{Fe}^{3+}: lose 2 from 4s and 1 from 3d → [Ar] 3d5[Ar]\, 3d^5
  • Fe2+\text{Fe}^{2+}: lose 2 from 4s → [Ar] 3d6[Ar]\, 3d^6

Fe3+\text{Fe}^{3+} (d5d^5): all five 3d electrons unpaired (Hund’s rule, half-filled subshell). n=5n = 5.

Fe2+\text{Fe}^{2+} (d6d^6): five orbitals, one pair, four unpaired. n=4n = 4.

μs=n(n+2) BM\mu_s = \sqrt{n(n+2)}\,\text{BM}

For Fe3+\text{Fe}^{3+}: μs=5×7=35≈5.92\mu_s = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 BM.

For Fe2+\text{Fe}^{2+}: μs=4×6=24≈4.90\mu_s = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 BM.

Final answer: Fe3+\text{Fe}^{3+} is more paramagnetic (μs≈5.92\mu_s \approx 5.92 BM > 4.90 BM)

Why This Works

Magnetic moment depends only on the number of unpaired electrons (assuming spin-only, no orbital contribution — generally true for first-row transition metals in weak fields). More unpaired electrons → more paramagnetic.

The half-filled d5d^5 configuration of Fe3+\text{Fe}^{3+} is especially stable (Hund’s rule + exchange energy) and gives the maximum unpaired count for any d-block 3+ ion.

Alternative Method

Some textbooks use μ=2S(S+1)\mu = 2\sqrt{S(S+1)} where S=n/2S = n/2 is the total spin quantum number. Same answer:

  • Fe3+\text{Fe}^{3+}: S=5/2S = 5/2, μ=2(5/2)(7/2)=235/4=35≈5.92\mu = 2\sqrt{(5/2)(7/2)} = 2\sqrt{35/4} = \sqrt{35} \approx 5.92 BM. ✓

Common Mistake

Students remove electrons from 3d before 4s when forming cations. Wrong direction — for atoms, you fill 4s before 3d, but you remove 4s first. Aufbau and ionisation follow different orders.

So Fe→Fe2+\text{Fe} \to \text{Fe}^{2+}: lose 4s² first, giving d6d^6. Not d44s2d^4 4s^2.

Maximum spin-only moment for first-row transition metals occurs at d5d^5 (Mn²⁺ or Fe³⁺): 5 unpaired electrons → 35≈5.92\sqrt{35} \approx 5.92 BM. NEET asks “which ion has highest magnetic moment?” — answer is almost always the d5d^5 candidate.

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