Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all three…

🧠 Lanthanide μ Comes From f-Electrons

For lanthanide $\mathrm{Ln^{n+}}$, the spin-only μ tracks the 4f electron count:

🗺️ Compute Each

(a) $\mathrm{(NH_4)_2[Ce(NO_3)_6]}$: $\mathrm{NH_4^+}$ × 2 + $\mathrm{NO_3^-}$ × 6 + Ce → charge balance: $2 + x + 6(-1) = 0 \Rightarrow x = +4$. Ce(IV) = $[\mathrm{Xe}] 4f^0$ → 0 unpaired → $\mu = 0$ BM.

(b) $\mathrm{Gd(NO_3)_3}$: Gd(III) = $[\mathrm{Xe}] 4f^7$ → half-filled f-shell → 7 unpaired → $\mu = \sqrt{63} \approx 7.94$ BM.

(c) $\mathrm{Eu(NO_3)_3}$: Eu(III) = $[\mathrm{Xe}] 4f^6$ → 6 unpaired → $\mu = \sqrt{48} \approx 6.93$ BM.

Order: $a (0) < c (6.93) < b (7.94)$ → option (4).

⚡ Lanthanides Are All "High-Spin"

The 4f orbitals are deeply buried — they don't interact strongly with ligand fields. So lanthanide complexes don't show CFT splitting and are always Hund-maximised. f-electron count = unpaired count.

⚠️ Spin-Only Underestimates Lanthanide μ

Real lanthanide μ values include orbital angular momentum contributions (because the 4f orbitals retain substantial L). The "true" formula is $\mu = g\sqrt{J(J+1)}$, but this question specifies spin-only, which uses $S$ only. So we just count unpaired electrons.

$\boxed{\text{Answer: (4) (a) < (c) < (b)}}$