Consider the following complex ions: P = [FeF6]3-, Q = [V(H2O)6]2+, R = [Fe(H2O)6]2+ The correct order of the complex…

🧠 Three Complexes, Three Magnetic Moments

Compute μ = $\sqrt{n(n+2)}$ for each.

🗺️ Walk Each

P = $[\mathrm{FeF_6}]^{3-}$: Fe³⁺ ($\mathrm{d^5}$). F⁻ is weak field → high-spin → 5 unpaired → $\mu = \sqrt{35} \approx 5.92$ BM.

Q = $[\mathrm{V(H_2O)_6}]^{2+}$: V²⁺ ($\mathrm{d^3}$). H₂O is mid-weak → high-spin (but for d³, configuration is $\mathrm{t_{2g}^3}$ regardless) → 3 unpaired → $\mu = \sqrt{15} \approx 3.87$ BM.

R = $[\mathrm{Fe(H_2O)_6}]^{2+}$: Fe²⁺ ($\mathrm{d^6}$). H₂O is weak field → high-spin → 4 unpaired → $\mu = \sqrt{24} \approx 4.90$ BM.

Order: $\mu_Q\,(3.87) < \mu_R\,(4.90) < \mu_P\,(5.92)$ → Q < R < P → option (3).

⚡ The "More d-Electrons (up to d⁵) = More Unpaired" Rule

For high-spin 3d: unpaired count rises monotonically d¹ → d⁵, then declines d⁵ → d¹⁰. Maximum unpaired (5) at d⁵ → maximum μ at d⁵.

⚠️ Don't Apply Low-Spin to F⁻ or H₂O

Both F⁻ and H₂O are weak-field — they almost never force pairing on 3d metals (except for very high-charge cations like Co³⁺, where even H₂O can occasionally give low-spin behaviour). For F⁻ and H₂O on Fe²⁺, Fe³⁺, V²⁺ → high-spin always.

$\boxed{\text{Answer: (3) Q < R < P}}$