In which of the following order are the given complex ions arranged correctly with respect to their decreasing spin…

🧠 Four Numbers, Sort Them Down

Compute the unpaired-electron count for each complex and sort largest → smallest. The complex with the most unpaired electrons has the highest spin-only μ.

| Complex | Metal | d-count | Geometry/Field | Unpaired | μ (BM) | |---|---|---|---|---|---| | (i) $[\mathrm{FeF_6}]^{3-}$ | $\mathrm{Fe^{3+}}$ | $\mathrm{d^5}$ | oct., $\mathrm{F^-}$ weak → HS | 5 | 5.92 | | (ii) $[\mathrm{Co(NH_3)_6}]^{3+}$ | $\mathrm{Co^{3+}}$ | $\mathrm{d^6}$ | oct., $\mathrm{NH_3}$ → LS | 0 | 0 | | (iii) $[\mathrm{NiCl_4}]^{2-}$ | $\mathrm{Ni^{2+}}$ | $\mathrm{d^8}$ | tetrahedral → 2 unpaired | 2 | 2.83 | | (iv) $[\mathrm{Cu(NH_3)_4}]^{2+}$ | $\mathrm{Cu^{2+}}$ | $\mathrm{d^9}$ | sq. planar → 1 unpaired | 1 | 1.73 |

Decreasing order: (i) 5.92 > (iii) 2.83 > (iv) 1.73 > (ii) 0 — option (3).

🗺️ Quick Justifications

• (i) Halide is weak field → no pairing on Fe(III) → all 5 electrons unpaired.
• (ii) NH₃ + Co(III) is the textbook low-spin pair → 0 unpaired.
• (iii) Tetrahedral d⁸ → high-spin (always) → 2 unpaired (one in each $\mathrm{e}$ orbital after pairing $\mathrm{t_2}$).
• (iv) d⁹ = always 1 unpaired regardless of geometry/field.

⚡ The "d⁹ Always Has 1 Unpaired" Anchor

A d⁹ ion has nine electrons in a five-orbital set — exactly one orbital must be singly occupied. So $\mathrm{Cu^{2+}}$ in any complex always shows 1 unpaired, μ = 1.73 BM.

⚠️ The "$\mathrm{Co^{3+}}$ + NH₃ = 4 Unpaired" Slip

If you miss that $\mathrm{NH_3}$ + $\mathrm{Co^{3+}}$ pairs all electrons (low-spin), you'd compute high-spin d⁶ → 4 unpaired and pick option (1) or (4). $\mathrm{Co^{3+}}$ + $\mathrm{NH_3}$, en, $\mathrm{CN^-}$, oxalate always low-spin for JEE.

$\boxed{\text{Answer: (3) (i) > (iii) > (iv) > (ii)}}$