The correct order of the number of unpaired electrons in the given complexes is: (A) [Fe(CN)6]3- (B) [FeF6]3- (C)…

🧠 Compute Unpaired for All Five

| Complex | Metal | d-count | Field | Unpaired | |---|---|---|---|---| | (A) $[\mathrm{Fe(CN)_6}]^{3-}$ | Fe(III) | d⁵ | strong → low-spin | 1 | | (B) $[\mathrm{FeF_6}]^{3-}$ | Fe(III) | d⁵ | weak → high-spin | 5 | | (C) $[\mathrm{CoF_6}]^{3-}$ | Co(III) | d⁶ | weak → high-spin | 4 | | (D) $[\mathrm{Cr(ox)_3}]^{3-}$ | Cr(III) | d³ | (any) | 3 | | (E) $[\mathrm{Ni(CO)_4}]$ | Ni(0) | d¹⁰ | tetrahedral | 0 |

Order: $E (0) < A (1) < D (3) < C (4) < B (5)$ → option (1).

🗺️ Why d⁵ Has Two Modes

d⁵ is the most spin-state-sensitive d-count. With strong field (CN⁻): all 5 electrons pair into $\mathrm{t_{2g}}$ → 1 unpaired. With weak field (F⁻): all 5 stay parallel → 5 unpaired (maximum). The same metal in the same OS goes from $\mu \approx 1.73$ BM to $\mu \approx 5.92$ BM just by switching ligand.

⚡ The "0–5" Spread for d⁵ Fe(III)

$\mathrm{Fe^{3+}}$ ($\mathrm{d^5}$) holds the record for largest spin-state range: 1 unpaired (with CN⁻, NO⁺) to 5 unpaired (with F⁻, halides, H₂O). Always check the ligand before assuming.

⚠️ Ni(0) Is Fully Paired

Don't list Ni(CO)₄ as "transition metal so unpaired > 0". It's $\mathrm{d^{10}}$ — completely paired. μ = 0.

$\boxed{\text{Answer: (1) E < A < D < C < B}}$