The calculated spin-only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2+ and [Fe(CN)6]4-…

🧠 Read "Anionic and Cationic" Carefully

The complexes given are $[\mathrm{Fe(H_2O)_6}]^{2+}$ (cationic) and $[\mathrm{Fe(CN)_6}]^{4-}$ (anionic). The question asks for the moments in that order: anionic first, cationic second. So compute:

• Anion $[\mathrm{Fe(CN)_6}]^{4-}$: $\mathrm{Fe^{2+}}\,(\mathrm{d^6})$ + strong-field $\mathrm{CN^-}$ → low-spin → 0 unpaired → μ = 0.
• Cation $[\mathrm{Fe(H_2O)_6}]^{2+}$: $\mathrm{Fe^{2+}}\,(\mathrm{d^6})$ + moderate-field $\mathrm{H_2O}$ → high-spin → 4 unpaired → μ = 4.9 BM.

So pair = (0, 4.9) — option (1).

🗺️ Verify Each

Anion $[\mathrm{Fe(CN)_6}]^{4-}$. Strong-field $\mathrm{CN^-}$ pairs the d⁶ to $\mathrm{t_{2g}^6\,e_g^0}$. All paired → diamagnetic → μ = 0.

Cation $[\mathrm{Fe(H_2O)_6}]^{2+}$. Moderate-field $\mathrm{H_2O}$ leaves d⁶ as high-spin $\mathrm{t_{2g}^4\,e_g^2}$. 4 unpaired → μ = $\sqrt{4 \cdot 6} = \sqrt{24} \approx 4.9$ BM.

⚡ The Same-Metal Twin Trick

JEE often pairs two complexes with the same metal and same d-count but different ligand strengths to test whether you understand that ligand identity flips the spin state. Here: identical $\mathrm{Fe^{2+}\,d^6}$ but $\mathrm{CN^-}$ → 0 unpaired vs $\mathrm{H_2O}$ → 4 unpaired. Always read the ligand before predicting.

⚠️ The Cation/Anion Mix-Up

If you read "cationic and anionic" as the order (option 3 inverts to "4.9 and 0"), you'll pick the wrong option. The question phrases it "anionic and cationic" — anion first. Read order labels carefully on every match-the-pair MCQ.

$\boxed{\text{Answer: (1) 0 and 4.9}}$