Arrange the following octahedral complexes in decreasing order of num…

The correct answer is Option (1) → B, A, C, D

In an octahedral complex, the five d orbitals split into a lower set t_2g (three orbitals) and a higher set e_g (two orbitals), separated by the energy Δ_o.

• Strong field ligands (\( \ce{CN^{-}} \); \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) give a large Δ_o, larger than the pairing energy, so electrons pair up in t_2g: a low spin complex with fewer unpaired electrons.

• Weak field ligands (\( \ce{F^{-}} \), \( \ce{H2O} \)) give a small Δ_o, so electrons enter e_g before pairing: a high spin complex with more unpaired electrons.

Spectrochemical series (weak to strong): \( \ce{I^{-}} \) < \( \ce{Br^{-}} \) < \( \ce{Cl^{-}} \) < \( \ce{F^{-}} \) < \( \ce{OH^{-}} \) < \( \ce{H2O} \) < \( \ce{NH3} \) < en < \( \ce{CN^{-}} \) < CO

• A. [Mn(CN)₆]³⁻: x + 6(−1) = −3, so x = +3. Mn (Z = 25) is [Ar]3d⁵4s²; removing two 4s and one 3d electron gives \( \ce{Mn^{3+}} \): 3d⁴. \( \ce{CN^{-}} \) is strong field: t_2g⁴e_g⁰, 2 unpaired.

• B. [CoF₆]³⁻: x + 6(−1) = −3, so x = +3. Co (Z = 27) is [Ar]3d⁷4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{F^{-}} \) is weak field: t_2g⁴e_g², 4 unpaired.

• C. [Fe(CN)₆]³⁻: x + 6(−1) = −3, so x = +3. Fe (Z = 26) is [Ar]3d⁶4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Fe^{3+}} \): 3d⁵. \( \ce{CN^{-}} \) is strong field: t_2g⁵e_g⁰, 1 unpaired.

• D. [Co(NH₃)₆]³⁺: \( \ce{NH3} \) is neutral, so x = +3, and \( \ce{Co^{3+}} \) is again 3d⁶. \( \ce{NH3} \) is strong field for \( \ce{Co^{3+}} \): t_2g⁶e_g⁰, 0 unpaired (diamagnetic).

    \( \displaystyle \text{B } 4 > \text{A } 2 > \text{C } 1 > \text{D } 0 \)

Shortcut: B is the only high spin complex and D is fully paired. Options (1) and (2) both start with B and end with D, so only A versus C decides: low spin d⁴ (2 unpaired) comes before low spin d⁵ (1 unpaired), which gives Option (1).