Arrange the following complex ions in increasing order of number of u…

The correct answer is Option (4) → C, B, A, D

The number of unpaired electrons in a complex depends on two things:

• The d electron count of the metal ion, found from its oxidation state.

• The ligand field: a weak field ligand (\( \ce{F^{-}} \), \( \ce{Cl^{-}} \), \( \ce{H2O} \)) gives high spin (electrons stay unpaired); a strong field ligand (\( \ce{NH3} \) with \( \ce{Co^{3+}} \), \( \ce{CN^{-}} \)) gives low spin (electrons pair in t_2g).

Working out each complex:

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

• B. \( \ce{[NiCl4]^{2-}} \): x + 4(−1) = −2, so Ni is +2. Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸. \( \ce{Cl^{-}} \) cannot pair the electrons; the complex is tetrahedral (sp³), with 2 unpaired.

• C. \( \ce{[Co(NH3)6]^{3+}} \): x + 6(0) = +3, so Co is +3, 3d⁶ as in A. \( \ce{NH3} \) acts as a strong field ligand for \( \ce{Co^{3+}} \): t_2g⁶e_g⁰, 0 unpaired (d²sp³, diamagnetic).

• D. \( \ce{[Fe(H2O)6]^{3+}} \): x + 6(0) = +3, so Fe is +3. Fe (Z = 26) is [Ar]3d⁶4s²; losing two 4s and one 3d electron gives \( \ce{Fe^{3+}} \): 3d⁵. \( \ce{H2O} \) is weak field: t_2g³e_g², 5 unpaired.

    0 < 2 < 4 < 5
    \( \ce{[Co(NH3)6]^{3+}} \) < \( \ce{[NiCl4]^{2-}} \) < \( \ce{[CoF6]^{3-}} \) < \( \ce{[Fe(H2O)6]^{3+}} \)

Shortcut: The diamagnetic complex C has no unpaired electrons and the d⁵ high spin complex D has the most. Only Option (4) starts with C and ends with D.