Arrange the following complexes in increasing order of spin only magn…

The correct answer is Option (2) → C, D, A, B

The spin-only magnetic moment depends only on the number of unpaired electrons, n:

    \( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)

So the more unpaired electrons a complex has, the higher its magnetic moment: n = 0, 1, 2, 3, 4, 5 give 0, 1.73, 2.83, 3.87, 4.90 and 5.92 BM.

To count the unpaired electrons, find the d-electron count of the metal ion and then decide whether the ligand pairs the electrons:

• Weak field ligands (F⁻, Cl⁻, Br⁻): the splitting Δ_o is smaller than the pairing energy. Electrons stay unpaired, giving a high spin complex.

• Strong field ligands (CN⁻, CO; NH₃ with Co³⁺): Δ_o is larger than the pairing energy. Electrons pair up in t_2g, giving a low spin complex.

• Tetrahedral complexes: the splitting is small (Δ_t ≈ 4/9 Δ_o), so they are always high spin.

Matching the given complexes:

• (A) \( \ce{[CoF6]^{3-}} \): x + 6(−1) = −3 gives Co(+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, so the complex is high spin (t_2g⁴e_g²) with 4 unpaired, μ = 4.90 BM.

• (B) \( \ce{[MnBr4]^{2-}} \): x + 4(−1) = −2 gives Mn(+2). Mn (Z = 25) is [Ar]3d⁵4s²; removing the two 4s electrons gives \( \ce{Mn^{2+}} \): 3d⁵. It is tetrahedral and high spin, with 5 unpaired, μ = 5.92 BM.

• (C) \( \ce{[Co(NH3)6]^{3+}} \): \( \ce{NH3} \) is neutral, so cobalt is +3, and \( \ce{Co^{3+}} \) is 3d⁶ as in (A). \( \ce{NH3} \) pairs the electrons of \( \ce{Co^{3+}} \) (t_2g⁶), giving 0 unpaired, μ = 0.

• (D) \( \ce{[Mn(CN)6]^{4-}} \): x + 6(−1) = −4 gives Mn(+2), 3d⁵ as in (B). \( \ce{CN^{-}} \) is strong field (t_2g⁵), giving 1 unpaired, μ = 1.73 BM.

Shortcut: C is diamagnetic, so it comes first. B has the largest possible number of unpaired electrons (5), so it comes last. Only Option (2) starts with C and ends with B.