Arrange the following complexes in increasing order of their magnetic…

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

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

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

So the order of magnetic moments is the order of unpaired electrons. To count them, find the oxidation state from the charge balance, write the d configuration of the ion from Z, and then decide how the d electrons are placed: a strong field ligand (\( \ce{CN^{-}} \); \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) pairs them up in the t_2g set (low spin), while a weak field ligand (\( \ce{Cl^{-}} \), \( \ce{H2O} \)) leaves them unpaired (high spin).

• (A) \( \ce{[Cr(H2O)6]^3+} \): \( \ce{H2O} \) is neutral, so Cr is +3. Cr (Z = 24) is [Ar]3d⁵4s¹; losing the 4s electron and two 3d electrons gives 3d³, that is t_2g³: n = 3, \( \mu = \sqrt{15} = 3.87 \) BM.

• (B) \( \ce{[Co(NH3)6]^3+} \): Co is +3. Co (Z = 27) is [Ar]3d⁷4s²; \( \ce{Co^{3+}} \) is 3d⁶. With \( \ce{NH3} \) the electrons pair up as t_2g⁶: n = 0, \( \mu = 0 \) (diamagnetic).

• (C) \( \ce{[FeCl6]^4-} \): x + 6(−1) = −4 gives x = +2. Fe (Z = 26) is [Ar]3d⁶4s²; \( \ce{Fe^{2+}} \) is 3d⁶. \( \ce{Cl^{-}} \) is weak field, so t_2g⁴e_g²: n = 4, \( \mu = \sqrt{24} = 4.90 \) BM.

• (D) \( \ce{[Mn(CN)6]^3-} \): x + 6(−1) = −3 gives x = +3. Mn (Z = 25) is [Ar]3d⁵4s²; \( \ce{Mn^{3+}} \) is 3d⁴. \( \ce{CN^{-}} \) is strong field, so t_2g⁴: n = 2, \( \mu = \sqrt{8} = 2.83 \) BM.

In increasing order of magnetic moment:

    \( \displaystyle \ce{[Co(NH3)6]^3+}\,(0) \lt \ce{[Mn(CN)6]^3-}\,(2.83) \lt \ce{[Cr(H2O)6]^3+}\,(3.87) \lt \ce{[FeCl6]^4-}\,(4.90) \)

Shortcut: The diamagnetic \( \ce{[Co(NH3)6]^3+} \) (B) must come first and the high-spin \( d^{6} \) complex (C), with four unpaired electrons, last. Only Option (2) starts with (B) and ends with (C).