Arrange the following ions in the increasing order of their magnetic moments:
(A) [Fe(CN)6]4−
(B) [Co(F)6]3−
(C) [Ni(H2O)6]2+
(D) [Cu(NH3)6]2+
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (D), (C), (B)
The magnetic moment of a complex comes from its unpaired electrons. By the spin-only formula, more unpaired electrons give a larger moment:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
To count n, find the oxidation state and the d-electron count, then decide the spin from the ligand. Strong field ligands (\( \ce{CN^{-}} \), \( \ce{NH3} \)) pair electrons up; weak field ligands (\( \ce{F^{-}} \), \( \ce{H2O} \)) leave them unpaired.
(A) \( \ce{[Fe(CN)6]^{4-}} \): x + 6(−1) = −4, so x = +2. Fe (Z = 26) is [Ar]3d64s2; losing both 4s electrons gives \( \ce{Fe^{2+}} \): 3d6. \( \ce{CN^{-}} \) is strong field, so low spin, t2g6: n = 0, μ = 0.
(B) \( \ce{[Co(F)6]^{3-}} \): x − 6 = −3, so x = +3. Co (Z = 27) is [Ar]3d74s2; losing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d6. \( \ce{F^{-}} \) is weak field, so high spin, t2g4eg2: n = 4, μ = √24 = 4.90 BM.
(C) \( \ce{[Ni(H2O)6]^{2+}} \): x + 6(0) = +2, so x = +2. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8, t2g6eg2: n = 2, μ = √8 = 2.83 BM.
(D) \( \ce{[Cu(NH3)6]^{2+}} \): x + 6(0) = +2, so x = +2. Cu (Z = 29) is [Ar]3d104s1; losing the 4s electron and one 3d electron gives \( \ce{Cu^{2+}} \): 3d9, t2g6eg3: n = 1, μ = √3 = 1.73 BM.
Shortcut: (A) is the only diamagnetic ion and (B) has the most unpaired electrons. Only Option (3) starts with (A) and ends with (B).