Arrange following complexes in decreasing order of number of unpaired electrons:
(A) [Co(H2O)6]2+
(B) [Fe(CN)6]3−
(C) [Co(H2O)6]3+
(D) [Fe(H2O)6]3+
Choose correct answer from options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (D), (A), (B), (C)
Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
The number of unpaired electrons in an octahedral complex depends on the d-electron count of the metal ion and on the crystal field splitting Δo compared with the pairing energy P. If Δo < P (weak field), electrons occupy eg before pairing: high spin. If Δo > P (strong field, e.g. \( \ce{CN^{-}} \)), they pair up in t2g: low spin. Δo also grows with the charge on the metal ion, so a +3 ion is split much more than a +2 ion by the same ligand.
(A) \( \ce{[Co(H2O)6]^{2+}} \) — Co (Z = 27) is [Ar]3d74s2; losing the two 4s electrons gives \( \ce{Co^{2+}} \): 3d7. With water it is high spin, t2g5eg2: 3 unpaired electrons.
(B) \( \ce{[Fe(CN)6]^{3-}} \) — x + 6(−1) = −3 gives 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, t2g5: 1 unpaired electron.
(C) \( \ce{[Co(H2O)6]^{3+}} \) — \( \ce{H2O} \) is neutral, so cobalt is +3; removing the two 4s electrons and one 3d electron from Co gives \( \ce{Co^{3+}} \): 3d6. For the highly charged \( \ce{Co^{3+}} \) ion, even water splits the d orbitals strongly enough to pair the electrons: low spin, t2g6, 0 unpaired electrons (diamagnetic). Even oxalate, a weaker ligand than water, gives the low-spin \( \ce{[Co(C2O4)3]^{3-}} \); only the very weak \( \ce{F^{-}} \) gives high-spin \( \ce{[CoF6]^{3-}} \). So (C) comes last.
(D) \( \ce{[Fe(H2O)6]^{3+}} \) — \( \ce{H2O} \) is neutral, so iron is +3, \( \ce{Fe^{3+}} \): 3d5 as in (B), high spin with water, t2g3eg2: 5 unpaired electrons.