A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with up to three electrons in the d orbitals, like Ti3+(d1); V3+(d2); Cr3+(d3); two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund’s rule). Thus, for d4 (Cr2+, Mn3+), d5 (Mn2+, Fe3+), d6 (Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications.
Match List-I with List-II.
| List-I (Complex ion) | List-II (Hybridisation; magnetic nature) |
|---|---|
| (A) \( \ce{[Co(C2O4)3]^3-} \) | (I) sp3d2; five unpaired electrons |
| (B) \( \ce{[CoF6]^3-} \) | (II) d2sp3; diamagnetic |
| (C) \( \ce{[FeF6]^3-} \) | (III) d2sp3; two unpaired electrons |
| (D) \( \ce{[Mn(CN)6]^3-} \) | (IV) sp3d2; four unpaired electrons |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
When the 3d electrons of the metal ion pair up, two inner 3d orbitals become free and the complex is an inner orbital complex (d2sp3); the magnetic data show this for the cyanido and oxalato complexes here.
With the weak ligand \( \ce{F^-} \) the electrons stay unpaired, the metal uses outer 4d orbitals and the complex is an outer orbital complex (sp3d2).
(A) \( \ce{[Co(C2O4)3]^3-} \) — This matches (II) d2sp3; diamagnetic. Co is +3 (d6); the six electrons pair up in three 3d orbitals, so no electron is unpaired.
(B) \( \ce{[CoF6]^3-} \) — This matches (IV) sp3d2; four unpaired electrons. With weak \( \ce{F^-} \) the 3d6 electrons of Co3+ are not paired up, so four stay unpaired.
(C) \( \ce{[FeF6]^3-} \) — This matches (I) sp3d2; five unpaired electrons. Fe is +3 (d5); with weak \( \ce{F^-} \) all five electrons stay unpaired.
(D) \( \ce{[Mn(CN)6]^3-} \) — This matches (III) d2sp3; two unpaired electrons. Mn is +3 (d4); pairing puts the four electrons in three 3d orbitals, leaving two unpaired.
NCERT (2026-27 edition):
5.5.2 Magnetic Properties of Coordination Compounds – “[Mn(CN)6]3– has magnetic moment of two unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic.” … “[Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridisation, the former two complexes are paramagnetic and the latter diamagnetic. On the other hand, [MnCl6]3–, [FeF6]3– and [CoF6]3– are outer orbital complexes involving sp3d2 hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.”