How many unpaired electrons are present in the high spin form of [CoF6]3- complex and which metal orbitals are used in bonding?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 unpaired electrons and 4s, 4p and 4d orbitals to give sp3d2 hybridisation.
Oxidation state of Co: in \( \ce{[CoF6]^3-} \), x + 6(−1) = −3, so x = +3.
Configuration: Co (Z = 27) is [Ar]3d74s2; losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): 3d6.
With \( \ce{F^-} \) ligands it is a high spin (spin free) complex, so the 3d electrons are not forced to pair: one 3d orbital holds a pair and four hold single electrons, giving 4 unpaired electrons.
With no empty 3d orbital, the six F– electron pairs go into hybrid orbitals made from one 4s, three 4p and two outer 4d orbitals: sp3d2. The complex is an outer orbital, high spin, paramagnetic complex.
Why the other options are incorrect
Option (1) 0 unpaired electrons and 4s, 4p and 4d orbitals to give sp3d2 hybridisation. — Incorrect. The orbitals are right, but zero unpaired electrons would need all six 3d electrons paired, which does not happen in this high spin complex.
Option (2) 0 unpaired electrons and 3d, 4s and 4p orbitals to give d2sp3 hybridisation. — Incorrect. This describes the inner orbital, low spin complex \( \ce{[Co(NH3)6]^3+} \), where pairing frees two 3d orbitals; it does not happen with \( \ce{F^-} \).
Option (4) 4 unpaired electrons and 3d, 4s and 4p orbitals to give d2sp3 hybridisation. — Incorrect. With 4 unpaired electrons all five 3d orbitals are occupied, so no 3d orbital is free for d2sp3 hybridisation.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “… The paramagnetic octahedral complex, [CoF6]3– uses outer orbital (4d) in hybridisation (sp3d2). It is thus called outer orbital or high spin or spin free complex.”
5.5.2 Magnetic Properties of Coordination Compounds – “… [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic.”