\([CoF_6]^{3-}\) using valence bond theory can be explained as
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Outer orbital, paramagnetic
In valence bond theory the metal ion uses hybrid orbitals to accept the ligand electron pairs. An octahedral complex uses either d2sp3 (inner 3d orbitals: inner orbital, low spin) when the ligand forces the d electrons to pair, or sp3d2 (outer 4d orbitals: outer orbital, high spin) when it does not.
Oxidation state of Co in \( \ce{[CoF6]^{3-}} \): x + 6(−1) = −3, so x = +3. Co (Z = 27) is [Ar] 3d74s2; removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): [Ar] 3d6.
\( \ce{F^{-}} \) is a weak field ligand and cannot force the 3d electrons to pair, so they stay spread out by Hund’s rule:
3d: [↑↓] [↑] [↑] [↑] [↑] 4s: [ ] 4p: [ ] [ ] [ ] 4d: [ ] [ ]
No 3d orbital is free, so the six \( \ce{F^{-}} \) pairs go into one 4s, three 4p and two 4d orbitals: sp3d2 hybridisation, an outer orbital (high spin) complex. Four 3d electrons remain unpaired, so the complex is paramagnetic (μ = √24 = 4.90 BM).
Why the other options are incorrect
Option (1) Inner orbital, paramagnetic — Incorrect. An inner orbital Co(III) complex needs two empty 3d orbitals, which means pairing all six d electrons; it would then have no unpaired electron. This combination does not apply to \( \ce{Co^{3+}} \).
Option (3) Inner orbital, diamagnetic — Incorrect. This describes \( \ce{[Co(NH3)6]^{3+}} \), where the strong field \( \ce{NH3} \) pairs the 3d6 electrons (d2sp3); \( \ce{F^{-}} \) is too weak to do this.
Option (4) Outer orbital, diamagnetic — Incorrect. An outer orbital complex keeps the 3d electrons unpaired; with 3d6 spread out there are four unpaired electrons, so it cannot be diamagnetic.
NCERT (2026-27 edition):
Valence Bond Theory – “… In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex. 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.”