The complex [CoF6]3- can be characterized as……..

The correct answer is Option (1) → sp³d², high spin

By valence bond theory, an octahedral complex uses six hybrid orbitals, of one of two kinds:

• d²sp³ (inner orbital complex): uses two inner 3d orbitals. They are freed when a strong field ligand pairs the d electrons, so the complex is low spin.

• sp³d² (outer orbital complex): uses two outer 4d orbitals. A weak field ligand does not pair the 3d electrons, so the complex is high spin.

In \( \ce{[CoF6]^{3-}} \), cobalt is +3 (x + 6(−1) = −3). Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron gives Co³⁺: 3d⁶. \( \ce{F^{-}} \) is a weak field ligand. The six 3d electrons stay spread out with four unpaired electrons, and the fluoride ions use the 4s, three 4p and two 4d orbitals:

    \( \ce{Co^{3+}} \): 3d⁶ (4 unpaired) + sp³d² from 4s, 4p, 4d

So \( \ce{[CoF6]^{3-}} \) is an outer orbital, sp³d², high spin complex and is paramagnetic.

Why the other options are incorrect

• Option (2) sp³d², low spin — Incorrect. sp³d² hybridisation is used because the electrons are not paired, so a \( \ce{Co^{3+}} \) outer orbital complex is high spin.

• Option (3) d²sp³, high spin — Incorrect. d²sp³ needs two empty 3d orbitals, which for d⁶ means pairing, so it cannot be high spin.

• Option (4) d²sp³, low spin — Incorrect. This describes [Co(NH₃)₆]³⁺, where the stronger \( \ce{NH3} \) ligand pairs the electrons.

NCERT (2026-27 edition):
Valence Bond Theory – …