The hybridisation and number of unpaired electrons in [CoF6]3- are:

🧠 The "Weak Field + d⁶" Combination

$[\mathrm{CoF_6}]^{3-}$ has $\mathrm{Co^{3+}}$ ($\mathrm{d^6}$) with fluoride — a weak-field ligand. Weak field means no electron pairing; the d-electrons stay distributed across all five 3d orbitals. So the inner 3d orbitals are full of unpaired electrons, and hybridisation must use the outer 4d set: $\mathrm{sp^3d^2}$ (outer-orbital, paramagnetic).

🗺️ Configuration Walk-Through

$\mathrm{Co^{3+}}$: $[\mathrm{Ar}]\,3\mathrm{d^6}$.

With $\mathrm{F^-}$ (weak field): no pairing → high-spin distribution across 3d. The standard high-spin $\mathrm{d^6}$ filling is $\mathrm{t_{2g}^4\,e_g^2}$ in CFT language; in VBT terms, all five 3d orbitals are populated, leaving only 4s, 4p, 4d to host the six F⁻ donors.

So hybridisation = $\mathrm{sp^3d^2}$ (uses 4d, outer-orbital).

Unpaired electrons: $\mathrm{t_{2g}^4\,e_g^2}$ → 1 paired + 3 unpaired in $\mathrm{t_{2g}}$, 2 unpaired in $\mathrm{e_g}$ → 4 unpaired.

⚡ The "Weak Field + Big Charge" Recipe

For $\mathrm{Co^{3+}}$ specifically:

• With weak ligand ($\mathrm{F^-}$, $\mathrm{H_2O}$ in some cases): $\mathrm{sp^3d^2}$, 4 unpaired.
• With strong ligand ($\mathrm{NH_3}$, $\mathrm{CN^-}$, $\mathrm{en}$, oxalate): $\mathrm{d^2sp^3}$, 0 unpaired.

This single mnemonic answers half the Co(III) hybridisation questions on JEE.

⚠️ Outer-Orbital ≠ Forbidden

Some students assume the inner d-orbitals are always used because they're lower in energy. With weak-field ligands, the d-electrons refuse to pair, so the inner d-set is unavailable for hybridisation — and the metal must use the outer 4d. That's normal physics, not a trick.

$\boxed{\text{Answer: (1) sp}^3\text{d}^2\text{, 4 unpaired electrons}}$