The number of species from the following that are involved in sp3d2 hybridization is: [Co(NH3)6]3+, SF6, [CrF6]3-,…

🧠 $sp^3 d^2$ = Outer-Orbital Octahedral

$sp^3 d^2$ uses outer (n)d orbitals — typical for HS transition metal complexes (where inner d orbitals are partially occupied by unpaired electrons).

Also $sp^3 d^2$ for main-group expanded octets (e.g., $\mathrm{SF_6}$).

🗺️ Evaluate Each

| Species | Metal/d-config | Field | Hybridisation | |---|---|---|---| | $[\mathrm{Co(NH_3)_6}]^{3+}$ | Co(III) d⁶ LS | strong | $d^2 sp^3$ | | $\mathrm{SF_6}$ | S, no d-electron in ground state | — | $sp^3 d^2$ ✓ | | $[\mathrm{CrF_6}]^{3-}$ | Cr(III) d³ | weak (F⁻) | $sp^3 d^2$ (HS, but d³ has only $t_{2g}$ filled — could go either way) | | $[\mathrm{CoF_6}]^{3-}$ | Co(III) d⁶ HS (F⁻) | weak | $sp^3 d^2$ ✓ | | $[\mathrm{Mn(CN)_6}]^{3-}$ | Mn(III) d⁴ LS | strong | $d^2 sp^3$ | | $[\mathrm{MnCl_6}]^{3-}$ | Mn(III) d⁴ HS | weak | $sp^3 d^2$ ✓ |

🗺️ Re-Examine $[\mathrm{CrF_6}]^{3-}$

For Cr(III) d³ ($t_{2g}^3$), all 3 electrons sit in $t_{2g}$. Inner 3d orbitals: 2 are empty ($e_g$), 3 are half-filled ($t_{2g}$). For $d^2 sp^3$, we need 2 inner d orbitals empty — but only the 2 $e_g$ orbitals are empty, AND those are upper-energy orbitals (not used for hybridisation in standard VBT).

For Cr(III) d³, VBT typically uses $d^2 sp^3$ (using two empty $e_g$ orbitals + 4s + 4p) — inner orbital.

So $[\mathrm{CrF_6}]^{3-}$ is $d^2 sp^3$, NOT $sp^3 d^2$.

🗺️ Final Count

$sp^3 d^2$ species: $\mathrm{SF_6}$, $[\mathrm{CoF_6}]^{3-}$, $[\mathrm{MnCl_6}]^{3-}$ = 3.

⚡ VBT for d³ Ions

For d³ HS configurations, the 2 empty $e_g$ orbitals are available for inner-orbital hybridisation → $d^2 sp^3$. So Cr(III) (d³) octahedral is universally $d^2 sp^3$.

⚠️ Main-Group $sp^3 d^2$

Main group atoms (P, S, Cl, etc.) use outer 3d orbitals for $sp^3 d^2$ hybridisation in expanded octets ($\mathrm{SF_6}$, $\mathrm{PCl_5}$ uses $sp^3 d$). These are still $sp^3 d^2$, not $d^2 sp^3$.

$\boxed{\text{Answer: (4) 3}}$