The number of species from the following that are involved in $$sp^3d^2$$ hybridization is : $$[Co(NH_3)_6]^{3+}$$, $$SF_6$$, $$[CrF_6]^{3-}$$, $$[CoF_6]^{3-}$$, $$[Mn(CN)_6]^{3-}$$, and $$[MnCl_6]^{3-}$$

We need to find how many of the given species are involved in $$sp^3d^2$$ hybridization.

The key distinction is between inner orbital complexes ($$d^2sp^3$$, which use inner $$(n-1)d$$ orbitals) and outer orbital complexes ($$sp^3d^2$$, which use outer $$nd$$ orbitals). Strong field ligands cause pairing and form inner orbital complexes, while weak field ligands form outer orbital complexes.

1. $$[Co(NH_3)_6]^{3+}$$

$$Co^{3+}$$: electronic configuration is $$[Ar] 3d^6$$.

$$NH_3$$ is a strong field ligand, so all 6 electrons pair up in the three $$3d$$ ($$t_{2g}$$) orbitals, leaving the two $$e_g$$ ($$3d$$) orbitals empty for bonding.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

2. $$SF_6$$

$$S$$ has 6 valence electrons and forms 6 bonds with fluorine atoms.

Sulfur uses one $$3s$$, three $$3p$$, and two $$3d$$ orbitals.

Hybridization: $$sp^3d^2$$. Yes.

3. $$[CrF_6]^{3-}$$

$$Cr^{3+}$$: electronic configuration is $$[Ar] 3d^3$$.

The three $$3d$$ electrons occupy the three $$t_{2g}$$ orbitals (one each), leaving the two $$e_g$$ ($$3d$$) orbitals empty.

Even though $$F^-$$ is a weak field ligand, since the $$e_g$$ orbitals are already empty (only 3 d-electrons), $$Cr^{3+}$$ uses these inner $$3d$$ orbitals for bonding.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

4. $$[CoF_6]^{3-}$$

$$Co^{3+}$$: electronic configuration is $$[Ar] 3d^6$$.

$$F^-$$ is a weak field ligand, so no pairing occurs. The 6 electrons are distributed as: $$t_{2g}^4 e_g^2$$ (with 4 unpaired electrons).

Since the $$e_g$$ orbitals are occupied, the inner $$3d$$ orbitals cannot be used. The complex uses outer $$4d$$ orbitals.

Hybridization: $$sp^3d^2$$ (outer orbital complex). Yes.

5. $$[Mn(CN)_6]^{3-}$$

$$Mn^{3+}$$: electronic configuration is $$[Ar] 3d^4$$.

$$CN^-$$ is a strong field ligand, so pairing occurs: the 4 electrons pair up into two orbitals in $$t_{2g}$$, leaving the $$e_g$$ orbitals empty.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

6. $$[MnCl_6]^{3-}$$

$$Mn^{3+}$$: electronic configuration is $$[Ar] 3d^4$$.

$$Cl^-$$ is a weak field ligand, so no pairing occurs. The 4 electrons are distributed as: $$t_{2g}^3 e_g^1$$.

Since one $$e_g$$ orbital is occupied, the inner $$3d$$ orbitals cannot both be used for bonding. The complex uses outer $$4d$$ orbitals.

Hybridization: $$sp^3d^2$$ (outer orbital complex). Yes.

Summary:

Species with $$sp^3d^2$$ hybridization: $$SF_6$$, $$[CoF_6]^{3-}$$, $$[MnCl_6]^{3-}$$ = 3 species.

Hence, the correct answer is Option D.