# Confusion in hybridisation of coordination compound

I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$.

In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$.

So there are $3$ more empty $\mathrm{d}$ orbitals.

Hence the hybridisation should be $\mathrm{d^3sp^2}$.

But it is $\mathrm{d^2sp^3}$ and paramagnetic .

I could not understand how .

Can you explain me?

Basically, it isn't really wise to use hybridization on coordination complexes to begin with. However, the hybridization of $d^2sp^3$ makes sense if you think of it as just describing the shape of the molecule. This should be octahedral, due to its 6 equivalent groups around the metal center, and $d^2sp^3$ is the valence bond description of the hybridization of an octahedron.
Using Ligand Field theory and MO diagrams is probably a better way to understand how the orbitals overlap and where electrons are in the final structure. This diagram is for $\ce{[Ni(H2O)6]^2+}$, but the diagram will look very similar for the $\ce{V}$ complex, just with fewer electrons. (Image source: dx.doi.org/10.1021/jp4100813 | J. Phys. Chem. B 2013, 117, 16512−16521)
• So actually it is not $d^2 sp^3$ . We write it as to denote the octahedral shape . – TrY iS CheM Mar 23 '17 at 2:11