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?